🧠:[<think>Okay, let's try to tackle this problem step by step. First, part (I) asks for the equation of the ellipse E given some conditions. Let me recall what I know about ellipses.An ellipse in standard form is given by (x²/a²) + (y²/b²) = 1, where a > b > 0. The eccentricity e is related to a and b by the formula e = c/a, where c is the distance from the center to each focus, and c² = a² - b². So, given that the eccentricity is 1/2, that tells me that c/a = 1/2, so c = a/2. Therefore, c² = a²/4. Also, since c² = a² - b², substituting gives a²/4 = a² - b². Let me write that down:a² - b² = a²/4So, subtracting a²/4 from both sides:b² = a² - a²/4 = (3/4)a²So, b = (√3/2)a. Got that.Now, the left focus F is at (-c, 0) which is (-a/2, 0). The problem says to draw a line perpendicular to the x-axis through F, which is a vertical line. Since F is at (-a/2, 0), the vertical line through F is x = -a/2. This line intersects the ellipse at points A and B, and the distance between A and B is 3.Let's find the points A and B. Substitute x = -a/2 into the ellipse equation:[ (-a/2)² / a² ] + [ y² / b² ] = 1Simplify:( a²/4 ) / a² + y² / b² = 1That's (1/4) + y² / b² = 1Subtract 1/4:y² / b² = 3/4Multiply both sides by b²:y² = (3/4)b²Therefore, y = ± (b√3)/2So the points A and B are (-a/2, (b√3)/2) and (-a/2, -(b√3)/2). The distance between A and B is the distance between these two points, which is the vertical distance since their x-coordinates are the same. So:|AB| = | (b√3)/2 - ( - (b√3)/2 ) | = | (b√3)/2 + (b√3)/2 | = | b√3 | = b√3But the problem states that |AB| = 3. Therefore:b√3 = 3 ⇒ b = 3 / √3 = √3But we have earlier that b² = (3/4)a². Since b = √3, then b² = 3. Therefore:3 = (3/4)a² ⇒ a² = 3 * (4/3) = 4 ⇒ a = 2Since a > 0, a = 2. Therefore, the equation of the ellipse is:x²/4 + y²/( (√3)² ) = x²/4 + y²/3 = 1So part (I) should be x²/4 + y²/3 = 1.Wait, let me confirm. If a = 2, then c = a/2 = 1. The foci are at (-1, 0) and (1, 0). The vertical line x = -1 intersects the ellipse. Substituting x = -1 into the ellipse equation:(-1)^2 /4 + y² /3 = 1 ⇒ 1/4 + y² /3 = 1 ⇒ y² = 3*(1 - 1/4) = 3*(3/4) = 9/4 ⇒ y = ±3/2. So points A and B are (-1, 3/2) and (-1, -3/2). The distance between them is |3/2 - (-3/2)| = 3, which matches the given |AB| = 3. So that seems correct.Therefore, part (I) is x²/4 + y²/3 = 1.Now part (II): We have a tangent line to the circle x² + y² = 12/7 at any point, and this tangent line intersects the ellipse E at points M and N. We need to determine if the dot product of vectors OM and ON is a constant value.First, let's recall that the equation of a tangent line to a circle centered at the origin at a point (x₀, y₀) on the circle is x*x₀ + y*y₀ = r². Here, the circle is x² + y² = 12/7, so radius squared is 12/7. Therefore, the tangent line at point (x₀, y₀) on the circle is x*x₀ + y*y₀ = 12/7.But the problem says "draw a tangent line to the circle... at any point", so this line will intersect the ellipse E at points M and N. Then, we need to check if OM ⋅ ON is a constant.First, let's parameterize the point of tangency on the circle. Since any point on the circle x² + y² = 12/7 can be written as (sqrt(12/7) cosθ, sqrt(12/7) sinθ) for some angle θ. Then, the tangent line at this point is x*sqrt(12/7) cosθ + y*sqrt(12/7) sinθ = 12/7.We can simplify this equation by dividing both sides by sqrt(12/7):x cosθ + y sinθ = sqrt(12/7)Alternatively, multiply numerator and denominator:Wait, sqrt(12/7) is equal to (2*sqrt(21))/7? Wait, sqrt(12/7) is sqrt(12)/sqrt(7) = 2*sqrt(3)/sqrt(7). Hmm, maybe it's better not to rationalize and just keep it as sqrt(12/7). Anyway, the tangent line is x cosθ + y sinθ = sqrt(12/7).Alternatively, perhaps parameterizing the tangent line with a parameter. Let me think. Maybe instead of using θ, I can use a general point (x₀, y₀) on the circle, but since we need to check for any point, it might be better to use θ as a parameter.So, the tangent line is x cosθ + y sinθ = sqrt(12/7). Let me confirm that. For a circle x² + y² = r², the tangent line at (x₀, y₀) is xx₀ + yy₀ = r². So here, (x₀, y₀) = (sqrt(12/7) cosθ, sqrt(12/7) sinθ). Then the tangent line is x*(sqrt(12/7) cosθ) + y*(sqrt(12/7) sinθ) = 12/7. Dividing both sides by sqrt(12/7):x cosθ + y sinθ = sqrt(12/7). Yes, that's correct.So, the equation of the tangent line is x cosθ + y sinθ = sqrt(12/7).We need to find the points M and N where this tangent line intersects the ellipse E: x²/4 + y²/3 = 1.To find the points of intersection, we can solve the system:1. x cosθ + y sinθ = sqrt(12/7)2. x²/4 + y²/3 = 1Let me solve equation 1 for one variable and substitute into equation 2. Let's solve for y:From equation 1: y sinθ = sqrt(12/7) - x cosθ ⇒ y = [sqrt(12/7) - x cosθ] / sinθNow substitute into equation 2:x²/4 + [ (sqrt(12/7) - x cosθ ) / sinθ ]² / 3 = 1This looks messy. Maybe instead, express x in terms of y. Alternatively, use parametric equations. Alternatively, use coordinates. Hmm. Alternatively, write the line equation parametrically.Alternatively, note that the line x cosθ + y sinθ = sqrt(12/7) can be parametrized. Let me think. Let me use a parameter t such that points on the line can be written as (sqrt(12/7)/cosθ - t sinθ, t cosθ) if cosθ ≠ 0. Wait, maybe that's more complicated.Alternatively, use substitution. Let me proceed step by step.Let me denote s = sqrt(12/7) to simplify notation. Then the line is x cosθ + y sinθ = s.So, y = (s - x cosθ)/sinθSubstitute into ellipse equation:x²/4 + [ (s - x cosθ)/sinθ ]² /3 = 1Multiply through by 12 sin²θ to eliminate denominators:3 sin²θ x² + 4 (s - x cosθ)^2 = 12 sin²θExpand the second term:4(s² - 2 s x cosθ + x² cos²θ) = 4s² - 8 s x cosθ + 4 x² cos²θTherefore, the equation becomes:3 sin²θ x² + 4s² - 8 s x cosθ + 4 x² cos²θ = 12 sin²θBring all terms to the left side:3 sin²θ x² + 4 x² cos²θ - 8 s x cosθ + 4s² - 12 sin²θ = 0Factor x² terms:x² (3 sin²θ + 4 cos²θ) - 8 s x cosθ + (4s² - 12 sin²θ) = 0This is a quadratic equation in x. Let me compute coefficients:A = 3 sin²θ + 4 cos²θB = -8 s cosθC = 4s² - 12 sin²θSo, the quadratic equation is A x² + B x + C = 0.Let me compute A, B, C with s = sqrt(12/7):First, s = sqrt(12/7), so s² = 12/7.Compute A:A = 3 sin²θ + 4 cos²θ = 3 sin²θ + 4 cos²θ = 3(sin²θ + cos²θ) + cos²θ = 3 + cos²θWait, 3 sin²θ + 4 cos²θ = 3(sin²θ + cos²θ) + cos²θ = 3 + cos²θ. Yes, that's correct.Similarly, B = -8 s cosθ = -8 sqrt(12/7) cosθC = 4s² - 12 sin²θ = 4*(12/7) - 12 sin²θ = 48/7 - 12 sin²θSo, substituting back:A = 3 + cos²θB = -8 sqrt(12/7) cosθC = 48/7 - 12 sin²θTherefore, the quadratic equation is:(3 + cos²θ)x² - 8 sqrt(12/7) cosθ x + (48/7 - 12 sin²θ) = 0This seems complicated. Let me see if we can find the product of the roots. For a quadratic equation Ax² + Bx + C = 0, the product of the roots is C/A. Therefore, if this quadratic has two solutions x1 and x2, then x1 x2 = C/A.But since the line intersects the ellipse at points M and N, their x-coordinates are the roots of this equation, so x1 x2 = C/A.Similarly, we can find y1 and y2 in terms of x1 and x2 using the line equation. Since y = (s - x cosθ)/sinθ, so y1 = (s - x1 cosθ)/sinθ and y2 = (s - x2 cosθ)/sinθ.Therefore, the points M and N are (x1, (s - x1 cosθ)/sinθ) and (x2, (s - x2 cosθ)/sinθ)We need to compute the dot product OM ⋅ ON, which is x1 x2 + y1 y2.Compute y1 y2:[(s - x1 cosθ)/sinθ] * [(s - x2 cosθ)/sinθ] = [s² - s cosθ (x1 + x2) + x1 x2 cos²θ] / sin²θTherefore, OM ⋅ ON = x1 x2 + [s² - s cosθ (x1 + x2) + x1 x2 cos²θ]/sin²θBut x1 + x2 = -B/A = [8 sqrt(12/7) cosθ]/(3 + cos²θ)And x1 x2 = C/A = (48/7 - 12 sin²θ)/(3 + cos²θ)Let me substitute these into the expression for OM ⋅ ON.First, let's compute x1 x2:C/A = (48/7 - 12 sin²θ)/(3 + cos²θ)Then, s² = 12/7, so:s² = 12/7Then, compute the numerator of the y1 y2 term:s² - s cosθ (x1 + x2) + x1 x2 cos²θ= 12/7 - sqrt(12/7) cosθ * [8 sqrt(12/7) cosθ / (3 + cos²θ)] + [ (48/7 - 12 sin²θ)/(3 + cos²θ) ] cos²θSimplify term by term.First term: 12/7Second term: - sqrt(12/7) * 8 sqrt(12/7) cos²θ / (3 + cos²θ )Compute sqrt(12/7)*sqrt(12/7) = 12/7, so:-8*(12/7) cos²θ / (3 + cos²θ) = -96/7 cos²θ / (3 + cos²θ)Third term: [ (48/7 - 12 sin²θ) cos²θ ] / (3 + cos²θ)So, putting all together:Numerator = 12/7 - 96/7 cos²θ / (3 + cos²θ) + [ (48/7 - 12 sin²θ) cos²θ ] / (3 + cos²θ)Combine the terms over the same denominator:= [12/7 (3 + cos²θ) - 96/7 cos²θ + (48/7 - 12 sin²θ) cos²θ ] / (3 + cos²θ)Let me compute numerator:First term: 12/7*(3 + cos²θ) = 36/7 + 12/7 cos²θSecond term: -96/7 cos²θThird term: (48/7 - 12 sin²θ) cos²θ = 48/7 cos²θ - 12 sin²θ cos²θSo combining all:36/7 + 12/7 cos²θ -96/7 cos²θ +48/7 cos²θ -12 sin²θ cos²θSimplify term by term:36/7 + (12 -96 +48)/7 cos²θ -12 sin²θ cos²θ= 36/7 + (-36/7) cos²θ -12 sin²θ cos²θFactor 36/7:36/7 (1 - cos²θ) -12 sin²θ cos²θBut 1 - cos²θ = sin²θ, so:36/7 sin²θ -12 sin²θ cos²θFactor 12 sin²θ:12 sin²θ (3/7 - cos²θ )Therefore, numerator = 12 sin²θ (3/7 - cos²θ )Thus, the entire y1 y2 term is [12 sin²θ (3/7 - cos²θ ) ] / [ sin²θ (3 + cos²θ ) ]Wait, the denominator was (3 + cos²θ), and we have divided by sin²θ because the expression was [numerator]/sin²θ. Wait, wait, let's retrace.Wait, the numerator of the y1 y2 term is [12 sin²θ (3/7 - cos²θ ) ] / (3 + cos²θ )And then the entire y1 y2 expression is numerator / sin²θ, right?Wait, no. Wait, original expression:OM ⋅ ON = x1 x2 + [numerator]/sin²θWhere numerator was:12/7 -96/7 cos²θ/(3 + cos²θ) + [ (48/7 -12 sin²θ) cos²θ ]/(3 + cos²θ )Which we simplified to [12 sin²θ (3/7 - cos²θ ) ] / (3 + cos²θ )Therefore, y1 y2 = [12 sin²θ (3/7 - cos²θ ) ] / [ (3 + cos²θ ) sin²θ ] = 12 (3/7 - cos²θ ) / (3 + cos²θ )So, OM ⋅ ON = x1 x2 + y1 y2 = [ (48/7 -12 sin²θ )/(3 + cos²θ ) ] + [12 (3/7 - cos²θ ) / (3 + cos²θ ) ]Since both terms have the denominator (3 + cos²θ ), combine numerators:[48/7 -12 sin²θ + 12*(3/7 - cos²θ ) ] / (3 + cos²θ )Compute numerator:48/7 -12 sin²θ + 36/7 -12 cos²θCombine constants:48/7 + 36/7 = 84/7 = 12Combine trigonometric terms:-12 sin²θ -12 cos²θ = -12(sin²θ + cos²θ ) = -12*1 = -12Therefore, numerator = 12 -12 = 0Wait, that's interesting. So the entire OM ⋅ ON = 0/(3 + cos²θ ) = 0. Therefore, OM ⋅ ON = 0.But wait, this can't be right. Because the problem is asking whether this dot product is a constant. If it's always zero, then yes, it's a constant. But is this correct?Wait, let me check my steps again because getting zero seems surprising.Starting from:OM ⋅ ON = x1 x2 + y1 y2We found x1 x2 = (48/7 -12 sin²θ )/(3 + cos²θ )And y1 y2 = [12 (3/7 - cos²θ ) ] / (3 + cos²θ )Adding them together:[ (48/7 -12 sin²θ ) + 12(3/7 - cos²θ ) ] / (3 + cos²θ )Compute numerator:48/7 -12 sin²θ + 36/7 -12 cos²θ= (48/7 + 36/7) -12(sin²θ + cos²θ )= 84/7 -12*1= 12 -12= 0So OM ⋅ ON = 0. Therefore, regardless of θ, the dot product is zero, which is a constant. Therefore, the answer is yes, it's a constant, and the value is 0.But let me verify this with a concrete example. Let's choose a specific θ where the calculation is easier.For example, take θ = 0. Then the tangent line is x cos0 + y sin0 = sqrt(12/7), which simplifies to x = sqrt(12/7). The line x = sqrt(12/7) intersects the ellipse x²/4 + y²/3 =1.Substitute x = sqrt(12/7):(12/7)/4 + y²/3 =1 ⇒ (3/7) + y²/3 =1 ⇒ y²/3 = 4/7 ⇒ y² =12/7 ⇒ y=±sqrt(12/7)Therefore, points M and N are (sqrt(12/7), sqrt(12/7)) and (sqrt(12/7), -sqrt(12/7))Compute OM ⋅ ON: (sqrt(12/7)*sqrt(12/7) ) + (sqrt(12/7)*(-sqrt(12/7)) ) = (12/7) - (12/7) = 0. So indeed, it's zero.Another example: θ = π/2. Then the tangent line is x cos(π/2) + y sin(π/2) = sqrt(12/7), which simplifies to y = sqrt(12/7). This line intersects the ellipse x²/4 + y²/3 =1.Substitute y = sqrt(12/7):x²/4 + (12/7)/3 =1 ⇒ x²/4 + 4/7 =1 ⇒ x²/4 = 3/7 ⇒ x²=12/7 ⇒ x=±sqrt(12/7)Therefore, points M and N are (sqrt(12/7), sqrt(12/7)) and (-sqrt(12/7), sqrt(12/7))Wait, but hold on: the line is y = sqrt(12/7), which intersects the ellipse at (sqrt(12/7), sqrt(12/7)) and (-sqrt(12/7), sqrt(12/7))? Wait, no. If we set y = sqrt(12/7), then x²/4 + (12/7)/3 = x²/4 + 4/7 =1 ⇒ x²/4 = 3/7 ⇒ x² = 12/7 ⇒ x=±sqrt(12/7). Therefore, points are (sqrt(12/7), sqrt(12/7)) and (-sqrt(12/7), sqrt(12/7)). Then OM ⋅ ON would be [sqrt(12/7)* -sqrt(12/7) ] + [sqrt(12/7)*sqrt(12/7) ] = -12/7 +12/7=0. So again, zero.Another example: θ = π/4. Let's see. The tangent line is x cos(π/4) + y sin(π/4) = sqrt(12/7). Which is (x + y)/sqrt(2) = sqrt(12/7). So x + y = sqrt(24/7). Let's find the points M and N.We can solve x + y = sqrt(24/7) and x²/4 + y²/3 =1.Let me express y = sqrt(24/7) - x, substitute into the ellipse equation:x²/4 + (sqrt(24/7) - x)^2 /3 =1Expand (sqrt(24/7) - x)^2 = 24/7 - 2x sqrt(24/7) + x²Therefore:x²/4 + (24/7 -2x sqrt(24/7) +x²)/3 =1Multiply through by 12 to eliminate denominators:3x² +4*(24/7 -2x sqrt(24/7) +x²) =12Compute each term:3x² + 96/7 -8x sqrt(24/7) +4x² =12Combine like terms:7x² -8x sqrt(24/7) +96/7 -12 =0Simplify 96/7 -12 =96/7 -84/7=12/7Thus:7x² -8x sqrt(24/7) +12/7=0Multiply through by 7 to eliminate denominators:49x² -56x sqrt(24/7) +12=0Let me compute sqrt(24/7) = (2*sqrt(42))/7. So sqrt(24/7)=2*sqrt(6/7). Therefore:56x *2*sqrt(6/7)=56x*(2 sqrt(6)/sqrt(7))=112x sqrt(6)/sqrt(7)But this is getting messy. Maybe instead, note that regardless of the quadratic equation, the product of the roots x1 x2 = C/A where in the original quadratic (after multiplying by 7):49x² -56x sqrt(24/7) +12=0A=49, B=-56 sqrt(24/7), C=12Thus x1 x2=12/49Similarly, y1 y2= [sqrt(24/7) -x1][sqrt(24/7)-x2] =24/7 -sqrt(24/7)(x1 +x2) +x1 x2x1 +x2=56 sqrt(24/7)/49= (56/49) sqrt(24/7)= (8/7) sqrt(24/7)Therefore:y1 y2=24/7 - sqrt(24/7)*(8/7 sqrt(24/7)) +12/49Compute sqrt(24/7)*sqrt(24/7)=24/7Thus:24/7 - (8/7)*(24/7) +12/49=24/7 -192/49 +12/49Convert 24/7 to 168/49:168/49 -192/49 +12/49= (168 -192 +12)/49= (-12)/49Therefore, OM ⋅ ON= x1 x2 + y1 y2=12/49 + (-12/49)=0So again, the dot product is zero. Therefore, even in this case, the result is zero.Therefore, regardless of θ, the dot product is zero. Therefore, the answer to part (II) is yes, it's a constant value of 0.But wait, the problem statement says "tangent line to the circle x²+y²=12/7 at any point, intersecting the ellipse E at points M and N". But in the case when the tangent line is horizontal or vertical, the points M and N are symmetric, leading to the dot product being zero. But even in the slanted case, the algebra shows that it's always zero. Therefore, the answer is yes, it's a constant zero.So, summarizing:(I) The equation of the ellipse E is x²/4 + y²/3 =1.(II) The dot product OM ⋅ ON is a constant value of 0.Final Answer(I) The equation of the ellipse E is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(II) boxed{0}</think>Given an ellipse ( E: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) with an eccentricity of ( frac{1}{2} ), and the left focus ( F ). A vertical line through ( F ) intersects the ellipse at points ( A ) and ( B ) with ( |AB| = 3 ).(I) To find the equation of the ellipse:1. The eccentricity ( e = frac{1}{2} ) gives ( c = frac{a}{2} ).2. Using ( c^2 = a^2 - b^2 ), we get ( left(frac{a}{2}right)^2 = a^2 - b^2 Rightarrow frac{a^2}{4} = a^2 - b^2 Rightarrow b^2 = frac{3}{4}a^2 ).3. The vertical line through ( F(-c, 0) ) is ( x = -frac{a}{2} ). Substituting this into the ellipse equation, we find ( y = pm frac{bsqrt{3}}{2} ).4. Given ( |AB| = 3 ), we have ( bsqrt{3} = 3 Rightarrow b = sqrt{3} ).5. Substituting ( b = sqrt{3} ) into ( b^2 = frac{3}{4}a^2 ), we get ( 3 = frac{3}{4}a^2 Rightarrow a^2 = 4 Rightarrow a = 2 ).Thus, the equation of the ellipse is ( frac{x^2}{4} + frac{y^2}{3} = 1 ).(II) To determine if ( overrightarrow{OM} cdot overrightarrow{ON} ) is a constant:1. The tangent line to the circle ( x^2 + y^2 = frac{12}{7} ) at point ( (sqrt{frac{12}{7}} costheta, sqrt{frac{12}{7}} sintheta) ) is ( x costheta + y sintheta = sqrt{frac{12}{7}} ).2. Substituting this line into the ellipse equation, solving for ( x ) and ( y ), we find the product of the roots ( x_1 x_2 ) and ( y_1 y_2 ).3. Using the properties of the quadratic equation, we show that the dot product ( overrightarrow{OM} cdot overrightarrow{ON} = x_1 x_2 + y_1 y_2 ) simplifies to 0 for all ( theta ).Thus, ( overrightarrow{OM} cdot overrightarrow{ON} ) is a constant value of 0. Final Answer(I) The equation of the ellipse ( E ) is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(II) boxed{0},<think>Okay, let's tackle this problem step by step. First, part (I) asks for the equation of the ellipse E given some conditions. Let's parse the information.We have an ellipse with the standard equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0). The eccentricity is given as (frac{1}{2}). The left focus is F, and a line perpendicular to the x-axis (which is a vertical line) passes through F, intersecting the ellipse at points A and B. The distance between A and B is 3. We need to find the equation of the ellipse, which means finding the values of (a) and (b).Starting with the eccentricity. For an ellipse, the eccentricity (e) is defined as (e = frac{c}{a}), where (c) is the distance from the center to each focus. Since the eccentricity is (frac{1}{2}), we have:[frac{c}{a} = frac{1}{2} implies c = frac{a}{2}]We also know that for an ellipse, (c^2 = a^2 - b^2). Substituting (c = frac{a}{2}) into this equation:[left(frac{a}{2}right)^2 = a^2 - b^2 implies frac{a^2}{4} = a^2 - b^2]Solving for (b^2):[b^2 = a^2 - frac{a^2}{4} = frac{3a^2}{4}]So, (b = frac{sqrt{3}a}{2}). Okay, so we have a relationship between (a) and (b). Now, moving on to the second part of the problem involving the vertical line through the left focus F.The left focus F is located at ((-c, 0)), which is ((- frac{a}{2}, 0)). The vertical line passing through F is (x = -frac{a}{2}). This line intersects the ellipse at points A and B. Let's find the coordinates of A and B by substituting (x = -frac{a}{2}) into the ellipse equation:[frac{(-frac{a}{2})^2}{a^2} + frac{y^2}{b^2} = 1 implies frac{frac{a^2}{4}}{a^2} + frac{y^2}{b^2} = 1 implies frac{1}{4} + frac{y^2}{b^2} = 1]Subtracting (frac{1}{4}) from both sides:[frac{y^2}{b^2} = frac{3}{4} implies y^2 = frac{3}{4}b^2 implies y = pm frac{sqrt{3}}{2}b]Therefore, points A and B are at (left(-frac{a}{2}, frac{sqrt{3}}{2}bright)) and (left(-frac{a}{2}, -frac{sqrt{3}}{2}bright)). The distance between A and B is the vertical distance, which is the difference in the y-coordinates. So:[|AB| = left(frac{sqrt{3}}{2}bright) - left(-frac{sqrt{3}}{2}bright) = sqrt{3}b]But the problem states that (|AB| = 3), so:[sqrt{3}b = 3 implies b = frac{3}{sqrt{3}} = sqrt{3}]So, (b = sqrt{3}). Earlier, we found that (b = frac{sqrt{3}a}{2}). Setting these equal:[frac{sqrt{3}a}{2} = sqrt{3} implies frac{a}{2} = 1 implies a = 2]Therefore, (a = 2) and (b = sqrt{3}). Plugging these back into the ellipse equation:[frac{x^2}{4} + frac{y^2}{3} = 1]So that should be the equation of the ellipse E. Let me just verify the steps again to ensure there are no mistakes.1. Eccentricity (e = frac{1}{2} = frac{c}{a}) gives (c = frac{a}{2}).2. Relationship (c^2 = a^2 - b^2) gives (frac{a^2}{4} = a^2 - b^2) leading to (b^2 = frac{3a^2}{4}).3. The vertical line through F at (x = -frac{a}{2}) intersects the ellipse, leading to (y = pm frac{sqrt{3}}{2}b).4. Distance AB is (sqrt{3}b = 3), so (b = sqrt{3}).5. Substitute (b) into the equation from step 2: (sqrt{3} = frac{sqrt{3}a}{2} implies a = 2).6. So, (a^2 = 4) and (b^2 = 3). The equation is (frac{x^2}{4} + frac{y^2}{3} = 1).Everything checks out. So part (I) is solved.Moving on to part (II). We need to determine whether (overrightarrow{OM} cdot overrightarrow{ON}) is a constant value when a tangent line to the circle (x^2 + y^2 = frac{12}{7}) is drawn at any point, and this tangent line intersects the ellipse E at points M and N. The origin is O. So, regardless of where the tangent is drawn on the circle, after intersecting the ellipse, the dot product of the vectors from the origin to M and N should be a constant.First, let's recall that the circle (x^2 + y^2 = frac{12}{7}) has a radius of (sqrt{frac{12}{7}}). The tangent lines to this circle at a point ((x_1, y_1)) on the circle have the equation (x x_1 + y y_1 = frac{12}{7}). Since the tangent line is drawn at any point on the circle, the equation of the tangent line is as above. Then, this tangent line intersects the ellipse (frac{x^2}{4} + frac{y^2}{3} = 1) at points M and N.We need to find the points M and N, compute vectors OM and ON, take their dot product, and see if it's a constant.Alternatively, perhaps there's a more straightforward way using properties of ellipses and circles, or using parametrization.Let me think. Let's parameterize a point on the circle. Let’s take a parameter θ such that the point on the circle is ((r cos theta, r sin theta)), where (r = sqrt{frac{12}{7}}). Then the tangent line at this point is (x cos theta + y sin theta = r), since the tangent line to the circle (x^2 + y^2 = r^2) at ((r cos theta, r sin theta)) is (x cos theta + y sin theta = r). Here, (r = sqrt{frac{12}{7}}), so the equation is (x cos theta + y sin theta = sqrt{frac{12}{7}}).But actually, let me check that. The standard tangent line to (x^2 + y^2 = r^2) at point ((x_1, y_1)) is (x x_1 + y y_1 = r^2). So if the circle here is (x^2 + y^2 = frac{12}{7}), then the tangent line at point ((x_1, y_1)) is (x x_1 + y y_1 = frac{12}{7}). If we parameterize ((x_1, y_1)) as ((sqrt{frac{12}{7}} cos theta, sqrt{frac{12}{7}} sin theta)), then the tangent line becomes:[x sqrt{frac{12}{7}} cos theta + y sqrt{frac{12}{7}} sin theta = frac{12}{7}]Divide both sides by (sqrt{frac{12}{7}}):[x cos theta + y sin theta = sqrt{frac{12}{7}} times frac{1}{sqrt{frac{12}{7}}} } quad text{Wait, that doesn't seem right. Let's do the division properly.}If we have (x sqrt{frac{12}{7}} cos theta + y sqrt{frac{12}{7}} sin theta = frac{12}{7}), then divide both sides by (sqrt{frac{12}{7}}):[x cos theta + y sin theta = frac{12}{7} div sqrt{frac{12}{7}} = frac{12}{7} times sqrt{frac{7}{12}} = sqrt{frac{12}{7}} times frac{12}{7} times sqrt{frac{7}{12}}} quad Hmm, maybe better to compute numerically.Wait, the equation is:[x x_1 + y y_1 = r^2]where (x_1 = r cos theta), (y_1 = r sin theta), and (r = sqrt{frac{12}{7}}).So substituting, the tangent line is:[x (r cos theta) + y (r sin theta) = r^2]Divide both sides by r:[x cos theta + y sin theta = r]Therefore, the tangent line is (x cos theta + y sin theta = sqrt{frac{12}{7}}).So the equation of the tangent line is (x cos theta + y sin theta = sqrt{frac{12}{7}}).Now, we need to find the points M and N where this tangent line intersects the ellipse (frac{x^2}{4} + frac{y^2}{3} = 1).To find the intersection points, we can solve the system of equations:1. (x cos theta + y sin theta = sqrt{frac{12}{7}}) (Equation of the tangent line)2. (frac{x^2}{4} + frac{y^2}{3} = 1) (Equation of the ellipse)We can express y from the tangent line equation in terms of x, or vice versa, and substitute into the ellipse equation. Let's solve for y:From the tangent line:[y sin theta = sqrt{frac{12}{7}} - x cos theta implies y = frac{sqrt{frac{12}{7}} - x cos theta}{sin theta}]Assuming (sin theta neq 0). If (sin theta = 0), the tangent line is vertical, but since the circle has radius (sqrt{frac{12}{7}}), the vertical tangent lines would be at (x = pm sqrt{frac{12}{7}}). However, since theta is a parameter, we can handle the case separately if needed, but let's proceed with (sin theta neq 0) for now.Substituting y into the ellipse equation:[frac{x^2}{4} + frac{left(frac{sqrt{frac{12}{7}} - x cos theta}{sin theta}right)^2}{3} = 1]This looks complicated. Let's simplify step by step.First, let's write ( sqrt{frac{12}{7}} = frac{2 sqrt{21}}{7} ). Wait, ( sqrt{12/7} = (2 sqrt{21}) /7 )? Let's check:[sqrt{frac{12}{7}} = frac{sqrt{12}}{sqrt{7}} = frac{2 sqrt{3}}{sqrt{7}} = frac{2 sqrt{21}}{7}]Yes, because multiplying numerator and denominator by (sqrt{7}):[frac{2 sqrt{3} cdot sqrt{7}}{7} = frac{2 sqrt{21}}{7}]So, ( sqrt{frac{12}{7}} = frac{2 sqrt{21}}{7} ). Let's use this for simplification.So, substituting into y:[y = frac{frac{2 sqrt{21}}{7} - x cos theta}{sin theta}]Now, substitute this into the ellipse equation:[frac{x^2}{4} + frac{left(frac{2 sqrt{21}}{7} - x cos thetaright)^2}{3 sin^2 theta} = 1]Multiply through by 12 to eliminate denominators:[3x^2 + 4 cdot frac{left(frac{2 sqrt{21}}{7} - x cos thetaright)^2}{sin^2 theta} = 12]This is still quite messy. Maybe expanding the squared term first:[left(frac{2 sqrt{21}}{7} - x cos thetaright)^2 = left(frac{2 sqrt{21}}{7}right)^2 - 2 cdot frac{2 sqrt{21}}{7} cdot x cos theta + x^2 cos^2 theta]Calculating each term:1. (left(frac{2 sqrt{21}}{7}right)^2 = frac{4 cdot 21}{49} = frac{84}{49} = frac{12}{7})2. (2 cdot frac{2 sqrt{21}}{7} cdot x cos theta = frac{4 sqrt{21}}{7} x cos theta)3. (x^2 cos^2 theta)So, substituting back:[frac{x^2}{4} + frac{frac{12}{7} - frac{4 sqrt{21}}{7} x cos theta + x^2 cos^2 theta}{3 sin^2 theta} = 1]Multiply each term by 3 sin²θ to eliminate denominators:Wait, no. Let's think again.Wait, in the original substitution, we had:[frac{x^2}{4} + frac{left(frac{2 sqrt{21}}{7} - x cos thetaright)^2}{3 sin^2 theta} = 1]So let's write each term:First term: ( frac{x^2}{4} )Second term: ( frac{1}{3 sin^2 theta} left( frac{12}{7} - frac{4 sqrt{21}}{7} x cos theta + x^2 cos^2 theta right) )So combining these:[frac{x^2}{4} + frac{12}{21 sin^2 theta} - frac{4 sqrt{21}}{21 sin^2 theta} x cos theta + frac{x^2 cos^2 theta}{3 sin^2 theta} = 1]Simplify the coefficients:1. (frac{12}{21 sin^2 theta} = frac{4}{7 sin^2 theta})2. (- frac{4 sqrt{21}}{21 sin^2 theta} x cos theta = - frac{4 sqrt{21}}{21 sin^2 theta} x cos theta)3. (frac{cos^2 theta}{3 sin^2 theta} x^2 = frac{cos^2 theta}{3 sin^2 theta} x^2)Combine the x² terms:[x^2 left( frac{1}{4} + frac{cos^2 theta}{3 sin^2 theta} right) - frac{4 sqrt{21}}{21 sin^2 theta} x cos theta + frac{4}{7 sin^2 theta} - 1 = 0]This is a quadratic equation in x. Let's denote the coefficients as A, B, C:A = ( frac{1}{4} + frac{cos^2 theta}{3 sin^2 theta} )B = ( - frac{4 sqrt{21}}{21 sin^2 theta} cos theta )C = ( frac{4}{7 sin^2 theta} - 1 )Let me compute A:First term: 1/4Second term: (frac{cos^2 theta}{3 sin^2 theta})So,A = ( frac{1}{4} + frac{cos^2 theta}{3 sin^2 theta} = frac{3 sin^2 theta + 4 cos^2 theta}{12 sin^2 theta} )Wait, let's find a common denominator:Multiply numerator and denominator appropriately:First term: (frac{1}{4} = frac{3 sin^2 theta}{12 sin^2 theta})Second term: (frac{cos^2 theta}{3 sin^2 theta} = frac{4 cos^2 theta}{12 sin^2 theta})So, adding them:A = ( frac{3 sin^2 theta + 4 cos^2 theta}{12 sin^2 theta} )Which can be written as:A = ( frac{3 sin^2 theta + 4 cos^2 theta}{12 sin^2 theta} )Similarly, term B:B = ( - frac{4 sqrt{21}}{21 sin^2 theta} cos theta )Term C:C = ( frac{4}{7 sin^2 theta} - 1 = frac{4 - 7 sin^2 theta}{7 sin^2 theta} )So now the quadratic equation is:[frac{3 sin^2 theta + 4 cos^2 theta}{12 sin^2 theta} x^2 - frac{4 sqrt{21} cos theta}{21 sin^2 theta} x + frac{4 - 7 sin^2 theta}{7 sin^2 theta} = 0]Multiply all terms by ( 84 sin^2 theta ) to eliminate denominators:- The coefficient A: ( frac{3 sin^2 theta + 4 cos^2 theta}{12 sin^2 theta} times 84 sin^2 theta = 7(3 sin^2 theta + 4 cos^2 theta) )- Coefficient B: ( - frac{4 sqrt{21} cos theta}{21 sin^2 theta} times 84 sin^2 theta = -16 sqrt{21} cos theta )- Coefficient C: ( frac{4 - 7 sin^2 theta}{7 sin^2 theta} times 84 sin^2 theta = 12(4 - 7 sin^2 theta) )So the equation becomes:[7(3 sin^2 theta + 4 cos^2 theta) x^2 - 16 sqrt{21} cos theta cdot x + 12(4 - 7 sin^2 theta) = 0]Simplify coefficients:First term:(7(3 sin^2 theta + 4 cos^2 theta) = 21 sin^2 theta + 28 cos^2 theta)Third term:(12(4 - 7 sin^2 theta) = 48 - 84 sin^2 theta)So the quadratic equation is:[(21 sin^2 theta + 28 cos^2 theta)x^2 - 16 sqrt{21} cos theta cdot x + (48 - 84 sin^2 theta) = 0]This is quite complicated, but maybe we can factor or find some relation here. Let's denote this quadratic as:[A x^2 + B x + C = 0]Where:A = (21 sin^2 theta + 28 cos^2 theta)B = (-16 sqrt{21} cos theta)C = (48 - 84 sin^2 theta)We can try solving for x using the quadratic formula:[x = frac{-B pm sqrt{B^2 - 4AC}}{2A}]But before doing that, let's consider that we need to find points M and N on the ellipse, and compute (overrightarrow{OM} cdot overrightarrow{ON}). The coordinates of M and N will be solutions (x1, y1) and (x2, y2) of the system, so their coordinates satisfy both the tangent line and the ellipse.But since the tangent line is from the circle (x^2 + y^2 = frac{12}{7}), and we're intersecting it with the ellipse, points M and N are intersection points. If we can find expressions for x1, x2, y1, y2 in terms of theta, then compute their dot product, maybe it simplifies to a constant.Alternatively, there's a property in conic sections that might help here. For instance, the concept of pole and polar lines. Maybe the tangent line to the circle is the polar line of the point with respect to the circle, and we can use properties of reciprocal polars or something else. But I'm not sure.Alternatively, think about the dot product. If M is (x1, y1) and N is (x2, y2), then (overrightarrow{OM} cdot overrightarrow{ON} = x1 x2 + y1 y2). So we need to compute x1 x2 + y1 y2 and see if it's a constant.From the quadratic equation above, the product of the roots x1 and x2 is C/A. Similarly, perhaps we can find expressions for y1 y2.Given that x1 and x2 are roots of the quadratic equation:x1 + x2 = -B/Ax1 x2 = C/ASo x1 x2 = (48 - 84 sin²θ) / (21 sin²θ + 28 cos²θ)Similarly, since y = [sqrt(12/7) - x cosθ]/sinθ, then y1 = [sqrt(12/7) - x1 cosθ]/sinθ and y2 = [sqrt(12/7) - x2 cosθ]/sinθ. Therefore,y1 y2 = [sqrt(12/7) - x1 cosθ][sqrt(12/7) - x2 cosθ] / sin²θExpand the numerator:= (12/7 - sqrt(12/7) cosθ (x1 + x2) + x1 x2 cos²θ) / sin²θSo:= (12/7 - sqrt(12/7) cosθ (-B/A) + (C/A) cos²θ) / sin²θBut since x1 + x2 = -B/A and x1 x2 = C/A.Let's compute each term step by step.First term: 12/7Second term: - sqrt(12/7) cosθ (x1 + x2) = - sqrt(12/7) cosθ (-B/A) = sqrt(12/7) cosθ (B/A)Third term: x1 x2 cos²θ = (C/A) cos²θTherefore, numerator:12/7 + sqrt(12/7) cosθ (B/A) + (C/A) cos²θTherefore,y1 y2 = [12/7 + sqrt(12/7) cosθ (B/A) + (C/A) cos²θ] / sin²θNow, let's compute this expression step by step.First, recall:B = -16 sqrt(21) cosθA = 21 sin²θ + 28 cos²θ = 7(3 sin²θ + 4 cos²θ)C = 48 - 84 sin²θ = 12(4 - 7 sin²θ)So, sqrt(12/7) is 2 sqrt(21)/7 as before.Let me compute sqrt(12/7) * cosθ * (B/A):sqrt(12/7) * cosθ * (-16 sqrt(21) cosθ / A )Wait, sqrt(12/7) is 2 sqrt(21)/7, so:= (2 sqrt(21)/7) * cosθ * (-16 sqrt(21) cosθ / A )Multiply the terms:= (2 sqrt(21) * -16 sqrt(21) ) / (7 A) * cos²θ= (-32 * 21) / (7 A) * cos²θ= (-672) / (7 A) * cos²θ= (-96)/A * cos²θSimilarly, the third term in the numerator:(C/A) cos²θ = (48 - 84 sin²θ)/A * cos²θSo putting together:Numerator = 12/7 + (-96/A) cos²θ + (48 - 84 sin²θ)/A cos²θ= 12/7 + [ (-96 + 48 - 84 sin²θ ) / A ] cos²θSimplify inside the brackets:-96 + 48 = -48So:= 12/7 + [ (-48 - 84 sin²θ ) / A ] cos²θFactor numerator:= 12/7 - [ (48 + 84 sin²θ ) / A ] cos²θNote that A = 21 sin²θ + 28 cos²θ = 7(3 sin²θ + 4 cos²θ)So:= 12/7 - [ (48 + 84 sin²θ ) / (7(3 sin²θ + 4 cos²θ)) ] cos²θFactor numerator in the bracket:48 + 84 sin²θ = 12*4 + 12*7 sin²θ = 12(4 + 7 sin²θ)So:= 12/7 - [ 12(4 + 7 sin²θ) / (7(3 sin²θ + 4 cos²θ)) ] cos²θSimplify:= 12/7 - (12 / 7) * [ (4 + 7 sin²θ) / (3 sin²θ + 4 cos²θ) ] cos²θFactor out 12/7:= (12/7) [ 1 - ( (4 + 7 sin²θ) cos²θ ) / (3 sin²θ + 4 cos²θ) ]Let’s compute the term inside the brackets:1 - [ (4 + 7 sin²θ) cos²θ / (3 sin²θ + 4 cos²θ) ]To combine these terms, express 1 as (3 sin²θ + 4 cos²θ)/(3 sin²θ + 4 cos²θ):= [ (3 sin²θ + 4 cos²θ) - (4 + 7 sin²θ) cos²θ ] / (3 sin²θ + 4 cos²θ)Expand numerator:3 sin²θ + 4 cos²θ - 4 cos²θ - 7 sin²θ cos²θSimplify:3 sin²θ - 7 sin²θ cos²θFactor sin²θ:= sin²θ (3 - 7 cos²θ)Therefore, the entire expression becomes:= (12/7) [ sin²θ (3 - 7 cos²θ) / (3 sin²θ + 4 cos²θ) ]Therefore, y1 y2:= [ (12/7) sin²θ (3 - 7 cos²θ) / (3 sin²θ + 4 cos²θ) ] / sin²θ= (12/7) (3 - 7 cos²θ) / (3 sin²θ + 4 cos²θ)So y1 y2 simplifies to:= (12/7) (3 - 7 cos²θ) / (3 sin²θ + 4 cos²θ)Therefore, combining x1 x2 and y1 y2:x1 x2 + y1 y2 = [ (48 - 84 sin²θ)/(21 sin²θ + 28 cos²θ) ] + [ (12/7)(3 - 7 cos²θ)/(3 sin²θ + 4 cos²θ) ]Note that denominator of the first term is 7*(3 sin²θ + 4 cos²θ), so:x1 x2 = (48 - 84 sin²θ) / [7(3 sin²θ + 4 cos²θ) ]Similarly, y1 y2 = (12/7)(3 - 7 cos²θ) / (3 sin²θ + 4 cos²θ)So write both terms over the same denominator:x1 x2 + y1 y2 = [ (48 - 84 sin²θ) + 12(3 - 7 cos²θ) ] / [7(3 sin²θ + 4 cos²θ) ]Compute numerator:48 - 84 sin²θ + 36 - 84 cos²θ = (48 + 36) - 84 sin²θ -84 cos²θ = 84 - 84 (sin²θ + cos²θ) = 84 - 84(1) = 0Wow! The numerator is zero. Therefore, x1 x2 + y1 y2 = 0 / [7(3 sin²θ + 4 cos²θ) ] = 0Therefore, regardless of θ, the dot product (overrightarrow{OM} cdot overrightarrow{ON} = x1 x2 + y1 y2 = 0). So it's a constant value, zero.Therefore, the answer to part (II) is yes, it is a constant value, and the value is 0.But let me verify this conclusion because it's surprising. If all the steps led us to x1 x2 + y1 y2 = 0, then the dot product is zero, meaning vectors OM and ON are perpendicular. However, this seems to hold for any tangent line to the circle, which is interesting.Alternatively, perhaps there is an underlying orthogonality condition here. Given that the tangent lines to the circle intersect the ellipse at points M and N such that OM and ON are perpendicular. This seems like a non-trivial result, but according to the algebra, it holds.Alternatively, maybe I made a mistake in the calculation. Let's check:We had:Numerator of x1 x2 + y1 y2 = 48 -84 sin²θ +36 -84 cos²θ = 84 -84(sin²θ + cos²θ) = 84 -84 =0. Yes, that's correct.Therefore, x1 x2 + y1 y2 =0, so the dot product is zero. So the answer is yes, it's a constant, and the value is zero.Hence, part (II) answer is 0.Final Answer(I) The equation of the ellipse E is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).(II) (overrightarrow{OM} cdot overrightarrow{ON}) is a constant value, and the value is (boxed{0}).</think>Given an ellipse ( E: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) (with ( a > b > 0 )) with an eccentricity of ( frac{1}{2} ), and a line perpendicular to the x-axis passing through the left focus ( F ), intersecting the ellipse at points ( A ) and ( B ) with ( |AB| = 3 ):(I) Find the equation of the ellipse E:1. The eccentricity ( e = frac{1}{2} ) gives ( c = frac{a}{2} ).2. Using the relationship ( c^2 = a^2 - b^2 ), we substitute ( c = frac{a}{2} ) to get ( frac{a^2}{4} = a^2 - b^2 ), leading to ( b^2 = frac{3a^2}{4} ).3. The vertical line through the left focus ( F(- frac{a}{2}, 0) ) intersects the ellipse at points ( A ) and ( B ). Solving for ( y ) when ( x = -frac{a}{2} ), we find ( y = pm frac{sqrt{3}}{2}b ).4. The distance ( |AB| = 3 ) gives ( sqrt{3}b = 3 ), so ( b = sqrt{3} ).5. Substituting ( b = sqrt{3} ) into ( b^2 = frac{3a^2}{4} ), we find ( a = 2 ).Thus, the equation of the ellipse is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).(II) Determine whether ( overrightarrow{OM} cdot overrightarrow{ON} ) is a constant value:1. The tangent line to the circle ( x^2 + y^2 = frac{12}{7} ) at any point ( (x_1, y_1) ) has the equation ( xx_1 + yy_1 = frac{12}{7} ).2. Parameterizing the point on the circle as ( (sqrt{frac{12}{7}} cos theta, sqrt{frac{12}{7}} sin theta) ), the tangent line equation becomes ( x cos theta + y sin theta = sqrt{frac{12}{7}} ).3. Solving the system of the tangent line and the ellipse equation, we find the coordinates of points ( M ) and ( N ).4. Using Vieta's formulas, the product of the roots ( x_1 x_2 ) and ( y_1 y_2 ) are calculated. The dot product ( overrightarrow{OM} cdot overrightarrow{ON} = x_1 x_2 + y_1 y_2 ) simplifies to 0.Thus, ( overrightarrow{OM} cdot overrightarrow{ON} ) is a constant value, and the value is (boxed{0}).]

🔑:(I) Given the eccentricity of the ellipse e = frac {c}{a}= sqrt {1- frac {b^{2}}{a^{2}}}= frac {1}{2}, we have 3a2=4b2.Also, |AB|= frac {2b^{2}}{a}=3, so a=2 and b= sqrt {3}.Thus, the equation of ellipse E is: frac {x^{2}}{4}+ frac {y^{2}}{3}=1.(II) We have two cases:a) When the slope of the tangent line does not exist, the tangent line equations are x= sqrt { frac {12}{7}} and x=- sqrt { frac {12}{7}}. In this case, overrightarrow {OM}cdot overrightarrow {ON}=0.b) When the slope exists, let the tangent line equation be y=kx+m. Solving the system of equations:begin{cases} x^2 + y^2 = frac{12}{7} y = kx + m end{cases}Substitute y and eliminate it, we get (3+4k2)x2+8kmx+4m2-12=0.Since the tangent line intersects the ellipse at M and N, the discriminant △ > 0, so 4k2 > m2-3.Let M(x1, y1) and N(x2, y2), then:x1 + x2 = -frac {8km}{3+4k^{2}} and x1x2 = frac {4m^{2}-12}{3+4k^{2}}.Thus, y1y2 = k2x1x2+km(x1+x2)+m2 = frac {3m^{2}-12k^{2}}{3+4k^{2}}.Now, overrightarrow {OM}cdot overrightarrow {ON} = x1x2 + y1y2 = frac {4m^{2}-12}{3+4k^{2}} + frac {3m^{2}-12k^{2}}{3+4k^{2}} = frac {7m^{2}-12k^{2}-12}{3+4k^{2}}.Since the line is tangent to the circle, we have frac {|m|}{ sqrt {k^{2}+1}}= sqrt { frac {12}{7}}, or 7m2-12k2=12.Substituting the value of 7m2-12k2 into the dot product, we get:overrightarrow {OM}cdot overrightarrow {ON} = 0.In conclusion, boxed{overrightarrow {OM}cdot overrightarrow {ON} = 0} is a constant value.

🧠:[<think>Okay, let me try to figure out this geometry problem. So, we have triangle ABC with medians AD and CF intersecting at point G. We know that angle AFG is 45 degrees and angle AGC is 60 degrees. We need to find the measure of angle ACF. The answer choices are 30°, 45°, 60°, or 75°.First, let me recall some properties of medians in a triangle. The point where the medians intersect is called the centroid, right? And the centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex. So, in this case, point G is the centroid of triangle ABC. That means AG is twice as long as GD, and CG is twice as long as GF.Given that AD and CF are medians, D is the midpoint of BC, and F is the midpoint of AB. So, AF = FB, and BD = DC.We are given two angles: angle AFG is 45°, and angle AGC is 60°. I need to relate these angles to find angle ACF. Let me try to sketch the triangle to visualize the problem.Let me draw triangle ABC. Let me label the vertices A, B, and C. Then, median AD connects vertex A to the midpoint D of BC, and median CF connects vertex C to the midpoint F of AB. The intersection of these two medians is centroid G.Since F is the midpoint of AB, AF = FB. Similarly, D is the midpoint of BC, so BD = DC.Now, angle AFG is 45°. Point F is the midpoint of AB, and point G is the centroid. So, angle AFG is the angle at point F between segments AF and FG. Wait, but AF is part of AB, and FG is part of the median CF? Wait, no. Let me think again. If F is the midpoint of AB, then CF is the median from C to AB. So, FG is the part of the median from F to G? Wait, but CF is the entire median from C to F (since F is the midpoint of AB). So, the median CF goes from C to F, and it is divided by centroid G into CG and GF, with CG:GF = 2:1.Similarly, median AD goes from A to D, and is divided by centroid G into AG:GD = 2:1.So, angle AFG is the angle at point F between points A, F, and G. So, triangle AFG has angle at F equal to 45°, and we need to relate this to angle AGC which is 60°, which is the angle at point G between points A, G, and C.Hmm. Maybe I should look at triangles AFG and AGC. Let me consider triangle AFG first. In triangle AFG, we know angle at F is 45°, and we might be able to find some side ratios if we consider the centroid properties. Similarly, in triangle AGC, we know angle at G is 60°, which might help.Alternatively, maybe using coordinate geometry? Assign coordinates to the triangle's vertices and use coordinates to find the angles. That might be a systematic approach.Let me try that. Let me place point A at (0, 0) for simplicity. Let me let AB be along the x-axis. Since F is the midpoint of AB, let me denote the coordinates of B as (2b, 0), so that F is at (b, 0). Then, point C can be at some coordinates (c, d). Then, midpoint D of BC would be at ((2b + c)/2, d/2).Since AD is a median from A to D, which is from (0,0) to ((2b + c)/2, d/2). The centroid G is the intersection of the medians AD and CF. The centroid divides each median into a 2:1 ratio, so coordinates of G can be calculated as the average of the vertices weighted by the centroid ratios.Wait, centroid coordinates are the average of the coordinates of the three vertices. So, if A is (0,0), B is (2b,0), and C is (c,d), then centroid G is at ((0 + 2b + c)/3, (0 + 0 + d)/3) = ((2b + c)/3, d/3). Hmm, that's another way to get G's coordinates.But maybe I can parameterize the problem. Let me let AB be of length 2 units, so that AF = FB = 1 unit. Then, setting coordinates: Let me place point A at (0,0), point B at (2,0), so midpoint F is at (1,0). Then, point C is somewhere in the plane, say (x,y). Then, midpoint D of BC is ((2 + x)/2, y/2).The centroid G is at ((0 + 2 + x)/3, (0 + 0 + y)/3) = ((2 + x)/3, y/3).Now, median CF connects point C (x,y) to point F (1,0). The centroid G is located along CF, 2/3 of the way from C to F. So, coordinates of G can also be found by moving from C towards F by 2/3 of the vector.So, G's coordinates would be (x + 2*(1 - x)/3, y + 2*(0 - y)/3) = (x + (2 - 2x)/3, y - (2y)/3) = ( (3x + 2 - 2x)/3 , (3y - 2y)/3 ) = ( (x + 2)/3 , y/3 ). Which matches the centroid formula. So, that checks out.Similarly, median AD connects A (0,0) to D ((2 + x)/2, y/2). The centroid G is 2/3 along AD from A to D. So, coordinates of G would be ( (2*( (2 + x)/2 ) + 0 ) / 3 , (2*( y/2 ) + 0 ) / 3 ) = ( ( (2 + x) + 0 ) / 3 , ( y + 0 ) / 3 ) = ( (2 + x)/3 , y/3 ). Again, same as centroid coordinates.So, coordinates are consistent. Now, angle AFG is 45°. Let's find angle at F (1,0) between points A (0,0), F (1,0), and G ((2 + x)/3, y/3).So, angle AFG is the angle at F between FA and FG.FA is the vector from F to A, which is (0 - 1, 0 - 0) = (-1, 0).FG is the vector from F to G, which is (( (2 + x)/3 - 1 ), ( y/3 - 0 )) = ( (2 + x - 3)/3, y/3 ) = ( (x -1)/3, y/3 ).The angle between vectors FA and FG is 45°. So, we can use the dot product formula to express this angle.The formula for the angle θ between two vectors u and v is:cosθ = (u · v) / (|u| |v|)Here, θ = 45°, so cosθ = √2/2.Vector FA is (-1, 0), vector FG is ( (x -1)/3, y/3 )Compute the dot product:u · v = (-1)*( (x -1)/3 ) + 0*( y/3 ) = - (x -1)/3The magnitude of u (FA) is sqrt( (-1)^2 + 0^2 ) = 1The magnitude of v (FG) is sqrt( ( (x -1)/3 )^2 + ( y/3 )^2 ) = (1/3) sqrt( (x -1)^2 + y^2 )So, putting into the formula:- (x -1)/3 = (1) * (1/3 sqrt( (x -1)^2 + y^2 )) * √2/2Multiply both sides by 3:- (x -1) = (1/ sqrt( (x -1)^2 + y^2 )) * √2/2 * 3Wait, no. Wait, the equation is:[ - (x -1)/3 ] = [ 1 * (1/3) sqrt( (x -1)^2 + y^2 ) ] * (√2 / 2 )Simplify both sides:Left side: - (x -1)/3Right side: ( √2 / 2 ) * (1/3) sqrt( (x -1)^2 + y^2 )Multiply both sides by 3:- (x -1) = ( √2 / 2 ) sqrt( (x -1)^2 + y^2 )Square both sides to eliminate the square root:Left side squared: (x -1)^2Right side squared: ( (√2 / 2 )^2 ) * ( (x -1)^2 + y^2 ) = (2 / 4 ) ( (x -1)^2 + y^2 ) = (1/2)( (x -1)^2 + y^2 )Set equal:(x -1)^2 = (1/2)( (x -1)^2 + y^2 )Multiply both sides by 2:2(x -1)^2 = (x -1)^2 + y^2Subtract (x -1)^2 from both sides:(x -1)^2 = y^2Take square roots:|x -1| = |y|So, either y = x -1 or y = -(x -1)But since point C is (x,y), and in a triangle, coordinates can be positive or negative depending on the position. However, since the centroid G has coordinates ( (2 + x)/3, y/3 ), and we have another condition involving angle AGC = 60°, we can use that to find which case applies.So, from angle AFG =45°, we get that |x -1| = |y|. So, either y = x -1 or y = 1 - x.Now, let's consider angle AGC =60°. Angle at G between points A, G, C.Point G is ((2 + x)/3, y/3). So, vectors GA and GC.Vector GA is from G to A: (0 - (2 + x)/3, 0 - y/3 ) = ( - (2 + x)/3, - y/3 )Vector GC is from G to C: (x - (2 + x)/3, y - y/3 ) = ( (3x - 2 - x)/3, (3y - y)/3 ) = ( (2x -2)/3, (2y)/3 )The angle between vectors GA and GC is 60°, so we can use the dot product formula again.Compute the dot product of GA and GC:[ - (2 + x)/3 ] * [ (2x -2)/3 ] + [ - y/3 ] * [ 2y /3 ]= [ - (2 + x)(2x -2) / 9 ] + [ -2y² / 9 ]= [ - ( (2)(2x -2) + x(2x -2) ) / 9 ] - 2y² /9First, expand (2 + x)(2x -2):= 2*(2x -2) + x*(2x -2) = 4x -4 + 2x² -2x = 2x² + 2x -4So, the first term becomes - (2x² + 2x -4)/9So, total dot product:= - (2x² + 2x -4)/9 - 2y²/9 = [ -2x² -2x +4 -2y² ] /9The magnitude of GA:|GA| = sqrt( [ - (2 + x)/3 ]² + [ - y/3 ]² ) = (1/3) sqrt( (2 + x)² + y² )Similarly, the magnitude of GC:|GC| = sqrt( [ (2x -2)/3 ]² + [ 2y /3 ]² ) = (1/3) sqrt( (2x -2)^2 + (2y)^2 ) = (1/3) sqrt(4(x -1)^2 +4y² ) = (2/3) sqrt( (x -1)^2 + y² )Therefore, the cosine of the angle between GA and GC is:[ (-2x² -2x +4 -2y²)/9 ] divided by [ (1/3 sqrt( (2 + x)^2 + y² )) * (2/3 sqrt( (x -1)^2 + y² )) ]Simplify denominator:(1/3)(2/3) sqrt( (2 + x)^2 + y² ) sqrt( (x -1)^2 + y² ) = (2/9) sqrt( [ (2 + x)^2 + y² ] [ (x -1)^2 + y² ] )So, the cosine of the angle (which is 60°, so cos60°=0.5) is:[ (-2x² -2x +4 -2y² ) /9 ] / [ (2/9) sqrt( [ (2 + x)^2 + y² ] [ (x -1)^2 + y² ] ) ]Simplify numerator divided by denominator:[ (-2x² -2x +4 -2y² ) /9 ] * [ 9 / (2 sqrt(...)) ] = [ (-2x² -2x +4 -2y² ) / (2 sqrt(...)) ] = [ - (2x² +2x -4 +2y² ) ] / (2 sqrt(...))This equals cos60° = 0.5So:[ - (2x² +2x -4 +2y² ) ] / (2 sqrt( [ (2 + x)^2 + y² ] [ (x -1)^2 + y² ] )) = 0.5Multiply both sides by 2 sqrt(...):- (2x² +2x -4 +2y² ) = sqrt( [ (2 + x)^2 + y² ] [ (x -1)^2 + y² ] )Square both sides:(2x² +2x -4 +2y² )² = [ (2 + x)^2 + y² ] [ (x -1)^2 + y² ]This seems complicated, but remember that from angle AFG, we have |x -1| = |y|. Let's substitute y² = (x -1)^2.Since y² = (x -1)^2, from the previous result.So, substituting y² = (x -1)^2 into the equation:Left side: (2x² +2x -4 +2(x -1)^2 )²First, expand 2(x -1)^2:2(x² -2x +1 ) = 2x² -4x +2So, left side becomes:[2x² +2x -4 +2x² -4x +2]² = [4x² -2x -2]²Right side: [ (2 + x)^2 + (x -1)^2 ] [ (x -1)^2 + (x -1)^2 ] = [ ( (2 + x)^2 + (x -1)^2 ) ] [ 2(x -1)^2 ]Compute (2 + x)^2 + (x -1)^2:= (4 +4x +x²) + (x² -2x +1) = 4 +4x +x² +x² -2x +1 = 2x² +2x +5So, right side is (2x² +2x +5) * 2(x -1)^2So, we have left side: (4x² -2x -2)^2Right side: (2x² +2x +5) * 2(x -1)^2Let me compute left side:(4x² -2x -2)^2 = (4x²)^2 + (-2x)^2 + (-2)^2 + 2*(4x²)(-2x) + 2*(4x²)(-2) + 2*(-2x)(-2)Wait, actually, better to expand as (a + b + c)^2, but maybe it's easier to factor 4x² -2x -2. Let's see.Alternatively, expand step by step:(4x² -2x -2)(4x² -2x -2) = 4x²*(4x²) + 4x²*(-2x) +4x²*(-2) + (-2x)*(4x²) + (-2x)*(-2x) + (-2x)*(-2) + (-2)*(4x²) + (-2)*(-2x) + (-2)*(-2)But this is tedious. Alternatively:Let me denote A = 4x² -2x -2Then, A^2 = (4x²)^2 + (-2x)^2 + (-2)^2 + 2*(4x²*(-2x)) + 2*(4x²*(-2)) + 2*(-2x*(-2))Wait, that's:16x^4 +4x² +4 + 2*(-8x³) + 2*(-8x²) + 2*(4x)=16x^4 +4x² +4 -16x³ -16x² +8xCombine like terms:16x^4 -16x³ + (4x² -16x²) +8x +4=16x^4 -16x³ -12x² +8x +4Left side:16x^4 -16x³ -12x² +8x +4Right side: (2x² +2x +5)*2(x -1)^2First compute (x -1)^2 =x² -2x +1Multiply by 2:2x² -4x +2Multiply by (2x² +2x +5):=2x²*(2x² +2x +5) + (-4x)*(2x² +2x +5) +2*(2x² +2x +5)=4x^4 +4x³ +10x² -8x³ -8x² -20x +4x² +4x +10Combine like terms:4x^4 + (4x³ -8x³) + (10x² -8x² +4x²) + (-20x +4x) +10=4x^4 -4x³ +6x² -16x +10So, right side is 4x^4 -4x³ +6x² -16x +10Set left side equal to right side:16x^4 -16x³ -12x² +8x +4 =4x^4 -4x³ +6x² -16x +10Bring all terms to left side:16x^4 -16x³ -12x² +8x +4 -4x^4 +4x³ -6x² +16x -10 =0Combine like terms:(16x^4 -4x^4) + (-16x³ +4x³) + (-12x² -6x²) + (8x +16x) + (4 -10)=012x^4 -12x³ -18x² +24x -6=0Divide both sides by 6:2x^4 -2x³ -3x² +4x -1=0So, now we have a quartic equation:2x^4 -2x³ -3x² +4x -1=0Hmm, solving quartic equations is complicated. Maybe we can factor this.Let me try rational roots. The possible rational roots are ±1, ±1/2.Test x=1:2(1)^4 -2(1)^3 -3(1)^2 +4(1) -1=2 -2 -3 +4 -1=0. So, x=1 is a root.So, (x -1) is a factor. Let's perform polynomial division or use synthetic division.Divide polynomial 2x^4 -2x³ -3x² +4x -1 by (x -1):Using synthetic division:Coefficients: 2 | -2 | -3 | 4 | -1Root x=1:Bring down 2.Multiply by 1: 2*1=2. Add to next coefficient: -2 +2=0.Multiply by1:0*1=0. Add to next coefficient: -3 +0=-3.Multiply by1:-3*1=-3. Add to next coefficient:4 +(-3)=1.Multiply by1:1*1=1. Add to last coefficient:-1 +1=0.So, the quotient is 2x³ +0x² -3x +1, so 2x³ -3x +1.Therefore, polynomial factors as (x -1)(2x³ -3x +1)=0Now, let's factor 2x³ -3x +1.Again, try rational roots. Possible roots: ±1, ±1/2.Test x=1:2(1)^3 -3(1) +1=2 -3 +1=0. So, x=1 is a root again.Therefore, (x -1) is a factor. Divide 2x³ -3x +1 by (x -1):Using synthetic division:Coefficients:2 | 0 | -3 |1Wait, note that the original polynomial is 2x³ +0x² -3x +1. So coefficients: 2,0,-3,1.Root x=1:Bring down 2.Multiply by1:2*1=2. Add to next coefficient:0 +2=2.Multiply by1:2*1=2. Add to next coefficient:-3 +2=-1.Multiply by1:-1*1=-1. Add to last coefficient:1 +(-1)=0.So, quotient is 2x² +2x -1.Thus, polynomial factors as (x -1)^2(2x² +2x -1)=0So, roots are x=1 (double root), and roots of 2x² +2x -1=0.Solve 2x² +2x -1=0:x = [-2 ± sqrt(4 +8)] /4 = [-2 ± sqrt(12)] /4 = [-2 ± 2*sqrt(3)] /4 = [ -1 ± sqrt(3) ] /2So, roots are x=1, x=( -1 + sqrt(3) )/2 ≈0.366, and x=( -1 - sqrt(3) )/2≈-1.366.Now, remember that in our coordinate system, point C is (x,y), and in the triangle, points A(0,0), B(2,0), C(x,y). To form a triangle, point C must not be on AB (unless degenerate). So, y must not be zero. From earlier, we have y²=(x -1)^2, so y= x -1 or y=1 -x.But let's check the possible x-values.First, x=1. If x=1, then y²=(1 -1)^2=0, so y=0. So, point C would be (1,0), but that's point F, which is the midpoint of AB. But then triangle ABC would be degenerate (all points colinear). So, x=1 is invalid.Next, x=( -1 + sqrt(3) )/2≈0.366. Then y²=(x -1)^2=( ( -1 + sqrt(3) )/2 -1 )^2=( ( -3 + sqrt(3) )/2 )^2=( ( sqrt(3) -3 )/2 )^2=( (3 - sqrt(3))^2 )/4=(9 -6sqrt(3) +3)/4=(12 -6sqrt(3))/4=(6 -3sqrt(3))/2So, y=±sqrt( (6 -3sqrt(3))/2 ). Let's compute sqrt( (6 -3sqrt(3))/2 ). Let's rationalize:sqrt( (6 -3sqrt(3))/2 )= sqrt(3*(2 -sqrt(3))/2 )= sqrt(3/2*(2 -sqrt(3)) )But this might not simplify nicely. Anyway, since we are dealing with coordinates, it's okay.Similarly, for x=( -1 - sqrt(3) )/2≈-1.366, then y²=(x -1)^2=( ( -1 - sqrt(3) )/2 -1 )^2=( ( -3 - sqrt(3) )/2 )^2=( (3 + sqrt(3))^2 )/4=(9 +6sqrt(3)+3)/4=(12 +6sqrt(3))/4=(6 +3sqrt(3))/2So, y=±sqrt( (6 +3sqrt(3))/2 ). Again, sqrt( (6 +3sqrt(3))/2 )=sqrt(3/2*(2 +sqrt(3)) )But we can check if these x-values make sense in the triangle.First, for x≈0.366, which is between 0 and 2, so point C is somewhere above or below the x-axis. Similarly, x≈-1.366 is left of point A (which is at 0,0). So, if point C is to the left of A, the triangle would still be valid as long as it's not colinear.But let's check angle AGC=60°. We need to see which of these roots satisfy the original equation.But solving this quartic equation was quite involved, and maybe there's a simpler approach.Alternatively, since we have |x -1| = |y|, so y = x -1 or y = 1 - x. Let's consider these two cases.Case 1: y = x -1Then, point C is (x, x -1). Let's see if this leads to angle AGC=60°.Coordinates of G would be ((2 +x)/3, (x -1)/3 )Vectors GA and GC:GA: from G to A: ( - (2 +x)/3, - (x -1)/3 )GC: from G to C: (x - (2 +x)/3, (x -1) - (x -1)/3 ) = ( (2x -2)/3, (2x -2)/3 )So, vectors GA = ( - (2 +x)/3, - (x -1)/3 )GC = ( (2x -2)/3, (2x -2)/3 )Compute the angle between GA and GC.Dot product: [ - (2 +x)/3 * (2x -2)/3 ] + [ - (x -1)/3 * (2x -2)/3 ]= [ - (2 +x)(2x -2) /9 - (x -1)(2x -2)/9 ]= [ - (2x -2)(2 +x +x -1) /9 ]Wait, factor out (2x -2):= [ - (2x -2)(2 +x +x -1) /9 ]= [ - (2x -2)(2x +1) /9 ]Alternatively, compute each term:First term: - (2 +x)(2x -2)/9Second term: - (x -1)(2x -2)/9Compute first term: - [2*(2x -2) +x*(2x -2) ] /9 = - [4x -4 +2x² -2x ] /9 = - [2x² +2x -4 ] /9Second term: - [x*(2x -2) -1*(2x -2) ] /9 = - [2x² -2x -2x +2 ] /9 = - [2x² -4x +2 ] /9Total dot product: (-2x² -2x +4 -2x² +4x -2)/9 = (-4x² +2x +2)/9Wait, no:Wait, first term: -2x² -2x +4 over 9Second term: -2x² +4x -2 over 9Add them:(-2x² -2x +4) + (-2x² +4x -2) = -4x² +2x +2So, total dot product: (-4x² +2x +2)/9The magnitudes:|GA| = sqrt( [ - (2 +x)/3 ]² + [ - (x -1)/3 ]² ) = (1/3) sqrt( (2 +x)^2 + (x -1)^2 )Compute (2 +x)^2 + (x -1)^2 =4 +4x +x² +x² -2x +1=2x² +2x +5So, |GA| = (1/3) sqrt(2x² +2x +5 )|GC| = sqrt( [ (2x -2)/3 ]² + [ (2x -2)/3 ]² ) = (1/3) sqrt( (2x -2)^2 + (2x -2)^2 ) = (1/3) sqrt(2*(2x -2)^2 ) = (1/3) * sqrt(2) |2x -2| = (sqrt(2)/3)|2x -2|So, |GC| = (sqrt(2)/3)|2x -2|The cosine of the angle between GA and GC is:[ (-4x² +2x +2)/9 ] / [ (1/3 sqrt(2x² +2x +5 )) * (sqrt(2)/3 |2x -2| ) ]Simplify denominator:(1/3)(sqrt(2)/3) sqrt(2x² +2x +5 ) |2x -2| = (sqrt(2)/9) sqrt(2x² +2x +5 ) |2x -2|So, cosine is:[ (-4x² +2x +2)/9 ] / [ (sqrt(2)/9) sqrt(2x² +2x +5 ) |2x -2| ) ] = [ (-4x² +2x +2) / sqrt(2) sqrt(2x² +2x +5 ) |2x -2| ) ]Set this equal to cos60°=0.5:[ (-4x² +2x +2) / (sqrt(2) sqrt(2x² +2x +5 ) |2x -2| ) ] = 0.5Multiply both sides by sqrt(2) sqrt(2x² +2x +5 ) |2x -2|:-4x² +2x +2 = 0.5 * sqrt(2) sqrt(2x² +2x +5 ) |2x -2|Square both sides:( -4x² +2x +2 )² = 0.25 * 2 * (2x² +2x +5 ) (2x -2)^2Simplify:Left side: ( -4x² +2x +2 )² = (4x² -2x -2 )² = same as before, 16x^4 -16x³ -12x² +8x +4Right side:0.25*2=0.5, so 0.5*(2x² +2x +5)(2x -2)^2Compute (2x -2)^2=4x² -8x +4Multiply by (2x² +2x +5):=4x²*(2x² +2x +5) -8x*(2x² +2x +5) +4*(2x² +2x +5)=8x^4 +8x³ +20x² -16x³ -16x² -40x +8x² +8x +20Combine terms:8x^4 + (8x³ -16x³) + (20x² -16x² +8x²) + (-40x +8x ) +20=8x^4 -8x³ +12x² -32x +20Multiply by 0.5:4x^4 -4x³ +6x² -16x +10Which matches the earlier right side. So, this equation is equivalent to the quartic we had earlier. Therefore, this case leads to the quartic equation. So, we need to check the roots.Previously, we found roots x=1 (invalid), x=( -1 ± sqrt(3) )/2.For case 1 (y =x -1), let's test x=( -1 + sqrt(3) )/2≈0.366.Then y=x -1≈0.366 -1≈-0.634. So, point C is at approximately (0.366, -0.634). Then, centroid G is at ((2 +x)/3, y/3)≈(2.366/3, -0.634/3 )≈(0.789, -0.211)Now, angle AGC is supposed to be 60°, let's see if this makes sense.Alternatively, let's check x=( -1 - sqrt(3) )/2≈-1.366. Then y=x -1≈-1.366 -1≈-2.366. So, point C is at (-1.366, -2.366). Then centroid G is at ((2 -1.366)/3, (-2.366)/3 )≈(0.634/3, -2.366/3 )≈(0.211, -0.789). Then, checking angle AGC here. However, since this is a mirror image, it might also satisfy the angle conditions, but we need to check which one is valid.Alternatively, maybe both roots are valid, but we need to see which one gives the required angle.But perhaps there's a simpler way. Since we are dealing with a triangle and angles, maybe using trigonometric properties or mass point geometry.Wait, another idea: use the properties of centroid and the given angles to apply the Law of Sines or Cosines in triangles AFG and AGC.Let me consider triangle AFG. In triangle AFG, angle at F is 45°, sides AF, FG, and AG.Since F is the midpoint of AB, AF = 1 (if AB=2). FG is a segment from F to centroid G. Since centroid divides the median CF into CG:GF=2:1, so FG = (1/3)CF. But we don't know the length of CF yet.Similarly, in triangle AGC, angle at G is 60°, and sides AG, GC, and AC.Given that AG is (2/3) of median AD, and GC is (2/3) of median CF.But perhaps we can assign variables to the lengths.Let me denote CF = 3m, so FG = m, CG =2m.Similarly, AD = 3n, so GD =n, AG=2n.In triangle AFG, sides are AF=1, FG=m, and AG=2n, with angle at F being 45°. Applying the Law of Cosines here:AG² = AF² + FG² - 2*AF*FG*cos(angle F)(2n)^2 = 1^2 + m^2 -2*1*m*cos45°4n² =1 +m² -2m*(√2/2 )4n² =1 +m² -√2 m ... (1)In triangle AGC, angle at G is 60°, sides AG=2n, GC=2m, and AC. Applying Law of Cosines here:AC² = AG² + GC² - 2*AG*GC*cos(angle G)AC² = (2n)^2 + (2m)^2 - 2*(2n)*(2m)*cos60°AC² =4n² +4m² -8nm*(0.5)AC² =4n² +4m² -4nm ... (2)Also, AC is a side of the triangle ABC, which can be related to other sides via coordinates or other means.Alternatively, since ABC is a triangle with AB=2, we can express AC and BC in terms of coordinates. But maybe it's better to find another relation between m and n.We need another equation relating m and n. Since AD and CF are medians, we can use Apollonius's theorem which states that the sum of the squares of the sides is equal to three times the sum of the squares of the medians.But in this case, we might need to relate the lengths of the medians AD and CF.Apollonius’s theorem: In any triangle, the sum of the squares of any two sides is equal to twice the square of the median to the third side plus twice the square of the half third side.For example, for median AD (to side BC):AB² + AC² = 2AD² + 2(BD² )Since BD = BC/2. But we don't know BC.Similarly, for median CF (to side AB):CA² + CB² = 2CF² + 2(AF² )Again, AF=1, so:CA² + CB² =2CF² +2(1)^2=2CF² +2But we might not have enough information. Alternatively, since we have two equations (1) and (2), and need another relation between m and n.But perhaps express AC² from equation (2) and relate it to another expression for AC².Alternatively, consider triangle AFC. Wait, point F is midpoint of AB, and we have some angles given.Alternatively, maybe using vectors or coordinate geometry again.Wait, perhaps in the coordinate system, since we have AF=1, AB=2, and point C(x,y) with y= x -1 or y=1 -x.Let me try with y= x -1.Then, as before, coordinates of G are ((2 +x)/3, (x -1)/3 )We need angle AGC=60°, which we translated into the quartic equation. The roots were x=( -1 ± sqrt(3) )/2.Let me take x=( -1 + sqrt(3) )/2≈0.366. Then, y=(x -1)= ( -1 + sqrt(3) )/2 -1=( -3 + sqrt(3) )/2≈-0.634.Now, compute angle ACF. Angle at point C between points A, C, F.Point A is (0,0), point C is (x,y), point F is (1,0).So, angle ACF is the angle at C between points A, C, F.To compute this angle, we can use vectors.Vector CA = A - C = (0 -x, 0 - y )=(-x, -y )Vector CF = F - C = (1 -x, 0 - y )=(1 -x, -y )The angle between vectors CA and CF at point C is angle ACF.Compute the cosine of the angle:cosθ=(CA · CF)/( |CA| |CF| )Compute the dot product:(-x)(1 -x) + (-y)(-y )= -x +x² + y²Compute |CA|=sqrt( x² + y² )|CF|=sqrt( (1 -x)^2 + y² )So,cosθ=( -x +x² + y² ) / ( sqrt(x² + y² ) sqrt( (1 -x)^2 + y² ) )Given that in this case, y= x -1, so y²=(x -1)^2.Substitute y²=(x -1)^2:cosθ=( -x +x² + (x -1)^2 ) / ( sqrt(x² + (x -1)^2 ) sqrt( (1 -x)^2 + (x -1)^2 ) )Simplify numerator:- x +x² +x² -2x +1=2x² -3x +1Denominator:sqrt(x² + (x -1)^2 ) * sqrt(2(1 -x)^2 )First, compute x² + (x -1)^2 =x² +x² -2x +1=2x² -2x +1So, sqrt(2x² -2x +1 )Then, sqrt(2(1 -x)^2 )=sqrt(2)|1 -x|Therefore,cosθ=(2x² -3x +1 ) / [ sqrt(2x² -2x +1 ) * sqrt(2)|1 -x| ]Simplify denominator:sqrt(2x² -2x +1 ) * sqrt(2)|1 -x|=sqrt(2) sqrt( (2x² -2x +1)(1 -x)^2 )Note that (1 -x)^2= (x -1)^2, so denominator becomes sqrt(2) * sqrt( (2x² -2x +1)(x -1)^2 )= sqrt(2) * |x -1| sqrt(2x² -2x +1 )So,cosθ=(2x² -3x +1 ) / [ sqrt(2) |x -1| sqrt(2x² -2x +1 ) ]Now, plug in x=( -1 + sqrt(3) )/2≈0.366. Let's compute numerator and denominator.First, compute x=( -1 + sqrt(3) )/2≈0.366.Numerator:2x² -3x +1Compute x²=( ( -1 + sqrt(3) )² )/4=(1 -2sqrt(3) +3)/4=(4 -2sqrt(3))/4=(2 -sqrt(3))/2≈(2 -1.732)/2≈0.134So, 2x²=2*(2 -sqrt(3))/2=2 -sqrt(3)Then, 2x² -3x +1= (2 -sqrt(3)) -3*( (-1 + sqrt(3))/2 ) +1=2 -sqrt(3) + (3 -3sqrt(3))/2 +1Convert to common denominator:= (4 -2sqrt(3))/2 + (3 -3sqrt(3))/2 +2/2= (4 -2sqrt(3) +3 -3sqrt(3) +2)/2=(9 -5sqrt(3))/2≈(9 -8.66)/2≈0.34/2≈0.17Denominator:sqrt(2)*|x -1|*sqrt(2x² -2x +1 )Compute |x -1|=1 -x=1 - ( -1 + sqrt(3) )/2=(2 +1 -sqrt(3))/2=(3 -sqrt(3))/2≈(3 -1.732)/2≈0.634Compute sqrt(2x² -2x +1 )First, compute 2x² -2x +1:We already have x²=(2 -sqrt(3))/2, so 2x²=2 -sqrt(3)Then, -2x= -2*( (-1 + sqrt(3))/2 )=1 -sqrt(3)Thus, 2x² -2x +1= (2 -sqrt(3)) + (1 -sqrt(3)) +1=4 -2sqrt(3)≈4 -3.464≈0.536So sqrt(2x² -2x +1 )≈sqrt(0.536 )≈0.732Therefore, denominator≈sqrt(2)*0.634*0.732≈1.414*0.634*0.732≈1.414*0.464≈0.658So, cosθ≈0.17/0.658≈0.258Which is approximately 75°, since cos75≈0.2588. So, angle ACF≈75°, which is option D.Alternatively, let's compute it exactly.Numerator: (2x² -3x +1 )With x=( -1 + sqrt(3) )/2:2x²=2*( (sqrt(3) -1 )² /4 )= (3 -2sqrt(3) +1 )/2=(4 -2sqrt(3))/2=2 -sqrt(3)-3x= -3*( (-1 + sqrt(3))/2 )=(3 -3sqrt(3))/2So, numerator=2 -sqrt(3) + (3 -3sqrt(3))/2 +1= (2 +1) + ( -sqrt(3) ) + (3 -3sqrt(3))/2=3 -sqrt(3) + (3/2 - (3sqrt(3))/2 )=3 +3/2 -sqrt(3) - (3sqrt(3))/2=4.5 - (5sqrt(3))/2Wait, this seems messy. Maybe there's an error.Wait, let's re-express numerator:2x² -3x +1=2x² -3x +1With x=(sqrt(3) -1)/2:Compute 2x²:x=(sqrt(3)-1)/2, x²=( (sqrt(3)-1)^2 )/4=(3 -2sqrt(3) +1)/4=(4 -2sqrt(3))/4=(2 -sqrt(3))/2Thus, 2x²=2*(2 -sqrt(3))/2=2 -sqrt(3)-3x= -3*(sqrt(3)-1)/2= (-3sqrt(3) +3)/2Thus, numerator=2 -sqrt(3) + (-3sqrt(3) +3)/2 +1Convert to common denominator of 2:= (4 -2sqrt(3) -3sqrt(3) +3 +2)/2= (4 +3 +2 -5sqrt(3))/2= (9 -5sqrt(3))/2Denominator:sqrt(2)*|x -1|*sqrt(2x² -2x +1 )|x -1|=1 -x=1 - (sqrt(3)-1)/2=(2 -sqrt(3)+1)/2=(3 -sqrt(3))/2sqrt(2x² -2x +1 ):Compute 2x² -2x +1=2x² -2x +1We have 2x²=2 -sqrt(3)-2x= -2*(sqrt(3)-1)/2= - (sqrt(3)-1 )= -sqrt(3)+1So, 2x² -2x +1=2 -sqrt(3) -sqrt(3) +1 +1=4 -2sqrt(3)So, sqrt(2x² -2x +1 )=sqrt(4 -2sqrt(3))Factor inside sqrt:4 -2sqrt(3). Let's see if this is a square.Suppose sqrt(a - b)=sqrt(c) -sqrt(d). Then, a -b= c +d -2sqrt(cd). Comparing:Let me try to write 4 -2sqrt(3) as (sqrt(3) -1)^2=3 +1 -2sqrt(3)=4 -2sqrt(3). Yes!So, sqrt(4 -2sqrt(3))=sqrt( (sqrt(3)-1)^2 )=sqrt(3)-1Therefore, denominator= sqrt(2)*(3 -sqrt(3))/2*(sqrt(3)-1 )Compute this:sqrt(2)/2*(3 -sqrt(3))(sqrt(3) -1 )First compute (3 -sqrt(3))(sqrt(3) -1 )=3*sqrt(3) -3*1 -sqrt(3)*sqrt(3) +sqrt(3)*1=3sqrt(3) -3 -3 +sqrt(3)=4sqrt(3) -6So, denominator= sqrt(2)/2*(4sqrt(3) -6 )=sqrt(2)*(2sqrt(3) -3 )Thus, cosθ= (9 -5sqrt(3))/2 divided by sqrt(2)*(2sqrt(3) -3 )Multiply numerator and denominator by sqrt(2):= (9 -5sqrt(3))/2 * sqrt(2) / (2sqrt(3) -3 )sqrt(2)*sqrt(2)Wait, no, to rationalize denominator:Wait, cosθ= (9 -5sqrt(3))/2 / [ sqrt(2)*(2sqrt(3) -3 ) ]Multiply numerator and denominator by sqrt(2):= (9 -5sqrt(3))/2 * sqrt(2) / [ (2sqrt(3) -3 ) *2 ]Wait, no:Wait, denominator is sqrt(2)*(2sqrt(3)-3). Multiply numerator and denominator by sqrt(2):Numerator becomes (9 -5sqrt(3))/2 * sqrt(2)Denominator becomes (2sqrt(3)-3)*2So,cosθ= [ (9 -5sqrt(3))sqrt(2) /2 ] / [ 2(2sqrt(3)-3) ]= (9 -5sqrt(3))sqrt(2) / [4(2sqrt(3)-3) ]This expression seems complicated, but maybe we can rationalize it further.Multiply numerator and denominator by (2sqrt(3)+3):= [ (9 -5sqrt(3))sqrt(2)(2sqrt(3)+3) ] / [4( (2sqrt(3)-3)(2sqrt(3)+3) ) ]Denominator:(2sqrt(3))² -3²=12 -9=3So, denominator=4*3=12Numerator:(9 -5sqrt(3))(2sqrt(3)+3)sqrt(2)First compute (9 -5sqrt(3))(2sqrt(3)+3):=9*2sqrt(3) +9*3 -5sqrt(3)*2sqrt(3) -5sqrt(3)*3=18sqrt(3) +27 -10*3 -15sqrt(3)=18sqrt(3) +27 -30 -15sqrt(3)=3sqrt(3) -3Thus, numerator= (3sqrt(3) -3 )sqrt(2)=3(sqrt(3)-1)sqrt(2)Therefore,cosθ=3(sqrt(3)-1)sqrt(2)/12= (sqrt(3)-1)sqrt(2)/4Simplify sqrt(2)(sqrt(3)-1)/4= (sqrt(6)-sqrt(2))/4Now, cosθ=(sqrt(6)-sqrt(2))/4≈(2.449 -1.414)/4≈1.035/4≈0.25875, which matches our approximate value earlier.Now, cosθ≈0.2588 corresponds to θ≈75°, since cos75°≈0.2588.Therefore, angle ACF is 75°, which is option D.But wait, let's check the other case where y=1 -x.Case 2: y=1 -x.Then, point C is (x,1 -x). Let's see if this leads to angle AGC=60°.Coordinates of G would be ((2 +x)/3, (1 -x)/3 )Vectors GA and GC:GA: from G to A: ( - (2 +x)/3, - (1 -x)/3 )GC: from G to C: (x - (2 +x)/3, (1 -x) - (1 -x)/3 ) = ( (2x -2)/3, (2 -2x)/3 )Compute the angle between GA and GC.Dot product:[ - (2 +x)/3 * (2x -2)/3 ] + [ - (1 -x)/3 * (2 -2x)/3 ]First term: - (2 +x)(2x -2)/9Second term: - (1 -x)(2 -2x)/9Expand both terms:First term: - [4x -4 +2x² -2x ]/9 = - [2x² +2x -4 ]/9Second term: - [2 -2x -2x +2x² ]/9 = - [2x² -4x +2 ]/9Total dot product: (-2x² -2x +4 -2x² +4x -2)/9 = (-4x² +2x +2)/9 (same as in case 1)The magnitudes:|GA|=sqrt( [ - (2 +x)/3 ]² + [ - (1 -x)/3 ]² )=(1/3)sqrt( (2 +x)^2 + (1 -x)^2 )=(1/3)sqrt(4 +4x +x² +1 -2x +x² )=(1/3)sqrt(2x² +2x +5 )|GC|=sqrt( [ (2x -2)/3 ]² + [ (2 -2x)/3 ]² )=(1/3)sqrt( (2x -2)^2 + (2 -2x)^2 )=(1/3)sqrt(4x² -8x +4 +4 -8x +4x² )=(1/3)sqrt(8x² -16x +8 )=(1/3)sqrt(8(x² -2x +1 ))=(1/3)sqrt(8(x -1)^2 )=(1/3)(2sqrt(2)|x -1| )=(2sqrt(2)/3)|x -1|Thus, the cosine of the angle is:[ (-4x² +2x +2)/9 ] / [ (1/3 sqrt(2x² +2x +5 )) * (2sqrt(2)/3 |x -1| ) ]Simplify:= [ (-4x² +2x +2)/9 ] / [ (2sqrt(2)/9 sqrt(2x² +2x +5 ) |x -1| ) ]= (-4x² +2x +2)/ (2sqrt(2) sqrt(2x² +2x +5 ) |x -1| )This is the same expression as in case 1, so setting it equal to 0.5 gives the same quartic equation. Therefore, the roots will be the same, x=( -1 ± sqrt(3) )/2. For case 2, y=1 -x.Take x=( -1 + sqrt(3) )/2≈0.366, then y=1 -x≈0.634. So, point C is (0.366,0.634). Compute angle ACF.Vectors CA = A - C = (-0.366, -0.634 )Vector CF = F - C = (1 -0.366, 0 -0.634 )=(0.634, -0.634 )Dot product CA · CF = (-0.366)(0.634) + (-0.634)(-0.634 )≈-0.232 +0.402≈0.17|CA|=sqrt(0.366² +0.634² )≈sqrt(0.134 +0.402 )≈sqrt(0.536 )≈0.732|CF|=sqrt(0.634² +0.634² )≈sqrt(0.402 +0.402 )≈sqrt(0.804 )≈0.897cosθ=0.17/(0.732*0.897 )≈0.17/0.656≈0.259, which is again approximately 75°, so angle ACF≈75°.Thus, regardless of whether y=x -1 or y=1 -x, angle ACF is approximately 75°, so the answer should be D) 75°.However, let me verify with the other root x=( -1 - sqrt(3) )/2≈-1.366. For case 2, y=1 -x≈1 -(-1.366 )≈2.366. Then, point C is (-1.366,2.366 ). Compute angle ACF.Vectors CA = (-(-1.366 ), -2.366 )=(1.366, -2.366 )Vector CF = F - C = (1 -(-1.366 ),0 -2.366 )=(2.366, -2.366 )Dot product:1.366*2.366 + (-2.366)*(-2.366 )≈3.232 +5.598≈8.83|CA|=sqrt(1.366² +2.366² )≈sqrt(1.866 +5.598 )≈sqrt(7.464 )≈2.732|CF|=sqrt(2.366² +2.366² )≈sqrt(5.598 +5.598 )≈sqrt(11.196 )≈3.346cosθ=8.83/(2.732*3.346 )≈8.83/9.15≈0.965, which corresponds to angle≈15°, which is not one of the options. So, this root likely doesn't satisfy the original angle conditions, or corresponds to a different configuration. Therefore, only the roots where x≈0.366 give valid triangles with the required angles, leading to angle ACF≈75°.Therefore, the answer is D) 75°.Final Answerboxed{D}</think>Given triangle ( triangle ABC ) with medians ( AD ) and ( CF ) intersecting at centroid ( G ). We know ( angle AFG = 45^circ ) and ( angle AGC = 60^circ ). We need to find the measure of ( angle ACF ).1. Centroid Properties: The centroid ( G ) divides each median into a ratio of 2:1. Thus, ( AG = 2GD ) and ( CG = 2GF ).2. Coordinate System: Place ( A ) at ( (0, 0) ), ( B ) at ( (2, 0) ), making ( F ) the midpoint of ( AB ) at ( (1, 0) ). Let ( C ) be ( (x, y) ).3. Angle AFG: Using the dot product for vectors ( overrightarrow{FA} ) and ( overrightarrow{FG} ), we find ( |x - 1| = |y| ). Thus, ( y = x - 1 ) or ( y = 1 - x ).4. Angle AGC: Using the dot product for vectors ( overrightarrow{GA} ) and ( overrightarrow{GC} ), we derive the quartic equation ( 2x^4 - 2x^3 - 3x^2 + 4x - 1 = 0 ). Solving this, we find valid roots ( x = frac{-1 pm sqrt{3}}{2} ).5. Valid Root Check: For ( x = frac{-1 + sqrt{3}}{2} approx 0.366 ), ( y = x - 1 approx -0.634 ) or ( y = 1 - x approx 0.634 ). Both cases lead to valid triangles.6. Angle Calculation: Using vectors ( overrightarrow{CA} ) and ( overrightarrow{CF} ), the cosine of ( angle ACF ) is found to be approximately ( 0.2588 ), corresponding to ( 75^circ ).Thus, the degree measure of ( angle ACF ) is (boxed{D}).,<think>Okay, so I need to solve this geometry problem about triangle ABC with medians AD and CF intersecting at point G. The given angles are angle AFG = 45 degrees and angle AGC = 60 degrees. I need to find the measure of angle ACF. The answer choices are 30°, 45°, 60°, or 75°. Hmm, let's start by drawing a diagram to visualize the problem.First, let me recall that in a triangle, the medians intersect at the centroid, which divides each median into a ratio of 2:1, with the longer segment being closer to the vertex. So, point G is the centroid here since it's the intersection of medians AD and CF. That means AG:GD = 2:1 and CG:GF = 2:1.Now, let's label the triangle ABC. Let me assume point A is at the top, B is at the bottom left, and C is at the bottom right. Then, median AD goes from A to the midpoint D of BC, and median CF goes from C to the midpoint F of AB. They intersect at centroid G.Given that angle AFG is 45°, and angle AGC is 60°, we need to find angle ACF. Let me label these angles on the diagram.First, angle AFG: point A, F, G. So, AFG is the angle at point F between points A, F, G. Wait, no. Angle AFG is at point F, right? Because the middle letter is the vertex. So angle at F between A, F, G. Similarly, angle AGC is at point G between A, G, C.Wait, but in the problem statement, it's written as angle AFG = 45°, so that's angle at F, and angle AGC = 60°, angle at G.Hmm. Let me confirm the notation: angle AFG is ∠AFG, so vertex at F, with sides FA and FG. Similarly, angle AGC is ∠AGC, vertex at G, sides AG and GC.Okay. So AFG is 45°, which is the angle at F between FA and FG. Since F is the midpoint of AB, AF is half of AB. Also, FG is a segment from F to centroid G. Since centroid divides the median CF into 2:1, so CG:GF = 2:1, which means FG is one-third of CF. But maybe I need to find some lengths or use trigonometry here.Similarly, angle AGC is 60°, which is at G between AG and GC. Since centroid divides the medians into 2:1, AG is twice as long as GD, and CG is twice as long as GF. Maybe we can apply the Law of Sines or Cosines in some triangles here.Let me consider triangle AFG first. In triangle AFG, we know angle at F is 45°, and we might need to relate the sides. But we don't know any side lengths. Similarly, in triangle AGC, angle at G is 60°, again, maybe relating sides using Law of Sines or Cosines.Alternatively, since medians are involved, perhaps coordinate geometry could be useful. Let me try setting coordinates for the triangle to compute the angles.Let's place point A at (0, 0) for simplicity. Let's let AB be along the x-axis, so point B is at (2b, 0), making F the midpoint at (b, 0). Then, point C is somewhere in the plane, say (c, d). Then, the midpoint D of BC would be at ((2b + c)/2, d/2). Then, centroid G is the intersection of AD and CF.But maybe choosing specific coordinates would make the problem easier. Let me choose coordinates such that point F is at (0, 0) to simplify calculations. Wait, but if I place point A at (-a, 0) and point B at (a, 0), then midpoint F is at (0, 0). Then point C can be at some (0, c) since if I place the midpoint at (0,0), but wait, no. If AB is from (-a, 0) to (a, 0), then midpoint F is at (0, 0). Then point C is somewhere else, say (d, e). Then, the midpoint D of BC would be at ((a + d)/2, (0 + e)/2) = ((a + d)/2, e/2). Then, centroid G is the intersection of AD and CF.But maybe this is getting too complicated. Alternatively, let's assign coordinates such that centroid G is at the origin. But maybe that complicates things. Alternatively, let's use vectors.Alternatively, let's consider the properties of medians and centroids. Since centroid divides the medians in 2:1 ratio, so in triangle AFG, FG is 1/3 of CF, and AF is 1/2 of AB.Wait, maybe I can consider mass point geometry. But I might not remember all the details. Alternatively, use coordinate geometry with specific values.Let me try coordinate geometry. Let me place point F at the origin (0,0). Then, since F is the midpoint of AB, let me let A be at (-a, 0) and B at (a, 0). Then, point C is somewhere else, say (0, c) to make it simple. Then, the midpoint D of BC is ((a + 0)/2, (0 + c)/2) = (a/2, c/2). Then, centroid G is the intersection of AD and CF.First, let's write equations for the medians AD and CF.Median AD goes from A (-a, 0) to D (a/2, c/2). The parametric equation for AD can be written as:x = -a + t( (a/2 + a) ) = -a + t(3a/2)y = 0 + t(c/2 - 0) = t(c/2)where t ranges from 0 to 1.Similarly, median CF goes from C (0, c) to F (0, 0). Wait, no. Wait, point F is the midpoint of AB, which is (0,0) in this coordinate system. So median CF is the line from C (0, c) to F (0, 0). Wait, that's a vertical line along the y-axis. Then centroid G is the intersection of AD and CF.But CF is the line x = 0 (since it goes from (0, c) to (0, 0)), and AD is parameterized as x = -a + (3a/2)t, y = (c/2)t.To find their intersection G, set x = 0:0 = -a + (3a/2)t => (3a/2)t = a => t = (a) / (3a/2) = 2/3.So t = 2/3, so y = (c/2)(2/3) = c/3. Therefore, centroid G is at (0, c/3).So in this coordinate system, centroid G is (0, c/3).Now, angle AFG is 45°, which is angle at F (0,0) between points A (-a,0), F (0,0), and G (0, c/3).So angle AFG is the angle between vectors FA and FG.Vector FA is from F to A: (-a, 0) - (0,0) = (-a, 0).Vector FG is from F to G: (0, c/3) - (0,0) = (0, c/3).Therefore, the angle between FA and FG is the angle between (-a, 0) and (0, c/3). The angle between these two vectors is 90°, because one is along the negative x-axis and the other is along the positive y-axis. But the problem states angle AFG is 45°, which contradicts this. Therefore, my coordinate system assumption must be wrong.Wait, so in this coordinate system, angle AFG is 90°, but the problem says it's 45°, so something is wrong with my setup.Hmm, so perhaps placing point C at (0, c) makes angle AFG 90°, but in the problem it's 45°, so maybe my coordinate system is not suitable. I need to adjust the coordinates so that angle AFG is 45°. Let me try a different approach.Let me instead place point F at the origin (0,0), and let me assign coordinates such that vector FA is along the positive x-axis and vector FG is in some direction making 45° with FA. So if FA is along the positive x-axis, then point A is at (a, 0), and since F is the midpoint of AB, point B would be at (-a, 0). Then, point F is at (0, 0). Now, angle AFG is 45°, so FG makes a 45° angle with FA. Since FA is from F(0,0) to A(a,0), which is along the positive x-axis. Therefore, FG is a vector making 45° above the x-axis. Let's denote point G as (g, g), since it makes a 45° angle with the x-axis, so the coordinates would be (g, g) for some g > 0.But G is also on the median CF. Let's figure out where point C is. Since CF is the median from C to F(0,0), midpoint of AB. Let me denote point C as (c, d). Then, the median CF is the line from C(c, d) to F(0,0). The centroid G is located at 2/3 of the median from C to F. Wait, centroid divides the median so that CG:GF = 2:1. So if the entire median CF is from C to F, then G is located 2/3 of the way from C to F. Therefore, coordinates of G would be:G_x = c + (0 - c)*(2/3) = c - (2c/3) = c/3G_y = d + (0 - d)*(2/3) = d - (2d/3) = d/3But we also have that G is at (g, g) from the previous consideration (since angle AFG is 45°). Therefore:c/3 = gd/3 = gThus, c = 3g, d = 3g. Therefore, point C is at (3g, 3g).Now, point G is at (g, g).Now, let's verify angle AFG is 45°. FA is from F(0,0) to A(a, 0), which is along the positive x-axis. FG is from F(0,0) to G(g, g). The angle between FA (x-axis) and FG (vector (g, g)) is indeed 45°, since the slope is 1. So that's correct.Now, we also know that angle AGC is 60°. Let's find the coordinates of point A. Since F is the midpoint of AB, and F is at (0,0), then point A is at (a, 0) and point B is at (-a, 0). So AB is from (-a, 0) to (a, 0), midpoint at F(0,0). Then, centroid G is at (g, g) as above, and point C is at (3g, 3g).Now, angle AGC is 60°. Let's find angle at G between points A, G, C.Point A is at (a, 0), G is at (g, g), and C is at (3g, 3g). So vectors GA and GC.Vector GA is from G to A: (a - g, 0 - g) = (a - g, -g)Vector GC is from G to C: (3g - g, 3g - g) = (2g, 2g)The angle between vectors GA and GC is 60°, so we can use the dot product formula:cos(theta) = (GA • GC) / (|GA| |GC|)theta = 60°, so cos(theta) = 0.5Compute GA • GC: (a - g)(2g) + (-g)(2g) = 2g(a - g) - 2g^2 = 2ga - 2g^2 - 2g^2 = 2ga - 4g^2|GA| = sqrt( (a - g)^2 + (-g)^2 ) = sqrt( (a - g)^2 + g^2 )|GC| = sqrt( (2g)^2 + (2g)^2 ) = sqrt(8g^2) = 2g*sqrt(2)So, putting into the formula:(2ga - 4g^2) / ( sqrt( (a - g)^2 + g^2 ) * 2g*sqrt(2) ) ) = 0.5Simplify numerator and denominator:Left-hand side (LHS):(2ga - 4g^2) / (2g*sqrt(2) * sqrt( (a - g)^2 + g^2 )) )Factor numerator: 2g(a - 2g)Denominator: 2g*sqrt(2)*sqrt( (a - g)^2 + g^2 )Cancel 2g in numerator and denominator:(a - 2g) / (sqrt(2)*sqrt( (a - g)^2 + g^2 )) ) = 0.5Multiply both sides by sqrt(2)*sqrt( (a - g)^2 + g^2 ):(a - 2g) = 0.5 * sqrt(2) * sqrt( (a - g)^2 + g^2 )Square both sides to eliminate square roots:(a - 2g)^2 = (0.5)^2 * 2 * ( (a - g)^2 + g^2 )Simplify right-hand side (RHS):0.25 * 2 * ( (a - g)^2 + g^2 ) = 0.5 * ( (a - g)^2 + g^2 )Left-hand side (LHS):(a - 2g)^2 = a^2 -4ag +4g^2RHS:0.5*( (a^2 - 2ag + g^2) + g^2 ) = 0.5*(a^2 -2ag + 2g^2) = 0.5a^2 - ag + g^2Set equal:a^2 -4ag +4g^2 = 0.5a^2 - ag + g^2Subtract RHS from both sides:a^2 -4ag +4g^2 -0.5a^2 + ag -g^2 = 0Simplify:0.5a^2 -3ag +3g^2 = 0Multiply both sides by 2 to eliminate decimal:a^2 -6ag +6g^2 = 0This is a quadratic equation in terms of a. Let's solve for a:a^2 -6g a +6g^2 =0Using quadratic formula:a = [6g ± sqrt( (6g)^2 -4*1*6g^2 ) ] / 2Compute discriminant:(36g^2) -24g^2 =12g^2So,a = [6g ± sqrt(12g^2)] /2 = [6g ± 2g*sqrt(3)] /2 = [6g ± 2g√3]/2 = 3g ± g√3Therefore, a = g(3 ± √3)But since point A is at (a, 0) and we have centroid G at (g, g), and point C at (3g, 3g). We need to ensure that the triangle is correctly oriented.Assuming a positive value for a, and since in our coordinate system, point A is at (a, 0) and centroid G is at (g, g). If a = g(3 + √3), then that's a larger value, or a = g(3 - √3). Let's check both possibilities.First, a = g(3 + √3). Then, point A is at (g(3 + √3), 0). Point G is at (g, g). Then, the vector GA is (g(3 + √3) - g, 0 - g) = (g(2 + √3), -g). The vector GC is (2g, 2g) as before.But if a = g(3 - √3), then point A is at (g(3 - √3), 0). Then, vector GA is (g(3 - √3) - g, -g) = (g(2 - √3), -g). Either way, we need to check which solution makes sense.But since angle AGC is 60°, both solutions may be possible, but we need to check consistency with the triangle.But in our coordinate system, point C is at (3g, 3g). Let's see if with a = g(3 - √3), then point A is at (g(3 - √3), 0). Let's compute the distance from A to C.AC distance would be sqrt( (3g - g(3 - √3))^2 + (3g - 0)^2 ) = sqrt( (3g -3g +g√3)^2 +9g^2 ) = sqrt( (g√3)^2 +9g^2 ) = sqrt(3g^2 +9g^2 )= sqrt(12g^2)= 2g√3.Similarly, with a =g(3 +√3), distance AC would be sqrt( (3g -g(3 +√3))^2 + (3g)^2 )= sqrt( (-g√3)^2 +9g^2 )= sqrt(3g^2 +9g^2)=sqrt(12g^2)=2g√3. So both give the same distance, which is okay.But we need to check if the angle at G is indeed 60°. Wait, but we already derived the angle condition, so both solutions should satisfy angle AGC =60°, but perhaps one leads to an impossible triangle.Alternatively, perhaps both are valid, but we can proceed with either.But perhaps the ratio between a and g will influence the other angles. However, since we need to find angle ACF, which is the angle at point C between points A, C, and F. Let's compute that angle.First, let's recall the coordinates:Point A: (a, 0) = either (g(3 +√3), 0) or (g(3 -√3), 0)Point C: (3g, 3g)Point F: (0,0) (midpoint of AB)So angle ACF is the angle at point C between points A, C, and F. So vectors CA and CF.Vector CA: from C to A: (a -3g, 0 -3g) = (a -3g, -3g)Vector CF: from C to F: (0 -3g, 0 -3g) = (-3g, -3g)We need to find the angle between vectors CA and CF at point C.So angle ACF is the angle between vectors CA and CF. Let's compute the cosine of that angle using the dot product.cos(theta) = (CA • CF) / (|CA| |CF| )First, compute the dot product:CA • CF = (a -3g)(-3g) + (-3g)(-3g) = -3g(a -3g) +9g^2 = -3ga +9g^2 +9g^2 = -3ga +18g^2Compute |CA|:|CA| = sqrt( (a -3g)^2 + (-3g)^2 ) = sqrt( (a -3g)^2 +9g^2 )Compute |CF|:|CF| = sqrt( (-3g)^2 + (-3g)^2 ) = sqrt(9g^2 +9g^2 )= sqrt(18g^2)= 3g√2Therefore,cos(theta) = [ -3ga +18g^2 ] / [ sqrt( (a -3g)^2 +9g^2 ) * 3g√2 ]Simplify numerator:Factor out 3g:3g( -a +6g )Denominator:3g√2 * sqrt( (a -3g)^2 +9g^2 )Cancel 3g:[ -a +6g ] / [ √2 * sqrt( (a -3g)^2 +9g^2 ) ]So,cos(theta) = (6g -a ) / [ √2 * sqrt( (a -3g)^2 +9g^2 ) ]Now, recall that a =3g ±g√3. Let's substitute both possibilities.First, take a =3g +g√3:cos(theta) = (6g -3g -g√3 ) / [ √2 * sqrt( (3g +g√3 -3g)^2 +9g^2 ) ]Simplify numerator:6g -3g -g√3 =3g -g√3 =g(3 -√3)Denominator:√2 * sqrt( (g√3)^2 +9g^2 ) =√2 * sqrt(3g^2 +9g^2 )=√2 * sqrt(12g^2 )=√2 * 2g√3=2g√6Therefore,cos(theta)= g(3 -√3)/ (2g√6 )= (3 -√3)/(2√6 )Rationalize denominator:Multiply numerator and denominator by √6:(3 -√3)√6 / (2*6 )= (3√6 -√18)/12= (3√6 -3√2)/12= [3(√6 -√2)]/12= (√6 -√2)/4 ≈ (2.449 -1.414)/4≈1.035/4≈0.25875Which corresponds to angle theta≈75°, since cos(75°)≈0.2588Alternatively, take a=3g -g√3:cos(theta)= (6g -3g +g√3 ) / [√2 * sqrt( (3g -g√3 -3g)^2 +9g^2 ) ]Simplify numerator:6g -3g +g√3=3g +g√3= g(3 +√3)Denominator:√2 * sqrt( (-g√3)^2 +9g^2 )=√2 * sqrt(3g^2 +9g^2 )= same as before=2g√6Therefore,cos(theta)=g(3 +√3)/(2g√6 )=(3 +√3)/(2√6 )Rationalize denominator:(3 +√3)√6 / (2*6 )= (3√6 +√18)/12= (3√6 +3√2)/12= [3(√6 +√2)]/12= (√6 +√2)/4≈(2.449 +1.414)/4≈3.863/4≈0.9659Which corresponds to angle theta≈15°, since cos(15°)≈0.9659But the answer choices are 30°,45°,60°,75°. So theta is either ≈75° or ≈15°. 15° is not among the options, so the correct answer must be 75°, which is option D.Wait, but the problem is asking for angle ACF. In our coordinate system, angle at C between points A, C, F. So if theta≈75°, then the answer is D)75°. However, let me verify with another approach.Alternatively, perhaps using mass point or properties of medians and angles.But according to the coordinate geometry approach, angle ACF is 75°, which is option D. However, the answer options are given, and 75° is an option. But let me check again the problem statement.Wait, the problem says "the degree measure of angle ACF". In the problem statement, angle ACF is the angle at point C between points A, C, and F. In our coordinate system, point C is at (3g, 3g), point A is at (a,0), point F is at (0,0).From the calculation, angle ACF is either approximately 75° or 15°, but only 75° is among the options, so the answer should be D)75°.But wait, the problem states angle AGC=60°, which we used in our coordinate system. But let me confirm again.Wait, in the problem statement, angle AGC=60°, which is the angle at G between points A, G, C. In our coordinate system, with G at (g, g), that angle was used to derive the value of a in terms of g. And in that case, we found two possibilities for a, leading to angle ACF being 75° or 15°, but only 75° is an option. So the answer is D)75°.But let me check another way. Suppose angle ACF is 75°, then the answer is D. However, I recall that in a triangle, the angles sum up to 180°, but since ACF is just one angle in triangle AFC, but maybe that's not helpful here.Alternatively, maybe using trigonometric identities. Given that cos(theta)= (√6 -√2)/4≈0.2588 corresponds to 75°, and cos(theta)= (√6 +√2)/4≈0.9659 corresponds to 15°, and since angle ACF is part of triangle ACF, which in our coordinate system, with points A(a,0), C(3g,3g), F(0,0). The side lengths are:AC: as we calculated earlier, 2g√3CF: distance from C(3g,3g) to F(0,0): sqrt(9g² +9g²)=sqrt(18g²)=3g√2AF: distance from A(a,0) to F(0,0): |a|.If a =g(3 -√3), then AF= a =g(3 -√3). If a =g(3 +√3), then AF= a= g(3 +√3). Then, using the Law of Sines in triangle ACF:sin(angle ACF)/AF = sin(angle AFC)/ACBut maybe this is getting too complicated. Alternatively, in our coordinate system, with the calculated angle ACF≈75°, which matches one of the answer choices. Since 75° is an option (D), that should be the answer.But wait, the answer options are:(A) 30°(B) 45°(C) 60°(D) 75°So 75° is option D, which we arrived at via coordinate geometry.But the problem seems to have a diagram, and maybe there's a simpler solution.Alternatively, consider using the properties of centroids and maybe some triangle similarity or special triangles.Given that angle AFG=45°, and FG is 1/3 of the median CF. Since AF is half of AB. If we can find the ratio of sides or some relationship, perhaps using trigonometric identities in triangle AFG and AGC.Alternatively, in triangle AFG, angle at F is 45°, sides AF, FG, and AG. In triangle AGC, angle at G is 60°, sides AG, GC, and AC.Since G is centroid, AG is 2/3 of median AD, and CG is 2/3 of median CF. But maybe not directly helpful.Alternatively, since in triangle AFG, we have angle at F=45°, and in triangle FGC, since FG is 1/3 of CF, and CG is 2/3 of CF. Maybe relating these.Alternatively, let's assign lengths. Let me let CF=3k, so FG=k and CG=2k. Since FG=k and angle AFG=45°, and AF is the length from A to F, which is half of AB. Let me denote AF=m.Then, in triangle AFG, sides are AF=m, FG=k, angle at F=45°. Using the Law of Sines in triangle AFG:AG/sin(45°) = AF/sin(angle at G) = FG/sin(angle at A)But I don't know angles at G or A in triangle AFG. Similarly, in triangle AGC, angle at G=60°, sides AG=?, CG=2k, and AC=?Alternatively, use coordinates again but more carefully. From earlier, we saw that angle ACF is 75°, which is an option. Alternatively, maybe there's a miscalculation.Wait, let's recap:Through coordinate geometry, we found that angle ACF is arccos( (√6 -√2)/4 )≈75°, which is option D. The problem gives that as an option, so I think that's the answer. Therefore, the degree measure of angle ACF is 75°, which is D.Final Answerboxed{D}</think>Given triangle ( triangle ABC ) with medians ( AD ) and ( CF ) intersecting at centroid ( G ). We know ( angle AFG = 45^circ ) and ( angle AGC = 60^circ ). We need to find the measure of ( angle ACF ).1. Coordinate Setup: - Place ( F ) at the origin ((0, 0)), midpoint of ( AB ). - Let ( A ) be at ((a, 0)) and ( B ) at ((-a, 0)). - Let ( C ) be at ((3g, 3g)) and centroid ( G ) at ((g, g)).2. Vectors and Angles: - Vector ( overrightarrow{FA} ) is ((a, 0)). - Vector ( overrightarrow{FG} ) is ((g, g)), forming a 45° angle with ( overrightarrow{FA} ).3. Centroid Properties: - ( G ) divides ( CF ) in the ratio 2:1, so ( C ) is at ((3g, 3g)). - Using the dot product to find the angle between vectors ( overrightarrow{GA} ) and ( overrightarrow{GC} ).4. Dot Product Calculation: - Vectors ( overrightarrow{GA} ) and ( overrightarrow{GC} ) lead to solving quadratic equations for ( a ) in terms of ( g ). - Solutions for ( a ) are ( g(3 pm sqrt{3}) ).5. Angle Calculation: - Using the dot product to find the angle between vectors ( overrightarrow{CA} ) and ( overrightarrow{CF} ). - Resultant angle ( angle ACF ) is found to be ( 75^circ ) using trigonometric identities and rationalization.Thus, the degree measure of ( angle ACF ) is (boxed{D}).]

🔑:5.D.As shown in Figure 5, draw C M perp A G at point M, and connect M F. Let G M = 1, then C G = 2, F G = 1. Therefore,F G = G M .Thus, begin{array}{l}angle M F C = angle G C M = 30^{circ},end{array}angle M F A = angle M A F = 15^{circ}.Hence, A M = F M = C M.Therefore, angle A C M = 45^{circ}.So, angle A C F = 75^{circ}.

🧠:[<think>Okay, let's try to figure out this problem step by step. So, we have a brigade with 10 members: 1 experienced brigadier and 9 young workers. Each young worker completed 15 instruments in a day. The brigadier completed 9 more instruments than the average of all ten members. We need to find the total number of instruments the brigade completed.First, let me note down the known information:- Number of young workers: 9- Each young worker's production: 15 instruments/day- Brigadier's production: 9 more than the average of all ten membersTotal instruments from young workers: 9 * 15. Let me calculate that first. 9 times 15 is 135. So, the young workers together made 135 instruments.Now, let the brigadier's production be B. According to the problem, the brigadier made 9 more than the average of all ten members. The average would be the total instruments divided by 10. But the total instruments are 135 (from young workers) plus B (from the brigadier). So, average is (135 + B)/10.Therefore, the brigadier's production B is equal to average plus 9. So:B = (135 + B)/10 + 9Hmm, okay, so this is an equation where B is on both sides. Let me solve for B.First, multiply both sides by 10 to eliminate the denominator:10B = 135 + B + 90Wait, because 9 multiplied by 10 is 90. So:10B = 135 + B + 90Combine like terms on the right side:10B = (135 + 90) + B10B = 225 + BSubtract B from both sides:9B = 225Divide both sides by 9:B = 225 / 9Calculating that: 225 divided by 9. Let's see, 9*25 is 225. So, B = 25.So, the brigadier made 25 instruments.Therefore, total instruments are 135 (from young workers) + 25 (brigadier) = 160.Wait, that seems straightforward. Let me check my steps again to make sure I didn't make a mistake.First, young workers: 9 * 15 = 135. Correct.Then, let B be the brigadier's instruments. The average of all ten is (135 + B)/10. The brigadier made 9 more than this average. So:B = (135 + B)/10 + 9Multiply both sides by 10:10B = 135 + B + 90Which simplifies to:10B = 225 + BSubtract B:9B = 225B = 25. Then total is 135 + 25 = 160.Wait, that seems okay. Let me check if 25 is indeed 9 more than the average.Average is (135 + 25)/10 = 160/10 = 16.25 is 16 + 9, which is correct. So, 25 = 16 + 9. Correct.Therefore, the total is 160. So, the answer should be 160.But let me just think again to make sure. Alternatively, maybe there's another way to approach it.Suppose the average of all ten workers is A. Then the brigadier's production is A + 9. The total production is 10A. But the total production is also 9*15 + (A + 9). Therefore:10A = 135 + A + 910A = 144 + ASubtract A:9A = 144A = 144 / 9 = 16Then the brigadier's production is 16 + 9 = 25. Total is 135 + 25 = 160. Same result.Yes, that confirms it. So, 160 instruments in total.Final AnswerThe total number of measuring instruments completed by the brigade is boxed{160}.</think>The brigade consists of 9 young workers and 1 experienced brigadier. Each young worker completed 15 instruments, so the total from the young workers is:[9 times 15 = 135]Let ( B ) be the number of instruments completed by the brigadier. The brigadier's production is 9 more than the average of all ten members. The average production of the ten members is:[frac{135 + B}{10}]Thus, the brigadier's production ( B ) can be expressed as:[B = frac{135 + B}{10} + 9]Multiplying both sides by 10 to clear the denominator:[10B = 135 + B + 90]Simplifying the right side:[10B = 225 + B]Subtracting ( B ) from both sides:[9B = 225]Solving for ( B ):[B = frac{225}{9} = 25]The total number of instruments completed by the brigade is the sum of the young workers' production and the brigadier's production:[135 + 25 = 160]Thus, the total number of measuring instruments completed by the brigade is (boxed{160}).,<think>Alright, let's see. So, the problem is about a brigade of workers who made measuring instruments. There's an experienced worker, the brigadier, and nine young workers. Each young worker completed 15 instruments in a day. The brigadier did 9 more than the average of all ten members. We need to find the total number of instruments the brigade made that day.Okay, let's break this down. First, the nine young workers each made 15 instruments. So, for the young workers alone, that would be 9 times 15. Let me calculate that: 9 * 15 = 135. So, the young workers contributed 135 instruments.Now, the brigadier's contribution is a bit trickier. It says the brigadier completed 9 more instruments than the average of all ten brigade members. Let's denote the average number of instruments made by all ten members as A. Then, the brigadier made A + 9 instruments.But the average A is the total number of instruments divided by 10, right? Let me denote the total instruments as T. Then, A = T / 10. So, the brigadier made (T / 10) + 9 instruments.But we also know that the total T is the sum of the young workers' instruments and the brigadier's instruments. So, T = 135 + (T / 10 + 9). Let me write that equation out:T = 135 + (T/10 + 9)Hmm, so I can solve for T here. Let's do that step by step. First, expand the equation:T = 135 + T/10 + 9Combine the constants: 135 + 9 = 144So, T = 144 + T/10Now, subtract T/10 from both sides to get:T - T/10 = 144Which is the same as (10T - T)/10 = 144Simplify numerator: 9T/10 = 144Multiply both sides by 10: 9T = 1440Divide both sides by 9: T = 1440 / 9Calculate that: 1440 divided by 9. Let me do the division. 9*160=1440, so T=160.Wait, so the total number of instruments is 160? Let me check that again.If T = 160, then the average A is 160 / 10 = 16. Therefore, the brigadier made 16 + 9 = 25 instruments. Then, the total is 135 (from the young workers) + 25 (brigadier) = 160. That adds up. Okay, so that seems correct.But let me verify each step again to be sure.First, young workers: 9 workers * 15 instruments = 135. That's straightforward.Brigadier's output: average of all ten plus 9. The average is total divided by ten. So, let the total be T. Then, the average is T/10, so the brigadier made T/10 + 9.Total T is sum of young workers and brigadier: 135 + (T/10 + 9) = T.Combine constants: 135 + 9 = 144. So, T = 144 + T/10.Subtract T/10: T - T/10 = 144. Which is (10T - T)/10 = 144 => 9T/10 = 144 => 9T = 1440 => T = 160.Yes, that seems correct. Then, check the brigadier's contribution: 160/10 = 16 average, so 16 + 9 = 25. Total is 135 + 25 = 160. Perfect.So, the total number of instruments completed is 160.Final AnswerThe total number of measuring instruments completed by the brigade is boxed{160}.</think>The brigade consists of nine young workers and one experienced brigadier. Each young worker completed 15 instruments, so the total contribution from the young workers is:[9 times 15 = 135]Let ( T ) be the total number of instruments completed by the entire brigade. The average number of instruments per brigade member is ( frac{T}{10} ). The brigadier completed 9 more instruments than this average, so the brigadier's contribution is:[frac{T}{10} + 9]The total number of instruments ( T ) is the sum of the contributions from the young workers and the brigadier:[T = 135 + left( frac{T}{10} + 9 right)]Simplifying this equation:[T = 135 + frac{T}{10} + 9][T = 144 + frac{T}{10}]Subtracting ( frac{T}{10} ) from both sides:[T - frac{T}{10} = 144][frac{9T}{10} = 144]Multiplying both sides by 10:[9T = 1440]Dividing by 9:[T = 160]Thus, the total number of measuring instruments completed by the brigade is boxed{160}.]

🔑:If all the devices completed by the nine young workers and 9 of the devices completed by the foreman are evenly distributed among the nine young workers, then each of the ten brigade members ends up with the mentioned average, meaning each member ends up with the same number of devices. Since each of the nine young workers ends up with exactly 16 devices, it follows that (10 cdot 16 = 160). Therefore, a total of 160 devices were completed on that day.Solutions of the I. Round 1981 adopted from [5] 5.23.2 II. Round 1981, Class 8

🧠:[<think>Okay, let me try to solve this problem step by step. So, the expression is [(5/(m - 2)) - m - 2] multiplied by (2m - 4)/(3 - m), and we need to evaluate it at m = -1/2. First, I should probably simplify the expression before plugging in the value of m. That way, it might be easier and less error-prone. Let me start by looking at the first part of the expression: (5/(m - 2)) - m - 2. Maybe I can combine those terms into a single fraction. To combine them, I need a common denominator. The first term has a denominator of (m - 2), and the other terms are just -m and -2, which can be considered as having a denominator of 1. So, the common denominator would be (m - 2). Let me rewrite the expression:5/(m - 2) - m - 2 = 5/(m - 2) - [m + 2]. To combine these, I can write [m + 2] as [(m + 2)(m - 2)]/(m - 2). Wait, but actually, since the denominator is (m - 2), I need to multiply both the numerator and denominator of (m + 2) by (m - 2). Let's see:= 5/(m - 2) - (m + 2)(m - 2)/(m - 2) Wait, hold on. If I write (m + 2) as (m + 2)/1, then to get the denominator (m - 2), I multiply numerator and denominator by (m - 2):= 5/(m - 2) - [(m + 2)(m - 2)]/(m - 2) - [2*(m - 2)]/(m - 2). Wait, no. Let me think again. The original expression is 5/(m - 2) - m - 2. So, the terms -m and -2 can be combined into -(m + 2). So, 5/(m - 2) - (m + 2). To combine into a single fraction, the common denominator is (m - 2). Therefore:= [5 - (m + 2)(m - 2)] / (m - 2)Is that right? Let me check. If I have 5/(m - 2) - (m + 2), then converting (m + 2) to have denominator (m - 2) would be (m + 2)(m - 2)/(m - 2). So:= [5 - (m + 2)(m - 2)] / (m - 2)Yes, that seems correct. So, expanding the numerator:First, compute (m + 2)(m - 2). That's a difference of squares: m² - (2)² = m² - 4. So:Numerator: 5 - (m² - 4) = 5 - m² + 4 = (5 + 4) - m² = 9 - m²Therefore, the first part simplifies to (9 - m²)/(m - 2)So now, the entire expression becomes:[(9 - m²)/(m - 2)] * [(2m - 4)/(3 - m)]Let me write that down:(9 - m²) (2m - 4)--------- * ---------- (m - 2) (3 - m)Now, let's see if we can factor each part to simplify further.First, 9 - m² is a difference of squares: 3² - m² = (3 - m)(3 + m). So:(9 - m²) = (3 - m)(3 + m)Then, 2m - 4 can be factored as 2(m - 2)So, substituting these into the expression:(3 - m)(3 + m) 2(m - 2)--------------- * ------------ (m - 2) (3 - m)Now, let's rearrange the terms:= [ (3 - m)(3 + m) / (m - 2) ] * [ 2(m - 2) / (3 - m) ]Looking at the numerators and denominators, there are common factors that can be canceled out.First, (3 - m) in the numerator of the first fraction cancels with (3 - m) in the denominator of the second fraction. But wait, (3 - m) in the first fraction's numerator and (3 - m) in the second fraction's denominator. Let's write it step by step.Multiply the numerators together: (3 - m)(3 + m) * 2(m - 2)Multiply the denominators together: (m - 2) * (3 - m)So, the entire expression is:[ (3 - m)(3 + m) * 2(m - 2) ] / [ (m - 2)(3 - m) ]Now, let's cancel out the common factors. First, (3 - m) in the numerator and denominator. There is one (3 - m) in the numerator and one in the denominator. So they cancel.Similarly, (m - 2) in the numerator and denominator. There is one (m - 2) in the numerator (from 2(m - 2)) and one in the denominator. So they cancel.What's left after canceling?Numerator: (3 + m) * 2Denominator: 1Therefore, the expression simplifies to 2(3 + m)So, the entire original expression simplifies to 2(3 + m). That's a big simplification!So instead of dealing with the complex fractions, it's just 2(3 + m). Let me verify that again.Original expression:[5/(m - 2) - m - 2] * (2m - 4)/(3 - m)After simplifying the first part, we had (9 - m²)/(m - 2). Then multiplied by (2m - 4)/(3 - m). Then factoring everything, we saw that terms canceled out, leading to 2(3 + m). Hmm, seems correct.Alternatively, maybe I can check with a different approach. Let's take m = -1/2 and compute both the original expression and the simplified expression to see if they match. Wait, but the problem asks to simplify first, then evaluate. So perhaps the answer is 2(3 + m). But let's proceed.So, simplified expression is 2(3 + m). Let's compute this when m = -1/2.First, compute 3 + m: 3 + (-1/2) = 3 - 1/2 = 5/2Then multiply by 2: 2*(5/2) = 5. So the answer should be 5.But let me verify this by plugging m = -1/2 into the original expression to make sure.Original expression: [5/(m - 2) - m - 2] * (2m - 4)/(3 - m)Compute each part step by step.First, compute m - 2: (-1/2) - 2 = (-1/2) - (4/2) = -5/2So 5/(m - 2) = 5/(-5/2) = 5 * (-2/5) = -2Next, compute -m - 2: -(-1/2) - 2 = 1/2 - 2 = 1/2 - 4/2 = -3/2So the first part [5/(m - 2) - m - 2] is (-2) - (3/2) = (-2) - 1.5 = -3.5 or -7/2Now compute the second part: (2m - 4)/(3 - m)2m - 4: 2*(-1/2) - 4 = -1 - 4 = -53 - m: 3 - (-1/2) = 3 + 1/2 = 7/2So the second part is (-5)/(7/2) = -5 * (2/7) = -10/7Multiply the two parts: (-7/2) * (-10/7) = (70/14) = 5. Which matches the simplified expression result. Therefore, the simplification is correct. So, after simplifying the expression to 2(3 + m), substituting m = -1/2 gives 5. But just to recap the simplification steps:Starting with [(5/(m - 2) - m - 2)] * [(2m - 4)/(3 - m)]First part simplifies to (9 - m²)/(m - 2). Then multiply by (2m - 4)/(3 - m). Factor numerator and denominator:9 - m² = (3 - m)(3 + m)2m - 4 = 2(m - 2)So, expression becomes [(3 - m)(3 + m)/ (m - 2)] * [2(m - 2)/(3 - m)]Cancel (3 - m) and (m - 2) terms, left with 2(3 + m). Therefore, when m = -1/2, 2*(3 + (-1/2)) = 2*(5/2) = 5.So the answer is 5. But since the problem says "put your final answer within boxed{}", the boxed answer should be boxed{5}.Final Answerboxed{5}</think>To simplify and evaluate the expression (left( frac{5}{m-2} - m - 2 right) cdot frac{2m-4}{3-m}) where (m = -frac{1}{2}):1. Simplify the first part of the expression: [ frac{5}{m-2} - m - 2 ] Combine the terms into a single fraction: [ frac{5 - (m+2)(m-2)}{m-2} = frac{5 - (m^2 - 4)}{m-2} = frac{9 - m^2}{m-2} ] Factor the numerator: [ frac{(3 - m)(3 + m)}{m - 2} ]2. Simplify the second part of the expression: [ frac{2m - 4}{3 - m} ] Factor the numerator: [ frac{2(m - 2)}{3 - m} ]3. Combine and simplify the entire expression: [ left( frac{(3 - m)(3 + m)}{m - 2} right) cdot left( frac{2(m - 2)}{3 - m} right) ] Cancel common factors: [ frac{(3 - m)(3 + m) cdot 2(m - 2)}{(m - 2)(3 - m)} = 2(3 + m) ]4. Substitute (m = -frac{1}{2}): [ 2 left( 3 + left( -frac{1}{2} right) right) = 2 left( frac{5}{2} right) = 5 ]Thus, the evaluated value of the expression is (boxed{5}).,<think>Okay, let me try to work through this problem step by step. So, I need to simplify the expression first and then evaluate it at m = -1/2. The expression is [(5/(m - 2)) - m - 2] multiplied by (2m - 4)/(3 - m). Hmm, let's see. Alright, breaking it down.First, let me look at the main components. There's a subtraction in the first part: 5/(m - 2) minus m minus 2. Then that entire result is multiplied by the fraction (2m - 4)/(3 - m). Maybe I can simplify each part before plugging in m = -1/2. That might make it easier than substituting right away.Starting with the first part: 5/(m - 2) - m - 2. To combine these terms, I need a common denominator. The first term is a fraction, and the others are polynomials. So maybe I can write m and 2 with the denominator (m - 2). Let me try that.So, 5/(m - 2) - m - 2 can be written as 5/(m - 2) - (m + 2). To combine these, I need to express (m + 2) as a fraction over (m - 2). That would be [5 - (m + 2)(m - 2)]/(m - 2). Wait, is that right? Let me check. If I have A - B, where A is 5/(m - 2) and B is (m + 2), then to subtract B from A, I need to express B as a fraction over the same denominator. So B = (m + 2)/1, so multiplying numerator and denominator by (m - 2) gives B = [(m + 2)(m - 2)]/(m - 2). Therefore, A - B becomes [5 - (m + 2)(m - 2)]/(m - 2).Let me compute (m + 2)(m - 2). That's a difference of squares: m^2 - 4. So the numerator becomes 5 - (m^2 - 4) = 5 - m^2 + 4 = (5 + 4) - m^2 = 9 - m^2. So the entire first part simplifies to (9 - m^2)/(m - 2).Wait, hold on, is that correct? Let's verify step by step. Original expression: 5/(m - 2) - m - 2. To combine terms, subtract (m + 2) from 5/(m - 2). So:5/(m - 2) - (m + 2) = 5/(m - 2) - m - 2.To combine these into a single fraction, yes, we need to get a common denominator. So, the common denominator is (m - 2). Therefore, rewrite m as [m(m - 2)]/(m - 2) and 2 as [2(m - 2)]/(m - 2). But actually, wait, that might not be the case. Let's think again.Wait, if we have 5/(m - 2) - m - 2, that is 5/(m - 2) - m - 2. To combine the terms -m and -2 with the fraction, we need to express them over the same denominator. So, -m can be written as -m(m - 2)/(m - 2), and -2 can be written as -2(m - 2)/(m - 2). Therefore, combining all terms:[5 - m(m - 2) - 2(m - 2)]/(m - 2). Let's compute the numerator:First term: 5Second term: -m(m - 2) = -m^2 + 2mThird term: -2(m - 2) = -2m + 4Combine all together: 5 - m^2 + 2m - 2m + 4 = 5 + 4 - m^2 + (2m - 2m) = 9 - m^2So yes, the numerator is 9 - m^2, so the first part simplifies to (9 - m^2)/(m - 2). That seems right.So now, the entire expression is [(9 - m^2)/(m - 2)] multiplied by (2m - 4)/(3 - m). Let me write that as:[(9 - m^2)/(m - 2)] * [(2m - 4)/(3 - m)]Now, let's look at each part. 9 - m^2 is a difference of squares, so that factors into (3 - m)(3 + m). Similarly, 2m - 4 can be factored as 2(m - 2). Let me note that.So, 9 - m^2 = (3 - m)(3 + m)2m - 4 = 2(m - 2)So substituting these into the expression:[(3 - m)(3 + m)/(m - 2)] * [2(m - 2)/(3 - m)]Now, let's rearrange the terms:[ (3 - m)(3 + m) * 2(m - 2) ] / [ (m - 2)(3 - m) )Notice that (3 - m) is in both the numerator and the denominator, so they can cancel out. Similarly, (m - 2) is in the denominator and numerator. Let me write that:The denominator has (m - 2)(3 - m)The numerator has (3 - m)(3 + m) * 2(m - 2)So canceling (3 - m) in numerator and denominator, and canceling (m - 2) in numerator and denominator:What remains in the numerator: (3 + m) * 2What remains in the denominator: 1Therefore, the entire expression simplifies to 2(3 + m)Wait, that seems too simplified. Let me check again.Original expression after factoring:[(3 - m)(3 + m)/(m - 2)] * [2(m - 2)/(3 - m)]Multiply numerator terms: (3 - m)(3 + m) * 2(m - 2)Multiply denominator terms: (m - 2)(3 - m)So, numerator: 2(3 - m)(3 + m)(m - 2)Denominator: (m - 2)(3 - m)Now, (3 - m) in the numerator and denominator can be canceled. Also, (m - 2) in numerator and denominator cancels. So:Numerator after cancels: 2(3 + m)Denominator: 1Therefore, the expression simplifies to 2(3 + m). So that's the simplified form. So instead of dealing with the original complex fraction, it's just 2*(3 + m). That seems correct.Alternatively, (3 - m)/(3 - m) cancels to 1, and (m - 2)/(m - 2) cancels to 1. So yeah, the simplified expression is 2*(3 + m). Nice, that makes the problem much easier.So then, the original expression simplifies to 2(3 + m). Therefore, when m = -1/2, substituting m into this simplified expression gives:2*(3 + (-1/2)) = 2*(3 - 1/2) = 2*(5/2) = 5.Wait, so the final answer is 5? Let me verify that step by step.But just to be thorough, let's check if my simplification is correct. Let's take the original expression, substitute m = -1/2, and see if it equals 5.Original expression:[(5/(m - 2) - m - 2)] * [(2m - 4)/(3 - m)]First, compute each part with m = -1/2.Compute m - 2: (-1/2) - 2 = (-1/2) - (4/2) = -5/2So 5/(m - 2) = 5 / (-5/2) = 5 * (-2/5) = -2Then compute -m - 2: -(-1/2) - 2 = 1/2 - 2 = 1/2 - 4/2 = -3/2So the first part of the expression is -2 - 3/2 = (-4/2 - 3/2) = -7/2Now, compute the second part: (2m - 4)/(3 - m)Compute 2m - 4: 2*(-1/2) - 4 = -1 - 4 = -5Compute 3 - m: 3 - (-1/2) = 3 + 1/2 = 7/2So the second part is -5 / (7/2) = -5 * (2/7) = -10/7Now multiply the two parts: (-7/2) * (-10/7) = (7/2)*(10/7) = (7*10)/(2*7) = 10/2 = 5Okay, so substituting directly gives 5, which matches the simplified expression's result. So that confirms that the simplification is correct. Therefore, the answer is 5.But let me just go through the simplification steps again to ensure I didn't make any errors.Original expression:[(5/(m - 2) - m - 2)] * [(2m - 4)/(3 - m)]Simplify the first bracket:5/(m - 2) - m - 2Combine terms over common denominator (m - 2):[5 - (m + 2)(m - 2)]/(m - 2) = [5 - (m^2 - 4)]/(m - 2) = (9 - m^2)/(m - 2)Then the entire expression becomes:[(9 - m^2)/(m - 2)] * [(2m - 4)/(3 - m)]Factor numerator terms:9 - m^2 = (3 - m)(3 + m)2m - 4 = 2(m - 2)So substituting:[(3 - m)(3 + m)/(m - 2)] * [2(m - 2)/(3 - m)]Cancel (3 - m) and (m - 2):Left with 2*(3 + m)Yes, so 2*(3 + m) is the simplified form. Then plugging in m = -1/2:2*(3 + (-1/2)) = 2*(5/2) = 5. Perfect.Alternatively, maybe I should check if there are any restrictions on m. For the original expression, the denominators cannot be zero. So m cannot be 2 (because m - 2 would be zero) and m cannot be 3 (because 3 - m would be zero). But since we're substituting m = -1/2, which is neither 2 nor 3, it's valid.Therefore, everything checks out. The simplified expression is 2*(3 + m), and substituting m = -1/2 gives 5. Both the direct substitution and simplification methods lead to the same answer, so I'm confident it's correct.Final Answerboxed{5}</think>To simplify and evaluate the expression (left( frac{5}{m-2} - m - 2 right) cdot frac{2m - 4}{3 - m}) where (m = -frac{1}{2}):1. Simplify the first part of the expression: [ frac{5}{m-2} - m - 2 ] Combine the terms over a common denominator: [ frac{5 - (m + 2)(m - 2)}{m - 2} = frac{5 - (m^2 - 4)}{m - 2} = frac{9 - m^2}{m - 2} ] Factor the numerator: [ frac{(3 - m)(3 + m)}{m - 2} ]2. Simplify the second part of the expression: [ frac{2m - 4}{3 - m} ] Factor the numerator: [ frac{2(m - 2)}{3 - m} ]3. Combine and simplify the entire expression: [ left( frac{(3 - m)(3 + m)}{m - 2} right) cdot left( frac{2(m - 2)}{3 - m} right) ] Cancel common factors: [ frac{(3 - m)(3 + m) cdot 2(m - 2)}{(m - 2)(3 - m)} = 2(3 + m) ]4. Substitute (m = -frac{1}{2}) into the simplified expression: [ 2 left( 3 + left( -frac{1}{2} right) right) = 2 left( frac{6}{2} - frac{1}{2} right) = 2 left( frac{5}{2} right) = 5 ]Thus, the final answer is (boxed{5}).]

🔑:To simplify and then evaluate the given expression ({frac{5}{{m-2}}-m-2})•frac{{2m-4}}{{3-m}} with m=-frac{1}{2}, we follow these steps:1. Simplify the Expression:First, we rewrite the original expression by combining like terms and factoring where possible:[begin{align*}Original expression &= frac{{5}}{{m-2}} - m - 2 cdot frac{{2m-4}}{{3-m}} &= frac{{5 - (m-2)(m+2)}}{{m-2}} cdot frac{{2(m-2)}}{{3-m}} &= frac{{5 - m^2 - 4}}{{m-2}} cdot frac{{2(m-2)}}{{3-m}} &= frac{{1 - m^2}}{{m-2}} cdot frac{{2(m-2)}}{{3-m}} &= frac{{(1-m)(1+m)}}{{m-2}} cdot frac{{2(m-2)}}{{3-m}} &= frac{{(3-m)(3+m)}}{{m-2}} cdot frac{{2(m-2)}}{{3-m}} &= (3+m) cdot frac{{2}}{{3-m}} cdot frac{{(m-2)}}{{m-2}} &= 2(3+m) &= 2m + 6.end{align*}]2. Evaluate the Expression for m=-frac{1}{2}:Next, we substitute m = -frac{1}{2} into the simplified expression:[begin{align*}2m + 6 &= 2left(-frac{1}{2}right) + 6 &= -1 + 6 &= 5.end{align*}]Therefore, after simplifying the given expression and evaluating it at m = -frac{1}{2}, we find that the value is boxed{5}.

🧠:[<think>Okay, so I need to simplify these three trigonometric sums. Let me start with the first one. Problem (1): The sum is from 45° to 133° where each term is 1 divided by the product of sine of n degrees and sine of (n+1) degrees. So it's like 1/(sin45 sin46) + 1/(sin47 sin48) + ... + 1/(sin133 sin134). Hmm. I remember that sometimes products of sines can be related to cotangent or tangent functions through some identity. Maybe there's a telescoping identity here. Let me recall the identity for cotangent differences. Wait, the formula for cot A - cot B is (sin(B - A))/(sin A sin B). Let me check that. If I have cot A - cot B, that's (cos A / sin A) - (cos B / sin B) = (cos A sin B - cos B sin A)/(sin A sin B) = sin(B - A)/sin A sin B. So, sin(B - A) = sin(- (A - B)) = -sin(A - B). So actually, cot A - cot B = - sin(A - B)/sin A sin B. Therefore, rearranged, 1/(sin A sin B) = (cot B - cot A)/sin(A - B). In the given problem, each term is 1/(sin n° sin(n+1)°). So here, A is n°, B is (n+1)°, so A - B = -1°, so sin(A - B) = sin(-1°) = -sin1°. Therefore, substituting into the identity: 1/(sin n° sin(n+1)°) = (cot(n+1)° - cot n°)/sin(-1°). Since sin(-1°) is -sin1°, this becomes (cot(n+1)° - cot n°)/(-sin1°) = (cot n° - cot(n+1)°)/sin1°. So each term in the sum can be written as (cot n° - cot(n+1)°)/sin1°. Therefore, the entire sum is [cot45° - cot46° + cot47° - cot48° + ... + cot133° - cot134°]/sin1°. Wait, but hold on. The original sum starts at n=45°, then skips to n=47°, right? Wait, no. Wait, the original problem says the terms are 1/(sin45 sin46) + 1/(sin47 sin48) + ... + 1/(sin133 sin134). Wait, is that stepping by 2 degrees each time? Because 45,46 then 47,48? Wait, that can't be. Wait, maybe the problem is written as (1)/(sin45 sin46) + (1)/(sin47 sin48) + ... + (1)/(sin133 sin134). So each term is consecutive degrees, but the denominators are consecutive angles? Wait, no. Wait, if it's 45 and 46, then next term is 47 and 48? Then the terms are stepping by 2. Wait, that would mean the sum skips every other term. But that seems odd. Wait, maybe the problem is written as a sequence where the first term is n=45°, the next term is n=46°, up to n=133°, such that each term is 1/(sin n° sin(n+1)°). So the sum is from n=45 to n=133 of 1/(sin n° sin(n+1)°). Then, in that case, the step is 1 each time, so the terms would be 1/(sin45 sin46) + 1/(sin46 sin47) + ... + 1/(sin133 sin134). But the problem as written says "..." between 1/(sin45 sin46) and 1/(sin47 sin48). Wait, maybe the user wrote it as starting at 45,46; then next is 47,48; so the step is 2 each time? That would mean the sum is over n=45, 47, 49,...,133. Then the denominators are sin45 sin46, sin47 sin48, etc. But that's a bit ambiguous. Wait, the problem is written as "1/(sin45° sin46°) + 1/(sin47° sin48°) + ... + 1/(sin133° sin134°)". The way it's written, the first term is 45,46; the second is 47,48; so each term increases by 2 degrees each time. So the sum is over terms where the angles in the denominators are consecutive odd degrees starting at 45 and 46 up to 133 and 134. Wait, but 45 to 134 in steps of 2 would require checking if 133 is the last term. Let's see: 45,46; 47,48; ...; 133,134. So the starting term is 45,46 and the ending term is 133,134. Then how many terms are there? From 45 to 133 inclusive, stepping by 2 each time: (133 -45)/2 +1 = (88)/2 +1=44+1=45 terms. But the original problem is written as (1) first sum: 1/(sin45 sin46) +1/(sin47 sin48)+...+1/(sin133 sin134). So if each term is two degrees apart, then each term is 1/(sin(45+2k) sin(46+2k)) for k from 0 to 44, since 45+2*44=45+88=133 and 46+2*44=134. So yes, 45 terms. But now, if each term is 1/(sin(45+2k) sin(46+2k)), then using the identity I found earlier, each term would be (cot(46+2k) - cot(45+2k))/sin1°. Therefore, each term is (cot(n+1) - cot n)/sin1°, where n=45+2k. So if we write out the sum, it would be [cot46 - cot45 + cot48 - cot47 + cot50 - cot49 + ... + cot134 - cot133]/sin1°. But notice that in the sum, every other term cancels. Let's write out the first few terms: (cot46 - cot45) + (cot48 - cot47) + (cot50 - cot49) + ... + (cot134 - cot133). So when we add all these up, the terms -cot45, then -cot47, -cot49, ..., -cot133, and the positive terms are cot46, cot48, ..., cot134. So the total sum would be [ (cot46 + cot48 + ... + cot134) - (cot45 + cot47 + ... + cot133) ] / sin1°. But this seems complicated. Maybe there's a telescoping if the terms were consecutive, but since they are every other term, the telescoping doesn't collapse as nicely. Wait, maybe there's another approach. Alternatively, if the problem was written as consecutive terms (without skipping degrees), then the sum would telescope completely to (cot46 - cot45 + cot47 - cot46 + ... + cot134 - cot133)/sin1° = (cot134 - cot45)/sin1°. But in that case, the sum would be straightforward. However, according to the problem's notation, the denominators are sin45 sin46, then sin47 sin48, etc., so the terms are not consecutive. So the sum is over every other term. But maybe there's a mistake here. Let me verify the problem statement again. It says: "Simplify: (1) 1/(sin45° sin46°) + 1/(sin47° sin48°) + ... + 1/(sin133° sin134°);"So the denominators are sin45 sin46, sin47 sin48,..., sin133 sin134. So each term is separated by 2 degrees. So each term is 1/(sin(n) sin(n+1)) where n starts at 45, increases by 2 each time, up to 133. Therefore, n = 45,47,49,...,133. Each term is 1/(sinn° sin(n+1)°). Therefore, each term can be expressed as (cot(n+1)° - cotn°)/sin1°, using the identity above. Therefore, the entire sum would be sum_{k=0}^{44} [cot(45 + 2k +1)° - cot(45 + 2k)°]/sin1° = [sum_{k=0}^{44} (cot(46 + 2k)° - cot(45 + 2k)°)] / sin1°. Expanding the sum, that would be [ (cot46 - cot45) + (cot48 - cot47) + (cot50 - cot49) + ... + (cot134 - cot133) ] / sin1°. Now, let's note that cot134° is cot(180° - 46°) = -cot46°, because cot(180° - x) = -cotx. Similarly, cot133° = cot(180° - 47°) = -cot47°, and so on. Similarly, cot90° is 0, but in this case, we don't reach 90°. Wait, 134° is 180° - 46°, right. So cot134° = -cot46°, cot133° = -cot47°, ..., cot91° = -cot89°, cot90° = 0. But in our sum, the last term is cot134° - cot133°, which is -cot46° - (-cot47°) = -cot46° + cot47°. Wait, but let's check this step-by-step. First, let's list all the terms in the sum: cot46 - cot45cot48 - cot47cot50 - cot49...cot134 - cot133So the total sum is [cot46 + cot48 + cot50 + ... + cot134] - [cot45 + cot47 + cot49 + ... + cot133]. But note that cot134 = cot(180 - 46) = -cot46Similarly, cot133 = cot(180 - 47) = -cot47cot132 = -cot48, but in our case, the angles in the positive terms go up to cot134, which is -cot46. Similarly, the negative terms go up to cot133, which is -cot47.Wait, perhaps it's better to pair terms.But let's think about the sum:Positive terms: cot46, cot48, cot50,..., cot134Negative terms: -cot45, -cot47, -cot49,..., -cot133So, positive terms: from cot46 to cot134, stepping by 2°. Since 134 = 46 + 88°, stepping by 2°, so number of terms: (134 -46)/2 +1 = 88/2 +1=44+1=45 terms.Negative terms: from cot45 to cot133, stepping by 2°, so (133 -45)/2 +1 =88/2 +1=44+1=45 terms.But cot134° = -cot46°, cot132° = -cot48°, etc. So cot134 = -cot46, cot132 = -cot48,..., cot90 = 0. Wait, but 134 is even, stepping down by 2°, so 134,132,...,90. But in our positive terms, it's cot46,48,...,134. So when we write cot134, that's -cot46. Similarly, cot132 is -cot48, but in our sum, cot134 is included as the last term. So the positive terms can be expressed as cot46 + cot48 + ... + cot88 + cot90 + ... + cot134. Wait, but 90° is cot90° = 0. So from cot46° to cot134°, stepping by 2°, we pass through 90°, which is 0. So cot90°=0. Therefore, the sum of the positive terms is cot46 + cot48 + ... + cot88 + 0 + cot92 + ... + cot134. But cot92 = -cot88, cot94 = -cot86, etc. Because cot(180 - x) = -cotx. So cot92 = cot(180 - 88) = -cot88, cot94 = -cot86, ..., cot134 = -cot46. Therefore, the positive terms: cot46 + cot48 + ... + cot88 + 0 + (-cot88) + (-cot86) + ... + (-cot46). So all terms cancel except 0. Similarly, the negative terms: -cot45 - cot47 - ... -cot133. Let's see, cot133 = -cot47, cot131 = -cot49,...,cot91 = -cot89. So -cot45 -cot47 - ... -cot89 - ... -cot133. Which is -cot45 -cot47 - ... -cot89 - (-cot89) - ... - (-cot47). Wait, similar cancellation? Let's check. Negative terms: from -cot45 to -cot133 stepping by 2°, which is -cot45, -cot47, ..., -cot89, -cot91, ..., -cot133. But cot91 = -cot89, cot93 = -cot87, etc. So -cot45 -cot47 - ... -cot89 - (-cot89) - ... - (-cot47). Wait, this would be -cot45 -cot47 - ... -cot89 + cot89 + ... + cot47. So again, terms cancel except for -cot45. Wait, maybe I need to elaborate:Positive terms:cot46 + cot48 + ... + cot88 + cot90 (which is 0) + cot92 + ... + cot134But cot92 = cot(180 - 88) = -cot88cot94 = -cot86...cot134 = -cot46So the positive terms become:cot46 + cot48 + ... + cot88 + 0 - cot88 - cot86 - ... - cot46This telescopes to 0, because each cotθ cancels with -cotθ. Similarly, the middle term is 0. So the entire sum of positive terms is 0.Negative terms:-cot45 - cot47 - ... - cot89 - cot91 - ... -cot133But cot91 = cot(180 -89) = -cot89cot93 = -cot87...cot133 = -cot47Therefore, negative terms become:-cot45 - cot47 - ... - cot89 + cot89 + ... + cot47Again, the terms from cot47 to cot89 cancel out, leaving only -cot45.Therefore, the total sum is [0 - (-cot45)] / sin1° = cot45° / sin1°. Since cot45° = 1, so the total sum is 1 / sin1°. Wait, that's neat! So the answer for part (1) is 1/sin1°, which is csc1°. But let me verify this again. Starting with the sum S = sum_{n=45,47,...,133} 1/(sinn° sin(n+1)°). Each term is (cot(n+1) - cotn)/sin1°. Sum becomes [sum (cot(n+1) - cotn)] / sin1°. Summing over n=45,47,...,133, the terms are (cot46 - cot45) + (cot48 - cot47) + ... + (cot134 - cot133). Then, when expanded, the sum is (cot46 + cot48 + ... + cot134) - (cot45 + cot47 + ... + cot133). But as we saw, the positive terms and negative terms collapse due to the cot(180 - x) = -cotx identity. So the positive terms sum to - (cot46 + cot48 + ... + cot88) and the negative terms sum to -cot45 + (cot47 + ... + cot89). Wait, maybe my previous reasoning was too hasty. Let me write out specific terms for a smaller case to check. Suppose instead of going up to 134°, let's take a smaller example. Let's say the sum is 1/(sin45 sin46) + 1/(sin47 sin48). Then according to the identity, this would be (cot46 - cot45)/sin1 + (cot48 - cot47)/sin1. Then the total sum would be [cot46 + cot48 - cot45 - cot47]/sin1. Then, using cot48 = -cot(180-48)= -cot132, but that's not helpful here. Wait, in this smaller case, how does it collapse? Maybe with more terms. But perhaps in the original problem, with all terms from 45 to 133, the positive cot terms from 46 to 134 (even steps) and negative terms from 45 to 133 (odd steps). Then, since 134 = 180 - 46, cot134 = -cot46. Similarly, cot132 = -cot48, etc. So the positive terms:cot46 + cot48 + ... + cot88 + cot90 + cot92 + ... + cot134But cot90 = 0, and cot92 = -cot88, cot94 = -cot86,..., cot134 = -cot46. Therefore, the positive terms from 46 to 88 cancel with those from 92 to 134, leaving only cot90 = 0. Similarly, the negative terms:-cot45 - cot47 - ... - cot89 - cot91 - ... -cot133But cot91 = -cot89, cot93 = -cot87,...,cot133 = -cot47. Therefore, the negative terms from 47 to 89 cancel with those from 91 to 133, leaving only -cot45. Hence, total sum is (0 - (-cot45)) / sin1° = cot45° / sin1° = 1 / sin1°, since cot45°=1. Therefore, the answer for part (1) is 1/sin1°, which is cosecant 1°, or csc1°. Moving on to problem (2): The sum is 1/(sin1° sin2°) + 1/(sin2° sin3°) + ... + 1/(sin89° sin90°). So this time, the terms are consecutive, starting from n=1° to n=89°, each term being 1/(sinn° sin(n+1)°). Using the same identity as before: 1/(sinn sin(n+1)) = (cot(n+1) - cotn)/sin1°. Therefore, the entire sum is sum_{n=1}^{89} (cot(n+1) - cotn)/sin1° = [cot90° - cot1°]/sin1°. Because it's a telescoping series where all intermediate terms cancel. But cot90° is 0, so the sum becomes [0 - cot1°]/sin1° = -cot1°/sin1°. But cot1° is cos1°/sin1°, so this is -cos1°/(sin1°)^2. However, this is negative, but the original sum is positive, so there must be an error here. Wait, let's check the identity again. Earlier, we had 1/(sinA sinB) = (cotB - cotA)/sin(B - A). In this case, for each term 1/(sinn° sin(n+1)°), A = n°, B = n+1°, so B - A = 1°. Therefore, the identity gives 1/(sinn sin(n+1)) = (cot(n+1) - cotn)/sin1°. Therefore, each term is (cot(n+1) - cotn)/sin1°, and the sum from n=1 to 89 would be [cot90 - cot1]/sin1°. Since cot90°=0, this gives (-cot1°)/sin1° = -cot1°/sin1°. But all terms in the original sum are positive, so this result being negative is a problem. Wait, perhaps I made a sign mistake in the identity. Let's rederive it. Start with cotB - cotA = (cosB/sinB - cosA/sinA) = (cosB sinA - cosA sinB)/(sinA sinB) = sin(A - B)/(sinA sinB). Therefore, cotB - cotA = sin(A - B)/(sinA sinB). Hence, 1/(sinA sinB) = (cotB - cotA)/sin(A - B). In our case, A = n°, B = n+1°, so A - B = -1°, so sin(A - B) = -sin1°. Therefore, 1/(sinn sin(n+1)) = (cot(n+1) - cotn)/(-sin1°) = (cotn - cot(n+1))/sin1°. Ah! So there was a sign error. The correct identity is 1/(sinn sin(n+1)) = (cotn - cot(n+1))/sin1°. Therefore, the sum becomes sum_{n=1}^{89} (cotn - cot(n+1))/sin1° = [cot1° - cot90°]/sin1°. Since cot90°=0, this becomes cot1°/sin1° = (cos1°/sin1°)/sin1° = cos1°/(sin1°)^2. But the original sum is a sum of positive terms, and this result is positive, so that makes sense. Therefore, the sum is cot1° / sin1° = cos1° / (sin1°)^2. But can this be simplified further? Alternatively, using the identity cotθ = cosθ / sinθ, so cot1° / sin1° = cos1° / (sin1°)^2. But perhaps expressing it in terms of cosecant or something else. However, maybe the problem expects the answer in terms of cot1° / sin1°, or perhaps simplifying further. Alternatively, note that cotθ / sinθ = cosθ / (sinθ)^2. But perhaps there's another telescoping identity or a known formula for this sum. Alternatively, let's recall that sum_{n=1}^{89} 1/(sinn sin(n+1)) = sum_{n=1}^{89} (cotn - cot(n+1))/sin1° = (cot1 - cot90)/sin1° = cot1 / sin1°, since cot90=0. So the answer is cot1° / sin1°, which is equal to cos1° / (sin1°)^2. But is there a better way to express this? For example, using the double-angle identity: 1/(sin1°) = csc1°, and cot1° = cos1° / sin1°, so cot1° / sin1° = cos1° / (sin1°)^2 = cot1° csc1°. Alternatively, using the identity for cosecant squared or something else. Alternatively, let's note that 1/(sinn sin(n+1)) can be written using another identity. Maybe using product-to-sum formulas. For example, sin(n+1) = sinn cos1 + cosn sin1. So 1/(sinn sin(n+1)) = 1/(sinn (sinn cos1 + cosn sin1)). But I don't see an immediate simplification here. Alternatively, maybe express 1/(sinn sin(n+1)) as 1/(sin1) [cotn - cot(n+1)], which is the identity we already used. Then the sum telescopes to cot1 / sin1. Therefore, the answer is cot1° / sin1°, which is equivalent to cos1° / (sin1°)^2. But maybe the problem expects a numerical value or a simpler form. But given that 1° is not a standard angle, and the answer is expressed in terms of cot and sin, perhaps this is the simplest form. Therefore, the answer is cot1° / sin1°, which can also be written as cot1° csc1°. Alternatively, in terms of cosecant squared: 2cot1° csc1° = csc1°(cot1° + cot1°), but not sure if that helps. Wait, perhaps the answer is cot1° / sin1°, which is equivalent to (cos1°/sin1°)/sin1° = cos1° / sin²1°. But maybe they want it in terms of cot and csc, but unless there's a further identity, this might be as simplified as it gets. Alternatively, since the problem says "simplify," and given part (1) simplified to 1/sin1°, maybe part (2) also simplifies to something similar. Let me check the numerical values. For example, cot1° / sin1° is approximately (57.28996) / 0.0174524 ≈ 57.29 / 0.01745 ≈ 3283. But 1/sin1° is approximately 57.298, so cot1° / sin1° is much larger. Alternatively, perhaps there's a different telescoping. Wait, but based on the identity, it's definitely telescoping to cot1° / sin1°, so unless there's a miscalculation here. Wait, let's recast the problem again. Each term 1/(sinn sin(n+1)) can be expressed as (cotn - cot(n+1))/sin1°, so the sum from n=1 to 89 is (cot1 - cot2 + cot2 - cot3 + ... + cot89 - cot90)/sin1° = (cot1 - cot90)/sin1° = (cot1 - 0)/sin1° = cot1 / sin1°. Therefore, the answer is cot1° / sin1°, which is equal to cos1° / sin²1°. Alternatively, using the identity 1 + cot²θ = csc²θ, so cotθ = sqrt(csc²θ -1), but that complicates it. Therefore, the simplest form is cot1° / sin1°, or cos1° / sin²1°. Maybe the answer is expressed as cot1° csc1°, but both are equivalent. Therefore, the answer is cot1° / sin1°, which is csc1° cot1°, but since they might prefer one form over another. Since part (1) was 1/sin1°, which is csc1°, and part (2) is cot1° / sin1°, which is csc1° cot1°, but unless there's a further simplification, I think this is acceptable. Therefore, problem (2)'s answer is cot1° / sin1°, which is cos1° / sin²1°. Now moving to problem (3): 2 sin2° + 4 sin4° + 6 sin6° + ... + 180 sin180°. So the sum is over even numbers multiplied by sine of that even degree. So the general term is 2k sin(2k)°, where k ranges from 1 to 90, since 2*90=180°. We need to compute S = sum_{k=1}^{90} 2k sin(2k)°. Hmm. This seems more challenging. Sums of k sinkθ can sometimes be evaluated using summation formulas in trigonometry. I recall that there is a formula for the sum_{k=1}^n k sin(kθ). Let me check if I can remember or derive it. One method is to use complex exponentials. Let me consider S = sum_{k=1}^n k sin(kθ). We can write sin(kθ) = (e^{ikθ} - e^{-ikθ})/(2i), so S = sum_{k=1}^n k * (e^{ikθ} - e^{-ikθ})/(2i) = 1/(2i) [sum_{k=1}^n k e^{ikθ} - sum_{k=1}^n k e^{-ikθ}]. Each sum is a standard arithmetic-geometric series. The sum sum_{k=1}^n k r^k is known to be r(1 - (n+1)r^n + n r^{n+1}) / (1 - r)^2. Therefore, if we let r = e^{iθ}, then sum_{k=1}^n k e^{ikθ} = e^{iθ}(1 - (n+1)e^{inθ} + n e^{i(n+1)θ}) / (1 - e^{iθ})^2. Similarly, sum_{k=1}^n k e^{-ikθ} = e^{-iθ}(1 - (n+1)e^{-inθ} + n e^{-i(n+1)θ}) / (1 - e^{-iθ})^2. Therefore, S = 1/(2i) [ e^{iθ}(1 - (n+1)e^{inθ} + n e^{i(n+1)θ}) / (1 - e^{iθ})^2 - e^{-iθ}(1 - (n+1)e^{-inθ} + n e^{-i(n+1)θ}) / (1 - e^{-iθ})^2 ]. This looks complicated, but perhaps can be simplified. Alternatively, maybe there's a real-valued formula. Alternatively, consider using the identity for sum_{k=1}^n k sin(kθ). I found a reference formula: sum_{k=1}^n k sin(kθ) = (sin(nθ/2) sin((n+1)θ/2)) / sin²(θ/2) - (n+1) cos((n+1/2)θ) / (2 sin(θ/2)). But I need to verify this. Alternatively, using recursion or generating functions. Alternatively, let's recall that the sum S = sum_{k=1}^n k sin(kθ) can be found by taking the imaginary part of sum_{k=1}^n k e^{ikθ}. So let's compute sum_{k=1}^n k e^{ikθ}. Let me denote z = e^{iθ}, then the sum is sum_{k=1}^n k z^k = z / (1 - z)^2 - (n + 1) z^{n + 1} / (1 - z)^2 + n z^{n + 2} / (1 - z)^2. Therefore, sum_{k=1}^n k z^k = z(1 - (n + 1) z^n + n z^{n + 1}) / (1 - z)^2. Therefore, the imaginary part of this sum is sum_{k=1}^n k sin(kθ) = Im [ z(1 - (n + 1) z^n + n z^{n + 1}) / (1 - z)^2 ]. This requires computing the imaginary part of a complex expression. Let's attempt it step by step. Let me denote z = e^{iθ} = cosθ + i sinθ. Then 1 - z = 1 - cosθ - i sinθ, so (1 - z)^2 = (1 - cosθ)^2 - (sinθ)^2 - 2i(1 - cosθ) sinθ. Alternatively, instead of expanding everything, maybe use trigonometric identities. Alternatively, consider using differentiation. Let me recall that sum_{k=1}^n sin(kθ) = (sin(nθ/2) sin((n + 1)θ/2)) / sin(θ/2). If we differentiate both sides with respect to θ, we can find sum_{k=1}^n k cos(kθ). But we need sum_{k=1}^n k sin(kθ). Alternatively, differentiate with respect to θ the sum of cos(kθ). Let me recall that sum_{k=1}^n cos(kθ) = [sin(nθ/2) cos((n + 1)θ/2)] / sin(θ/2). If we take the derivative with respect to θ of both sides, we get:sum_{k=1}^n (-k sin(kθ)) = [ (n/2 cos(nθ/2) cos((n + 1)θ/2) - sin(nθ/2) ( (n +1)/2 sin((n +1)θ/2) ) ) / sin(θ/2) - [sin(nθ/2) cos((n + 1)θ/2)] * ( (1/2) cos(θ/2) ) / sin²(θ/2) ]This is quite complicated, but perhaps manageable. However, the left side is -sum_{k=1}^n k sin(kθ). Therefore, solving for the sum gives:sum_{k=1}^n k sin(kθ) = - derivative of sum_{k=1}^n cos(kθ). But this might not be the most straightforward approach. Alternatively, refer to standard formulas. According to some references, the formula for the sum is:sum_{k=1}^n k sin(kθ) = (sin(nθ) - (n + 1) sin((n +1)θ) + n sin((n +2)θ)) / (4 sin²(θ/2)). Alternatively, this might be derived via summation by parts. Alternatively, let's look up the formula. Upon checking, the formula for sum_{k=1}^n k sin(kθ) is:[sin(nθ/2) ( (n +1) sin((n +1)θ/2) - n sin((n +1)θ/2 - θ) ) ] / (4 sin²(θ/2))But this is getting too convoluted. Alternatively, let's consider using complex numbers step-by-step. Let’s define S = sum_{k=1}^{n} k sin(kθ). Let’s express sin(kθ) as the imaginary part of e^{ikθ}, so S = Im [ sum_{k=1}^n k e^{ikθ} ].Let’s compute sum_{k=1}^n k z^k where z = e^{iθ}. The formula for this sum is z(1 - (n +1)z^n + n z^{n+1}) / (1 - z)^2. Therefore, sum_{k=1}^n k e^{ikθ} = e^{iθ}(1 - (n +1)e^{inθ} + n e^{i(n+1)θ}) / (1 - e^{iθ})^2.Let’s compute the denominator first: (1 - e^{iθ})^2 = (1 - cosθ -i sinθ)^2 = (1 - cosθ)^2 - (sinθ)^2 - 2i(1 - cosθ) sinθ. Expand (1 - cosθ)^2 - (sinθ)^2 = 1 - 2cosθ + cos²θ - sin²θ = 1 - 2cosθ + cos2θ. But 1 + cos2θ = 2cos²θ, so this becomes 2cos²θ - 2cosθ = 2cosθ(cosθ -1). Alternatively, perhaps express 1 - e^{iθ} in terms of sine: 1 - e^{iθ} = 1 - cosθ -i sinθ = 2 sin²(θ/2) - 2i sin(θ/2) cos(θ/2) = 2 sin(θ/2) [ sin(θ/2) - i cos(θ/2) ] = -2i sin(θ/2) e^{iθ/2}. Therefore, (1 - e^{iθ})^2 = (-2i sin(θ/2) e^{iθ/2})^2 = (-2i)^2 sin²(θ/2) e^{iθ} = -4 sin²(θ/2) e^{iθ}. Therefore, denominator is -4 sin²(θ/2) e^{iθ}. Now, the numerator is e^{iθ}(1 - (n +1)e^{inθ} + n e^{i(n+1)θ}) = e^{iθ} - (n +1)e^{i(n+1)θ} + n e^{i(n+2)θ}. Therefore, the entire sum is [e^{iθ} - (n +1)e^{i(n+1)θ} + n e^{i(n+2)θ}] / (-4 sin²(θ/2) e^{iθ}) = [1 - (n +1)e^{inθ} + n e^{i(n+1)θ}] / (-4 sin²(θ/2)). Therefore, sum_{k=1}^n k e^{ikθ} = [1 - (n +1)e^{inθ} + n e^{i(n+1)θ}] / (-4 sin²(θ/2)). Now, take the imaginary part of this to get S = sum_{k=1}^n k sin(kθ). Let’s compute the numerator's imaginary part:Im [1 - (n +1)e^{inθ} + n e^{i(n+1)θ}] = Im[1] - (n +1) Im[e^{inθ}] + n Im[e^{i(n+1)θ}] = 0 - (n +1) sin(nθ) + n sin((n +1)θ). Therefore, the imaginary part is - (n +1) sin(nθ) + n sin((n +1)θ). Hence, S = [ - (n +1) sin(nθ) + n sin((n +1)θ) ] / (-4 sin²(θ/2)) = [ (n +1) sin(nθ) - n sin((n +1)θ) ] / (4 sin²(θ/2)). Therefore, the formula is:sum_{k=1}^n k sin(kθ) = [ (n +1) sin(nθ) - n sin((n +1)θ) ] / (4 sin²(θ/2)). Let me check this formula with n=1: sum_{k=1}^1 k sin(kθ) = sinθ. The formula gives [2 sinθ - 1 sin2θ ] / (4 sin²(θ/2)). Compute numerator: 2 sinθ - sin2θ = 2 sinθ - 2 sinθ cosθ = 2 sinθ (1 - cosθ). Denominator: 4 sin²(θ/2) = 4 * ( (1 - cosθ)/2 ) = 2(1 - cosθ). Therefore, numerator / denominator = 2 sinθ (1 - cosθ) / [2(1 - cosθ)] = sinθ. Correct. Good, so the formula works for n=1. Let's try n=2: sum_{k=1}^2 k sin(kθ) = sinθ + 2 sin2θ. Formula: [3 sin2θ - 2 sin3θ] / (4 sin²(θ/2)). Let's compute:Numerator: 3 sin2θ - 2 sin3θ. Using trig identities: sin3θ = 3 sinθ - 4 sin³θ. So 3 sin2θ - 2(3 sinθ - 4 sin³θ) = 3 sin2θ - 6 sinθ + 8 sin³θ. Denominator: 4 sin²(θ/2) = 2(1 - cosθ). Not sure if this simplifies to sinθ + 2 sin2θ, but let's compute for θ=30°, for example. Let θ=30°, then sum is sin30 + 2 sin60 = 0.5 + 2*(√3/2) = 0.5 + √3 ≈ 0.5 + 1.732 ≈ 2.232. Formula: [3 sin60° - 2 sin90°]/[4 sin²15°] = [3*(√3/2) - 2*1]/[4*( (sqrt(2 - sqrt(3))/2 )² )] = [ (3√3/2 - 2) ] / [4*( (2 - sqrt(3))/4 ) ] = (3√3/2 - 2) / (2 - sqrt(3)). Rationalize denominator: multiply numerator and denominator by (2 + sqrt(3)):Numerator: (3√3/2 - 2)(2 + sqrt(3)) = (3√3/2)(2) + (3√3/2)(sqrt(3)) - 2*2 - 2*sqrt(3) = 3√3 + (9/2) -4 - 2√3 = (3√3 -2√3) + (9/2 -4) = √3 + 1/2 ≈ 1.732 + 0.5 ≈ 2.232. Denominator: (2 - sqrt(3))(2 + sqrt(3)) = 4 - 3 =1. Therefore, the formula gives √3 + 1/2 ≈ 2.232, which matches the direct calculation. So the formula works. Therefore, the formula is valid. Returning to problem (3): sum_{k=1}^{90} 2k sin(2k)°. Here, θ=2°, and n=90. But in the formula, θ is in radians, but since we are working in degrees, we need to be cautious. Wait, but trigonometric functions are the same in terms of degrees or radians in terms of identities, but differentiation and series depend on the unit. However, since we are using degrees, we need to ensure that θ=2°, and apply the formula. But let's confirm: In the formula, θ is in whatever unit we are using. Since our angles are in degrees, θ=2°, and n=90. Therefore, applying the formula:sum_{k=1}^{90} k sin(2k°) = [ (90 +1) sin(90*2°) - 90 sin(91*2°) ] / (4 sin²(2°/2)) = [91 sin180° - 90 sin182°] / (4 sin²1°). But sin180°=0, sin182°=sin(180°+2°)= -sin2°. Therefore, the numerator becomes 0 - 90*(-sin2°) = 90 sin2°. Denominator is 4 sin²1°. Therefore, the sum is 90 sin2° / (4 sin²1°) = (90/4) * [sin2° / sin²1°]. But the problem's sum is sum_{k=1}^{90} 2k sin(2k°) = 2 * sum_{k=1}^{90} k sin(2k°). Therefore, the total sum is 2 * [90 sin2° / (4 sin²1°)] = (180 sin2°) / (4 sin²1°) = (45 sin2°)/ sin²1°. Now, simplify sin2° in terms of sin1°. Using double-angle formula: sin2°=2 sin1° cos1°. Therefore:45 * 2 sin1° cos1° / sin²1° = 90 cos1° / sin1° = 90 cot1°. Therefore, problem (3)'s sum is 90 cot1°. Let me verify the steps again:1. For sum_{k=1}^{n} k sin(kθ), formula gives [ (n +1) sin(nθ) - n sin((n +1)θ) ] / (4 sin²(θ/2)).2. In problem (3), θ=2°, n=90.3. Plugging into formula: [91 sin(180°) -90 sin(182°)] / (4 sin²(1°)) = [0 -90*(-sin2°)] / (4 sin²1°) = 90 sin2° / (4 sin²1°).4. Multiply by 2: 2*(90 sin2°)/(4 sin²1°) = (180 sin2°)/(4 sin²1°) = (45 sin2°)/ sin²1°.5. sin2°=2 sin1° cos1°, so substitute: 45*(2 sin1° cos1°)/ sin²1° = 90 cos1°/ sin1° =90 cot1°.Therefore, the answer is 90 cot1°. But wait, the problem states "2 sin2° +4 sin4° +6 sin6° + ... +180 sin180°", which is indeed sum_{k=1}^{90} 2k sin(2k°). Therefore, the calculation is correct. Therefore, the answer for part (3) is 90 cot1°. To recap:1. The first problem telescopes due to cotangent identities, collapsing to 1/sin1°.2. The second problem also telescopes, but since it's a full consecutive sum from 1° to 89°, it results in cot1° / sin1°.3. The third problem uses the summation formula for k sin(kθ), leading to 90 cot1°.All answers involve cot1° or csc1°, which are known to be large values due to the small angle. Double-checking problem (3):Given the sum is over 2k sin2k° from k=1 to 90, which is 2*(1 sin2° +2 sin4° +...+90 sin180°). We applied the formula for sum_{k=1}^{n} k sin(kθ) with n=90 and θ=2°, obtaining 90 cot1°, which seems correct. Yes, all steps check out. Final Answer(1) boxed{dfrac{1}{sin 1^{circ}}}(2) boxed{dfrac{cot 1^{circ}}{sin 1^{circ}}}(3) boxed{90 cot 1^{circ}}</think>To simplify the given trigonometric sums, we use trigonometric identities and telescoping series properties. Problem (1)The sum is (frac{1}{sin 45^{circ} sin 46^{circ}} + frac{1}{sin 47^{circ} sin 48^{circ}} + cdots + frac{1}{sin 133^{circ} sin 134^{circ}}).1. Use the identity (frac{1}{sin A sin B} = frac{cot A - cot B}{sin(B - A)}).2. For each term (frac{1}{sin n^{circ} sin (n+1)^{circ}}), this becomes (frac{cot n^{circ} - cot (n+1)^{circ}}{sin 1^{circ}}).3. The series telescopes, leaving (frac{cot 45^{circ} - cot 134^{circ}}{sin 1^{circ}}).4. Since (cot 134^{circ} = -cot 46^{circ}) and other terms cancel out, the result simplifies to (frac{1}{sin 1^{circ}}).[boxed{dfrac{1}{sin 1^{circ}}}] Problem (2)The sum is (frac{1}{sin 1^{circ} sin 2^{circ}} + frac{1}{sin 2^{circ} sin 3^{circ}} + cdots + frac{1}{sin 89^{circ} sin 90^{circ}}).1. Use the same identity as in Problem (1).2. Each term (frac{1}{sin n^{circ} sin (n+1)^{circ}}) becomes (frac{cot n^{circ} - cot (n+1)^{circ}}{sin 1^{circ}}).3. The series telescopes, leaving (frac{cot 1^{circ} - cot 90^{circ}}{sin 1^{circ}}).4. Since (cot 90^{circ} = 0), the result is (frac{cot 1^{circ}}{sin 1^{circ}}).[boxed{dfrac{cot 1^{circ}}{sin 1^{circ}}}] Problem (3)The sum is (2 sin 2^{circ} + 4 sin 4^{circ} + 6 sin 6^{circ} + cdots + 180 sin 180^{circ}).1. Use the formula for the sum (sum_{k=1}^{n} k sin(ktheta)).2. For (theta = 2^{circ}) and (n = 90), the formula simplifies to (frac{(91 sin 180^{circ} - 90 sin 182^{circ})}{4 sin^2 1^{circ}}).3. Since (sin 180^{circ} = 0) and (sin 182^{circ} = -sin 2^{circ}), the result is (frac{90 sin 2^{circ}}{4 sin^2 1^{circ}}).4. Simplifying using (sin 2^{circ} = 2 sin 1^{circ} cos 1^{circ}) gives (90 cot 1^{circ}).[boxed{90 cot 1^{circ}}],<think>Okay, so I need to solve these three trigonometric simplification problems. Let me take them one at a time. Starting with problem (1): The expression is a sum of terms like 1/(sin a° sin (a+1)°), starting from a = 45° up to a = 133°, since the last term is 1/(sin133° sin134°). So that's from 45 to 133, but each term increases by 2 degrees? Wait, no, let's check. The first term is 45 and 46, the next is 47 and 48, so each term's angles increase by 2 degrees each time. So the difference between the first angle of consecutive terms is 2°, which would mean the number of terms is (133 - 45)/2 + 1 = (88)/2 +1 = 44 +1=45 terms? Wait, 133 -45 is 88, divided by 2 is 44, plus 1 is 45. Hmm, okay, 45 terms in total. But how do I approach summing 1/(sin a° sin (a+1)°)? Maybe there's a telescoping identity here. Let me recall some trigonometric identities. There's an identity for cotangent differences or sums. Let me think. The formula for cot x - cot y in terms of sine functions. Yes, I remember that cot x - cot y = (sin(y - x))/(sin x sin y). Let me verify this. Let's start with cot x - cot y:cot x - cot y = (cos x / sin x) - (cos y / sin y) = (cos x sin y - cos y sin x)/(sin x sin y) = sin(y - x)/sin x sin y. Wait, sin(y - x) = sin y cos x - cos y sin x. So yes, that's exactly the numerator. Therefore, cot x - cot y = sin(y - x)/(sin x sin y). So rearranged, 1/(sin x sin y) = (cot x - cot y)/sin(y - x). In our case, each term is 1/(sin a° sin (a+1)°). So here, x = a°, y = a+1°, so y - x = 1°, so sin(y - x) = sin1°. Therefore, 1/(sin a sin (a+1)) = (cot a - cot (a+1))/sin1°. Therefore, each term in the sum can be written as (cot a° - cot (a+1)°)/sin1°. Therefore, the entire sum is [ (cot45° - cot46°) + (cot47° - cot48°) + ... + (cot133° - cot134°) ] / sin1°. Wait, hold on. Wait, the first term is a=45°, so (cot45 - cot46)/sin1°, next term a=47°, so (cot47 - cot48)/sin1°, etc., up to a=133°, which gives (cot133 - cot134)/sin1°. So the sum becomes [ (cot45 - cot46) + (cot47 - cot48) + ... + (cot133 - cot134) ] / sin1°. Hmm, this seems like a telescoping series, but there are gaps between the terms. Because after cot46, the next term is cot47, so the negative cot46 and positive cot47 don't cancel. Similarly, between each pair of terms, there's a gap. Wait, unless there's a pattern here. Let me see how many terms there are. Starting at a=45°, then a=47°, ..., up to a=133°. So the angles a are 45,47,49,...,133. How many terms? Let me compute. From 45 to 133 inclusive, stepping by 2. The number of terms is (133 -45)/2 +1 = (88)/2 +1=44+1=45 terms. So 45 terms. So each term is (cot a - cot(a+1)) where a is odd degrees from 45 to 133. Then when we add them up, the sum is [ (cot45 - cot46) + (cot47 - cot48) + ... + (cot133 - cot134) ] / sin1°. If I can pair the terms cleverly or see if there is cancellation. Let me write out the first few and the last few terms:First few terms: cot45 - cot46 + cot47 - cot48 + cot49 - cot50 + ... + cot133 - cot134.So all the inner terms are -cot46 + cot47 - cot48 + cot49 - ... -cot134. But since every other term is negative and positive, but the intermediate cotangent terms don't cancel because they are separated by one degree. Wait, unless there's a symmetry here. Let me check if cot(180 - x) = -cotx. Because 180 - x is in degrees. Since cot(180 - x) = cos(180 -x)/sin(180 -x) = (-cosx)/sinx = -cotx. Yes, that's correct. So cot(180 -x) = -cotx. So cot134° = cot(180 -46°) = -cot46°. Similarly, cot133° = cot(180 -47°) = -cot47°, cot132° = -cot48°, etc. Wait, let me check:134° = 180° -46°, so cot134° = cot(180° -46°) = -cot46°.Similarly, cot133° = cot(180° -47°) = -cot47°, and cot135° would be cot(180 -45) = -cot45°, but 135 is beyond our range. Wait, in the sum, the last term is cot133 - cot134. Let's substitute cot134 = -cot46, so cot133 - (-cot46) = cot133 + cot46. But cot133 = cot(180 -47) = -cot47. So cot133 + cot46 = -cot47 + cot46. Hmm, that's similar to the first term, which was cot45 - cot46. Wait, perhaps if I group the terms from the start and end towards the middle? Let's see. Let's pair the first term and the last term:First term: cot45 - cot46Last term: cot133 - cot134 = -cot47 - (-cot46) = -cot47 + cot46Wait, adding these two: (cot45 - cot46) + (-cot47 + cot46) = cot45 - cot47Wait, but that's not telescoping further. Maybe pair the first term with the term that would cancel the -cot46? But the next term after the first term is (cot47 - cot48). So the first term is cot45 - cot46, then cot47 - cot48. So if I group them as (cot45 - cot46 + cot47 - cot48), that becomes cot45 - cot48 + ( -cot46 + cot47 ). Hmm, not helpful.Alternatively, maybe consider the entire sum:Sum = [cot45 - cot46 + cot47 - cot48 + ... + cot133 - cot134]But notice that cot45, cot47, cot49,..., cot133 are all the cotangents of odd degrees from 45 to 133. And the negative terms are cot46, cot48,..., cot134, which are even degrees from 46 to 134. But maybe if I express cot45 + cot47 + ... + cot133 and subtract (cot46 + cot48 + ... + cot134). Then perhaps there is a symmetry here. Given that cot(180 - x) = -cotx, so for example, cot133 = cot(180 -47) = -cot47, cot131 = -cot49, etc. Let me see:The positive terms are from 45°, 47°, 49°, ..., 133°. Let me note that 133° is 180° -47°, 131° is 180° -49°, ..., 91° is 180° -89°, but 89° is less than 90°, but our terms start at 45°, so 180° -45°=135°, which is beyond the last term. Wait, the last term is 133°, which is 180°-47°, so if we look at cot45° + cot47° + ... + cot133°, each term from 45° to 133° in steps of 2°, which is 45 terms. But since 133°=180°-47°, cot133°= -cot47°, similarly cot131°= -cot49°, etc. So perhaps the sum of cot45° + cot47° + ... + cot133° is equal to cot45° + sum_{k=47}^{133} cotk°, which can be written as cot45° + sum_{k=47}^{89} cotk° + sum_{k=91}^{133} cotk°. Wait, but 91 to 133 can be expressed as 180 - (89 to 47). So sum_{k=91}^{133} cotk° = sum_{m=47}^{89} cot(180 -m)° = sum_{m=47}^{89} (-cotm)° = -sum_{m=47}^{89} cotm°. Therefore, the total sum cot45 + sum_{47}^{89} cotk + sum_{91}^{133} cotk = cot45 + sum_{47}^{89} cotk - sum_{47}^{89} cotk = cot45°. Similarly, the sum of the positive terms in the sum is cot45°. Similarly, the negative terms are cot46° + cot48° + ... + cot134°. Let's check the same idea. The sum from cot46° to cot134° in steps of 2°. 134° is 180° -46°, so cot134°= -cot46°, similarly cot132°= -cot48°, etc. So split the sum into two parts: cot46° + cot48° + ... + cot88° + cot90° + ... + cot134°. Wait, but 134° is even. Let's see:Starting from 46°, stepping by 2°, up to 134°. Let's see how many terms: (134 -46)/2 +1 = (88)/2 +1=44 +1=45 terms. So 45 terms. Now, let's split this sum at 90°. The terms from 46° to 88° (stepping by 2°), then 90°, then from 92° to 134°. But 90°, cot90° = 0, since cos90°=0. So the sum becomes (cot46° + cot48° + ... + cot88°) + 0 + (cot92° + ... + cot134°). Now, cot92° = cot(180 -88°)= -cot88°, cot94°= -cot86°, ..., up to cot134°= -cot46°. So the sum from 92° to134° is - (cot88° + cot86° + ... + cot46°). Therefore, the total negative sum is [cot46 + cot48 + ... + cot88] + [ - (cot88 + ... + cot46) ] = 0. Wait, but that can't be. Wait, but the sum from 46 to88 stepping by 2° is reversed and negated in the sum from 92 to134. Therefore, the sum cancels out, except for the middle term which is cot90°=0. Therefore, the total negative sum is zero. Wait, but wait, let's check with specific numbers. Suppose from 46° to134°, stepping by2°: 46,48,...,88,90,92,...,134. Split into 46-88 and 92-134. Each term in 92-134 is 180 - (88-46). So cot92° = cot(180-88°)= -cot88°, cot94°=cot(180-86°)= -cot86°, ..., cot134°=cot(180-46°)= -cot46°. Therefore, the sum from 92° to134° is -cot88° -cot86° - ... -cot46°. But the sum from46° to88° is cot46° + cot48° + ... + cot88°. Therefore, adding them together: cot46° + ... + cot88° - (cot88° + ... + cot46°) =0. So the sum is zero. Therefore, the negative terms sum up to zero. Therefore, the entire sum in the numerator is cot45° -0 = cot45°. Therefore, the total expression is cot45° / sin1°. But cot45° is 1, so the sum is 1/sin1°. Wait, let me verify that again. So the positive terms sum to cot45°, the negative terms sum to zero, so the total sum is cot45° - 0 = cot45°, divided by sin1°, so 1/sin1°. Therefore, the answer to part (1) is 1/sin1°, which is csc1°. Hmm, that seems too straightforward. Let me check with a small example. Suppose instead of starting at 45°, let's take a smaller sum. For example, 1/(sin1°sin2°) + 1/(sin3°sin4°). According to the formula, this would be [cot1 - cot2 + cot3 - cot4]/sin1°. Then, using the same logic, positive terms cot1 + cot3, negative terms -cot2 -cot4. Then using cot(180 -x) = -cotx, but in this small case, angles are not near 180. So maybe this approach only works when the terms are symmetric around 90°. Wait, in problem (1), the terms go from 45° to133°, which is symmetric around 90°, since 45+133=178, 178/2=89°, close to 90°. Wait, not exactly symmetric. Wait, 45° to 133° is 89° on either side? Wait, 45° to 90° is 45°, 90° to135° is 45°, but 133° is 43° from 90°. Wait, no. Maybe not exactly symmetric. But when we considered the sum of cotangents from45° to133°, stepping by2°, each term in the upper half (above90°) cancels with a term in the lower half. Wait, but 133°=180-47°, so cot133°=-cot47°, and since 47° is part of the lower half. Similarly, cot131°=-cot49°, and so on. Therefore, the sum of cot45° + cot47° + ... + cot133° becomes cot45° + (cot47° + ... + cot89°) + (cot91° + ... + cot133°). But cot91°=cot(180-89°)=-cot89°, cot93°=-cot87°, etc., up to cot133°=-cot47°. So the sum from91° to133° is - (cot89° + ... + cot47°). Therefore, the total sum is cot45° + (cot47° + ... + cot89°) - (cot47° + ... + cot89°) = cot45°. Similarly, the negative terms sum from46° to134°, stepping by2°, which includes angles from46° up, and due to the same symmetry, they cancel out. Therefore, the sum is indeed cot45° / sin1°, which is 1/sin1°, so the answer is csc1°, which is 1/sin1°. Therefore, problem (1) simplifies to 1/sin1°. Moving on to problem (2): The expression is the sum from1° to89° of 1/(sin a° sin(a+1)°). So similar to problem (1), but starting at1°, ending at89°, each term is 1/(sin a sin(a+1)). So here, a goes from1 to89, but each term is consecutive. Wait, in problem (1), each term was separated by2°, but here the terms are consecutive. So the sum is 1/(sin1°sin2°) +1/(sin2°sin3°)+...+1/(sin89°sin90°). Wait, similar to problem (1), but the angles are consecutive. Let's try using the same identity. As before, 1/(sin a sin(a+1)) = (cot a - cot(a+1)) / sin1°. Because as before, cot a - cot(a+1) = sin((a+1)-a)/sin a sin(a+1) = sin1° / sin a sin(a+1). Therefore, 1/(sin a sin(a+1)) = (cot a - cot(a+1))/sin1°. Therefore, the entire sum becomes [ (cot1° - cot2°) + (cot2° - cot3°) + ... + (cot89° - cot90°) ] / sin1°. This is a telescoping series! All the intermediate terms cancel out. The first term is cot1°, then -cot2° + cot2° cancels, -cot3° + cot3° cancels, ..., up until -cot89° + cot89° cancels, leaving cot1° - cot90°. But cot90° is 0, since cos90°=0. Therefore, the sum simplifies to (cot1° -0)/sin1° = cot1° / sin1°. But cot1° is cos1°/sin1°, so cot1° / sin1° = cos1° / sin²1°. However, let me check if there's a simpler form. Alternatively, maybe there is another identity. Let's see:cot1° / sin1° = (cos1° / sin1°) / sin1° = cos1° / sin²1°. But is there a better way to write this? Alternatively, using the identity 1 + cot²θ = csc²θ, so cotθ = √(csc²θ -1). But I don't think that helps here. Alternatively, express in terms of double angles. Wait, perhaps another approach. Let me compute cot1° / sin1°:cot1° / sin1° = (cos1° / sin1°) / sin1° = cos1° / sin²1°. Alternatively, using the identity for cosecant squared: 1/sin²1° = 1 + cot²1°. Therefore, cot1° / sin1° = cot1° * sqrt(1 + cot²1°). But this seems more complicated. Alternatively, recall that cotθ = tan(90° -θ), so cot1° = tan89°. Therefore, cot1° / sin1° = tan89° / sin1°. But tan89° is very large, but not sure if helpful. Alternatively, maybe there's another telescoping identity that can lead to a simpler expression. Wait, but in the previous problem, the telescoping resulted in a single term. Here, the telescoping gives cot1° - cot90°, which is cot1°, and then divided by sin1°. So the answer is cot1° / sin1°. But maybe we can write this in terms of another trigonometric function. Let's compute:cot1° / sin1° = (cos1° / sin1°) / sin1° = cos1° / sin²1°. Alternatively, using double angle identity: sin2θ = 2 sinθ cosθ. But not sure. Alternatively, use the identity 1/sin²θ = 1 + cot²θ. Therefore, cos1°/sin²1° = cot1°(1 + cot²1°). Hmm, but that might not be helpful. Alternatively, use the identity for cotθ in terms of sine and cosine. Alternatively, express in terms of csc. Since 1/sinθ = cscθ, so 1/sin²1° = csc²1°, so cos1° / sin²1° = cos1° csc²1°. But that's still the same. Alternatively, check if this is equal to something like 2 cot2°, but let's compute 2 cot2°:2 cot2° = 2*(cos2°/sin2°) = (2 cos2°)/sin2°. Is this equal to cos1°/sin²1°? Let's compute:Using double angle identity: sin2° = 2 sin1° cos1°, so sin2° = 2 sin1° cos1°. Then 2 cos2° / sin2° = 2 cos2° / (2 sin1° cos1°) )= cos2° / (sin1° cos1°). But cos2° = 2cos²1° -1, so:cos2° / (sin1° cos1°) = (2cos²1° -1)/(sin1° cos1°). Compare this to cos1° / sin²1°. Let me compute cos1° / sin²1°:= cos1° / (sin1° * sin1°) = cot1° / sin1°. Hmm, not sure if they are equal. Let me compute numerical values. Let's compute cot1° / sin1°:cot1° ≈ 57.28996163sin1° ≈ 0.017452406Thus cot1° / sin1° ≈ 57.28996163 / 0.017452406 ≈ 3283. But 2 cot2°:cot2° ≈ 28.636253282 cot2° ≈ 57.27250656So different. Therefore, they are not equal. Therefore, perhaps the answer is simply cot1° / sin1°, but maybe the problem expects it written in terms of another trigonometric function. Alternatively, since the answer is cos1° / sin²1°, perhaps leave it as that, but maybe there's another telescoping step. Wait, but I don't see an immediate identity here. Wait, but perhaps the answer is intended to be cot1° / sin1°, which is equivalent to cos1° / sin²1°, and that's the simplest form. So maybe that's the answer. Alternatively, maybe there is an identity that can express this in terms of another function. Wait, alternatively, using product-to-sum identities? Let me try:We have 1/(sin a sin(a+1)) = 2/(cos1° - cos(2a+1)°). Wait, is that a valid identity? Let me recall that 1/(sin a sin b) can be expressed using cos(a - b) - cos(a + b). Yes, indeed, using the identity:cos(A - B) - cos(A + B) = 2 sin A sin B.Therefore, 2/(cos(A - B) - cos(A + B)) = 1/(sin A sin B). In our case, for each term, A = a°, B = (a+1)°, so A - B = -1°, and A + B = 2a +1°. Therefore,1/(sin a sin(a+1)) = 2 / [cos(-1°) - cos(2a +1°)] = 2 / [cos1° - cos(2a +1°)].But this might not be helpful for telescoping unless there is a way to telescope the sum. Let me check:If each term is 2 / [cos1° - cos(2a +1°)], then summing over a from1° to89°, we get sum_{a=1}^{89} 2 / [cos1° - cos(2a +1°)]. Hmm, not sure how that would telescope. Alternatively, perhaps integrating another approach. Wait, but since we already have the telescoping sum from the previous method leading to cot1° / sin1°, which is about 3283. However, maybe the problem expects an exact form, like in terms of cot or csc, so the answer would be cot1° / sin1°, or written as cot1° csc1°, since 1/sin1° is csc1°. Alternatively, express as cot1° * csc1°. But in terms of the problem statement, it just says "Simplify", so perhaps either form is acceptable. However, since in part (1) the answer was 1/sin1°, which is csc1°, and here it's cot1° / sin1°, which is cot1° csc1°, but maybe there's a different simplification. Wait, let me check with an example. Suppose I take a smaller sum, like sum_{a=1}^{2} 1/(sin a sin(a+1)). Then according to the telescoping sum, it would be [cot1 - cot3]/sin1°. But computing directly:1/(sin1 sin2) +1/(sin2 sin3). Compute using the identity: [cot1 - cot2 + cot2 - cot3]/sin1 = [cot1 - cot3]/sin1. But if we compute numerically, let's see:Approximate values:sin1 ≈0.017452406, sin2≈0.0348995, sin3≈0.0523359561/(0.017452406 *0.0348995) ≈1/(0.0006091)≈1642.41/(0.0348995 *0.052335956)≈1/(0.001827)≈547.2Total ≈1642.4 +547.2≈2189.6On the other hand, [cot1 - cot3]/sin1:cot1≈57.28996163, cot3≈19.081136657.28996163 -19.0811366≈38.20882538.208825 /0.017452406≈2189.6, which matches. So the formula works. So for the total sum from1 to89, it's [cot1 - cot90]/sin1° = cot1 / sin1°, as cot90=0. Therefore, the answer is indeed cot1° / sin1°, which is approximately 57.29 /0.01745≈3283. But since the problem says "Simplify", maybe we can write this in terms of a single trigonometric function. cot1° / sin1° = cos1° / sin²1°, which is as simplified as it gets. Unless there's another identity. Alternatively, perhaps use the double-angle identity. Let's see:We have sin2° = 2 sin1° cos1°, so sin1° = sin2°/(2 cos1°). Therefore, sin²1° = sin2°/(2 cos1°) * sin1°. Wait, not helpful. Alternatively, using the identity:sin(A) sin(B) = [cos(A - B) - cos(A + B)] / 2But this brings us back to the previous approach. Alternatively, if we consider the sum from problem (2) and problem (1), they both have similar structures but different ranges, leading to different results. But in problem (1), the sum was 1/sin1°, and here it's cot1°/sin1°, which is different. So perhaps we just leave it as cot1° divided by sin1°, or cos1° divided by sin squared1°, but in the problem statement, it's acceptable. Alternatively, the problem might have a typo or I might have made a mistake in applying the telescoping. Let me check again. The sum in problem (2) is sum_{a=1}^{89} 1/(sin a° sin(a+1)°). Applying the identity, each term is (cot a - cot(a+1))/sin1°. Therefore, the entire sum is [cot1 - cot2 + cot2 - cot3 + ... + cot89 - cot90]/sin1° = [cot1 - cot90]/sin1° = cot1/sin1°, since cot90=0. Yes, that's correct. Therefore, the answer is cot1° divided by sin1°, which is cot1° * csc1°, but written as cot1° / sin1°. Alternatively, expressed as 2 cot2° + 1, but let me check. Wait, let's see:cot1° / sin1° = cos1° / sin²1°But using the identity:cos1° = 1 - 2 sin²0.5°, but that might not help. Alternatively, using the double-angle identity:sin2° = 2 sin1° cos1°, so cos1° = sin2°/(2 sin1°). Therefore,cos1° / sin²1° = (sin2°/(2 sin1°)) / sin²1° = sin2° / (2 sin³1°). But that seems more complicated. Alternatively, perhaps rationalizing or something. But I think the answer is supposed to be cot1° / sin1°, which is equivalent to cos1° / sin²1°, and that's the simplified form. Therefore, problem (2) simplifies to cot1° / sin1°, or cos1° / sin²1°. Either form is acceptable, but since cot1° / sin1° is more compact, that might be the preferred answer. Now moving on to problem (3): The expression is the sum 2 sin2° +4 sin4° +6 sin6° + ... +180 sin180°. This is a sum of terms where each term is 2n sin2n°, starting from n=1 (2*1 sin2°) up to n=90 (2*90 sin180°). So the sum is sum_{n=1}^{90} 2n sin(2n°). We need to compute this sum. I recall that there is a formula for the sum of k sin kθ, but here it's sum_{k=1}^N k sin kθ. Let me recall the identity. The general formula for the sum S = sum_{k=1}^n k sin(kθ). There is a formula for this. Yes, using the identity:sum_{k=1}^n k sin(kθ) = (sin(nθ) - (n+1) sin((n+1)θ) + n sin((n+2)θ)) / (4 sin²(θ/2))But I need to verify this. Alternatively, another approach is to use complex exponentials or generating functions.Alternatively, use the identity:sum_{k=1}^n k sin(kθ) = ( sinθ + 2 sin2θ + 3 sin3θ + ... + n sin nθ )There's a formula for this. Let me check.Yes, the formula can be derived using summation by parts or using differentiation. Let me recall that:sum_{k=1}^n sin(kθ) = (sin(nθ/2) sin((n+1)θ/2)) / sin(θ/2)But that's the sum of sines. For the sum of k sin(kθ), let's differentiate with respect to θ. Let S = sum_{k=1}^n sin(kθ) = Im( sum_{k=1}^n e^{i kθ} ) = Im( e^{iθ} (1 - e^{i nθ}) / (1 - e^{iθ}) )But differentiating S with respect to θ gives sum_{k=1}^n k cos(kθ) = dS/dθ. But we need sum_{k=1}^n k sin(kθ). Hmm. Alternatively, integrating or differentiating another function. Alternatively, consider that the sum sum_{k=1}^n k sin(kθ) is the imaginary part of sum_{k=1}^n k e^{i kθ}. Let's compute that sum. Let me denote z = e^{iθ}, then sum_{k=1}^n k z^k = z (1 - (n+1) z^n + n z^{n+1}) / (1 - z)^2. Therefore, the imaginary part would be the imaginary part of that expression. Let me write this out:sum_{k=1}^n k sin(kθ) = Im [ z (1 - (n+1) z^n + n z^{n+1}) / (1 - z)^2 ]But this seems complicated. Let me try to compute it step by step. Let me first compute the sum sum_{k=1}^n k z^k = z / (1 - z)^2 - (n+1) z^{n+1} / (1 - z)^2 + n z^{n+2} / (1 - z)^2. But this is getting messy. Maybe there's a better approach. Alternatively, I remember that there is a formula:sum_{k=1}^n k sin(kθ) = [ sinθ - (n +1) sin(nθ) + n sin((n +1)θ) ] / (4 sin²(θ/2))But I need to verify this. Let me test for small n.Let n=1: sum is 1*sinθ. Formula gives [ sinθ - 2 sinθ +1 sin2θ ] / (4 sin²(θ/2)). Wait, [ sinθ -2 sinθ + sin2θ ] / (4 sin²(θ/2)) = [ -sinθ + sin2θ ] / (4 sin²(θ/2)). But sin2θ = 2 sinθ cosθ, so:[ -sinθ + 2 sinθ cosθ ] / (4 sin²(θ/2)) = sinθ ( -1 + 2 cosθ ) / (4 sin²(θ/2))But using identity: cosθ = 1 - 2 sin²(θ/2). So:-1 + 2 cosθ = -1 + 2(1 - 2 sin²(θ/2)) = -1 +2 -4 sin²(θ/2) =1 -4 sin²(θ/2)Thus, sinθ (1 -4 sin²(θ/2)) / (4 sin²(θ/2)). But for n=1, the sum is sinθ, so this formula does not give sinθ. Hence, the formula must be incorrect. Alternatively, maybe another formula. Another approach: Use recursion or derive the formula. Let me denote S = sum_{k=1}^n k sin(kθ). Multiply both sides by 2 sin(θ/2):2 sin(θ/2) S = sum_{k=1}^n k [ 2 sin(θ/2) sin(kθ) ]Using identity 2 sin A sin B = cos(A - B) - cos(A + B). So,2 sin(θ/2) sin(kθ) = cos(kθ - θ/2) - cos(kθ + θ/2) = cos((k - 0.5)θ) - cos((k +0.5)θ)Therefore,2 sin(θ/2) S = sum_{k=1}^n k [ cos((k -0.5)θ) - cos((k +0.5)θ) ]This telescopes when we expand the sum. Let's write out the terms:For k=1: 1 [ cos(0.5θ) - cos(1.5θ) ]For k=2: 2 [ cos(1.5θ) - cos(2.5θ) ]For k=3: 3 [ cos(2.5θ) - cos(3.5θ) ]...For k=n: n [ cos((n -0.5)θ) - cos((n +0.5)θ) ]Now, summing these terms:= cos0.5θ - cos1.5θ + 2 cos1.5θ - 2 cos2.5θ + 3 cos2.5θ -3 cos3.5θ + ... +n cos(n -0.5θ) -n cos(n +0.5θ)Now, group like terms:cos0.5θ + (-cos1.5θ + 2 cos1.5θ) + (-2 cos2.5θ +3 cos2.5θ) + ... + [ - (n -1) cos(n -0.5θ) + n cos(n -0.5θ) ] -n cos(n +0.5θ)Each group simplifies to:cos0.5θ + (1)cos1.5θ + (1)cos2.5θ + ... + (1)cos(n -0.5θ) -n cos(n +0.5θ)Therefore, the entire sum is sum_{m=0.5θ}^{(n -0.5)θ} cos(mθ) with step θ, but actually, it's sum_{k=1}^{n} cos((k -0.5)θ) -n cos((n +0.5)θ)Wait, no. Wait, the terms are:Starting with cos0.5θ, then for each k from1 to n, we have an additional term (k - (k-1)) cos((k -0.5)θ). Wait, actually, each coefficient is 1 for cos((k -0.5)θ), so total sum is sum_{k=1}^n cos((k -0.5)θ) + cos0.5θ -n cos((n +0.5)θ). Wait, no, wait:Wait, when expanding:First term: k=1 gives cos0.5θ -cos1.5θSecond term: k=2 gives 2cos1.5θ -2cos2.5θThird term: k=3 gives 3cos2.5θ -3cos3.5θ...k=n term: n cos(n -0.5θ) -n cos(n +0.5θ)So combining all terms:cos0.5θ + (-cos1.5θ +2cos1.5θ) + (-2cos2.5θ +3cos2.5θ) + ... + (- (n-1)cos(n -0.5θ) +n cos(n -0.5θ)) -n cos(n +0.5θ)Each of the middle terms simplifies to cos(k -0.5θ):For the first bracket: -cos1.5θ +2cos1.5θ = cos1.5θSecond bracket: -2cos2.5θ +3cos2.5θ=cos2.5θ...Last bracket before the final term: - (n-1)cos(n -0.5θ) +n cos(n -0.5θ)=cos(n -0.5θ)Therefore, the total sum is cos0.5θ + cos1.5θ + cos2.5θ + ... +cos(n -0.5θ) -n cos(n +0.5θ)So we have:2 sin(θ/2) S = sum_{k=0}^{n-1} cos((k +0.5)θ) -n cos((n +0.5)θ)Therefore, sum_{k=0}^{n-1} cos((k +0.5)θ) is another trigonometric sum. Let's compute that. The sum sum_{m=0}^{n-1} cos(a + m d), where a=0.5θ, d=θ. There's a formula for this:sum_{m=0}^{n-1} cos(a + m d) = [sin(n d /2) / sin(d /2)] cos(a + (n -1)d /2)In our case, a=0.5θ, d=θ, so:sum_{m=0}^{n-1} cos(0.5θ + mθ) = [sin(nθ /2)/sin(θ/2)] * cos(0.5θ + (n -1)θ/2 )Simplify the argument of the cosine:0.5θ + (n -1)θ/2 = 0.5θ + (nθ/2 - θ/2) = nθ/2Therefore, the sum becomes:[sin(nθ/2)/sin(θ/2)] * cos(nθ/2) = [ sin(nθ/2) cos(nθ/2) ] / sin(θ/2) = [ (1/2) sin(nθ) ] / sin(θ/2 )Therefore, sum_{k=0}^{n-1} cos((k +0.5)θ) = (1/2) sin(nθ) / sin(θ/2)Therefore, going back to our previous equation:2 sin(θ/2) S = (1/2) sin(nθ)/ sin(θ/2) -n cos((n +0.5)θ)Multiply both sides by sin(θ/2):2 sin²(θ/2) S = (1/2) sin(nθ) -n sin(θ/2) cos((n +0.5)θ)Therefore, solving for S:S = [ (1/2) sin(nθ) -n sin(θ/2) cos((n +0.5)θ) ] / (2 sin²(θ/2))Simplify:S = [ sin(nθ) - 2n sin(θ/2) cos((n +0.5)θ) ] / (4 sin²(θ/2))This seems complicated, but perhaps we can use another identity on the term 2n sin(θ/2) cos((n +0.5)θ). Using the identity: 2 sin A cos B = sin(A + B) + sin(A - B). Let A=θ/2, B=(n +0.5)θ:2 sin(θ/2) cos((n +0.5)θ) = sin(θ/2 + (n +0.5)θ) + sin(θ/2 - (n +0.5)θ) = sin((n +1)θ) + sin(-nθ) = sin((n +1)θ) - sin(nθ)Therefore, substituting back:S = [ sin(nθ) -n ( sin((n +1)θ) - sin(nθ) ) ] / (4 sin²(θ/2))= [ sin(nθ) -n sin((n +1)θ) +n sin(nθ) ] / (4 sin²(θ/2))= [ (1 +n) sin(nθ) -n sin((n +1)θ) ] / (4 sin²(θ/2))Therefore, the formula for the sum S = sum_{k=1}^n k sin(kθ) is:S = [ (n +1) sin(nθ) -n sin((n +1)θ) ] / (4 sin²(θ/2))Let me verify this formula with a small n. Let's take n=1, θ=θ. Then:S = [2 sinθ -1 sin2θ]/(4 sin²(θ/2))But sin2θ =2 sinθ cosθ, so:[2 sinθ -2 sinθ cosθ]/(4 sin²(θ/2)) = [2 sinθ(1 - cosθ)] / (4 sin²(θ/2))Using the identity 1 - cosθ = 2 sin²(θ/2), so:[2 sinθ *2 sin²(θ/2)] / (4 sin²(θ/2)) = [4 sinθ sin²(θ/2)] / [4 sin²(θ/2)] = sinθ, which is correct for n=1. Another test with n=2, θ=θ. Then S=1 sinθ +2 sin2θ. The formula:[3 sin2θ -2 sin3θ]/(4 sin²(θ/2))Let me compute for θ=90°, n=2:S=sin90 +2 sin180=1 +0=1Formula: [3 sin180 -2 sin270]/(4 sin²45) = [0 -2*(-1)]/(4*(√2/2)^2)= [2]/(4*(0.5))=2 /2=1. Correct. Therefore, the formula seems valid. Therefore, applying this formula to problem (3). Problem (3): sum_{k=1}^{90} 2k sin(2k°). Wait, in our formula, the sum is sum_{k=1}^n k sin(kθ). Here, each term is 2k sin(2k°). So if we let θ=2°, then the term is 2k sin(k*2°)=2k sin(kθ). So the sum is 2 * sum_{k=1}^{90} k sin(kθ), where θ=2°. Therefore, using the formula, sum_{k=1}^n k sin(kθ) = [ (n +1) sin(nθ) -n sin((n +1)θ) ] / (4 sin²(θ/2))Here, n=90, θ=2°, so:sum_{k=1}^{90} k sin(2k°) = [91 sin(90*2°) -90 sin(91*2°)] / (4 sin²(1°))But sin(180°)=0, sin(182°)=sin(180°+2°)= -sin2°. Therefore:sum_{k=1}^{90} k sin(2k°) = [91*0 -90*(-sin2°)] / (4 sin²1°)= [90 sin2°] / (4 sin²1°)= [45 sin2°]/(2 sin²1°)But the original problem's sum is 2*sum_{k=1}^{90}k sin(2k°), so multiply by 2:Total sum = 2 * [45 sin2° / (2 sin²1°)] = 45 sin2° / sin²1°Therefore, the total sum is 45 sin2° / sin²1°. Now, simplify this expression. First, note that sin2°=2 sin1° cos1°, so substituting:45 * 2 sin1° cos1° / sin²1° = 90 sin1° cos1° / sin²1° =90 cos1° / sin1°=90 cot1°Therefore, the sum simplifies to 90 cot1°. But let me verify this:sin2°=2 sin1° cos1°, so sin2° / sin²1°= 2 cos1° / sin1°, therefore, 45 sin2° / sin²1°=45*2 cos1° / sin1°=90 cot1°. Yes, correct. Therefore, problem (3) simplifies to 90 cot1°. Therefore, the three answers are:(1) 1/sin1°=csc1°,(2) cot1° / sin1°=cot1° csc1°,(3) 90 cot1°.But let me check if there's a way to express answer (2) differently. Wait, earlier, for problem (2), we had the sum equal to cot1° / sin1°, which is equivalent to 90 cot1° when n=89, but no, they are different. But problem (3)'s answer is 90 cot1°, which is a different expression. Alternatively, let me verify problem (3) with a small n. Suppose n=1: sum is 2*1 sin2°, which is 2 sin2°. Using the formula, sum_{k=1}^1 2k sin2k°=2 sin2°. According to the formula, 45 cot1°, but wait for n=1 in problem (3), n=1 would correspond to the upper limit being 2°, but in the original problem, n=90. Therefore, the formula applies correctly for n=90. Therefore, the final answers are:(1) boxed{dfrac{1}{sin 1^circ}},(2) boxed{dfrac{cot 1^circ}{sin 1^circ}},(3) boxed{90 cot 1^circ}Final Answer(1) boxed{dfrac{1}{sin 1^circ}};(2) boxed{dfrac{cot 1^circ}{sin 1^circ}};(3) boxed{90 cot 1^circ}</think>To solve the given trigonometric simplification problems, we use trigonometric identities and telescoping series techniques. Problem (1)The expression is:[frac{1}{sin 45^{circ} sin 46^{circ}} + frac{1}{sin 47^{circ} sin 48^{circ}} + cdots + frac{1}{sin 133^{circ} sin 134^{circ}}]Using the identity:[frac{1}{sin a sin (a+1)} = frac{cot a - cot (a+1)}{sin 1^{circ}}]Each term can be written as:[frac{cot a - cot (a+1)}{sin 1^{circ}}]Summing these terms from (a = 45^{circ}) to (a = 133^{circ}) (in steps of (2^{circ})) results in a telescoping series. The intermediate terms cancel out, leaving:[frac{cot 45^{circ} - cot 134^{circ}}{sin 1^{circ}}]Using (cot 45^{circ} = 1) and (cot 134^{circ} = -cot 46^{circ}) (due to symmetry), the series simplifies to:[frac{1}{sin 1^{circ}}] Problem (2)The expression is:[frac{1}{sin 1^{circ} sin 2^{circ}} + frac{1}{sin 2^{circ} sin 3^{circ}} + cdots + frac{1}{sin 89^{circ} sin 90^{circ}}]Using the same identity as above, each term can be written as:[frac{cot a - cot (a+1)}{sin 1^{circ}}]Summing these terms from (a = 1^{circ}) to (a = 89^{circ}) results in a telescoping series. The intermediate terms cancel out, leaving:[frac{cot 1^{circ} - cot 90^{circ}}{sin 1^{circ}}]Since (cot 90^{circ} = 0), the series simplifies to:[frac{cot 1^{circ}}{sin 1^{circ}}] Problem (3)The expression is:[2 sin 2^{circ} + 4 sin 4^{circ} + 6 sin 6^{circ} + cdots + 180 sin 180^{circ}]This sum can be written as:[sum_{k=1}^{90} 2k sin (2k^{circ})]Using the formula for the sum of (k sin(ktheta)) and simplifying, we find that the sum telescopes to:[90 cot 1^{circ}] Final Answers(1) (boxed{dfrac{1}{sin 1^circ}})(2) (boxed{dfrac{cot 1^circ}{sin 1^circ}})(3) (boxed{90 cot 1^circ})]

🔑:begin{array}{l}text { 7. Since } frac{sin 1^{circ}}{sin k^{circ} sin (k+1)^{circ}}=frac{sin left[(k+1)^{circ}-k^{circ}right]}{sin k^{circ} sin (k+1)^{circ}}=frac{sin (k+1)^{circ} cos k^{circ}-sin k^{circ} cos (k+1)^{circ}}{sin k^{circ} sin (k+1)^{circ}}=cot k^{circ}- cos (k+1)^{circ}, text { hence } frac{1}{sin k^{circ} sin (k+1)^{circ}}=frac{1}{sin 1^{circ}}left[cot k^{circ}-cot (k+1)^{circ}right] text { (1) } frac{1}{sin 45^{circ} sin 46^{circ}}+frac{1}{sin 47^{circ} sin 48^{circ}}+cdots+frac{1}{sin 133^{circ} sin 134^{circ}} =frac{1}{sin 1^{circ}}left[left(cot 45^{circ}-cot 46^{circ}right)+left(cot 47^{circ}-cot 48^{circ}right)+cdots+left(cot 133^{circ}-cot 134^{circ}right)right] =frac{1}{sin 1^{circ}}left[left(cot 45^{circ}-cot 46^{circ}right)+left(cot 47^{circ}-cot 48^{circ}right)+cdots+left(cot 89^{circ}-cot 90^{circ}right)+left(cot 88^{circ}-cot 89^{circ}right)+left(cot 86^{circ}right.right. left.left.-cot 87^{circ}right)+cdots+left(cot 46^{circ}-cot 47^{circ}right)right]=frac{1}{sin 1^{circ}}left(cot 45^{circ}-cot 90^{circ}right)=frac{1}{sin 1^{circ}} text { (2) } frac{1}{sin 1^{circ} sin 2^{circ}}+frac{1}{sin 2^{circ} sin 3^{circ}}+cdots+frac{1}{sin 89^{circ} sin 90^{circ}}=frac{1}{sin 1^{circ}}left[left(cot 1^{circ}-cos 2^{circ}right)+left(cot 2^{circ}-cot 3^{circ}right)+cdots+right. left.left(cot 89^{circ}-cot 90^{circ}right)right]=frac{1}{sin 1^{circ}}left(cot 1^{circ}-cot 90^{circ}right)=frac{cos 1^{circ}}{sin 1^{circ}}end{array}begin{array}{l}text { (3) Since } 2 sin 2 k^{circ} cdot sin 1^{circ}=cos (2 k-1)^{circ}-cos (2 k+1)^{circ} text { Hence } 2 sin 2^{circ} cdot sin 1^{circ}+2left(2 sin 4^{circ} cdot sin 1^{circ}right)+3left(2 sin 6^{circ} cdot sin 1^{circ}right)+cdots+89 cdotleft(2 sin 178^{circ} cdot sin 1^{circ}right) =left(cos 1^{circ}-cos 3^{circ}right)+2left(cos 3^{circ}-cos 5^{circ}right)+3left(cos 5^{circ}-cos 7^{circ}right)+cdots+89left(cos 177^{circ}-cos 179^{circ}right) =cos 1^{circ}+cos 3^{circ}+cos 5^{circ}+cdots+cos 177^{circ}-89 cos 179^{circ} =cos 1^{circ}+89 cos 1^{circ}=90 cos 1^{circ}end{array}Therefore, 2 sin 2^{circ}+4 sin 4^{circ}+6 sin 6^{circ}+cdots+180 sin 180^{circ}=90 cot 1^{circ}.

🧠:[<think>Alright, let's try to tackle this geometry problem. Hmm, it's about a rhombus and a triangle both circumscribed around a circle with radius 1. The triangle has two sides parallel to the diagonals of the rhombus and the third side is parallel to one of the sides of the rhombus and is equal to 5. We need to find the side of the rhombus.First, let's recall some properties of rhombuses and circumscribed figures. A rhombus has all sides equal, and its diagonals are perpendicular bisectors of each other. Also, if a circle is inscribed in a rhombus, the radius of the circle is equal to the height of the rhombus divided by 2. Wait, but actually, in a rhombus, the area can be calculated as (diagonal1 * diagonal2)/2 or as side * height. Since the circle is inscribed, the radius is related to the area and the perimeter. For any tangential quadrilateral (which a rhombus is), the area is equal to the product of the semiperimeter and the radius of the inscribed circle. So, for the rhombus, the semiperimeter is (4 * side)/2 = 2 * side. Therefore, area = 2 * side * radius. Since the radius is 1, the area would be 2 * side * 1 = 2 * side.But also, area of the rhombus can be expressed as (d1 * d2)/2, where d1 and d2 are the diagonals. So, we have 2 * side = (d1 * d2)/2 => d1 * d2 = 4 * side. That's one equation involving the diagonals.Now, the problem also mentions a triangle that is circumscribed around the same circle. So, this triangle is also tangential, meaning it has an incircle tangent to all three sides. The radius of the incircle is 1. For a triangle, the area is equal to the semiperimeter multiplied by the inradius. So, area of triangle = (a + b + c)/2 * 1 = (a + b + c)/2, where a, b, c are the sides of the triangle. But we know that two sides of the triangle are parallel to the diagonals of the rhombus, and the third side is parallel to a side of the rhombus and has length 5. Let me visualize this. Let's denote the rhombus as ABCD with diagonals AC and BD intersecting at O. The diagonals are perpendicular, and they bisect each other. The sides of the rhombus are all equal, let's say each side is 's'. The triangle in question has two sides parallel to the diagonals AC and BD, and the third side is parallel to, say, side AB and is of length 5.Wait, but how exactly is the triangle positioned? Since it's circumscribed around the same circle, which is tangent to all three sides. The triangle's sides are related to the rhombus's diagonals and a side. Let me try to think step by step.First, maybe it's helpful to assign coordinates to the rhombus. Let's place the rhombus so that its center is at the origin (0,0), and its diagonals are along the x-axis and y-axis. Let the diagonals be AC along the x-axis and BD along the y-axis. Then, the vertices of the rhombus would be at (d1/2, 0), (-d1/2, 0), (0, d2/2), (0, -d2/2). Wait, no, actually, since diagonals intersect at the center, each vertex is a combination of half-diagonals. So, vertices would be (d1/2, 0), (-d1/2, 0), (0, d2/2), (0, -d2/2). Wait, that seems incorrect. Wait, no, in a rhombus, the diagonals split each other into halves. So, if the diagonals are of length d1 and d2, then the vertices are at (d1/2, 0), (0, d2/2), (-d1/2, 0), (0, -d2/2). But then connecting these points would form a rhombus with sides of length sqrt[(d1/2)^2 + (d2/2)^2], which is indeed the formula for the side length of a rhombus: s = sqrt[(d1/2)^2 + (d2/2)^2] = (sqrt(d1^2 + d2^2))/2.But we also know from earlier that d1 * d2 = 4 * s. So, maybe we can set up equations here. Let's note that s = sqrt(d1^2 + d2^2)/2, and d1 * d2 = 4s. Let me square both sides of the s equation: s^2 = (d1^2 + d2^2)/4. So, 4s^2 = d1^2 + d2^2. Also, d1*d2 = 4s. Hmm, so we have two equations:1) d1^2 + d2^2 = 4s^22) d1*d2 = 4sWe can perhaps use these to find a relationship. Let me square the second equation: (d1*d2)^2 = 16s^2. So, d1^2*d2^2 = 16s^2. But from the first equation, d1^2 + d2^2 = 4s^2. Hmm, maybe express (d1^2 + d2^2)^2 = d1^4 + 2d1^2*d2^2 + d2^4 = (4s^2)^2 = 16s^4. But we also have d1^2*d2^2 = 16s^2. Let me see if that's helpful. Alternatively, perhaps we can use substitution. Let me denote x = d1 and y = d2. Then, we have:x^2 + y^2 = 4s^2xy = 4sLet me try to express x^2 + y^2 in terms of (x + y)^2. We know (x + y)^2 = x^2 + 2xy + y^2. So, x^2 + y^2 = (x + y)^2 - 2xy. Plugging into the first equation:(x + y)^2 - 2xy = 4s^2But from the second equation, xy = 4s. So,(x + y)^2 - 2*(4s) = 4s^2=> (x + y)^2 = 4s^2 + 8sHmm, but we don't know x + y. Alternatively, maybe express x in terms of y from the second equation: x = 4s/y. Then plug into the first equation:(4s/y)^2 + y^2 = 4s^216s^2/y^2 + y^2 = 4s^2Multiply both sides by y^2:16s^2 + y^4 = 4s^2*y^2Bring all terms to one side:y^4 - 4s^2*y^2 + 16s^2 = 0Let me set z = y^2, so equation becomes:z^2 - 4s^2*z + 16s^2 = 0This is a quadratic in z. Let's apply quadratic formula:z = [4s^2 ± sqrt(16s^4 - 64s^2)] / 2= [4s^2 ± sqrt(16s^2(s^2 - 4))]/2= [4s^2 ± 4s*sqrt(s^2 - 4)]/2= 2s^2 ± 2s*sqrt(s^2 - 4)But since z = y^2 must be positive, and s is positive, the discriminant must be non-negative, so s^2 - 4 ≥ 0 => s ≥ 2. Hmm, so the side length must be at least 2. Interesting.But maybe this approach is getting too complicated. Let's see if we can find another way using the triangle.The triangle is circumscribed around the same circle (radius 1), with two sides parallel to the diagonals of the rhombus and the third side parallel to a side of the rhombus and equal to 5. Let's try to model this triangle.First, since the triangle is tangential and has an inradius of 1, its area is equal to its semiperimeter. Let's denote the sides of the triangle as a, b, c. The semiperimeter is (a + b + c)/2, so area = (a + b + c)/2 * 1 = (a + b + c)/2.But we know that two sides are parallel to the diagonals of the rhombus, and the third side is parallel to a side of the rhombus and is 5. Let's denote the two sides parallel to the diagonals as m and n, corresponding to lengths parallel to d1 and d2. Wait, but lengths in which proportion?Wait, if two sides of the triangle are parallel to the diagonals of the rhombus, then their lengths would be proportional to the diagonals. But how exactly? Let's think. Suppose the triangle has sides parallel to d1, d2, and a side of the rhombus. Let me consider the triangle. If two sides are parallel to the diagonals, which are perpendicular to each other (since rhombus diagonals are perpendicular), then the triangle must be a right triangle? Because two sides are perpendicular, so the triangle would have a right angle where those two sides meet. Then, the third side would be the hypotenuse. But the problem states the third side is parallel to a side of the rhombus and is equal to 5. Hmm, but in a rhombus, the sides are not perpendicular unless it's a square. So, maybe the triangle is a right triangle with legs parallel to the diagonals of the rhombus and hypotenuse parallel to a side of the rhombus, with length 5.Alternatively, maybe the triangle is not right-angled, but has two sides parallel to the diagonals (which are perpendicular), so those sides would be perpendicular? Wait, if two sides are parallel to the diagonals, which are perpendicular, then those sides of the triangle are perpendicular. So, the triangle is right-angled. So, the triangle is a right-angled triangle with legs parallel to the diagonals of the rhombus and hypotenuse parallel to a side of the rhombus, length 5. That seems plausible.So, if the triangle is right-angled, with legs m and n (parallel to d1 and d2) and hypotenuse 5 (parallel to the rhombus side). Let's denote the legs as m and n. Then, by Pythagoras, m^2 + n^2 = 5^2 = 25.Also, since the triangle is circumscribed around a circle with radius 1, its area is equal to the semiperimeter. The area of the right-angled triangle is (m * n)/2. The semiperimeter is (m + n + 5)/2. Therefore, according to the formula for the area of a tangential polygon:Area = semiperimeter * inradiusSo,(m * n)/2 = (m + n + 5)/2 * 1Multiply both sides by 2:m * n = (m + n + 5)/1Thus,m * n = m + n + 5So, we have two equations:1) m^2 + n^2 = 252) m * n = m + n + 5We need to solve these equations for m and n.Let me denote S = m + n, P = m * n. Then, from equation 2, P = S + 5. From equation 1, m^2 + n^2 = 25. But m^2 + n^2 = (m + n)^2 - 2mn = S^2 - 2P = 25. Substitute P = S + 5:S^2 - 2(S + 5) = 25S^2 - 2S - 10 = 25S^2 - 2S - 35 = 0Solve this quadratic equation:S = [2 ± sqrt(4 + 140)] / 2 = [2 ± sqrt(144)] / 2 = [2 ± 12]/2Discarding the negative solution because S = m + n must be positive:S = (2 + 12)/2 = 14/2 = 7Thus, m + n = 7, and m * n = 7 + 5 = 12. So, we have:m + n = 7m * n = 12Therefore, m and n are the roots of x^2 - 7x + 12 = 0, which factors as (x - 3)(x - 4) = 0. So, m = 3, n = 4 or vice versa.So, the legs of the triangle are 3 and 4, hypotenuse 5. That's a classic 3-4-5 right triangle. Makes sense.Now, we need to relate this triangle to the rhombus. The legs of the triangle (3 and 4) are parallel to the diagonals of the rhombus. So, the lengths 3 and 4 correspond to the directions of the diagonals, but not necessarily their actual lengths. Since the triangle is circumscribed around the same circle as the rhombus, but with sides parallel to the diagonals. Wait, but how exactly are the sides of the triangle related to the diagonals?Wait, perhaps the sides of the triangle are scaled versions of the diagonals. Since they are parallel, they could be similar in some way. Let me think. If two sides of the triangle are parallel to the diagonals of the rhombus, then the triangle's sides are in the same direction as the diagonals. However, the lengths of these sides (3 and 4) might be proportional to the lengths of the diagonals. Let's consider that.Suppose the diagonals of the rhombus are d1 and d2. The sides of the triangle parallel to these diagonals are 3 and 4. Since the triangle is a right triangle with legs 3 and 4, those legs are aligned with the diagonals. However, the actual lengths of the diagonals may be different. But how?Wait, maybe the sides of the triangle are scaled by some factor relative to the diagonals. Since they are parallel, the ratio of lengths could be related to the scaling factor. Let's think about homothety or similar figures.Alternatively, perhaps the lengths of the triangle's sides (3 and 4) are related to the rhombus's diagonals through the inradius. Wait, but both the rhombus and the triangle are circumscribed around the same circle. The inradius is 1 for both. For the rhombus, as we saw earlier, the inradius is equal to area / semiperimeter. For the triangle, same formula. But the problem is that the inradius is given, so the formulas apply.Wait, but for the rhombus, the inradius is equal to (d1 * d2) / (2 * perimeter). Wait, no. Wait, area of rhombus is (d1 * d2)/2, and semiperimeter is (4 * s)/2 = 2s. So, inradius r = area / semiperimeter = (d1 * d2)/2 / (2s) = (d1 * d2)/(4s). But we are given that the inradius is 1. So,(d1 * d2)/(4s) = 1 => d1 * d2 = 4s. Which is the same as we had before.For the triangle, the inradius is 1, and we found that it's a 3-4-5 triangle. Let's check the inradius for a 3-4-5 triangle. The area is (3*4)/2 = 6. The semiperimeter is (3+4+5)/2 = 6. So, inradius r = area / semiperimeter = 6 / 6 = 1. Perfect, that matches the given inradius.So, the triangle is indeed a 3-4-5 triangle with inradius 1.Now, the key is to relate the sides of the triangle (3,4,5) to the rhombus. The legs 3 and 4 are parallel to the diagonals of the rhombus, and the hypotenuse 5 is parallel to a side of the rhombus. So, the hypotenuse is parallel to the rhombus's sides and has length 5. So, the side of the rhombus must be related to this hypotenuse. Since they are parallel, maybe the rhombus's side is a scaled version of this hypotenuse? But how?Wait, in the rhombus, the sides are vectors that are combinations of the diagonals. Let me recall that in a rhombus, the sides can be considered as the vectors resulting from the sum of half of each diagonal. Wait, perhaps we need to use vector analysis here.Let me consider the coordinate system again. Let’s place the rhombus with diagonals along the x and y axes. Then, the sides of the rhombus are vectors from the center to the vertices. Wait, no. The vertices are at (d1/2, 0), (0, d2/2), (-d1/2, 0), (0, -d2/2). So, the sides of the rhombus connect these vertices. For example, from (d1/2, 0) to (0, d2/2) is one side. The vector for this side is (-d1/2, d2/2). Similarly, the next side goes from (0, d2/2) to (-d1/2, 0), which is (-d1/2, -d2/2), etc. So, each side of the rhombus is a vector of magnitude s, which is sqrt[(d1/2)^2 + (d2/2)^2].But the triangle's hypotenuse is parallel to a side of the rhombus and has length 5. Let's consider the direction of the hypotenuse. The hypotenuse is parallel to the rhombus's side, so its direction vector is the same as the rhombus's side vector, which is (-d1/2, d2/2) or scaled.But the hypotenuse is length 5. So, the hypotenuse vector is a scalar multiple of the rhombus's side vector. Let's denote the hypotenuse vector as k*(-d1, d2) where k is a scalar. The length of this vector is 5. The length of the hypotenuse vector is sqrt[(k*d1)^2 + (k*d2)^2] = k*sqrt(d1^2 + d2^2) = 5. But from the rhombus's side length, s = sqrt[(d1/2)^2 + (d2/2)^2] = sqrt(d1^2 + d2^2)/2. Therefore, sqrt(d1^2 + d2^2) = 2s. Therefore, hypotenuse length 5 = k*2s => k = 5/(2s). Therefore, the hypotenuse vector is ( -5/(2s)*d1, 5/(2s)*d2 ). But the hypotenuse vector is also the sum of the two legs of the triangle. Since the triangle is a right triangle with legs parallel to the diagonals (d1 and d2), the legs are vectors along the x and y axes. So, the legs are vectors (3,0) and (0,4) scaled by some factors.Wait, but since the legs are parallel to the diagonals, which are along the x and y axes in our coordinate system, the legs of the triangle would be horizontal and vertical. But the hypotenuse is parallel to the rhombus's side. Wait, but in our coordinate system, the rhombus's side has a direction that is a combination of x and y components. So, if the hypotenuse is parallel to the rhombus's side, which is a diagonal direction, then the hypotenuse of the triangle must be oriented at the same angle as the rhombus's side.But the triangle in question is circumscribed around the circle. Wait, but how is the triangle positioned relative to the circle? Since both the rhombus and the triangle are circumscribed around the same circle, the circle is tangent to all sides of both the rhombus and the triangle.But that seems impossible unless the rhombus and triangle share the same incircle, which is possible if they are both tangential polygons with the same inradius and center. However, the problem states that both the rhombus and the triangle are circumscribed around the circle, meaning the circle is inscribed in both shapes. However, having two different polygons inscribed around the same circle with the same center is possible, but their positions can vary. However, the problem states that two sides of the triangle are parallel to the diagonals of the rhombus, and the third is parallel to a side of the rhombus. Therefore, the triangle is oriented in the same way as the rhombus, with sides aligned to the rhombus's diagonals and one side.But perhaps the triangle is inscribed around the same circle, meaning the circle is tangent to all three sides of the triangle and all four sides of the rhombus. Since the circle is the incircle for both, their centers must coincide. Therefore, the inradius is 1 for both, and the center of the circle is the same for both polygons.Given that, let's try to model the triangle in the coordinate system. Let's assume the circle is centered at (0,0), the same as the rhombus. The triangle is a right triangle with legs parallel to the diagonals (x and y axes) and hypotenuse parallel to a side of the rhombus.So, the right triangle would have its right angle at some point, and the legs along the x and y axes. The hypotenuse would then be a line from (a,0) to (0,b), where a and b are the lengths of the legs. Wait, but in our case, the legs are 3 and 4, so the triangle's vertices would be at (3,0), (0,4), and (0,0), forming a right triangle. But the hypotenuse in this case would be from (3,0) to (0,4), which has length 5, which matches. But then, the hypotenuse is the line connecting (3,0) to (0,4). The direction of this hypotenuse is given by the vector (-3,4), which should be parallel to a side of the rhombus.Wait, the rhombus's sides have direction vectors like (-d1, d2) as we considered earlier. So, the direction of the hypotenuse (-3,4) should be a scalar multiple of the rhombus's side direction (-d1, d2). So, (-3,4) = k*(-d1, d2) for some scalar k. Therefore, we have:-3 = -k*d1 => 3 = k*d14 = k*d2Therefore, from these two equations:k = 3/d1 = 4/d2 => 3/d1 = 4/d2 => 3d2 = 4d1 => d2 = (4/3)d1So, the diagonals are in a 4:3 ratio. From earlier, we had the product of the diagonals d1*d2 = 4s. Substituting d2 = (4/3)d1:d1*(4/3)d1 = 4s => (4/3)d1^2 = 4s => d1^2 = 3s => d1 = sqrt(3s)But we also have from the side length formula:s = sqrt( (d1/2)^2 + (d2/2)^2 )Substitute d2 = (4/3)d1:s = sqrt( (d1^2)/4 + ( (16/9)d1^2 )/4 )= sqrt( (d1^2)/4 + (4d1^2)/9 )= sqrt( (9d1^2 + 16d1^2)/36 )= sqrt(25d1^2/36)= (5d1)/6Therefore, s = (5d1)/6 => d1 = (6s)/5But earlier, we had d1^2 = 3s. Substitute d1 = (6s)/5:( (6s)/5 )^2 = 3s(36s^2)/25 = 3sMultiply both sides by 25:36s^2 = 75sDivide both sides by s (assuming s ≠ 0):36s = 75Therefore, s = 75/36 = 25/12 ≈ 2.083...So, the side length of the rhombus is 25/12.Wait, let's check this. Let me verify step by step.First, we found that d2 = (4/3)d1.Then, from the product of diagonals:d1*d2 = 4s => d1*(4/3)d1 = 4s => (4/3)d1^2 = 4s => d1^2 = 3s.From the side length formula:s = sqrt( (d1/2)^2 + (d2/2)^2 ) = sqrt( (d1^2 + d2^2)/4 )But since d2 = (4/3)d1:s = sqrt( (d1^2 + (16/9)d1^2)/4 ) = sqrt( (25/9 d1^2)/4 ) = sqrt(25d1^2 / 36 ) = (5d1)/6Therefore, s = (5d1)/6 => d1 = (6s)/5Substitute into d1^2 = 3s:( (6s)/5 )^2 = 3s => 36s^2/25 = 3s => 36s = 75 => s = 75/36 = 25/12Yes, that seems correct. So, the side of the rhombus is 25/12. Let's check if this makes sense.25/12 is approximately 2.083, which is greater than 2, which matches our earlier conclusion that s ≥ 2. Also, the diagonals would be d1 = 6*(25/12)/5 = (150/12)/5 = (150/60) = 2.5. So, d1 = 2.5, d2 = (4/3)*2.5 = 10/3 ≈ 3.333. Then, product d1*d2 = 2.5*(10/3) = 25/3 ≈ 8.333. Then, 4s = 4*(25/12) = 100/12 = 25/3 ≈ 8.333, so that matches. The product of diagonals equals 4s. So that's good.Additionally, the side length s = sqrt( (2.5/2)^2 + (10/3 / 2)^2 ) = sqrt( (1.25)^2 + (1.666...)^2 ) = sqrt(1.5625 + 2.777...) ≈ sqrt(4.34) ≈ 2.083, which is 25/12. So, that checks out.Therefore, the side of the rhombus is 25/12. Let me write this as a boxed fraction.Final AnswerThe side of the rhombus is boxed{dfrac{25}{12}}.</think>Given a rhombus and a triangle circumscribed around a circle with a radius of 1, where two sides of the triangle are parallel to the diagonals of the rhombus and the third side is parallel to one of the sides of the rhombus and is equal to 5, we need to find the side of the rhombus.1. Properties of the Rhombus: - The diagonals of the rhombus are perpendicular bisectors of each other. - The area of the rhombus can be expressed as ( frac{d_1 cdot d_2}{2} ) where ( d_1 ) and ( d_2 ) are the diagonals. - The semiperimeter of the rhombus is ( 2s ), and the area is also given by ( 2s cdot 1 = 2s ). - Therefore, ( frac{d_1 cdot d_2}{2} = 2s ) implies ( d_1 cdot d_2 = 4s ).2. Properties of the Triangle: - The triangle is a right triangle with legs parallel to the diagonals of the rhombus and hypotenuse parallel to a side of the rhombus with length 5. - The triangle is a 3-4-5 right triangle, confirming the legs are 3 and 4, and the hypotenuse is 5. - The inradius of the triangle is 1, confirming the area is 6 and the semiperimeter is 6.3. Relating the Triangle to the Rhombus: - The legs of the triangle (3 and 4) are parallel to the diagonals of the rhombus. - The hypotenuse (5) is parallel to a side of the rhombus. - The direction vector of the hypotenuse is proportional to the side vector of the rhombus.4. Calculations: - The direction vector of the hypotenuse (-3, 4) is proportional to the side vector of the rhombus. - Solving for the diagonals ( d_1 ) and ( d_2 ) in the ratio 3:4, we find ( d_2 = frac{4}{3}d_1 ). - Using the product of the diagonals ( d_1 cdot d_2 = 4s ) and the side length formula ( s = sqrt{left(frac{d_1}{2}right)^2 + left(frac{d_2}{2}right)^2} ), we solve for ( s ).5. Final Solution: - Substituting ( d_2 = frac{4}{3}d_1 ) into the equations, we find ( d_1 = frac{6s}{5} ). - Solving the quadratic equation, we find ( s = frac{25}{12} ).Thus, the side of the rhombus is boxed{dfrac{25}{12}}.,<think>Okay, so I have this geometry problem here. Let me try to parse it out step by step. The problem says:"A rhombus and a triangle are circumscribed around a circle with a radius of 1, two sides of which are parallel to the diagonals of the rhombus, and the third side is parallel to one of the sides of the rhombus and is equal to 5. Find the side of the rhombus."Hmm, first, let's make sure I understand all the terms here. A rhombus is a quadrilateral with all sides equal, and its diagonals are perpendicular bisectors of each other. A triangle circumscribed around a circle means that the circle is the incircle of the triangle, tangent to all three sides. Similarly, a rhombus circumscribed around a circle would have the circle as its incircle, tangent to all four sides. The radius of the circle is given as 1.Now, the triangle has two sides parallel to the diagonals of the rhombus, and the third side is parallel to one of the sides of the rhombus and is equal to 5. We need to find the side length of the rhombus.Let me start by recalling some properties:1. For a rhombus circumscribed around a circle (i.e., a tangential rhombus), all sides are tangent to the circle. Since all sides of a rhombus are equal, the radius of the incircle can be related to the area and perimeter. The formula for the radius r is r = Area / Perimeter. Since the radius is 1, we have Area = Perimeter * 1 = Perimeter. But wait, for a rhombus, the area can also be calculated as (d1 * d2)/2, where d1 and d2 are the lengths of the diagonals. Also, the perimeter is 4 * side (since all sides are equal). Let me note that.So for the rhombus:- Let the side length be 'a'.- Diagonals are d1 and d2.- Then, area = (d1 * d2)/2.- Perimeter = 4a.- Since it's tangential, the inradius r = 1 = Area / Perimeter => (d1 * d2)/2 / (4a) = 1 => (d1 * d2) = 8a.So, we have that the product of the diagonals is 8a.Also, in a rhombus, the diagonals are perpendicular bisectors, so they form four right triangles with legs d1/2 and d2/2, and hypotenuse 'a'. Therefore, by Pythagoras:(d1/2)^2 + (d2/2)^2 = a^2Which simplifies to:(d1^2 + d2^2)/4 = a^2 => d1^2 + d2^2 = 4a^2.So now, we have two equations:1. d1 * d2 = 8a2. d1^2 + d2^2 = 4a^2Let me note these down. Maybe we can solve for d1 and d2 in terms of a? Let's see.If we let x = d1 and y = d2, then:1. xy = 8a2. x^2 + y^2 = 4a^2We can express (x + y)^2 = x^2 + y^2 + 2xy = 4a^2 + 16a = 4a^2 + 16a. So x + y = sqrt(4a^2 + 16a). Not sure if that helps immediately.Alternatively, perhaps we can express one variable in terms of the other. For example, from xy = 8a, we have y = 8a / x. Substitute into the second equation:x^2 + (8a / x)^2 = 4a^2Multiply through by x^2:x^4 + 64a^2 = 4a^2 x^2Bring all terms to one side:x^4 - 4a^2 x^2 + 64a^2 = 0This is a quadratic in x^2. Let me set z = x^2:z^2 - 4a^2 z + 64a^2 = 0Solving for z:z = [4a^2 ± sqrt(16a^4 - 256a^2)] / 2Hmm, discriminant is sqrt(16a^4 - 256a^2) = sqrt(16a^2(a^2 - 16)) = 4a sqrt(a^2 - 16). Wait, but sqrt(a^2 - 16) would require a^2 >= 16, which would mean a >= 4. But let's check if that makes sense.But wait, this seems complicated. Maybe there's a different approach.Alternatively, for a rhombus with side length 'a', the inradius is given by r = (d1 * d2) / (2 * perimeter / 2). Wait, actually, for a tangential quadrilateral, the area is equal to the inradius multiplied by the semiperimeter. Wait, hold on, inradius formula is r = Area / semiperimeter.Wait, maybe I made a mistake earlier. Let me verify that.Yes, in general, for any tangential polygon, the area is equal to the inradius multiplied by the semiperimeter. So for a quadrilateral, the semiperimeter is (4a)/2 = 2a. Therefore, the area should be r * semiperimeter = 1 * 2a = 2a. But earlier, I had Area = (d1 * d2)/2. So setting (d1 * d2)/2 = 2a => d1 * d2 = 4a. Wait, but earlier I had d1 * d2 = 8a. So there must be an error in my initial calculation.Wait, this is critical. Let me redo that.The formula for the area of a rhombus is (d1 * d2)/2. The inradius formula for a tangential quadrilateral is r = Area / semiperimeter. So, here, semiperimeter is (4a)/2 = 2a. Therefore, Area = r * semiperimeter = 1 * 2a = 2a. Therefore, (d1 * d2)/2 = 2a => d1 * d2 = 4a. So my initial equation was wrong; it should be d1 * d2 = 4a, not 8a.That's a crucial mistake. Let me correct that.So, correct equations are:1. d1 * d2 = 4a2. d1^2 + d2^2 = 4a^2 (from Pythagoras)Therefore, let's write x = d1 and y = d2. Then:x * y = 4ax^2 + y^2 = 4a^2Again, let's compute (x + y)^2 = x^2 + y^2 + 2xy = 4a^2 + 8aSo x + y = sqrt(4a^2 + 8a). Not sure yet.Alternatively, solving for x and y.From x * y = 4a, so y = 4a / x. Substitute into the second equation:x^2 + (16a^2)/(x^2) = 4a^2Multiply through by x^2:x^4 + 16a^2 = 4a^2 x^2Bring all terms to left:x^4 - 4a^2 x^2 + 16a^2 = 0Let z = x^2:z^2 - 4a^2 z + 16a^2 = 0Quadratic in z: solutions are [4a^2 ± sqrt(16a^4 - 64a^2)] / 2Simplify discriminant:sqrt(16a^4 - 64a^2) = sqrt(16a^2(a^2 - 4)) = 4a sqrt(a^2 - 4)Thus, z = [4a^2 ± 4a sqrt(a^2 - 4)] / 2 = 2a^2 ± 2a sqrt(a^2 - 4)Since z = x^2 must be positive, both solutions are positive, but we need to check if sqrt(a^2 - 4) is real, which requires a^2 >=4 => a >= 2. Which makes sense as the side length of a rhombus with inradius 1.But maybe this approach is getting too complicated. Perhaps instead of solving for diagonals, think about angles?In a rhombus, the inradius can also be expressed in terms of the side length and the angles. Since all sides are equal, and the inradius is the distance from the center to any side, which is equal to the radius of the inscribed circle. For a rhombus, the inradius r is given by r = a * sin(theta), where theta is any of the interior angles. Wait, is that correct?Wait, let me recall. In a rhombus, the area is also equal to base * height. The height is the distance between two opposite sides, which is equal to 2r, since the inradius is the distance from the center to a side, but the height would be twice that. Wait, no. Wait, in a rhombus circumscribed around a circle, the circle is tangent to all four sides, so the diameter of the circle (which is 2*1=2) would be equal to the height of the rhombus. Wait, but actually, no. Wait, the inradius is the distance from the center to the side, which is the same as the radius, so the height (distance between two opposite sides) is 2r, which would be 2*1=2.But the area of the rhombus is also equal to base * height = a * 2. So area is 2a, which matches the previous result (since we had area = 2a). So this is consistent.Alternatively, the area can also be expressed as a^2 sin(theta), where theta is one of the interior angles. Therefore, equating:a^2 sin(theta) = 2a => sin(theta) = 2/a.Since sin(theta) must be <=1, this implies that 2/a <=1 => a >=2. Which again matches the previous conclusion that a must be at least 2.But how does this relate to the triangle?The problem states that there's a triangle circumscribed around the same circle (radius 1), with two sides parallel to the diagonals of the rhombus, and the third side parallel to a side of the rhombus and equal to 5. We need to relate this triangle to the rhombus.Let me try to visualize this. The triangle has an incircle of radius 1. Two of its sides are parallel to the diagonals of the rhombus, and the third side is parallel to a side of the rhombus and has length 5.First, since the triangle is circumscribed around the circle, it is a tangential triangle, meaning it has an incircle tangent to all three sides. The radius of this incircle is 1.In a tangential triangle, the area is equal to the inradius multiplied by the semiperimeter: Area = r * s, where s is the semiperimeter. Therefore, in this case, Area = 1 * s = s.But we need to find the sides of the triangle. Wait, but the sides are related to the rhombus's diagonals and side.Let me denote the triangle. Let's call the triangle ABC, with sides parallel to the rhombus's diagonals and one side parallel to the rhombus's side. Let me suppose that side BC is parallel to a side of the rhombus and has length 5. Then sides AB and AC are parallel to the diagonals of the rhombus.Wait, but which diagonals? The rhombus has two diagonals, one longer and one shorter, unless it's a square (but a square is a rhombus with equal diagonals). But since the rhombus is not necessarily a square, the diagonals are of different lengths. So the triangle has two sides parallel to these diagonals, and one side parallel to a side of the rhombus.Since the rhombus's sides are all equal and the triangle's side is 5, which is parallel to a side of the rhombus, maybe there's a scaling factor involved.But how do we relate the triangle's sides to the rhombus's diagonals?Alternatively, since the triangle's sides are parallel to the diagonals, perhaps the triangle is similar in some way to the rhombus's diagonals? Wait, not necessarily.Alternatively, maybe the triangle is formed by lines parallel to the diagonals and a side of the rhombus? But I need to think more clearly.Let me consider coordinate geometry. Let's place the rhombus in a coordinate system such that its center is at the origin, and its diagonals are along the axes. Since the diagonals are perpendicular and bisect each other, this is a common setup.Let’s suppose the rhombus has diagonals of lengths d1 and d2 along the x-axis and y-axis, respectively. Then the vertices of the rhombus would be at (d1/2, 0), (-d1/2, 0), (0, d2/2), and (0, -d2/2). The sides of the rhombus can be calculated as the distance between these points. For example, between (d1/2, 0) and (0, d2/2), which is sqrt((d1/2)^2 + (d2/2)^2) = a, the side length. So as before, (d1/2)^2 + (d2/2)^2 = a^2.Now, the triangle is circumscribed around the same circle (radius 1), with two sides parallel to the diagonals of the rhombus (i.e., the x-axis and y-axis), and the third side is parallel to a side of the rhombus and has length 5.Wait, if the third side is parallel to a side of the rhombus, which in this coordinate system, a side goes from (d1/2, 0) to (0, d2/2). The slope of this side is (d2/2 - 0)/(0 - d1/2) = -d2/d1. So the side of the triangle that is parallel to the rhombus's side will have the same slope. However, the problem states that the third side is parallel to one of the sides of the rhombus and is equal to 5. So this side of the triangle has slope -d2/d1 (or d2/d1 if considering the opposite direction) and length 5.But how can we model this triangle? Let's try to define the triangle's sides.Since two sides are parallel to the diagonals (x-axis and y-axis), those sides must be horizontal and vertical. So the triangle has one horizontal side, one vertical side, and one slanted side with slope -d2/d1 and length 5.Wait, but in that case, the triangle would be a right triangle with legs along the axes and hypotenuse of length 5. But the hypotenuse would have a slope of -1 if the legs are equal, but here the hypotenuse has slope -d2/d1. Hmm, maybe not a right triangle? Wait, if two sides are parallel to the diagonals (which are perpendicular), then the triangle would have two sides perpendicular to each other, making it a right triangle. But the third side is the hypotenuse with length 5 and slope -d2/d1.Therefore, the triangle is a right triangle with legs parallel to the diagonals of the rhombus (i.e., along the x and y axes) and hypotenuse parallel to a side of the rhombus with length 5.Wait, in this case, the legs of the triangle would be horizontal and vertical, with lengths say 'p' and 'q', and the hypotenuse would have length 5 and slope -d2/d1.But since the hypotenuse is parallel to a side of the rhombus, its slope must be equal to the slope of the rhombus's side. The rhombus's side has slope (d2/2)/( -d1/2) = -d2/d1, so the hypotenuse of the triangle has slope -d2/d1. Also, the length of the hypotenuse is 5.Given that the legs are 'p' and 'q', the hypotenuse length is sqrt(p^2 + q^2) = 5. So sqrt(p^2 + q^2) = 5.Additionally, the slope of the hypotenuse is -q/p (since it goes from (p, 0) to (0, q), the slope is (q - 0)/(0 - p) = -q/p). This slope must be equal to -d2/d1. Therefore, -q/p = -d2/d1 => q/p = d2/d1 => q = (d2/d1) p.So, substituting into the hypotenuse length equation:sqrt(p^2 + ( (d2/d1) p )^2 ) = 5sqrt(p^2 (1 + (d2/d1)^2 )) = 5p * sqrt(1 + (d2/d1)^2 ) = 5Therefore,p = 5 / sqrt(1 + (d2/d1)^2 )But from the rhombus, we know that (d1/2)^2 + (d2/2)^2 = a^2 => (d1^2 + d2^2)/4 = a^2 => d1^2 + d2^2 = 4a^2.Therefore, (d2/d1)^2 = ( (sqrt(4a^2 - d1^2))/d1 )^2 = (4a^2 - d1^2)/d1^2 = (4a^2)/d1^2 - 1.Wait, maybe not helpful. Alternatively, let's express (d2/d1) in terms of a. Let’s denote k = d2/d1. Then, from the rhombus:d1^2 + d2^2 = 4a^2 => d1^2 (1 + k^2) = 4a^2 => d1 = 2a / sqrt(1 + k^2)Similarly, d2 = 2a k / sqrt(1 + k^2)But maybe this is getting too abstract.Alternatively, since the triangle is a right triangle with legs p and q, hypotenuse 5, and q = (d2/d1) p.Also, since the triangle is circumscribed around a circle of radius 1, its inradius is 1. For a right triangle, the inradius is given by r = (p + q - hypotenuse)/2. Let me verify that formula.Yes, in a right triangle, the inradius is r = (p + q - c)/2, where c is the hypotenuse. Therefore, here:1 = (p + q - 5)/2 => p + q -5 = 2 => p + q = 7.But we already have q = (d2/d1) p, and from the hypotenuse, sqrt(p^2 + q^2) = 5.So, substituting q = (d2/d1) p into these equations:1. p + (d2/d1) p = 7 => p(1 + d2/d1) = 72. sqrt(p^2 + ( (d2/d1) p )^2 ) = 5 => p sqrt(1 + (d2/d1)^2 ) = 5Let me denote m = d2/d1. Then:1. p(1 + m) = 7 => p = 7 / (1 + m)2. p sqrt(1 + m^2 ) = 5Substitute p from 1 into 2:(7 / (1 + m)) * sqrt(1 + m^2 ) = 5Multiply both sides by (1 + m):7 sqrt(1 + m^2 ) = 5 (1 + m)Square both sides:49 (1 + m^2 ) = 25 (1 + 2m + m^2 )Expand:49 + 49m^2 = 25 + 50m + 25m^2Bring all terms to left:49 + 49m^2 -25 -50m -25m^2 = 0Simplify:24m^2 -50m +24 =0Divide equation by 2:12m^2 -25m +12 =0Now, solve quadratic equation:m = [25 ± sqrt(625 - 576)] / 24 = [25 ± sqrt(49)] /24 = [25 ±7]/24Thus, two solutions:m = (25 +7)/24 =32/24=4/3≈1.333...or m=(25-7)/24=18/24=3/4=0.75So m=4/3 or m=3/4.But m= d2/d1. So depending on which diagonal is longer, m could be greater than 1 or less than 1. But in a rhombus, the ratio of diagonals can be either way, depending on the angles. So both possibilities are valid.Therefore, we have two cases:Case 1: m =4/3 => d2/d1=4/3 => d2= (4/3) d1Case 2: m=3/4 => d2/d1=3/4 => d2=(3/4) d1We need to relate this back to the rhombus.From the rhombus, we have two equations:1. d1 * d2 =4a (from area)2. d1^2 + d2^2 =4a^2 (from Pythagoras)Let’s substitute d2 in terms of d1 for each case.Case 1: d2 = (4/3)d1Substitute into equation 1:d1*(4/3 d1) =4a => (4/3)d1^2=4a => d1^2=3a => d1= sqrt(3a)But wait, d1 is a length, so positive. Then d2=(4/3)sqrt(3a)Then substitute into equation 2:d1^2 + d2^2 =4a^2=> 3a + (16/9)*3a =4a^2Compute:3a + (16/9 *3a)= 3a + (16/3)a= (9/3 +16/3)a=25/3 aSet equal to 4a^2:25/3 a =4a^2 => 25/3 =4a => a=25/(12)=25/12≈2.083...Case 2: d2=(3/4)d1Substitute into equation 1:d1*(3/4 d1)=4a => (3/4)d1^2=4a => d1^2= (16/3)a => d1=4/sqrt(3) *sqrt(a)Wait, that seems messy. Wait, let me check:Wait, (3/4)d1^2=4a => d1^2= (4a)*(4/3)=16a/3 => d1=4*sqrt(a/3)Hmm, maybe not helpful. Then substitute into equation 2:d1^2 +d2^2=4a^2d1^2 + (9/16)d1^2=4a^2 => (25/16)d1^2=4a^2 => d1^2= (64/25)a^2 => d1=(8/5)aBut from equation 1: d1^2=16a/3. So:(8/5 a)^2=16a/3 => 64/25 a^2=16a/3Multiply both sides by 75 (LCM of 25 and 3):64*3 a^2=16*25 a192a^2=400aDivide both sides by a (assuming a ≠0):192a=400 => a=400/192=25/12≈2.083...Same as in Case1.Wait, so both cases lead to a=25/12? That’s interesting.Wait, let me check again.Case1:d2=(4/3)d1From equation1:d1*d2=4a => d1*(4/3 d1)=4a => (4/3)d1^2=4a => d1^2=3a => d1=sqrt(3a)From equation2:d1^2 + d2^2=3a + (16/9)d1^2=3a + (16/9)(3a)=3a + 16/3 a= (9a +16a)/3=25a/3=4a^2Thus, 25a/3=4a^2 =>25/3=4a =>a=25/(12)Case2:d2=(3/4)d1From equation1:d1*d2=4a => d1*(3/4 d1)=4a => (3/4)d1^2=4a => d1^2=16a/3 => d1=4*sqrt(a/3)From equation2:d1^2 + d2^2= d1^2 + (9/16)d1^2= (25/16)d1^2=4a^2 => d1^2= (4a^2)*(16/25)=64a^2/25But from equation1, d1^2=16a/3. Thus,64a^2/25=16a/3 =>64a/25=16/3 =>a=(16/3)*(25/64)= (16*25)/(3*64)= (25)/(3*4)=25/12Same result. So regardless of the case, a=25/12.Therefore, the side of the rhombus is 25/12, which is approximately 2.083...But let me check if this answer makes sense.First, a=25/12≈2.083. The inradius is 1, which for a rhombus should satisfy r = (d1*d2)/(2*2a) [Wait, no, earlier we had Area = (d1*d2)/2 = 2a. Then r = Area / semiperimeter = 2a / (2a) = 1. Wait, semiperimeter is (4a)/2=2a. So Area / semiperimeter = (2a)/(2a)=1, which matches the given inradius. So that's consistent.Also, for the triangle, let's check if the inradius is indeed 1.In the triangle, the legs p and q, hypotenuse 5. From above, in either case, p + q=7. The inradius is 1, which is (p + q -5)/2=(7-5)/2=1. Correct.Also, the legs p and q can be calculated.From case1: m=4/3. So p=7/(1 +4/3)=7/(7/3)=3. Then q=4/3 *3=4.Therefore, legs 3 and4, hypotenuse5. This is the classic 3-4-5 triangle, which has inradius (3+4-5)/2=2/2=1. Correct.Similarly, case2: m=3/4. Then p=7/(1 +3/4)=7/(7/4)=4. Then q=3/4*4=3. Same triangle, legs4 and3, hypotenuse5. Same result.Therefore, the triangle is a 3-4-5 triangle, which has inradius1. That checks out.So, all the conditions are satisfied with a=25/12.Hence, the side of the rhombus is 25/12.But let me confirm once again with another approach.Alternatively, since the triangle is a 3-4-5 triangle with sides 3,4,5, and inradius1. The sides parallel to the rhombus's diagonals are 3 and4, which are the legs. So these sides are parallel to the diagonals of the rhombus. The hypotenuse is parallel to the rhombus's side and has length5.Now, in the coordinate system where the rhombus has diagonals along the axes, the sides of the rhombus have slope -d2/d1. The hypotenuse of the triangle has slope -q/p. In our case, the hypotenuse has slope -4/3 (if legs are3 and4) or -3/4 (if legs are4 and3). But the rhombus's sides have slope -d2/d1. In our calculation, d2/d1 was either4/3 or3/4. Therefore, the slope of the rhombus's side is either -4/3 or -3/4, which matches the slope of the triangle's hypotenuse. Therefore, the side of the triangle is indeed parallel to the rhombus's side.Moreover, the length of the triangle's hypotenuse is5, which is given. So everything matches.Therefore, the answer is25/12, which is approximately2.083..., but expressed as a fraction25/12.Before finalizing, let me check if25/12 reduces, but25 and12 are coprime, so that's the simplest form.Thus, the side of the rhombus is25/12.Final AnswerThe side of the rhombus is boxed{dfrac{25}{12}}.</think>Given a rhombus and a triangle circumscribed around a circle with a radius of 1, we need to find the side length of the rhombus. The triangle has two sides parallel to the diagonals of the rhombus and a third side parallel to one of the sides of the rhombus with length 5.1. Properties of the Rhombus: - The rhombus is circumscribed around a circle (incircle with radius 1). - The area of the rhombus is given by ( text{Area} = 2a ) where ( a ) is the side length. - The product of the diagonals ( d_1 ) and ( d_2 ) is ( d_1 cdot d_2 = 4a ). - The sum of the squares of the diagonals is ( d_1^2 + d_2^2 = 4a^2 ).2. Properties of the Triangle: - The triangle is a right triangle with legs parallel to the diagonals of the rhombus and hypotenuse parallel to a side of the rhombus with length 5. - The inradius of the triangle is 1, leading to the semiperimeter ( s = 7 ) (since ( r = frac{text{Area}}{s} ) and ( text{Area} = r cdot s )). - The legs of the triangle are 3 and 4, forming a 3-4-5 triangle.3. Relating the Triangle to the Rhombus: - The legs of the triangle (3 and 4) are proportional to the diagonals of the rhombus. - The ratio of the diagonals ( frac{d_2}{d_1} ) is found to be ( frac{4}{3} ) or ( frac{3}{4} ).4. Solving for the Side Length ( a ): - Using the equations ( d_1 cdot d_2 = 4a ) and ( d_1^2 + d_2^2 = 4a^2 ), we find that both cases lead to the same side length ( a = frac{25}{12} ).Thus, the side length of the rhombus is (boxed{dfrac{25}{12}}).]

🔑:Let the side KM of the triangle KLM be parallel to the diagonal BD of the rhombus ABCD and touch the given circle at point P, the side KL of the triangle be parallel to the diagonal AC of the rhombus and touch the circle at point Q, and the side LM of the triangle be parallel to the side AB of the rhombus and touch the circle at point T. Since the sides KM and KL are parallel to the diagonals of the rhombus, and the diagonals of the rhombus are perpendicular, the triangle KLM is a right triangle, with angle LKM = 90^circ. If O is the center of the circle, then the quadrilateral KPOQ is a square with a side equal to the radius of the circle, i.e., 1. Let MP = MD = x. ThenMT = MP = x, LQ = LT = ML - MT = 5 - x, KL = KQ + LQ = 1 + (5 - x) = 6 - x.By the Pythagorean theorem, ML^2 = KM^2 + KL^2, or25 = (1 + x)^2 + (6 - x)^2, x^2 - 5x + 6 = 0.Assume that KM < KL. Then we take the smaller root of the obtained equation: x = 2. Thus, KM = 1 + x = 1 + 2 = 3, KL = 6 - x = 6 - 2 = 4. Let angle KLM = alpha. Thensin alpha = frac{KM}{ML} = frac{3}{5}, cos alpha = frac{4}{5}, sin 2alpha = 2 sin alpha cos alpha = 2 cdot frac{3}{5} cdot frac{4}{5} = frac{24}{25}.Let H be the projection of the vertex B of the rhombus onto its side AD. Then BH is the height of the rhombus, so BH is twice the radius of the circle inscribed in the rhombus, i.e., BH = 2. From the right triangle AHB, we find thatAB = frac{BH}{sin angle BAD} = frac{BH}{sin 2 angle BAC} = frac{BC}{sin 2 angle KLM} = frac{2}{sin 2 alpha} = frac{2}{frac{24}{25}} = frac{25}{12}.## Answerfrac{25}{12}.