🧠:[<think>Okay, so I have this problem here where I need to find the value of b/a given that (a sin π/7 + b cos π/7) divided by (a cos π/7 - b sin π/7) equals tan(10π/21). Alright, let's try to break this down step by step.First, let me recall that tan θ = sin θ / cos θ, so the given equation is a ratio of two expressions equal to another tangent. Maybe I can rewrite the left side in terms of tangent? Let me see.The left-hand side (LHS) is [a sin(π/7) + b cos(π/7)] / [a cos(π/7) - b sin(π/7)]. Hmm, this looks similar to the formula for tangent of a sum or difference. Remember that tan(A + B) = (tan A + tan B) / (1 - tan A tan B). If I can express the numerator and denominator in terms of a single sine and cosine, maybe I can relate it to a tangent function.Alternatively, maybe I can divide numerator and denominator by cos(π/7) to introduce tan(π/7). Let me try that.Dividing numerator and denominator by cos(π/7):Numerator becomes a sin(π/7)/cos(π/7) + b = a tan(π/7) + b.Denominator becomes a - b sin(π/7)/cos(π/7) = a - b tan(π/7).So the LHS simplifies to [a tan(π/7) + b] / [a - b tan(π/7)]. Let me set this equal to tan(10π/21). So:[a tan(π/7) + b] / [a - b tan(π/7)] = tan(10π/21).Let me denote x = b/a. Since a and b are non-zero, x is defined. Then, divide numerator and denominator by a:[tan(π/7) + x] / [1 - x tan(π/7)] = tan(10π/21).So now, if I let x = b/a, then the equation becomes:(tan(π/7) + x) / (1 - x tan(π/7)) = tan(10π/21).This looks like the tangent addition formula! Because tan(A + B) = (tan A + tan B)/(1 - tan A tan B). So comparing this to the left side, which is (tan(π/7) + x)/(1 - x tan(π/7)), it's like tan(A + B) where A = π/7 and tan B = x. Therefore, this expression is tan(A + B) = tan(π/7 + B). Therefore, set this equal to tan(10π/21).Therefore, π/7 + B = 10π/21 + kπ for some integer k. But since we are dealing with real numbers and x is a real number, B would be arctan(x). But we can ignore the periodicity for now and just take the principal value.So, π/7 + B = 10π/21. Solving for B:B = 10π/21 - π/7 = 10π/21 - 3π/21 = 7π/21 = π/3.Therefore, B = π/3, so tan B = x = tan(π/3) = √3.Therefore, x = √3, so b/a = √3. Let me check this.Wait, but let me verify the steps again to make sure I didn't make a mistake.Starting from:(tan(π/7) + x)/(1 - x tan(π/7)) = tan(10π/21)If this is tan(A + B) where A = π/7 and B is such that tan B = x, then the equation becomes tan(A + B) = tan(10π/21). Therefore, A + B = 10π/21 + kπ. Since A = π/7 = 3π/21, then 3π/21 + B = 10π/21 + kπ. Therefore, B = 7π/21 + kπ = π/3 + kπ. Therefore, tan B = tan(π/3 + kπ) = tan(π/3) = √3. Since tangent has a period of π, all solutions for B would give tan B = √3, so x = √3. Therefore, regardless of k, x is √3. Therefore, b/a = √3.So the answer is √3. Let me check with actual numbers to be sure.Let’s suppose that b/a = √3, so let’s take a = 1, b = √3. Then compute the LHS:[sin(π/7) + √3 cos(π/7)] / [cos(π/7) - √3 sin(π/7)].Let me compute numerator and denominator.First, note that sin(π/7) + √3 cos(π/7) can be written as 2*( (1/2) sin(π/7) + (√3/2) cos(π/7) ) = 2 sin(π/7 + π/3), because sin(A + B) = sin A cos B + cos A sin B. Wait, but (√3/2) cos(π/7) + (1/2) sin(π/7) is sin(π/7 + π/6) ?Wait, wait. Let's recall that sin(A + B) = sin A cos B + cos A sin B. So if I have C sin x + D cos x, I can write it as R sin(x + φ), where R = sqrt(C² + D²), and tan φ = D/C. Wait, actually, more precisely:C sin x + D cos x = R sin(x + φ), where R = sqrt(C² + D²), and φ = arctan(D/C). Let me verify:R sin(x + φ) = R sin x cos φ + R cos x sin φ. Therefore, matching coefficients:C = R cos φD = R sin φTherefore, tan φ = D/C. So if C and D are coefficients in front of sin x and cos x, respectively, then φ = arctan(D/C).In the numerator, sin(π/7) + √3 cos(π/7): here, C = 1, D = √3. Therefore, R = sqrt(1 + 3) = 2, and tan φ = √3 / 1 = √3, so φ = π/3. Therefore, numerator = 2 sin(π/7 + π/3). Similarly, the denominator is cos(π/7) - √3 sin(π/7). Let me do the same for denominator.Denominator: cos(π/7) - √3 sin(π/7) = 1 * cos(π/7) + (-√3) sin(π/7). So here, C = -√3, D = 1. Therefore, R = sqrt( (√3)^2 + 1^2 ) = sqrt(4) = 2. The angle φ is arctan(D/C) = arctan(-1/√3) = -π/6. So the denominator can be written as 2 cos(π/7 + π/6). Wait, because:Wait, using the identity C cos x + D sin x = R cos(x - φ), where R = sqrt(C² + D²), and tan φ = D/C. Alternatively, maybe I need to adjust the formula.Wait, let's consider the denominator: cos(π/7) - √3 sin(π/7). Let me factor out the 2:2*( (1/2) cos(π/7) - (√3/2) sin(π/7) )Recognize that 1/2 = cos(π/3) and √3/2 = sin(π/3). Therefore:= 2*( cos(π/3) cos(π/7) - sin(π/3) sin(π/7) )Which is 2*cos(π/3 + π/7) by the cosine addition formula.Similarly, the numerator was 2 sin(π/7 + π/3).Therefore, the entire expression becomes [2 sin(π/7 + π/3)] / [2 cos(π/7 + π/3)] = tan(π/7 + π/3).Compute π/7 + π/3: common denominator is 21, so 3π/21 + 7π/21 = 10π/21. Therefore, the expression is tan(10π/21), which matches the right-hand side (RHS). So this checks out. Therefore, when b/a = √3, the equation holds.Therefore, the value of b/a is √3.Alternatively, if I didn't recognize the tangent addition formula initially, I could have cross-multiplied and solved for x. Let's try that approach as a verification.Starting from [tan(π/7) + x] / [1 - x tan(π/7)] = tan(10π/21). Cross-multiplying gives:tan(π/7) + x = tan(10π/21) * [1 - x tan(π/7)]Expanding the RHS:= tan(10π/21) - x tan(10π/21) tan(π/7)Bring all terms to the left side:tan(π/7) + x - tan(10π/21) + x tan(10π/21) tan(π/7) = 0Factor out x:x[1 + tan(10π/21) tan(π/7)] + [tan(π/7) - tan(10π/21)] = 0Solve for x:x = [tan(10π/21) - tan(π/7)] / [1 + tan(10π/21) tan(π/7)]Again, this is the tangent subtraction formula! Because tan(A - B) = [tan A - tan B] / [1 + tan A tan B]. So here, if we let A = 10π/21 and B = π/7, then x = tan(A - B) = tan(10π/21 - π/7). Compute 10π/21 - π/7 = 10π/21 - 3π/21 = 7π/21 = π/3. Therefore, x = tan(π/3) = √3. So same result. Therefore, x = √3, so b/a = √3.Either approach works. So regardless of whether I use the addition formula or cross-multiplied and used the subtraction formula, I end up with the same answer. Therefore, I'm confident that the value of b/a is √3.Final AnswerThe value of dfrac{b}{a} is boxed{sqrt{3}}.</think>Given the equation:[frac{a sin frac{pi}{7} + b cos frac{pi}{7}}{a cos frac{pi}{7} - b sin frac{pi}{7}} = tan frac{10pi}{21}]we start by letting ( x = frac{b}{a} ). Dividing the numerator and denominator of the left-hand side by ( cos frac{pi}{7} ) gives:[frac{a tan frac{pi}{7} + b}{a - b tan frac{pi}{7}} = tan frac{10pi}{21}]Substituting ( x = frac{b}{a} ) into the equation, we get:[frac{tan frac{pi}{7} + x}{1 - x tan frac{pi}{7}} = tan frac{10pi}{21}]Recognizing this as the tangent addition formula ( tan(A + B) = frac{tan A + tan B}{1 - tan A tan B} ), we identify ( A = frac{pi}{7} ) and ( B ) such that ( tan B = x ). Therefore, we have:[tanleft( frac{pi}{7} + B right) = tan frac{10pi}{21}]Solving for ( B ):[frac{pi}{7} + B = frac{10pi}{21} implies B = frac{10pi}{21} - frac{pi}{7} = frac{10pi}{21} - frac{3pi}{21} = frac{7pi}{21} = frac{pi}{3}]Thus, ( tan B = tan frac{pi}{3} = sqrt{3} ), so ( x = sqrt{3} ). Therefore, the value of ( frac{b}{a} ) is:[boxed{sqrt{3}}],<think>Okay, so I need to find the value of b/a given the equation:[frac{a sin frac{pi}{7} + b cos frac{pi}{7}}{a cos frac{pi}{7} - b sin frac{pi}{7}} = tan frac{10pi}{21}]Hmm, let's see. First, I remember that when you have a fraction involving sine and cosine in the numerator and denominator, it might resemble the tangent of some angle. The right side is already a tangent, so maybe I can rewrite the left side as a tangent of an angle, which would then let me set up an equation.The left side looks like (A sin x + B cos x)/(A cos x - B sin x). If I recall correctly, there's a formula for this. Let me think. If you have something like (sin θ cos x + cos θ sin x)/(cos θ cos x - sin θ sin x), that would be sin(θ + x)/cos(θ + x) which is tan(θ + x). Wait, but here it's A sin x + B cos x over A cos x - B sin x. So maybe factor out something?Alternatively, maybe divide numerator and denominator by cos(π/7) to express everything in terms of tan(π/7). Let's try that.Divide numerator and denominator by cos(π/7):Numerator becomes a tan(π/7) + bDenominator becomes a - b tan(π/7)So the left side becomes [a tan(π/7) + b] / [a - b tan(π/7)]Let me denote tan(π/7) as t for simplicity.Then the equation becomes:(a t + b)/(a - b t) = tan(10π/21)Let me compute tan(10π/21). Since π/3 is 7π/21, 10π/21 is π/3 + π/7. Wait, π/3 is 7π/21, π/7 is 3π/21, so 10π/21 is indeed π/3 + π/7. So using tan(A + B) formula.tan(π/3 + π/7) = [tan(π/3) + tan(π/7)] / [1 - tan(π/3) tan(π/7)]We know tan(π/3) is √3, and tan(π/7) is t. So tan(10π/21) = (√3 + t)/(1 - √3 t)So substituting back into the equation:(a t + b)/(a - b t) = (√3 + t)/(1 - √3 t)Cross-multiplying:(a t + b)(1 - √3 t) = (a - b t)(√3 + t)Let me expand both sides.Left side:a t * 1 - a t * √3 t + b * 1 - b * √3 t= a t - a √3 t² + b - b √3 tRight side:a * √3 + a t - b t * √3 - b t²= a √3 + a t - b √3 t - b t²So set left = right:a t - a √3 t² + b - b √3 t = a √3 + a t - b √3 t - b t²Hmm, let's subtract the right side from the left side to see what's left:Left - Right:(a t - a √3 t² + b - b √3 t) - (a √3 + a t - b √3 t - b t²)= a t - a √3 t² + b - b √3 t - a √3 - a t + b √3 t + b t²Simplify term by term:a t - a t = 0- a √3 t² + b t² = t² (-a √3 + b)b - a √3- b √3 t + b √3 t = 0So overall:t² (-a √3 + b) + (b - a √3) = 0Factor out (b - a √3):(b - a √3)(t² + 1) = 0Since t = tan(π/7), which is a real number, so t² + 1 is always positive (since t² is non-negative and we add 1). Therefore, t² + 1 ≠ 0. Hence, the equation simplifies to:(b - a √3) = 0Therefore:b = a √3Thus, the ratio b/a is √3.Wait, so the answer is √3? Let me check if I made any mistakes in the algebra.Starting again:After cross-multiplying:(a t + b)(1 - √3 t) = (a - b t)(√3 + t)Expanding:Left: a t (1) + a t (-√3 t) + b (1) + b (-√3 t)= a t - a √3 t² + b - b √3 tRight: a √3 + a t - b t √3 - b t²= a √3 + a t - b √3 t - b t²Subtracting right from left:Left - Right = (a t - a √3 t² + b - b √3 t) - (a √3 + a t - b √3 t - b t²)= a t - a √3 t² + b - b √3 t - a √3 - a t + b √3 t + b t²= (-a √3 t² + b t²) + (b - a √3) + (a t - a t) + (-b √3 t + b √3 t)= t² ( -a √3 + b ) + (b - a √3 )So yes, that's correct. Then since t² + 1 ≠ 0, then (b - a √3) must be zero, so b = a √3. Therefore, b/a = √3.But wait, let me verify with specific values. Suppose a = 1, then b = √3. Let's check if the original equation holds.Compute numerator: sin(π/7) + √3 cos(π/7)Denominator: cos(π/7) - √3 sin(π/7)Compute the ratio:[sin(π/7) + √3 cos(π/7)] / [cos(π/7) - √3 sin(π/7)]This resembles the formula for sin(π/7 + π/3) / cos(π/7 + π/3) because:sin(A + B) = sin A cos B + cos A sin Bcos(A + B) = cos A cos B - sin A sin BBut here, in the numerator: sin(π/7) * 1 + cos(π/7) * √3Which is similar to sin(π/7) cos(π/3) + cos(π/7) sin(π/3) since cos(π/3) = 0.5 and sin(π/3) = √3/2. But we have coefficients 1 and √3. So actually, if we factor out 2, it's 2*( sin(π/7) * 0.5 + cos(π/7) * (√3/2) ) = 2 sin(π/7 + π/3). Similarly, the denominator:cos(π/7) * 1 - sin(π/7) * √3= 2*( cos(π/7) * 0.5 - sin(π/7) * (√3/2) ) = 2 cos(π/7 + π/3)Therefore, the ratio is [2 sin(π/7 + π/3)] / [2 cos(π/7 + π/3)] = tan(π/7 + π/3) = tan(10π/21), which matches the given right-hand side. So yes, that works. Hence, the answer is √3.Final Answerboxed{sqrt{3}}</think>Given the equation:[frac{a sin frac{pi}{7} + b cos frac{pi}{7}}{a cos frac{pi}{7} - b sin frac{pi}{7}} = tan frac{10pi}{21}]we start by rewriting the left-hand side. Let ( t = tan frac{pi}{7} ). Dividing the numerator and denominator by ( cos frac{pi}{7} ), we get:[frac{a tan frac{pi}{7} + b}{a - b tan frac{pi}{7}} = frac{a t + b}{a - b t}]The right-hand side is ( tan frac{10pi}{21} ). We know that ( frac{10pi}{21} = frac{pi}{3} + frac{pi}{7} ), so using the tangent addition formula:[tan left( frac{pi}{3} + frac{pi}{7} right) = frac{tan frac{pi}{3} + tan frac{pi}{7}}{1 - tan frac{pi}{3} tan frac{pi}{7}} = frac{sqrt{3} + t}{1 - sqrt{3} t}]Equating the two expressions, we have:[frac{a t + b}{a - b t} = frac{sqrt{3} + t}{1 - sqrt{3} t}]Cross-multiplying gives:[(a t + b)(1 - sqrt{3} t) = (a - b t)(sqrt{3} + t)]Expanding both sides:Left side:[a t - a sqrt{3} t^2 + b - b sqrt{3} t]Right side:[a sqrt{3} + a t - b sqrt{3} t - b t^2]Subtracting the right side from the left side:[-a sqrt{3} t^2 + b t^2 + b - a sqrt{3} = 0]Factoring out ( (b - a sqrt{3}) ):[(b - a sqrt{3})(t^2 + 1) = 0]Since ( t^2 + 1 neq 0 ), we have ( b - a sqrt{3} = 0 ), leading to ( b = a sqrt{3} ). Thus, the value of ( frac{b}{a} ) is:[boxed{sqrt{3}}]]

🔑:Given that frac{asinfrac{pi}{7} + bcosfrac{pi}{7}}{acosfrac{pi}{7} - bsinfrac{pi}{7}} = tanfrac{10pi}{21}, we havetanfrac{10pi}{21} = frac{asinfrac{pi}{7} + bcosfrac{pi}{7}}{acosfrac{pi}{7} - bsinfrac{pi}{7}} = frac{tanfrac{pi}{7} + frac{b}{a}}{1 - frac{b}{a}tanfrac{pi}{7}}.Let tantheta = frac{b}{a}. Then,tanfrac{10pi}{21} = frac{tanfrac{pi}{7} + tantheta}{1 - tanfrac{pi}{7}tantheta} = tanleft(frac{pi}{7} + thetaright).This implies thatfrac{10pi}{21} + kpi = frac{pi}{7} + theta, where k is an integer.Thus, theta = kpi + frac{pi}{3}.It follows that frac{b}{a} = tantheta = tanleft(kpi + frac{pi}{3}right) = tanfrac{pi}{3} = sqrt{3}.Therefore, the answer is boxed{frac{b}{a} = sqrt{3}}.

🧠:[<think>Okay, let me try to work through this problem step by step. I need to solve both parts (1) and (2). Let's start with part (1).Problem (1): Find the value of sin A.We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. The vectors m and n are given as:- m = (cos(A - B), sin(A - B))- n = (cos B, -sin B)And their dot product is -3/5. So, the dot product of vectors m and n is equal to -3/5. First, recall that the dot product of two vectors (a, b) and (c, d) is ac + bd. So applying that here:m · n = cos(A - B) * cos B + sin(A - B) * (-sin B) = -3/5Let me compute this expression step by step.Compute the first term: cos(A - B) * cos BCompute the second term: sin(A - B) * (-sin B) = -sin(A - B) * sin BSo, the entire expression becomes:cos(A - B)cos B - sin(A - B)sin B = -3/5Wait a minute, that looks like a trigonometric identity. Let me recall: cos(X + Y) = cos X cos Y - sin X sin Y. So, if we let X = (A - B) and Y = B, then:cos((A - B) + B) = cos A = cos(A - B)cos B - sin(A - B)sin BTherefore, the left side of the equation simplifies to cos A. Therefore:cos A = -3/5So, cos A = -3/5. Then, they want sin A. Since we know cos A, we can find sin A using the Pythagorean identity:sin² A + cos² A = 1Therefore:sin² A = 1 - cos² A = 1 - (9/25) = 16/25Thus, sin A = ±4/5. However, in a triangle, all angles are between 0 and π radians (0° and 180°). The sine of any angle in this range is non-negative. Therefore, sin A = 4/5.Wait, but angle A could be in the second quadrant (between 90° and 180°), where cosine is negative and sine is positive. Since cos A is -3/5, which is negative, so angle A is indeed in the second quadrant. Therefore, sin A is positive. So yes, sin A = 4/5.Therefore, the answer to part (1) is 4/5. That seems straightforward once I recognized the trigonometric identity. Let me double-check my steps.1. Calculated the dot product using components: correct.2. Recognized the identity for cos((A - B) + B) = cos A: correct.3. Solved for sin A using Pythagorean identity: correct, and considered the sign based on the angle's quadrant: correct. Okay, part (1) seems done. Now, moving on to part (2).Problem (2): If a = 4√2, b = 5, find the measure of angle B and the projection of vector AB onto vector BC.First, let's note the sides:In triangle ABC,- a is the side opposite angle A: so side BC = a = 4√2- b is the side opposite angle B: so side AC = b = 5- c is the side opposite angle C: so side AB = c (unknown)We need to find angle B and the projection of vector AB onto vector BC.First, let's find angle B. We know sides a and b, and from part (1) we know sin A = 4/5. Maybe we can use the Law of Sines or Law of Cosines here.Law of Sines states that a/sin A = b/sin B = c/sin CWe know a = 4√2, b = 5, and sin A = 4/5. Let's try using the Law of Sines to find angle B.From Law of Sines:a / sin A = b / sin BSo, 4√2 / (4/5) = 5 / sin BCompute left side:4√2 divided by (4/5) is equal to 4√2 * (5/4) = 5√2Therefore:5√2 = 5 / sin BDivide both sides by 5:√2 = 1 / sin BTherefore:sin B = 1 / √2 = √2 / 2 ≈ 0.7071So sin B = √2 / 2, which corresponds to angles of 45° or 135° (since sine is positive in first and second quadrants). However, in a triangle, all angles are less than 180°, so possible angles are 45° or 135°.But we need to check which one is valid. Since we already have angle A with cosine -3/5, which is in the second quadrant (since cosine is negative there). Let's compute angle A.cos A = -3/5, so angle A is arccos(-3/5). Let's compute that in degrees:cos A = -3/5, so angle A is approximately 126.87° (since cos 120° = -0.5, cos 135° ≈ -0.707, so -0.6 is between 90° and 180°, around 126.87°). Let me confirm with a calculator:cos⁻¹(-3/5) ≈ 126.87°So angle A is approximately 126.87°, which is more than 90°, so the sum of angles A and B must be less than 180°, so angle B must be less than 180° - 126.87° ≈ 53.13°. Therefore, angle B can't be 135°, which is more than 53.13°, so angle B must be 45°.Therefore, angle B is 45°, which is π/4 radians.Alternatively, perhaps we can use the Law of Cosines to verify.But let's see. Alternatively, since we know angle A and sides a and b, maybe using Law of Cosines.Wait, but the Law of Sines gave us angle B as 45°, and that seems consistent with angle A being 126.87°, because angles A + B + C = 180°, so angle C would be 180 - 126.87 - 45 ≈ 8.13°, which is possible.Alternatively, let's compute angle B using Law of Cosines.Law of Cosines: c² = a² + b² - 2ab cos C. But we don't know angle C or side c. Alternatively, perhaps Law of Cosines on angle B.Wait, Law of Cosines for angle B:For angle B, which is opposite side b (AC = 5). Wait, no: in standard notation, side a is opposite angle A, side b opposite angle B, side c opposite angle C.Wait, angle B is opposite side b = 5. So Law of Cosines:b² = a² + c² - 2ac cos BBut we don't know side c. So maybe not helpful yet. Alternatively, using Law of Sines seems okay.But let's check if angle B is 45°, then using Law of Sines, sin A = 4/5, so angle A is arcsin(4/5). Wait, but wait, we already found cos A = -3/5, so angle A is in the second quadrant. So sin A is 4/5, as given. So angle A is arcsin(4/5) in the second quadrant, which is π - arcsin(4/5). Wait, but arcsin(4/5) is approximately 53.13°, so angle A is 180 - 53.13 = 126.87°, which matches previous calculation.So with angle B as 45°, and angle A as 126.87°, then angle C is 180 - 126.87 - 45 ≈ 8.13°. Then, using Law of Sines, we can find side c.Alternatively, if we use Law of Sines:a / sin A = c / sin CBut maybe that's not necessary unless needed. Since the problem only asks for angle B and the projection, which we can handle.But to confirm angle B, since sin B = √2 / 2, so B is 45°, which is valid given the other angles. Therefore, angle B is 45°, which is π/4 radians. So that's the measure of angle B.Now, the second part of problem (2): Find the projection of vector AB onto vector BC.First, let's clarify the vectors. In triangle ABC, vector AB and vector BC.But need to be precise with the points. Let me sketch triangle ABC:- Let’s denote the vertices as A, B, C.- Side opposite angle A is BC = a = 4√2- Side opposite angle B is AC = b = 5- Side opposite angle C is AB = cSo, points:- A is opposite side a (BC)- B is opposite side b (AC)- C is opposite side c (AB)Therefore, the triangle has vertices A, B, C with sides:- BC = a = 4√2- AC = b = 5- AB = cVectors:- Vector AB is from point A to point B.- Vector BC is from point B to point C.We need the projection of vector AB onto vector BC.Projection of vector AB onto vector BC is given by the formula:proj_{BC}(AB) = |AB| * cos θ, where θ is the angle between vectors AB and BC.But wait, vectors AB and BC. The angle between them is not necessarily an angle of the triangle. Let's see.Wait, in triangle ABC, the angle at point B is angle B, which is 45°. But the angle between vectors AB and BC is different. Let me think.Vector AB goes from A to B, and vector BC goes from B to C. So, the angle between AB and BC is the angle at point B between BA and BC. Wait, but vector AB is from A to B, and vector BC is from B to C.Wait, to find the angle between vector AB and vector BC, we need to consider their directions. Vector AB is from A to B, and vector BC is from B to C. So, if we place point B at the origin, vector AB would be pointing towards B from A, but actually, vector AB is from A to B. Wait, perhaps it's easier to model the coordinates.Alternatively, maybe it's better to compute the projection using the dot product formula.The projection of vector AB onto vector BC is equal to (AB · BC) / |BC|.But we need to compute AB · BC, then divide by |BC|.But to compute the dot product, we need the vectors in component form. Maybe we can assign coordinates to the triangle to compute the vectors.Let me try to assign coordinates to the triangle to make this easier.Let's place point B at the origin (0,0).Let’s place point C along the positive x-axis, so point C is at (a, 0) where a = BC = 4√2.Point A is somewhere in the plane. We need to find its coordinates.We know side AC = b = 5, and angle B is 45°, which is at point B. Wait, angle B is 45°, which is the angle between sides AB and BC. Wait, angle at point B is between BA and BC. Since we placed point B at (0,0) and point C at (4√2, 0), then BA is the vector from B to A, and BC is the vector from B to C. The angle between BA and BC is angle B, which is 45°.So, with point B at (0,0), point C at (4√2, 0), and angle at B is 45°, so point A is somewhere in the plane. Let's find coordinates of point A.We can use Law of Cosines to find coordinates of A.Wait, in triangle ABC, with points B(0,0), C(4√2, 0), angle at B is 45°, and AC = 5. Let me use coordinates.Let’s denote coordinates:- B: (0, 0)- C: (4√2, 0)Let’s denote point A as (x, y). Then, BA is the vector from B to A: (x, y). BC is the vector from B to C: (4√2, 0).The angle between BA and BC is 45°, so the cosine of the angle between BA and BC is equal to the dot product of BA and BC divided by (|BA| |BC|).So:cos 45° = (BA · BC) / (|BA| |BC|)Compute BA · BC = x * 4√2 + y * 0 = 4√2 x|BA| = √(x² + y²)|BC| = 4√2Therefore:√2 / 2 = (4√2 x) / (√(x² + y²) * 4√2 )Simplify:√2 / 2 = (4√2 x) / (4√2 √(x² + y²))Cancel out 4√2:√2 / 2 = x / √(x² + y²)Multiply both sides by √(x² + y²):√2 / 2 * √(x² + y²) = xSquare both sides to eliminate the square root:( (√2 / 2)^2 ) * (x² + y²) = x²Compute (√2 / 2)^2 = (2 / 4) = 1/2So:(1/2)(x² + y²) = x²Multiply both sides by 2:x² + y² = 2x²Therefore:y² = x² => y = ±xBut in the triangle, point A cannot be on the x-axis because angle B is 45°, and points B and C are on the x-axis. Since angle at B is 45°, point A must be above the x-axis (assuming the triangle is drawn in standard position). So, y = x.Therefore, coordinates of A are (x, x). Now, we also know that AC = 5. The distance from A(x, x) to C(4√2, 0) is 5.Compute AC:√[(4√2 - x)^2 + (0 - x)^2] = 5Square both sides:(4√2 - x)^2 + x² = 25Expand (4√2 - x)^2:(4√2)^2 - 2*4√2*x + x² + x² = 25Compute (4√2)^2 = 16 * 2 = 32So:32 - 8√2 x + x² + x² = 25Combine like terms:2x² - 8√2 x + 32 - 25 = 0Simplify:2x² - 8√2 x + 7 = 0Divide both sides by 1:2x² - 8√2 x + 7 = 0Let me solve this quadratic equation for x.Quadratic in the form ax² + bx + c = 0, where:a = 2b = -8√2c = 7Discriminant D = b² - 4ac = ( -8√2 )² - 4*2*7 = 128 - 56 = 72Square root of D is √72 = 6√2Solutions:x = [8√2 ± 6√2]/(2*2) = [ (8√2 ± 6√2) ] / 4 = [√2(8 ± 6)] / 4So:x = (√2 * 14)/4 = (14√2)/4 = (7√2)/2 ≈ 4.95orx = (√2 * 2)/4 = (2√2)/4 = (√2)/2 ≈ 0.707Now, we need to check which solution is valid. Let's consider the coordinates.If x = (7√2)/2 ≈ 4.95, then since point C is at 4√2 ≈ 5.656, so 4.95 is less than 5.656, so that's possible.If x = √2 / 2 ≈ 0.707, then point A would be closer to point B. Let's check the distances.But we also know from the Law of Sines that side AB (which is c) can be found using a/sin A = 4√2 / (4/5) = 5√2, so c = 5√2 * sin C.But angle C is approximately 8.13°, so sin C ≈ sin 8.13° ≈ 0.1414, so c ≈ 5√2 * 0.1414 ≈ 5 * 1.414 * 0.1414 ≈ 1.0. But this seems inconsistent with x ≈ 0.707. Wait, perhaps I need to check.Alternatively, maybe both solutions are possible? Wait, but in a triangle, given two sides and the included angle, there could be two possible triangles, but in this case, since we have two sides and a non-included angle, maybe there are two solutions. Wait, but in our case, we have side AC = 5, angle at B = 45°, and side BC = 4√2.Wait, maybe there are two possible positions for point A. However, since angle A is fixed at 126.87°, which is greater than 90°, then there can only be one triangle. Wait, but how?Wait, if angle A is fixed (from part 1), then there is only one possible triangle. Therefore, only one solution for x. Therefore, we need to determine which of the two x solutions corresponds to angle A being 126.87°.Given that angle A is at point A, which is opposite side BC. If point A is at (x, x), then the position of A affects the angles. Let's compute coordinates for both x values and check the angles.First, x = (7√2)/2 ≈ 4.95. Then, coordinates of A are (4.95, 4.95). Point C is at (5.656, 0). Then, vector BA is (4.95, 4.95), vector BC is (5.656, 0). The angle at B is 45°, which is consistent.Alternatively, if x = √2 / 2 ≈ 0.707, then coordinates of A are (0.707, 0.707). Then, vector BA is (0.707, 0.707), vector BC is (5.656, 0). The angle between BA and BC would still be 45°, but in this case, the distance AC would be:Distance between (0.707, 0.707) and (5.656, 0):√[(5.656 - 0.707)^2 + (0 - 0.707)^2] ≈ √[(4.949)^2 + (-0.707)^2] ≈ √[24.49 + 0.5] ≈ √25 ≈ 5. So that's correct. So both points satisfy AC = 5. But which one gives angle A as 126.87°?Wait, angle A is at point A. So we need to compute angle at point A in both cases.First, let's compute angle at A for x = (7√2)/2.Coordinates:A: ( (7√2)/2 , (7√2)/2 )B: (0,0)C: (4√2, 0)Vectors:AB = B - A = (- (7√2)/2, - (7√2)/2 )AC = C - A = (4√2 - (7√2)/2, - (7√2)/2 ) = ( (8√2 - 7√2)/2, - (7√2)/2 ) = (√2/2, -7√2/2 )To find angle at A, we can compute the angle between vectors AB and AC.Vectors AB and AC:AB = (-7√2/2, -7√2/2 )AC = (√2/2, -7√2/2 )The angle between AB and AC can be found using the dot product:cos θ = (AB · AC) / (|AB| |AC| )Compute AB · AC:(-7√2/2)(√2/2) + (-7√2/2)(-7√2/2 )First term: (-7√2/2)(√2/2) = (-7 * 2 / 4 ) = -7/2Second term: (-7√2/2)(-7√2/2 ) = (49 * 2)/4 = 49/2Sum: -7/2 + 49/2 = 42/2 = 21Compute |AB|:|AB| = sqrt( (-7√2/2)^2 + (-7√2/2)^2 ) = sqrt( 49*2/4 + 49*2/4 ) = sqrt( 98/4 + 98/4 ) = sqrt(196/4) = sqrt(49) = 7Compute |AC|:|AC| = sqrt( (√2/2)^2 + (-7√2/2)^2 ) = sqrt( (2/4) + (49*2)/4 ) = sqrt( (2 + 98)/4 ) = sqrt(100/4) = sqrt(25) = 5So cos θ = 21 / (7 * 5) = 21/35 = 3/5Thus, angle θ = arccos(3/5) ≈ 53.13°, but angle at A is supposed to be ≈126.87°, which is 180° - 53.13°. Wait, that's confusing.Wait, angle at A is the angle between vectors AB and AC. Wait, but in the coordinate system, when we compute the angle between AB and AC, we have to consider the direction of the vectors. Since AB is from A to B and AC is from A to C, the angle between them is indeed the angle at A. Wait, but when we computed the vectors AB and AC as vectors from A, but in the calculation above, vectors AB and AC are from point A, but we took vectors as AB = B - A and AC = C - A, which is correct. So the angle between vectors AB and AC is angle at A.But arccos(3/5) ≈53.13°, but we know from part (1) that angle A is ≈126.87°, which is in the second quadrant. This inconsistency suggests an error.Wait, but cos(theta) = 3/5, so theta is 53.13°, but we expected angle A to be 126.87°. Therefore, this suggests that the angle at A in this coordinate system is 53.13°, which contradicts part (1). Therefore, this solution x = (7√2)/2 is invalid. Let's check the other solution.For x = √2 / 2 ≈0.707, coordinates of A are (√2/2, √2/2).Vectors:AB = B - A = (-√2/2, -√2/2 )AC = C - A = (4√2 - √2/2, -√2/2 ) = ( (8√2 - √2)/2, -√2/2 ) = (7√2/2, -√2/2 )Compute angle at A between vectors AB and AC.Compute AB · AC:(-√2/2)(7√2/2) + (-√2/2)(-√2/2 )First term: (-√2/2)(7√2/2) = (-7*2)/4 = -14/4 = -7/2Second term: (-√2/2)(-√2/2 ) = (2)/4 = 1/2Sum: -7/2 + 1/2 = -6/2 = -3Compute |AB|:|AB| = sqrt( (-√2/2)^2 + (-√2/2)^2 ) = sqrt( (2/4) + (2/4) ) = sqrt(1) = 1Wait, but in reality, side AB is opposite angle C, and if |AB| = 1, but from Law of Sines earlier, a/sin A = 4√2 / (4/5) =5√2, so c = AB = 5√2 * sin C. Since angle C is ≈8.13°, sin C ≈0.1414, so c≈5√2*0.1414≈1.0. So this matches. So AB length is 1.But then |AC| is given as 5, but from coordinates:Distance from A(√2/2, √2/2) to C(4√2, 0):sqrt( (4√2 - √2/2)^2 + (0 - √2/2)^2 )= sqrt( ( (8√2 - √2)/2 )^2 + ( -√2/2 )^2 )= sqrt( (7√2/2)^2 + (√2/2)^2 )= sqrt( (49*2)/4 + (2)/4 )= sqrt(98/4 + 2/4) = sqrt(100/4) = sqrt(25) =5, which is correct.So in this case, vectors AB and AC are:AB = (-√2/2, -√2/2 ), AC = (7√2/2, -√2/2 )Compute angle between them:Dot product AB · AC = (-√2/2)(7√2/2) + (-√2/2)(-√2/2 )= (-7*2)/4 + (2)/4 = (-14 + 2)/4 = -12/4 = -3|AB| = 1, |AC| =5Thus, cos(theta) = (-3)/(1*5) = -3/5Therefore, theta = arccos(-3/5) ≈126.87°, which matches part (1). Therefore, this is the correct position for point A. Therefore, x = √2/2 is the valid solution.Therefore, coordinates of A are (√2/2, √2/2), B is (0,0), C is (4√2, 0).Now, we need to find the projection of vector AB onto vector BC.First, let's find vector AB and vector BC.Vector AB is from point A to point B: B - A = (0 - √2/2, 0 - √2/2) = (-√2/2, -√2/2 )But wait, in vector terms, the projection formula is:proj_u v = (v · u / |u|² ) uBut the projection length is |v| cos(theta), which is (v · u)/|u|.But we need to confirm the direction. The projection of vector AB onto vector BC can be a scalar or a vector. The problem says "the projection of vector AB onto vector BC". Depending on the convention, it might be the scalar projection (the magnitude) or the vector projection. The problem mentions "the projection", but since it's asking for a measure, perhaps the scalar projection.But let's check.First, compute vectors AB and BC.Vector AB: from A to B: B - A = (-√2/2, -√2/2 )Vector BC: from B to C: C - B = (4√2, 0 )But the projection can be computed as:proj_{BC}(AB) = (AB · BC / |BC|² ) * BC (vector projection)But if they ask for the projection as a scalar, it's (AB · BC)/|BC|But the problem says "the projection of vector AB onto vector BC". Depending on the context, in physics, sometimes projection refers to the scalar projection. However, in math, it can refer to the vector projection. Let me check the problem statement again.Original problem: "the projection of vector AB onto vector BC". It just says "the projection", but in Chinese problems, sometimes it refers to the scalar projection. However, since in part (1) they used vectors and the dot product, perhaps they expect the scalar projection. But to be safe, let's compute both and see.But first, compute AB and BC vectors.Wait, in the problem statement: "the projection of vector AB onto vector BC".Vector AB is from A to B: as per coordinates, that's (-√2/2, -√2/2 )Vector BC is from B to C: (4√2, 0 )The scalar projection of AB onto BC is equal to the dot product of AB and the unit vector in the direction of BC.Which is (AB · BC)/|BC|.Alternatively, the scalar projection is |AB| cos(theta), where theta is the angle between AB and BC.Compute AB · BC:(-√2/2)(4√2) + (-√2/2)(0) = (-√2/2)(4√2) + 0 = (-4*2)/2 = -8/2 = -4|BC| = 4√2Therefore, scalar projection is (-4)/ (4√2) = -1/√2 = -√2/2 ≈ -0.7071But projections can be negative, indicating direction opposite to BC.But if the problem asks for the projection as a scalar, it's -√2/2. If they want the magnitude, it's √2/2. But we need to check.Alternatively, in some contexts, the projection is considered as a scalar value, considering the sign. So if the projection is in the opposite direction of BC, it's negative.But let's verify with the angle.The angle between AB and BC. Vector AB is from A to B, vector BC is from B to C.But angle between AB and BC can be calculated.But since AB is (-√2/2, -√2/2 ), and BC is (4√2, 0 )The angle theta between them can be found via:cos(theta) = (AB · BC) / (|AB| |BC| )We have AB · BC = -4|AB| = sqrt( (-√2/2)^2 + (-√2/2)^2 ) = sqrt( (2/4 + 2/4 )) = sqrt(1) =1|BC| =4√2Thus, cos(theta) = -4 / (1 * 4√2 ) = -1/√2Thus, theta = 135°, which makes sense. So the angle between AB and BC is 135°, so the projection of AB onto BC is |AB| cos(theta) = 1 * cos(135°) = -√2/2Therefore, the scalar projection is -√2/2.But let's confirm with coordinates.Alternatively, since point A is at (√2/2, √2/2), B is at (0,0), so vector AB is (-√2/2, -√2/2). Vector BC is (4√2, 0 ).The projection of AB onto BC is:( AB · BC ) / |BC| = (-4)/ (4√2 ) = -1/√2 = -√2/2Therefore, the projection is -√2/2. However, depending on the problem's requirement, if they want the magnitude, it's √2/2, but since the problem says "the projection of vector AB onto vector BC", and given the vectors, the projection is a scalar which can be negative, indicating direction. Since the answer is likely a scalar with sign, so the answer is -√2/2.But let me check if there's a different interpretation.Alternatively, perhaps the problem considers vectors AB and BC as vectors from point A to B and from point B to C, but in the projection, sometimes the vectors are considered geometrically.But in standard terms, the projection of vector AB onto vector BC is calculated as above. Therefore, the answer should be -√2/2.But let me check with another method.Alternatively, since we know the coordinates:Vector AB is from A(√2/2, √2/2) to B(0,0): (-√2/2, -√2/2 )Vector BC is from B(0,0) to C(4√2,0): (4√2, 0 )The projection of AB onto BC is:The component of AB in the direction of BC. Since BC is along the x-axis from B to C, the projection is simply the x-component of vector AB. But vector AB is (-√2/2, -√2/2 ). The x-component is -√2/2. However, since vector BC is in the positive x-direction, the projection is negative, which matches our previous result.Therefore, the projection is -√2/2.But the problem might expect a positive value since it's a length, but in vector projection, it can be negative. Let me check the problem statement again:"the projection of vector AB onto vector BC".If it's a scalar projection, it's -√2/2. If it's the vector projection, it's (-√2/2, 0 ). But given that in part (1) they used vectors and the dot product, which resulted in a scalar, maybe here they want the scalar projection. However, the problem says "the projection of vector AB onto vector BC", which is technically a vector. However, in Chinese contexts, sometimes "projection" refers to the scalar projection. But given that in the problem statement part (1) mentions vectors and their dot product, which is scalar, perhaps here they also expect a scalar.But in the problem's original Chinese context (if this is a translation), sometimes projection is scalar. Alternatively, the problem might want the signed scalar projection.But let's see. In the problem statement, part (2) says:"find the measure of angle B and the projection of vector AB onto vector BC".Since angle B is a measure (45°), the projection is likely a scalar. Therefore, the answer is -√2/2. However, sometimes projection is given as a magnitude, but since the problem mentions "projection of vector AB onto vector BC", and in vector terms, the scalar projection can be negative. Therefore, I think the answer is -√2/2.But let me confirm with another approach.Using the Law of Cosines in triangle ABC, we can find the length of AB, which is c.From part (1), we have sin A = 4/5, cos A = -3/5.Using Law of Sines:a / sin A = b / sin B = c / sin CFrom earlier, a =4√2, sin A =4/5, so 4√2 / (4/5) =5√2.Therefore, the common ratio is 5√2.Thus, side c = AB =5√2 * sin C.But angle C = 180° - A - B = 180° - 126.87° -45° ≈8.13°, so sin C ≈0.1414, so c≈5√2 *0.1414≈1.0, which matches our previous result where |AB| =1.Therefore, AB has length 1. The projection of AB onto BC is |AB| cos(theta), where theta is the angle between AB and BC, which is 135°, so cos(theta)= -√2/2, so the projection is 1*(-√2/2)= -√2/2.Therefore, the projection is -√2/2.But in some contexts, especially in geometry, projection lengths are given as absolute values. However, since the problem involves vectors and their projection considering direction, the answer should include the sign. Therefore, the projection is -√2/2.But let me check once more with coordinates:Coordinates:A: (√2/2, √2/2)B: (0,0)C: (4√2,0)Vector AB: from A to B: (-√2/2, -√2/2 )Vector BC: from B to C: (4√2, 0 )The scalar projection is AB · (BC / |BC| ) = (-√2/2, -√2/2 ) · (4√2,0 ) / |BC|Wait, compute unit vector in direction of BC: (4√2,0 ) / (4√2 ) = (1,0 )Then, the scalar projection is AB · (1,0 ) = (-√2/2 )*1 + (-√2/2 )*0 = -√2/2.Yes, that's consistent.Therefore, the projection is -√2/2.Thus, the answers are:(1) sin A = 4/5(2) Angle B = 45°, projection is -√2/2But before finalizing, let's check if the problem might want the magnitude of the projection. If so, it's √2/2, but given the problem uses vectors and the dot product leading to a negative value, I think the negative sign is important. However, in some geometrical contexts, projection might refer to the absolute value. Wait, the problem says "the projection of vector AB onto vector BC". In vector terms, the scalar projection can be negative. However, in geometry, sometimes the term "projection" refers to the length, hence absolute value. But in the context of vectors and the given information leading to a negative dot product, the answer should retain the sign. Therefore, I will go with -√2/2.Final check:- Angle B: 45° (π/4 radians), which is correct.- Projection of AB onto BC: -√2/2.Therefore, the answers are:(1) boxed{dfrac{4}{5}}(2) Angle B: boxed{dfrac{pi}{4}} or boxed{45^circ}, and projection: boxed{-dfrac{sqrt{2}}{2}}But since the problem says "the measure of angle B", it's better to present it in radians or degrees. Given the original problem doesn't specify, but in Chinese textbooks, commonly degrees are used. However, since π/4 is exact, maybe radians is better. But it's safer to check both. But given the first part is numerical, perhaps degrees. Alternatively, both are correct.Wait, the problem says "find the measure of angle B". "Measure" can be in degrees or radians. Given the context is trigonometry with triangles, degrees are common. However, in higher mathematics, radians are standard. But since the problem didn't specify, either is acceptable, but likely degrees. However, in Chinese textbooks, often degrees are used. But given that in the calculation, we found angle B as 45°, which is π/4 radians, so writing both is possible, but to be precise, the exact value is π/4, which is 45 degrees. So both are correct, but since the problem doesn't specify, I can write 45°.Final Answer(1) boxed{dfrac{4}{5}}(2) Angle B is boxed{45^circ} and the projection is boxed{-dfrac{sqrt{2}}{2}}</think>Given the problem, we need to solve two parts:(1) Find the value of (sin A).We start with the vectors (overrightarrow{m} = (cos(A - B), sin(A - B))) and (overrightarrow{n} = (cos B, -sin B)). Their dot product is given as (-frac{3}{5}). The dot product is calculated as:[overrightarrow{m} cdot overrightarrow{n} = cos(A - B)cos B - sin(A - B)sin B]Using the trigonometric identity (cos(X + Y) = cos X cos Y - sin X sin Y), we recognize this as (cos(A - B + B) = cos A). Therefore:[cos A = -frac{3}{5}]Using the Pythagorean identity:[sin^2 A + cos^2 A = 1 implies sin^2 A = 1 - left(-frac{3}{5}right)^2 = 1 - frac{9}{25} = frac{16}{25} implies sin A = frac{4}{5}]Since (A) is an angle in a triangle, (sin A) is positive. Thus, (sin A = frac{4}{5}).(2) Find the measure of angle (B) and the projection of vector (overrightarrow{AB}) onto (overrightarrow{BC}).Given sides (a = 4sqrt{2}) (opposite angle (A)) and (b = 5) (opposite angle (B)), we use the Law of Sines:[frac{a}{sin A} = frac{b}{sin B} implies frac{4sqrt{2}}{frac{4}{5}} = frac{5}{sin B} implies 5sqrt{2} = frac{5}{sin B} implies sin B = frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]Thus, (B = 45^circ) (since it must be less than (90^circ) given angle (A) is obtuse).For the projection of vector (overrightarrow{AB}) onto (overrightarrow{BC}), we place points (B) at ((0, 0)), (C) at ((4sqrt{2}, 0)), and find coordinates of (A). Using the Law of Cosines and solving quadratic equations, we determine coordinates of (A) as ((frac{sqrt{2}}{2}, frac{sqrt{2}}{2})).Vectors:- (overrightarrow{AB} = (-frac{sqrt{2}}{2}, -frac{sqrt{2}}{2}))- (overrightarrow{BC} = (4sqrt{2}, 0))The scalar projection is:[text{proj}_{overrightarrow{BC}} overrightarrow{AB} = frac{overrightarrow{AB} cdot overrightarrow{BC}}{|overrightarrow{BC}|} = frac{(-frac{sqrt{2}}{2})(4sqrt{2}) + (-frac{sqrt{2}}{2})(0)}{4sqrt{2}} = frac{-4}{4sqrt{2}} = -frac{sqrt{2}}{2}]Final answers:(1) (boxed{dfrac{4}{5}})(2) Angle (B) is (boxed{45^circ}) and the projection is (boxed{-dfrac{sqrt{2}}{2}}),<think>Okay, let's tackle this problem step by step. First, let's read the problem again to make sure I understand what's being asked.We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. There are two vectors: vector m is (cos(A - B), sin(A - B)) and vector n is (cos B, -sin B). Their dot product is given as -3/5. Part (1) asks for sin A. Part (2) gives specific values for sides a and b (4√2 and 5 respectively) and asks for the measure of angle B and the projection of vector AB onto vector BC.Starting with part (1). The key here is the dot product of vectors m and n. Let's recall that the dot product of two vectors (x1, y1) and (x2, y2) is x1x2 + y1y2. So applying that here:m · n = cos(A - B) * cos B + sin(A - B) * (-sin B) = -3/5Let's compute this expression:cos(A - B) * cos B - sin(A - B) * sin BHmm, this looks like the formula for cosine of a sum. Remember that cos(X + Y) = cos X cos Y - sin X sin Y. Comparing this to our expression, if we set X = (A - B) and Y = B, then the expression becomes cos((A - B) + B) = cos A. Therefore, the left-hand side simplifies to cos A. So:cos A = -3/5Therefore, the value of cos A is -3/5. Then, since we need sin A, we can use the Pythagorean identity sin² A + cos² A = 1. Therefore:sin² A = 1 - cos² A = 1 - (9/25) = 16/25Thus, sin A = ±4/5. However, in a triangle, all angles are between 0 and π radians (0° and 180°). The cosine of A is negative, which implies that angle A is in the second quadrant (between 90° and 180°). In the second quadrant, sine is positive, so sin A = 4/5. So part (1) is 4/5.Wait, let me check that again. If cos A is -3/5, then angle A is obtuse? Wait, cos A = -3/5. So A is greater than 90°, but less than 180°, so yes, in that case, sin A is positive. Therefore, sin A = 4/5. That makes sense. So part (1)'s answer is 4/5.Moving on to part (2). Given a = 4√2, b = 5, find angle B and the projection of vector AB onto vector BC.First, we know sides a and b. In triangle ABC, side a is opposite angle A, and side b is opposite angle B. Since we found sin A = 4/5 in part (1), perhaps we can use the Law of Sines to find angle B.Law of Sines states that a / sin A = b / sin B = c / sin C.We have a = 4√2, b = 5, sin A = 4/5. Let's compute a / sin A:(4√2) / (4/5) = (4√2) * (5/4) = 5√2Therefore, the Law of Sines ratio is 5√2. Then, b / sin B = 5 / sin B = 5√2. Therefore:5 / sin B = 5√2 => sin B = 5 / (5√2) = 1/√2So sin B = √2 / 2. Therefore, angle B is either 45° or 135° (since sine is positive in first and second quadrants). However, since angle A is already greater than 90°, the sum of angles A and B would be greater than 90° + 45° = 135°, so angle C would be less than 45°, which is possible. Alternatively, if angle B were 135°, then angle A + angle B would be greater than 90° + 135° = 225°, which would make angle C negative, which is impossible. Therefore, angle B must be 45°, i.e., π/4 radians.Therefore, angle B is 45 degrees.Now, the projection of vector AB onto vector BC.First, let's recall what the projection of a vector onto another vector is. The projection of vector v onto vector u is (v · u) / |u|. However, sometimes the definition is given as (v · u) / |u|² * u, but here, since the problem says "the projection of vector AB onto vector BC", it's likely asking for the scalar projection, which is |AB| cos θ, where θ is the angle between AB and BC.Alternatively, depending on the context, maybe we need to compute the vector projection. But since the problem mentions "the projection", and given that it's in a triangle, we need to clarify.But let's recall the projection of AB onto BC. Let's model the triangle. Let's denote the points:Point A, B, C.Vector AB goes from point A to point B. Vector BC goes from point B to point C. So the projection of AB onto BC would be the scalar projection, which is |AB| cos φ, where φ is the angle between AB and BC.But wait, in the triangle, AB is a side, and BC is another side. The angle between AB and BC at point B is angle B. Wait, angle at point B is angle B, which is 45 degrees. Wait, but AB and BC are vectors. Let me visualize the triangle.In triangle ABC, point A is opposite side a (BC), point B is opposite side b (AC), and point C is opposite side c (AB). Wait, maybe I should clarify the notation. In standard notation, side a is BC, side b is AC, and side c is AB. So:- Side a = BC = 4√2 (Wait, no. Wait, the problem states: "In triangle ABC, a, b, c are the sides opposite to angles A, B, C respectively." So side a is opposite angle A, which is BC. Side b is opposite angle B, which is AC. Side c is opposite angle C, which is AB.Therefore, given a = 4√2 (BC = 4√2), b = 5 (AC = 5).But the projection of AB onto BC. Let's think about vectors.Vector AB is from point A to point B. Vector BC is from point B to point C. The angle between AB and BC is the angle at point B, but since AB is going from A to B, and BC is going from B to C, the angle between AB and BC is 180° - angle B. Because if you have vector AB and vector BC, the angle between them is the angle outside the triangle at point B.Wait, let me draw this mentally. If you have triangle ABC, with AB, BC, and AC. Vector AB is from A to B, and vector BC is from B to C. The angle between AB and BC is the angle between the vector AB and the vector BC, which is not the internal angle at B. The internal angle at B is between BA and BC. So BA is the opposite direction of AB. Therefore, the angle between AB and BC is 180° - angle at B.Since angle B is 45°, the angle between AB and BC is 180° - 45° = 135°. Therefore, the projection of AB onto BC is |AB| * cos(135°).But |AB| is the length of AB, which is side c. Wait, side c is opposite angle C, so AB is side c. But we don't know side c yet. Wait, but maybe we can compute it using the Law of Cosines or Law of Sines.Alternatively, since we know angle A (from part (1)), we can compute angle C, then use Law of Sines.Wait, from part (1), we know sin A = 4/5, cos A = -3/5. Therefore, angle A is arccos(-3/5). Let me compute that. Since cos A = -3/5, angle A is in the second quadrant. The reference angle is arccos(3/5). Let's compute that angle. arccos(3/5) is approximately 53.13°, so angle A is 180° - 53.13° = 126.87°. But maybe we can keep it exact.But perhaps we don't need the exact measure. Let's see.We have angle A and angle B. Since the sum of angles in a triangle is 180°, angle C = 180° - A - B. So angle C = 180° - angle A - 45°. Since angle A is arccos(-3/5), but maybe we can use sine instead. Since sin A = 4/5, angle A is arcsin(4/5). However, since angle A is in the second quadrant, we have angle A = π - arcsin(4/5). But maybe we don't need to calculate it explicitly.Alternatively, using the Law of Sines again. We have sides a, b, and c.We know a = 4√2, b = 5, angle B = 45°, sin A = 4/5. From Law of Sines:a / sin A = b / sin B = c / sin CWe have a / sin A = 4√2 / (4/5) = 5√2, as calculated before. Then, c = sin C * 5√2.But perhaps we can compute angle C first. Since angle A + angle B + angle C = 180°, angle C = 180° - angle A - angle B. Since angle B is 45°, angle C = 135° - angle A.But angle A is arccos(-3/5). Let's compute angle A:cos A = -3/5, so angle A = arccos(-3/5). Let's compute this angle:Using a calculator, arccos(-3/5) is approximately 126.87°, as I mentioned before. So angle C ≈ 180° - 126.87° - 45° ≈ 8.13°. So angle C is approximately 8.13°, but let's keep it exact.Alternatively, we can use the Law of Cosines to find side c. Let's see:Law of Cosines: c² = a² + b² - 2ab cos C. But we don't know angle C. Alternatively, using Law of Sines:sin C = sin(180° - A - B) = sin(A + B) because sin(180° - x) = sin x.Wait, no. sin(180° - x) = sin x. So angle C = 180° - A - B, so sin C = sin(A + B). Wait, sin(180° - (A + B)) = sin(A + B). Wait, no: sin(π - θ) = sin θ. So sin C = sin(A + B). But A + B is 180° - C, so sin(A + B) = sin(180° - C) = sin C. Hmm, that's circular.Alternatively, maybe we can compute sin C using angle addition formula. Since angle C = 180° - A - B, then sin C = sin(A + B). Wait, no, angle C = 180° - A - B, so sin C = sin(π - (A + B)) = sin(A + B). But sin(A + B) = sin A cos B + cos A sin B.We know sin A = 4/5, cos A = -3/5, sin B = √2/2, cos B = √2/2 (since angle B is 45°). Therefore:sin C = sin(A + B) = sin A cos B + cos A sin B = (4/5)(√2/2) + (-3/5)(√2/2) = [4/5 - 3/5] * √2/2 = (1/5)(√2/2) = √2/10Therefore, sin C = √2/10. Then, using Law of Sines:c = sin C * (a / sin A) = (√2/10) * 5√2 = (√2 * 5√2) / 10 = (5 * 2) / 10 = 10 / 10 = 1So side c = AB = 1. Wait, really? That seems very small. Let me check that again.Law of Sines: a / sin A = 4√2 / (4/5) = 4√2 * 5/4 = 5√2. Then, c = 5√2 * sin C. Since sin C = √2/10, so c = 5√2 * √2/10 = 5*(2)/10 = 10/10 = 1. Yes, that's correct. So side c is 1. So AB = 1. Interesting.Therefore, the length of AB is 1. So |AB| = 1.Now, the projection of AB onto BC. As mentioned earlier, the projection is |AB| * cos θ, where θ is the angle between vectors AB and BC. The angle between AB and BC is 180° - angle B, which is 135°, as angle B is 45°. Therefore, cos θ = cos 135° = -√2/2.Therefore, the projection is |AB| * cos 135° = 1 * (-√2/2) = -√2/2.But wait, is this the scalar projection or the vector projection? The problem says "the projection of vector AB onto vector BC". The scalar projection is the magnitude times the cosine of the angle, which can be negative, indicating direction. The vector projection would be that scalar times the unit vector in the direction of BC. However, since the problem mentions "the projection", and given that it's a geometry problem, it might just want the scalar projection. However, sometimes in physics, projection can refer to the scalar value with sign. Let's check the problem statement again.It says: "the projection of vector AB onto vector BC". The term "projection" in vector context can mean either scalar or vector projection, but since it's asked for "the projection" without specifying vector, it's possible they want the scalar projection. However, let's make sure.Alternatively, maybe the problem is referring to the projection in the geometric sense, meaning the length of the shadow of AB onto BC, which would take into account the direction. Since the angle is 135°, the projection would be negative, indicating the direction opposite to BC. However, in geometric terms, sometimes projection length is given as absolute value. But given the problem mentions vectors, it's more likely they want the scalar projection with sign.But let's cross-verify. Let's compute the projection using vector components.Vector AB: from point A to point B. The coordinates are a bit tricky since we don't have a coordinate system. Let's perhaps assign coordinates to the triangle to compute vectors.Let me place point B at the origin (0, 0). Let's orient the triangle such that point B is at (0, 0), point C is along the positive x-axis, so point C is at (c, 0), where c is the length of BC, which is side a = 4√2. Wait, no: in the problem, side a is opposite angle A, which is BC. Therefore, BC = a = 4√2, so the length of BC is 4√2.Wait, earlier, we found side AB = c = 1. Wait, side AB is c, which is opposite angle C. Wait, let's recap:In standard triangle notation:- Side a is opposite angle A: BC = a = 4√2- Side b is opposite angle B: AC = b = 5- Side c is opposite angle C: AB = c = 1Therefore, BC = 4√2, AC = 5, AB = 1.So if we place point B at (0, 0), point C at (4√2, 0). Now, point A is somewhere in the plane. We need to find coordinates of A such that AB = 1 and AC = 5. Let's find coordinates of point A.Let me denote coordinates:- Point B: (0, 0)- Point C: (4√2, 0)- Point A: (x, y)Then, distance from A to B is √(x² + y²) = 1, so x² + y² = 1.Distance from A to C is √((x - 4√2)² + y²) = 5, so (x - 4√2)² + y² = 25.Subtract the first equation from the second:(x - 4√2)² + y² - x² - y² = 25 - 1 => x² - 8√2 x + (4√2)² - x² = 24Simplify:-8√2 x + 32 = 24-8√2 x = -8x = (-8) / (-8√2) = 1/√2Therefore, x = √2/2.Then, from x² + y² = 1:( (√2/2)² ) + y² = 1 => (2/4) + y² = 1 => (1/2) + y² = 1 => y² = 1/2 => y = ±√(1/2) = ±√2/2But in the triangle, how is point A positioned? Since angle B is 45°, and angle A is obtuse, point A is above the x-axis. Let's assume y is positive. So y = √2/2.Therefore, coordinates of point A are (√2/2, √2/2).Therefore, vector AB is from A to B: B - A = (0 - √2/2, 0 - √2/2) = (-√2/2, -√2/2)Vector BC is from B to C: C - B = (4√2 - 0, 0 - 0) = (4√2, 0)Now, the projection of vector AB onto vector BC is given by the dot product of AB and the unit vector in the direction of BC.First, compute the unit vector in the direction of BC. Since BC is (4√2, 0), its magnitude is 4√2. Therefore, the unit vector is (1, 0).Wait, the vector BC is (4√2, 0), so the unit vector in the direction of BC is (1, 0), since it's along the x-axis. Therefore, the projection of AB onto BC is the dot product of AB and the unit vector in the direction of BC.But AB vector is (-√2/2, -√2/2). The unit vector in BC direction is (1, 0). So the dot product is (-√2/2)(1) + (-√2/2)(0) = -√2/2.Therefore, the projection is -√2/2, which matches our earlier calculation. So the projection is indeed -√2/2. The negative sign indicates that the projection is in the direction opposite to BC.Alternatively, if we use the formula for scalar projection: |AB| cos θ = 1 * cos 135° = -√2/2, which is the same.Therefore, the projection is -√2/2.But let's confirm once more. Since vector AB is (-√2/2, -√2/2), and vector BC is (4√2, 0). The projection scalar is (AB · BC) / |BC|.Compute AB · BC = (-√2/2)(4√2) + (-√2/2)(0) = (-√2/2)(4√2) = - (4*2)/2 = -4.Then, |BC| = 4√2, so the projection is -4 / (4√2) = -1/√2 = -√2/2.Yes, that's correct. So the projection is indeed -√2/2.Therefore, the answers are:(1) sin A = 4/5(2) Angle B = 45°, projection = -√2/2.But wait, angle B is 45°, which is π/4 radians, so depending on the required format, but the problem says "measure of angle B", so it's likely expected in degrees, so 45°. However, sometimes problems prefer radians, but since the answer is a common angle, 45° is safe.But let's check if angle B is 45°. Earlier, we used Law of Sines: a / sin A = b / sin B. So with a = 4√2, sin A = 4/5, b = 5.So 4√2 / (4/5) = 5 / sin B => 5√2 = 5 / sin B => sin B = 1/√2, which is 45°, correct. So yes, angle B is 45°.Thus, the answers are:1. sin A = 4/52. Angle B = 45°, projection = -√2/2.Final Answer(1) boxed{dfrac{4}{5}}(2) Angle ( B ) is boxed{45^circ} and the projection is boxed{-dfrac{sqrt{2}}{2}}.</think>Given the problem, we need to find the value of (sin A) and solve for angle (B) and the projection of vector (overrightarrow{AB}) onto (overrightarrow{BC}) in (triangle ABC). Part (1)The vectors (overrightarrow{m} = (cos(A - B), sin(A - B))) and (overrightarrow{n} = (cos B, -sin B)) have a dot product of (-frac{3}{5}). Using the dot product formula:[overrightarrow{m} cdot overrightarrow{n} = cos(A - B)cos B - sin(A - B)sin B = cos(A) = -frac{3}{5}]This simplifies to (cos A = -frac{3}{5}). Using the Pythagorean identity:[sin^2 A + cos^2 A = 1 implies sin^2 A = 1 - left(-frac{3}{5}right)^2 = 1 - frac{9}{25} = frac{16}{25} implies sin A = frac{4}{5}]Since (A) is in the second quadrant, (sin A) is positive. Thus, (sin A = frac{4}{5}). Part (2)Given (a = 4sqrt{2}) and (b = 5), we use the Law of Sines:[frac{a}{sin A} = frac{b}{sin B} implies frac{4sqrt{2}}{frac{4}{5}} = frac{5}{sin B} implies 5sqrt{2} = frac{5}{sin B} implies sin B = frac{1}{sqrt{2}} implies B = 45^circ]For the projection of (overrightarrow{AB}) onto (overrightarrow{BC}), we first determine the coordinates of points. Placing (B) at ((0, 0)) and (C) at ((4sqrt{2}, 0)), we find coordinates of (A) using distances (AB = 1) and (AC = 5). Solving, (A) is ((frac{sqrt{2}}{2}, frac{sqrt{2}}{2})). Vector (overrightarrow{AB}) is ((- frac{sqrt{2}}{2}, - frac{sqrt{2}}{2})) and vector (overrightarrow{BC}) is ((4sqrt{2}, 0)). The projection is:[text{Projection} = frac{overrightarrow{AB} cdot overrightarrow{BC}}{|overrightarrow{BC}|} = frac{-4}{4sqrt{2}} = -frac{sqrt{2}}{2}] Final Answers(1) (boxed{dfrac{4}{5}})(2) Angle (B) is (boxed{45^circ}) and the projection is (boxed{-dfrac{sqrt{2}}{2}}).]

🔑:(1) From overrightarrow{m} cdot overrightarrow{n} = -frac{3}{5}, we have cos(A-B)cos(B) - sin(A-B)sin(B) = -frac{3}{5}. Simplifying, we get cos(A) = -frac{3}{5}. Since 0 < A < pi, we find sin(A) = sqrt{1 - cos^2(A)} = boxed{frac{4}{5}}.(2) Applying the Sine Law, we have frac{a}{sin(A)} = frac{b}{sin(B)}, which leads to sin(B) = frac{sqrt{2}}{2} and thus B = boxed{frac{pi}{4}}.Next, applying the Cosine Law, we have a^2 = b^2 + c^2 - 2bccos(A), which simplifies to (4sqrt{2})^2 = 5^2 + c^2 - 2 cdot 5 cdot c cdot (-frac{3}{5}). Solving this quadratic equation for c, we find c = 1 or c = -7 (we discard the negative value as length cannot be negative).Therefore, the projection of vector overrightarrow{AB} onto overrightarrow{BC} is boxed{frac{overrightarrow{AB} cdot overrightarrow{BC}}{lvert overrightarrow{BC} rvert} = -ccos(B) = -frac{sqrt{2}}{2}}.

🧠:[<think>Okay, let's see. I need to find the maximum value of |OC| where O is the origin, and C is part of an isosceles right triangle ABC with A being the right angle. Point A is given as (2, 0), and point B is on the curve y = sqrt(1 - x²), which is the upper half of the unit circle. So, point B must lie somewhere on the semicircle centered at the origin with radius 1.First, let me visualize this. Point A is at (2, 0), which is two units to the right of the origin on the x-axis. Point B is somewhere on the unit semicircle above the x-axis. The triangle ABC is an isosceles right triangle with the right angle at A. That means the legs of the triangle are AAAB and AAAC, right? Wait, no. In a right triangle, the legs are the two sides that form the right angle. So, since the right angle is at A, the legs must be segments AB and AC. Since it's isosceles, those two legs must be of equal length. So AB = AC, and the hypotenuse is BC.Therefore, the problem is to find points B on the unit semicircle and point C such that triangle ABC is an isosceles right triangle with right angle at A, and then find the maximum distance from the origin O to point C.Hmm. So first, I need to express the coordinates of point C in terms of point B, then find |OC|, and then maximize that. Let's see.Given that ABC is an isosceles right triangle with right angle at A, the coordinates of C can be determined based on the coordinates of B. Since AB and AC are the legs and are equal in length, we can use rotation or some coordinate transformations to find C.Let me recall that in a coordinate system, if you have a right angle at a point, the other two points can be found by moving along perpendicular directions from that point. Since it's an isosceles right triangle, moving equal distances along two perpendicular directions.But here, the legs AB and AC must be equal in length and perpendicular. Wait, but in an isosceles right triangle with right angle at A, the legs are AB and AC, so they must be perpendicular and equal in length. Therefore, vectors AB and AC must satisfy (AB) · (AC) = 0 (since they are perpendicular) and |AB| = |AC|.Alternatively, since we know point A is (2, 0), and point B is (x, y) on the unit circle (since y = sqrt(1 - x²)), then we can find point C such that ABC is an isosceles right triangle with right angle at A.One method to find point C is to realize that C can be obtained by rotating point B around A by 90 degrees, either clockwise or counterclockwise. Because in a right isosceles triangle, rotating one leg around the right angle vertex by 90 degrees gives the other leg. So, if we take vector AB and rotate it 90 degrees around A, we get vector AC, which would give point C.Therefore, to find C, we can take the coordinates of B, subtract the coordinates of A to get the vector from A to B, rotate that vector by 90 degrees, and then add it back to A to get the coordinates of C.Let's formalize this. Let point B be (x, y). Then vector AB is (x - 2, y - 0) = (x - 2, y). Rotating this vector by 90 degrees counterclockwise would give (-y, x - 2). Therefore, adding this rotated vector to point A gives point C. So:C = A + rotated vector ABSo coordinates of C would be:(2 + (-y), 0 + (x - 2)) = (2 - y, x - 2)Alternatively, rotating clockwise would be (y, -(x - 2)), leading to C = (2 + y, 0 - (x - 2)) = (2 + y, 2 - x). So there are two possible positions for C depending on the direction of rotation.Therefore, point C can be either (2 - y, x - 2) or (2 + y, 2 - x). Since we need the maximum |OC|, we might need to consider both possibilities and see which gives the larger distance.But let's check this rotation approach. Let me verify with an example. Suppose point B is (1, 0), which is on the unit circle. Then vector AB is (1 - 2, 0 - 0) = (-1, 0). Rotating this 90 degrees counterclockwise around A would give (0, 1), so adding to A gives (2 + 0, 0 + 1) = (2, 1). Wait, but according to the formula above, (2 - y, x - 2) would be (2 - 0, 1 - 2) = (2, -1). Hmm, that's different. So perhaps I made a mistake in the rotation direction.Wait, rotation matrices: rotating a vector (a, b) 90 degrees counterclockwise gives (-b, a). So if vector AB is (x - 2, y), then rotating it 90 degrees counterclockwise would give (-y, x - 2). Then adding this to point A (2, 0) gives (2 - y, 0 + x - 2) = (2 - y, x - 2). But in the example where B is (1, 0), the rotated vector would be (-0, 1 - 2) = (0, -1). Adding to A gives (2 + 0, 0 - 1) = (2, -1). Which makes sense: the triangle with points A(2,0), B(1,0), and C(2, -1) is an isosceles right triangle with legs of length 1 (AB is length 1, AC is length 1, and BC is sqrt(2)). So that works.Alternatively, rotating clockwise would be rotating the vector AB by -90 degrees, which gives (y, -(x - 2)). Adding to A gives (2 + y, 0 - (x - 2)) = (2 + y, 2 - x). In the example, rotating vector AB (-1, 0) by 90 degrees clockwise would give (0, 1), adding to A gives (2 + 0, 0 + 1) = (2, 1). So that triangle would be A(2,0), B(1,0), C(2,1), which is also an isosceles right triangle with legs of length 1.Therefore, depending on the rotation direction, we get two different points C. So for each point B, there are two possible Cs. Therefore, when trying to maximize |OC|, we need to consider both possibilities.Therefore, we can model point C as either (2 - y, x - 2) or (2 + y, 2 - x), where point B is (x, y) on the unit circle, so x² + y² = 1 and y ≥ 0.So our goal is to maximize |OC| for both cases and find the maximum between them.First, let's write expressions for |OC| in both cases.Case 1: C = (2 - y, x - 2)Then |OC| is sqrt[(2 - y)^2 + (x - 2)^2]Case 2: C = (2 + y, 2 - x)Then |OC| is sqrt[(2 + y)^2 + (2 - x)^2]Given that B is on the unit circle, x² + y² = 1, so we can parametrize B as (cosθ, sinθ) where θ ∈ [0, π] because y ≥ 0.Therefore, substituting x = cosθ and y = sinθ.Therefore, in Case 1: C = (2 - sinθ, cosθ - 2)Then |OC| = sqrt[(2 - sinθ)^2 + (cosθ - 2)^2]Similarly, in Case 2: C = (2 + sinθ, 2 - cosθ)Then |OC| = sqrt[(2 + sinθ)^2 + (2 - cosθ)^2]Therefore, we can write these distances squared to make it easier:Case 1: (2 - sinθ)^2 + (cosθ - 2)^2Case 2: (2 + sinθ)^2 + (2 - cosθ)^2Let me compute both expressions.Starting with Case 1:(2 - sinθ)^2 + (cosθ - 2)^2Expand both terms:= (4 - 4 sinθ + sin²θ) + (cos²θ - 4 cosθ + 4)Combine terms:= 4 - 4 sinθ + sin²θ + cos²θ - 4 cosθ + 4Since sin²θ + cos²θ = 1:= 4 - 4 sinθ + 1 - 4 cosθ + 4= 4 + 4 + 1 - 4 sinθ - 4 cosθ= 9 - 4(sinθ + cosθ)Similarly, for Case 2:(2 + sinθ)^2 + (2 - cosθ)^2Expand both terms:= (4 + 4 sinθ + sin²θ) + (4 - 4 cosθ + cos²θ)Combine terms:= 4 + 4 sinθ + sin²θ + 4 - 4 cosθ + cos²θAgain, sin²θ + cos²θ = 1:= 4 + 4 sinθ + 1 + 4 - 4 cosθ= 4 + 4 + 1 + 4 sinθ - 4 cosθ= 9 + 4(sinθ - cosθ)So, the squared distances for Case 1 and Case 2 are 9 - 4(sinθ + cosθ) and 9 + 4(sinθ - cosθ), respectively.Therefore, our problem reduces to finding the maximum of sqrt(9 - 4(sinθ + cosθ)) and sqrt(9 + 4(sinθ - cosθ)) over θ ∈ [0, π], and then taking the larger of the two maxima.But since sqrt is an increasing function, maximizing the square root is equivalent to maximizing the expression inside. Therefore, we can instead maximize 9 - 4(sinθ + cosθ) and 9 + 4(sinθ - cosθ) separately, find their maximum values, then take the square roots and compare.First, let's handle Case 1: 9 - 4(sinθ + cosθ)To find the maximum of this expression, we need to minimize the term 4(sinθ + cosθ). Since 4 is positive, minimizing sinθ + cosθ will maximize the entire expression.Similarly, for Case 2: 9 + 4(sinθ - cosθ)To maximize this, we need to maximize sinθ - cosθ.Therefore, let's find the extrema for sinθ + cosθ and sinθ - cosθ over θ ∈ [0, π].Starting with Case 1: minimize sinθ + cosθ.The function sinθ + cosθ can be written as sqrt(2) sin(θ + 45°). The maximum value of this is sqrt(2), and the minimum is -sqrt(2). However, since θ ∈ [0, π], θ + 45° ∈ [45°, 225°]. The sine function reaches its maximum at 90°, which is 1, and minimum at 270°, but 225° is the upper limit here. The minimum value in this interval would be at θ + 45° = 225°, which is θ = 180°, so sin(180° + 45°) = sin(225°) = -sqrt(2)/2. However, sinθ + cosθ at θ = 180° is sin(180°) + cos(180°) = 0 - 1 = -1. Wait, this seems conflicting.Wait, let's compute sinθ + cosθ for θ ∈ [0, π]. Let's compute derivative:d/dθ (sinθ + cosθ) = cosθ - sinθ. Setting to zero, cosθ - sinθ = 0 => tanθ = 1 => θ = π/4 or 5π/4. But 5π/4 is outside [0, π], so critical point at θ = π/4.Compute sin(π/4) + cos(π/4) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2) ≈ 1.414.At θ = 0: sin0 + cos0 = 0 + 1 = 1At θ = π/2: sin(π/2) + cos(π/2) = 1 + 0 = 1At θ = π: sinπ + cospi = 0 - 1 = -1Therefore, the maximum of sinθ + cosθ on [0, π] is sqrt(2) at θ = π/4, and the minimum is -1 at θ = π.Therefore, for Case 1, 9 - 4(sinθ + cosθ) will be maximized when sinθ + cosθ is minimized, which is -1. Therefore, maximum value is 9 - 4*(-1) = 9 + 4 = 13. Therefore, the maximum |OC| for Case 1 is sqrt(13) ≈ 3.605.For Case 2: 9 + 4(sinθ - cosθ)To maximize this, we need to maximize sinθ - cosθ over θ ∈ [0, π].Similarly, let's analyze sinθ - cosθ.Compute derivative: d/dθ (sinθ - cosθ) = cosθ + sinθ. Setting to zero: cosθ + sinθ = 0 => tanθ = -1 => θ = 3π/4 (since θ ∈ [0, π]).Therefore, critical points at θ = 3π/4.Compute sinθ - cosθ at critical points and endpoints.At θ = 0: 0 - 1 = -1At θ = π/2: 1 - 0 = 1At θ = 3π/4: sin(3π/4) - cos(3π/4) = sqrt(2)/2 - (-sqrt(2)/2) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2) ≈ 1.414At θ = π: 0 - (-1) = 1Therefore, maximum of sinθ - cosθ on [0, π] is sqrt(2) at θ = 3π/4.Therefore, maximum value for Case 2 is 9 + 4*sqrt(2) ≈ 9 + 5.656 ≈ 14.656. Therefore, maximum |OC| for Case 2 is sqrt(14.656) ≈ 3.828.Comparing both cases, Case 2 gives a larger maximum |OC|. Therefore, the maximum |OC| is sqrt(9 + 4*sqrt(2)).Wait, let me verify the calculations again.For Case 1, the maximum of 9 - 4*(sinθ + cosθ) occurs when sinθ + cosθ is minimized, which is -1. So 9 -4*(-1) = 13. Therefore, sqrt(13) ≈ 3.605.For Case 2, the maximum of 9 +4*(sinθ - cosθ) occurs when sinθ - cosθ is maximized, which is sqrt(2). So 9 +4*sqrt(2). Then the squared distance is 9 +4*sqrt(2). Wait, but 4*sqrt(2) is approximately 5.656, so 9 +5.656 ≈14.656. Then sqrt(14.656)≈3.828.Therefore, the maximum |OC| is sqrt(9 +4*sqrt(2)). Let me compute that exactly.sqrt(9 +4*sqrt(2)). Is this the final answer? Let me see.Alternatively, maybe we can write it in a simplified form. Let's compute (sqrt(9 +4*sqrt(2))).But perhaps it can be expressed as sqrt(a) + sqrt(b) for some a and b. Let me check.Suppose sqrt(9 +4*sqrt(2)) = sqrt(a) + sqrt(b). Then squaring both sides:9 +4*sqrt(2) = a + b + 2*sqrt(ab)Equating the rational and irrational parts:a + b = 92*sqrt(ab) = 4*sqrt(2) => sqrt(ab)=2*sqrt(2) => ab = 8So solving:a + b =9ab =8This is a quadratic equation: x² -9x +8=0, solutions x=(9±sqrt(81-32))/2=(9±sqrt(49))/2=(9±7)/2. Thus x= (16)/2=8 or x=2/2=1. Therefore, a=8, b=1. Therefore, sqrt(9 +4*sqrt(2))=sqrt(8)+sqrt(1)=2*sqrt(2)+1.Let me verify:(2*sqrt(2) +1)^2 = (2*sqrt(2))^2 +2*(2*sqrt(2))*1 +1^2=8 +4*sqrt(2)+1=9 +4*sqrt(2). Yes, that's correct.Therefore, sqrt(9 +4*sqrt(2))=1 +2*sqrt(2). So |OC| maximum is 1 +2*sqrt(2). Which is approximately 1 +2.828=3.828, which matches the decimal approximation earlier.Therefore, the maximum value of |OC| is 1 +2*sqrt(2), which is approximately 3.828.Therefore, the answer is 1 +2*sqrt(2), which in boxed form is boxed{1 + 2sqrt{2}}.But wait, let me verify once more. Let's check if point C when θ=3π/4 gives this distance.If θ=3π/4, then point B is (cos(3π/4), sin(3π/4))=(-sqrt(2)/2, sqrt(2)/2). Then, for Case 2: C=(2 + sinθ, 2 - cosθ)=2 + sqrt(2)/2, 2 - (-sqrt(2)/2)=2 + sqrt(2)/2, 2 + sqrt(2)/2.Therefore, coordinates of C are (2 + sqrt(2)/2, 2 + sqrt(2)/2). Then |OC|=sqrt[(2 + sqrt(2)/2)^2 + (2 + sqrt(2)/2)^2] = sqrt(2*(2 + sqrt(2)/2)^2) = sqrt(2)*(2 + sqrt(2)/2).Compute this:sqrt(2)*(2 + sqrt(2)/2)=2*sqrt(2) + (sqrt(2)*sqrt(2))/2=2*sqrt(2) + 2/2=2*sqrt(2) +1. Which is indeed 1 +2*sqrt(2). So that's correct.Alternatively, computing the squared distance:(2 + sqrt(2)/2)^2 + (2 + sqrt(2)/2)^2 =2*(4 + 2*sqrt(2) + (2)/4)=2*(4 + 2*sqrt(2) + 0.5)=2*(4.5 +2*sqrt(2))=9 +4*sqrt(2). Then sqrt(9 +4*sqrt(2))=1 +2*sqrt(2).Therefore, this is correct.Another check: if we take θ=3π/4, which is 135 degrees, point B is in the second quadrant. Rotating vector AB 90 degrees clockwise gives point C in the first quadrant with coordinates as above, leading to maximum distance from O.Hence, the maximum |OC| is indeed 1 +2*sqrt(2). So I think this is the correct answer.Final Answerboxed{1 + 2sqrt{2}}</think>Given point ( A(2, 0) ) and point ( B ) on the curve ( y = sqrt{1 - x^2} ), we need to find the maximum value of ( |OC| ) where ( triangle ABC ) is an isosceles right triangle with ( A ) as the right-angle vertex.1. Coordinates of Point C: - For an isosceles right triangle with right angle at ( A ), point ( C ) can be found by rotating vector ( overrightarrow{AB} ) by 90 degrees around ( A ). - If ( B ) has coordinates ( (x, y) ), then vector ( overrightarrow{AB} ) is ( (x - 2, y) ). - Rotating ( overrightarrow{AB} ) 90 degrees counterclockwise gives ( (-y, x - 2) ), leading to ( C = (2 - y, x - 2) ). - Rotating ( overrightarrow{AB} ) 90 degrees clockwise gives ( (y, -(x - 2)) ), leading to ( C = (2 + y, 2 - x) ).2. Distance Calculations: - For ( C = (2 - y, x - 2) ), the distance ( |OC| ) is ( sqrt{(2 - y)^2 + (x - 2)^2} ). - For ( C = (2 + y, 2 - x) ), the distance ( |OC| ) is ( sqrt{(2 + y)^2 + (2 - x)^2} ).3. Parametrization: - Point ( B ) on the unit circle can be parametrized as ( (costheta, sintheta) ) where ( theta in [0, pi] ). - Substituting ( x = costheta ) and ( y = sintheta ), we get: - For ( C = (2 - sintheta, costheta - 2) ), ( |OC|^2 = 9 - 4(sintheta + costheta) ). - For ( C = (2 + sintheta, 2 - costheta) ), ( |OC|^2 = 9 + 4(sintheta - costheta) ).4. Maximizing the Distances: - Minimizing ( sintheta + costheta ) over ( theta in [0, pi] ) gives the minimum value ( -1 ), leading to ( |OC| = sqrt{13} approx 3.605 ). - Maximizing ( sintheta - costheta ) over ( theta in [0, pi] ) gives the maximum value ( sqrt{2} ), leading to ( |OC| = sqrt{9 + 4sqrt{2}} approx 3.828 ).5. Simplification: - ( sqrt{9 + 4sqrt{2}} ) simplifies to ( 1 + 2sqrt{2} ).Thus, the maximum value of ( |OC| ) is (boxed{1 + 2sqrt{2}}).,<think>Okay, so I need to find the maximum value of |OC| where point C is part of an isosceles right triangle ABC with A being the right angle. Point A is given as (2,0), and point B is on the curve y = sqrt(1 - x²), which I recognize as the upper half of the unit circle. So point B lies somewhere on the semicircle above the x-axis with radius 1 centered at the origin.First, let's visualize this. Point A is at (2,0), which is outside the unit circle. The triangle ABC is an isosceles right triangle with the right angle at A. That means the legs of the triangle, which are the sides from A to B and from A to C, must be equal in length, and the hypotenuse is BC. Since it's a right triangle at A, the legs are perpendicular to each other. So vectors AB and AC should be perpendicular and of equal length.So, if I can find the coordinates of points B and C such that AB and AC are perpendicular and equal in length, then compute |OC|, which is the distance from the origin to point C, and then find the maximum of that distance.But how do I approach this? Let's break it down step by step.First, point B is on the unit circle's upper half, so its coordinates can be parameterized. Let's use a parameter θ such that B is (cosθ, sinθ) where θ is between 0 and π because it's the upper half. But wait, the curve is y = sqrt(1 - x²), which is indeed the upper unit circle, so yes, B can be written as (x, sqrt(1 - x²)) where x is between -1 and 1. Alternatively, using θ as a parameter might make the calculations easier.So let's parametrize point B as (cosθ, sinθ). Then, since triangle ABC is an isosceles right triangle with right angle at A, we need to find point C such that vectors AB and AC are perpendicular and have the same length.Let me recall that in an isosceles right triangle with legs of length L, the hypotenuse is L*sqrt(2). So, the distance from B to C should be L*sqrt(2) where L is the length of AB (and AC). But maybe working with vectors would be better here.Vector AB is from point A(2,0) to point B(cosθ, sinθ), so AB = (cosθ - 2, sinθ - 0) = (cosθ - 2, sinθ).Similarly, vector AC is from A(2,0) to point C. Let's denote point C as (c_x, c_y). Then AC = (c_x - 2, c_y - 0) = (c_x - 2, c_y).Since triangle ABC is a right triangle at A with legs AB and AC, vectors AB and AC must be perpendicular. Therefore, their dot product should be zero:AB · AC = 0So:(cosθ - 2)(c_x - 2) + (sinθ)(c_y) = 0Also, since the triangle is isosceles, |AB| = |AC|. So:sqrt[(cosθ - 2)^2 + (sinθ)^2] = sqrt[(c_x - 2)^2 + (c_y)^2]Squaring both sides:(cosθ - 2)^2 + sin²θ = (c_x - 2)^2 + c_y²Expand the left side:cos²θ - 4cosθ + 4 + sin²θ = (c_x - 2)^2 + c_y²But cos²θ + sin²θ = 1, so:1 - 4cosθ + 4 = (c_x - 2)^2 + c_y²Which simplifies to:5 - 4cosθ = (c_x - 2)^2 + c_y²So, the equation for |AC| squared is 5 - 4cosθ.Now, we have two equations:1) (cosθ - 2)(c_x - 2) + sinθ * c_y = 02) (c_x - 2)^2 + c_y² = 5 - 4cosθThese two equations can be used to solve for c_x and c_y in terms of θ.Alternatively, since we need to express point C in terms of point B, maybe there's a transformation involved. In an isosceles right triangle with right angle at A, point C can be obtained by rotating point B around A by 90 degrees, either clockwise or counterclockwise. Because rotating one leg by 90 degrees around the right-angle vertex would give the other leg.Yes, that's a good point. Since AB and AC are perpendicular and equal in length, point C can be obtained by rotating vector AB by 90 degrees around point A. The direction of rotation (clockwise or counterclockwise) would give two possible positions for C.So, if we rotate vector AB by 90 degrees, we can find point C. The rotation of a vector (u, v) by 90 degrees counterclockwise is (-v, u), and clockwise is (v, -u). Therefore, depending on the direction of rotation, we can get two different points C1 and C2.Therefore, point C can be either:C1 = A + rotation of AB by 90 degrees counterclockwiseorC2 = A + rotation of AB by 90 degrees clockwise.So let's compute both possibilities.First, vector AB is (cosθ - 2, sinθ). Rotating this 90 degrees counterclockwise gives (-sinθ, cosθ - 2). Adding this to point A(2,0):C1 = (2,0) + (-sinθ, cosθ - 2) = (2 - sinθ, 0 + cosθ - 2) = (2 - sinθ, cosθ - 2)Similarly, rotating AB 90 degrees clockwise gives (sinθ, - (cosθ - 2)) = (sinθ, 2 - cosθ). Adding this to point A(2,0):C2 = (2,0) + (sinθ, 2 - cosθ) = (2 + sinθ, 0 + 2 - cosθ) = (2 + sinθ, 2 - cosθ)Therefore, depending on the direction of rotation, point C can be either (2 - sinθ, cosθ - 2) or (2 + sinθ, 2 - cosθ).Therefore, for each point B on the unit circle, there are two possible points C. To find the maximum |OC|, we need to compute |OC| for both possibilities and find the maximum over θ.Therefore, we can compute |OC1| and |OC2|, then find the maximum of these two over θ.So let's compute |OC1| and |OC2|.First, for C1: (2 - sinθ, cosθ - 2)|OC1|² = (2 - sinθ)^2 + (cosθ - 2)^2Similarly, for C2: (2 + sinθ, 2 - cosθ)|OC2|² = (2 + sinθ)^2 + (2 - cosθ)^2We need to compute both and find their maximum.Let's compute |OC1|²:(2 - sinθ)^2 + (cosθ - 2)^2 = (4 - 4 sinθ + sin²θ) + (cos²θ - 4 cosθ + 4)= 4 - 4 sinθ + sin²θ + cos²θ - 4 cosθ + 4= (sin²θ + cos²θ) + (4 + 4) + (-4 sinθ -4 cosθ)= 1 + 8 - 4(sinθ + cosθ)= 9 - 4(sinθ + cosθ)Similarly, |OC2|²:(2 + sinθ)^2 + (2 - cosθ)^2 = (4 + 4 sinθ + sin²θ) + (4 - 4 cosθ + cos²θ)= 4 + 4 sinθ + sin²θ + 4 - 4 cosθ + cos²θ= (sin²θ + cos²θ) + (4 + 4) + (4 sinθ -4 cosθ)= 1 + 8 + 4(sinθ - cosθ)= 9 + 4(sinθ - cosθ)Therefore, we have:|OC1|² = 9 - 4(sinθ + cosθ)|OC2|² = 9 + 4(sinθ - cosθ)Now, we need to find the maximum of sqrt(|OC1|²) and sqrt(|OC2|²). Since sqrt is a monotonically increasing function, the maximum of |OC| will occur at the maximum of |OC1|² and |OC2|². Therefore, we can just find the maximum of |OC1|² and |OC2|² over θ, then take the square root.So, first, analyze |OC1|² = 9 - 4(sinθ + cosθ)To find its maximum, we need to minimize (sinθ + cosθ), since it's subtracted.Similarly, for |OC2|² = 9 + 4(sinθ - cosθ), to find its maximum, we need to maximize (sinθ - cosθ).So, let's compute the extrema for both expressions.First, for |OC1|²:We need to minimize (sinθ + cosθ). The minimum of sinθ + cosθ is -sqrt(2). Wait, actually:The expression sinθ + cosθ can be rewritten as sqrt(2) sin(θ + 45°), which has a maximum of sqrt(2) and a minimum of -sqrt(2). Therefore, sinθ + cosθ ∈ [-sqrt(2), sqrt(2)]Therefore, the minimum value of (sinθ + cosθ) is -sqrt(2). Therefore, the minimum of -4(sinθ + cosθ) is -4*(-sqrt(2)) = 4 sqrt(2). Therefore, the maximum value of |OC1|² is 9 + 4 sqrt(2). Wait, no:Wait, |OC1|² = 9 - 4(sinθ + cosθ). To maximize |OC1|², since it's 9 minus 4 times (sinθ + cosθ), we need to minimize (sinθ + cosθ). Since the minimum of sinθ + cosθ is -sqrt(2), then:Maximum |OC1|² = 9 - 4*(-sqrt(2)) = 9 + 4 sqrt(2)Similarly, for |OC2|² = 9 + 4(sinθ - cosθ). To maximize this, we need to maximize (sinθ - cosθ). Let's analyze sinθ - cosθ.Similarly, sinθ - cosθ can be written as sqrt(2) sin(θ - 45°), which also has a maximum of sqrt(2) and a minimum of -sqrt(2). Therefore, the maximum value of (sinθ - cosθ) is sqrt(2). Therefore:Maximum |OC2|² = 9 + 4*(sqrt(2)) = 9 + 4 sqrt(2)So both |OC1|² and |OC2|² have the same maximum value of 9 + 4 sqrt(2). Therefore, the maximum |OC| is sqrt(9 + 4 sqrt(2)).Therefore, the answer should be sqrt(9 + 4 sqrt(2)).But let me verify that.Wait, is that correct? Because when we rotate in both directions, both C1 and C2 can reach the same maximum |OC|. So regardless of the direction of rotation, the maximum |OC| is the same. Therefore, the maximum value of |OC| is sqrt(9 + 4 sqrt(2)).But let's compute sqrt(9 + 4 sqrt(2)) to check if that simplifies further.Compute sqrt(9 + 4 sqrt(2)). Let's see if this can be written as sqrt(a) + sqrt(b):Suppose sqrt(9 + 4 sqrt(2)) = sqrt(a) + sqrt(b)Then squaring both sides:9 + 4 sqrt(2) = a + b + 2 sqrt(ab)Therefore, we have:a + b = 92 sqrt(ab) = 4 sqrt(2) => sqrt(ab) = 2 sqrt(2) => ab = 8So solving:a + b = 9ab = 8The solutions are the roots of x² - 9x + 8 = 0, which factors as (x - 1)(x - 8) = 0. Therefore, a = 1, b = 8 or vice versa.Therefore, sqrt(9 + 4 sqrt(2)) = sqrt(1) + sqrt(8) = 1 + 2 sqrt(2). Let's check:(1 + 2 sqrt(2))² = 1 + 4 sqrt(2) + 8 = 9 + 4 sqrt(2). Yes, correct.Therefore, sqrt(9 + 4 sqrt(2)) = 1 + 2 sqrt(2).Therefore, the maximum |OC| is 1 + 2 sqrt(2).Therefore, the answer should be 1 + 2 sqrt(2), which is approximately 3.828.But let me check again if I made a mistake in the rotation direction.Wait, when we rotate vector AB by 90 degrees counterclockwise, we get (-sinθ, cosθ - 2). Then adding to point A(2,0) gives (2 - sinθ, cosθ - 2). Similarly, rotating clockwise gives (sinθ, 2 - cosθ), so adding to A gives (2 + sinθ, 2 - cosθ). Then, when we calculated |OC1|² and |OC2|², both ended up with maximums of 9 + 4 sqrt(2). Therefore, their square roots are 1 + 2 sqrt(2). Therefore, that's correct.But let me verify with a specific example. Let's take θ where sinθ + cosθ is minimized, which is at θ = 225°, but since point B is on the upper semicircle, θ is between 0 and π. So θ cannot be 225°, which is 5π/4, which is in the lower semicircle. Therefore, the minimum of sinθ + cosθ over θ ∈ [0, π] is actually at θ = 5π/4, but since 5π/4 is not in [0, π], the minimum on [0, π] occurs at θ = π, where sinθ + cosθ = 0 + (-1) = -1.Wait, hold on. Wait, sinθ + cosθ in θ ∈ [0, π]. Let's compute its minimum.The derivative of sinθ + cosθ is cosθ - sinθ. Setting derivative to zero: cosθ - sinθ = 0 => tanθ = 1 => θ = π/4 or 5π/4. But 5π/4 is not in [0, π], so critical point is at θ = π/4.Evaluating sinθ + cosθ at θ = 0: 0 + 1 = 1At θ = π/4: sqrt(2)/2 + sqrt(2)/2 = sqrt(2) ≈ 1.414At θ = π/2: 1 + 0 = 1At θ = π: 0 + (-1) = -1Therefore, on θ ∈ [0, π], the minimum of sinθ + cosθ is -1 at θ = π, and maximum is sqrt(2) at θ = π/4.Wait, so my previous assumption that the minimum was -sqrt(2) was incorrect because θ is restricted to [0, π]. Therefore, the actual minimum of sinθ + cosθ over [0, π] is -1, and maximum is sqrt(2).Similarly, for sinθ - cosθ, let's compute its maximum over θ ∈ [0, π].Compute derivative: cosθ + sinθ. Setting derivative to zero: cosθ + sinθ = 0 => tanθ = -1 => θ = 3π/4, which is in [0, π].So critical point at θ = 3π/4.Evaluate sinθ - cosθ at θ=0: 0 - 1 = -1At θ=3π/4: sqrt(2)/2 - (-sqrt(2)/2) = sqrt(2)At θ=π/2: 1 - 0 = 1At θ=π: 0 - (-1) = 1Therefore, the maximum of sinθ - cosθ over [0, π] is sqrt(2) at θ = 3π/4, and minimum is -1 at θ=0.Therefore, going back, for |OC1|² = 9 -4(sinθ + cosθ). The maximum of |OC1|² occurs when sinθ + cosθ is minimized, which is -1. Therefore, |OC1|²_max = 9 -4*(-1) = 9 +4 =13. Therefore, |OC1|_max = sqrt(13) ≈3.605.But wait, this contradicts the previous conclusion. Wait, this is a problem.Wait, because earlier I considered θ over all real numbers, but in reality, θ is restricted to [0, π] since point B is on the upper semicircle. Therefore, sinθ + cosθ can only go down to -1 (at θ=π), not to -sqrt(2). Similarly, sinθ - cosθ can go up to sqrt(2) (at θ=3π/4), which is within [0, π].Therefore, recalculating:For |OC1|² = 9 -4(sinθ + cosθ). The minimum value of (sinθ + cosθ) is -1, so maximum |OC1|² is 9 -4*(-1) = 13, so |OC1|_max = sqrt(13) ≈3.605.For |OC2|² = 9 +4(sinθ - cosθ). The maximum value of (sinθ - cosθ) is sqrt(2) (at θ=3π/4), so maximum |OC2|² is 9 +4*sqrt(2) ≈9 +5.656=14.656, so |OC2|_max ≈sqrt(14.656)=3.827, which is approximately 1 + 2 sqrt(2) ≈1 +2.828≈3.828, which matches.Therefore, I made a mistake earlier by not considering the domain of θ. Therefore, the maximum of |OC| is sqrt(9 +4 sqrt(2)) ≈3.828, which is achieved when sinθ - cosθ is maximized at sqrt(2) (i.e., at θ=3π/4). However, we need to check if point C in that case is valid (i.e., whether point B at θ=3π/4 is on the curve y=sqrt(1 -x²)).Wait, θ=3π/4 is in [0, π], so point B is (cos(3π/4), sin(3π/4))=(-sqrt(2)/2, sqrt(2)/2), which is on the upper semicircle y=sqrt(1 -x²). Yes, because x=-sqrt(2)/2 ≈-0.707, which is within [-1,1], and y=sqrt(2)/2≈0.707, which is sqrt(1 - (sqrt(2)/2)^2)=sqrt(1 -0.5)=sqrt(0.5)=sqrt(2)/2. So yes, point B is valid.Therefore, the maximum |OC| is sqrt(9 +4 sqrt(2))=1 + 2 sqrt(2). Wait, but sqrt(9 +4 sqrt(2)) is equal to 1 + 2 sqrt(2)?Let's verify:(1 + 2 sqrt(2))² = 1 + 4 sqrt(2) + 8 =9 +4 sqrt(2). Yes, correct. Therefore, sqrt(9 +4 sqrt(2))=1 + 2 sqrt(2).Therefore, the maximum |OC| is 1 + 2 sqrt(2).But why in the previous calculation, |OC2|²_max is 9 +4 sqrt(2) which gives sqrt(9 +4 sqrt(2))≈3.828, which is indeed 1 + 2 sqrt(2)≈1+2.828≈3.828. So that's correct.But earlier, when considering |OC1|², the maximum was 13, but that is less than 9 +4 sqrt(2)≈14.656. Therefore, the overall maximum is achieved by |OC2|.Therefore, the answer is 1 + 2 sqrt(2).Therefore, the maximum value of |OC| is 1 + 2 sqrt(2).But let me check with θ=3π/4, what is point C?Point B is (cos(3π/4), sin(3π/4))=(-sqrt(2)/2, sqrt(2)/2). Then vector AB is (-sqrt(2)/2 -2, sqrt(2)/2 -0)= (-2 -sqrt(2)/2, sqrt(2)/2)Rotating this vector 90 degrees clockwise gives (sqrt(2)/2, 2 + sqrt(2)/2). Therefore, point C2 is A + rotated vector: (2,0) + (sqrt(2)/2, 2 + sqrt(2)/2)= (2 + sqrt(2)/2, 2 + sqrt(2)/2)Therefore, point C is (2 + sqrt(2)/2, 2 + sqrt(2)/2). The distance from the origin is sqrt[(2 + sqrt(2)/2)^2 + (2 + sqrt(2)/2)^2] = sqrt{2*(2 + sqrt(2)/2)^2}Compute (2 + sqrt(2)/2)^2 =4 + 2*sqrt(2) + (sqrt(2)/2)^2 =4 + 2 sqrt(2) + (2)/4=4 + 2 sqrt(2) +0.5=4.5 + 2 sqrt(2)Multiply by 2:9 +4 sqrt(2)Then sqrt(9 +4 sqrt(2)) as before, which is 1 +2 sqrt(2). So this checks out.Alternatively, compute the coordinates:C is (2 + sqrt(2)/2, 2 + sqrt(2)/2). So distance from origin is sqrt{(2 + sqrt(2)/2)^2 + (2 + sqrt(2)/2)^2} = sqrt{2*(2 + sqrt(2)/2)^2} = sqrt{2}*(2 + sqrt(2)/2) = 2 sqrt(2) + (sqrt(2)*sqrt(2))/2 =2 sqrt(2) + 2/2=2 sqrt(2) +1. Which is 1 +2 sqrt(2). Yes, correct.Therefore, the maximum |OC| is indeed 1 + 2 sqrt(2).Therefore, the answer is 1 + 2 sqrt(2). So boxed, it's boxed{1 + 2sqrt{2}}.Final Answerboxed{1 + 2sqrt{2}}</think>Given point ( A(2,0) ) and point ( B ) on the curve ( y = sqrt{1 - x^2} ), we need to find the maximum value of ( |OC| ) where ( triangle ABC ) is an isosceles right triangle with ( A ) as the right-angle vertex.1. Parameterization of Point B: - Point ( B ) can be parameterized as ( (costheta, sintheta) ) where ( theta in [0, pi] ).2. Vectors and Rotation: - Vector ( overrightarrow{AB} = (costheta - 2, sintheta) ). - To form an isosceles right triangle, point ( C ) can be obtained by rotating ( overrightarrow{AB} ) by 90 degrees around ( A ).3. Rotation Directions: - Rotating ( overrightarrow{AB} ) 90 degrees counterclockwise gives ( (-sintheta, costheta - 2) ), leading to point ( C_1 = (2 - sintheta, costheta - 2) ). - Rotating ( overrightarrow{AB} ) 90 degrees clockwise gives ( (sintheta, 2 - costheta) ), leading to point ( C_2 = (2 + sintheta, 2 - costheta) ).4. Distance Calculations: - For ( C_1 ): [ |OC_1|^2 = (2 - sintheta)^2 + (costheta - 2)^2 = 9 - 4(sintheta + costheta) ] - For ( C_2 ): [ |OC_2|^2 = (2 + sintheta)^2 + (2 - costheta)^2 = 9 + 4(sintheta - costheta) ]5. Maximizing the Distances: - The minimum value of ( sintheta + costheta ) over ( [0, pi] ) is -1, but the maximum value of ( |OC_1|^2 ) is 13. - The maximum value of ( sintheta - costheta ) over ( [0, pi] ) is ( sqrt{2} ), leading to: [ |OC_2|^2_{text{max}} = 9 + 4sqrt{2} ] - Therefore, the maximum value of ( |OC| ) is ( sqrt{9 + 4sqrt{2}} ).6. Simplification: - Recognizing ( sqrt{9 + 4sqrt{2}} = 1 + 2sqrt{2} ).Thus, the maximum value of ( |OC| ) is (boxed{1 + 2sqrt{2}}).]

🔑:3. 2 sqrt{2}+1.As shown in Figure 2, the trajectory of point C is a semicircular arc with center D(2,2) (in fact, by rotating triangle A O B 90^{circ} clockwise around point A).text { Therefore, } |O C|_{max }=O D+1=2 sqrt{2}+1 text {. }

🧠:[<think>Okay, let's tackle this problem step by step. So, we need to select 3 people from 5 boys and 5 girls, making sure there's at least one girl in the team. The answer choices are A. 100, B. 120, C. 110, and D. 180. Hmm, let me think about how to approach this.First, I remember that when dealing with combinations, especially with restrictions like "at least one," it's often easier to calculate the total number of possible combinations without any restrictions and then subtract the ones that don't meet the requirement. In this case, the requirement is having at least one girl. So, the opposite of that would be having no girls, which means all boys. So, if I can find the total number of ways to choose 3 people from all 10 (5 boys + 5 girls) and then subtract the number of all-boy teams, that should give me the answer.Let me write that down to visualize:Total number of possible teams without restrictions: C(10, 3)Number of teams with all boys: C(5, 3)Number of valid teams (with at least one girl) = C(10, 3) - C(5, 3)Now, let's compute these values. First, C(10, 3). The combination formula is C(n, k) = n! / (k! * (n - k)! )So, C(10, 3) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.Wait, that seems right. 10 choose 3 is 120. But wait, one of the answer choices is 120. Hmm. But then we need to subtract the all-boy teams.C(5, 3) is 5 choose 3. Let's calculate that: 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10.So, 120 total teams minus 10 all-boy teams gives 110. So the answer should be 110, which is option C. Hmm, that seems straightforward.But let me verify if there's another way to approach this to double-check. Sometimes, when you do problems in different ways, you can catch mistakes.Another method would be to consider the cases where there is exactly 1 girl, exactly 2 girls, or exactly 3 girls in the team, and add those up. So:Case 1: 1 girl and 2 boysCase 2: 2 girls and 1 boyCase 3: 3 girls and 0 boysCalculating each case:Case 1: C(5, 1) * C(5, 2)C(5,1) is 5, and C(5,2) is (5*4)/(2*1) = 10. So 5*10 = 50.Case 2: C(5, 2) * C(5, 1)Same as above, since it's symmetric. 10 * 5 = 50.Case 3: C(5, 3) * C(5, 0)C(5,3) is 10, and C(5,0) is 1 (since there's only one way to choose nothing). So 10*1 = 10.Adding them up: 50 + 50 + 10 = 110. That matches the previous result. So that's good. Both methods give 110. Therefore, the answer must be C. 110.But let me check once more to ensure I didn't make any calculation errors. Sometimes when you're multiplying numbers quickly, you can slip up.For total combinations C(10,3): 10*9*8 / 6. 10*9 is 90, 90*8 is 720. Divide by 6: 720/6 = 120. That's correct.C(5,3): 5*4*3 / 6 = 60/6 = 10. Correct.Subtracting 10 from 120 gives 110. So that's correct.Alternatively, the second method:Case 1: 1 girl: 5 choices. 2 boys: 5 choose 2. Which is 10. 5*10=50.Case 2: 2 girls: 5 choose 2 is 10. 1 boy: 5 choices. 10*5=50.Case 3: 3 girls: 5 choose 3 is 10. 0 boys: 1. 10*1=10.Total: 50+50+10=110. Correct.So both methods confirm the answer is 110, which is option C. Therefore, the correct answer is C. 110.Final Answerboxed{C}</think>To determine the number of possible team combinations with at least one girl, we can use two methods:1. Total combinations minus all-boy teams: - Total number of ways to choose 3 people from 10 (5 boys + 5 girls): [ binom{10}{3} = frac{10!}{3!(10-3)!} = frac{10 times 9 times 8}{3 times 2 times 1} = 120 ] - Number of ways to choose 3 boys from 5: [ binom{5}{3} = frac{5!}{3!(5-3)!} = frac{5 times 4}{2 times 1} = 10 ] - Number of valid teams (with at least one girl): [ 120 - 10 = 110 ]2. Sum of combinations for each valid case: - Case 1 (1 girl and 2 boys): [ binom{5}{1} times binom{5}{2} = 5 times 10 = 50 ] - Case 2 (2 girls and 1 boy): [ binom{5}{2} times binom{5}{1} = 10 times 5 = 50 ] - Case 3 (3 girls and 0 boys): [ binom{5}{3} times binom{5}{0} = 10 times 1 = 10 ] - Total number of valid teams: [ 50 + 50 + 10 = 110 ]Both methods confirm that the number of possible team combinations is (boxed{C}).,<think>Okay, let's see. The problem is about selecting 3 people from 5 boys and 5 girls to form a team, with the condition that there has to be at least one girl in the team. We need to find the number of possible team combinations, and the options are A. 100, B. 120, C. 110, D. 180.First, I need to recall how combinations work. The basic idea is that if we want to choose k people from a group of n without considering the order, the formula is n choose k, which is n! / (k! (n - k)!).But here, there's a condition: at least one girl. So, maybe the way to approach this is to first calculate the total number of possible teams without any restrictions and then subtract the number of teams that don't meet the condition (i.e., teams with no girls, which would be all boys).Let me check if that makes sense. The total number of teams is choosing 3 people from 10 (since there are 5 boys + 5 girls = 10 people). Then, subtract the number of teams that are all boys. That should give the number of teams with at least one girl. Yeah, that seems like a standard approach. Sometimes called the complementary counting method.So, total teams: C(10, 3). Then subtract the all-boys teams: C(5, 3). Then the result is the answer.Let me compute that. First, C(10,3). That's 10! / (3! * 7!) = (10*9*8)/(3*2*1) = 120. Then, C(5,3) is 10. Because 5!/(3!2!)=10. So, 120 - 10 = 110. Hmm, so is the answer 110? That's option C.Wait, but let me verify. Maybe I made a mistake in the calculation. Let me recalculate C(10,3). 10*9*8=720, divided by 6 (since 3!=6) gives 120. That's correct. Then, C(5,3): 5*4*3=60, divided by 6=10. So 120-10=110. That seems right.Alternatively, maybe the question is interpreted differently. Let me think again. Another way could be to count the number of teams with 1 girl and 2 boys, plus teams with 2 girls and 1 boy, plus teams with 3 girls. Then sum them up. Let's see if that gives the same answer.First, 1 girl and 2 boys: C(5,1)*C(5,2). That is 5 * 10 = 50.Then, 2 girls and 1 boy: C(5,2)*C(5,1). That is 10 * 5 = 50.Then, 3 girls and 0 boys: C(5,3)*C(5,0). That is 10 *1=10.Adding those up: 50 + 50 +10=110. So same answer. So that's good. So the answer is 110, option C. So that must be the correct answer.Wait, just to make sure, maybe the problem is in Chinese or has some translation issues? The problem says "Select 3 people from 5 boys and 5 girls". So total 10 people. So the total combinations are C(10,3). Then subtract the all-boys teams. So 120 -10=110. Both methods give the same result. So answer is C.110.Alternatively, perhaps someone might have thought that the teams have positions or something else, but the problem says "team combinations", which usually implies that the order doesn't matter. So combinations are the right approach. So yes, 110 is correct. Therefore, option C.Final Answerboxed{C}</think>To determine the number of possible team combinations of 3 people from 5 boys and 5 girls with at least one girl, we can use complementary counting. First, calculate the total number of ways to choose 3 people from 10 (5 boys + 5 girls):[binom{10}{3} = frac{10!}{3!(10-3)!} = frac{10 times 9 times 8}{3 times 2 times 1} = 120]Next, subtract the number of ways to choose 3 people with no girls (all boys):[binom{5}{3} = frac{5!}{3!(5-3)!} = frac{5 times 4 times 3}{3 times 2 times 1} = 10]Subtracting the all-boys teams from the total teams gives:[120 - 10 = 110]We can verify this by considering the valid teams directly:1. 1 girl and 2 boys: (binom{5}{1} times binom{5}{2} = 5 times 10 = 50)2. 2 girls and 1 boy: (binom{5}{2} times binom{5}{1} = 10 times 5 = 50)3. 3 girls and 0 boys: (binom{5}{3} times binom{5}{0} = 10 times 1 = 10)Summing these combinations:[50 + 50 + 10 = 110]Thus, the number of possible team combinations is (boxed{C}).]

🔑:AnswerThe total number of ways to select 3 people from 10 (5 boys and 5 girls) without any restrictions is binom{10}{3} = 120. However, this includes the combinations where all selected members are boys. The number of ways to select 3 boys from the 5 available is binom{5}{3} = 10. Therefore, the number of combinations where at least one girl is selected is the total combinations minus the all-boys combinations, which is 120 - 10 = 110.So, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to figure out how to inscribe a cone in a sphere such that the cone has the maximum possible volume. Hmm, let me start by visualizing this. A sphere, and inside it, there's a cone touching the sphere at certain points. The cone has a base that's a circle, and a height from the base to the tip of the cone. Since it's inscribed, I guess the tip of the cone and the edge of its base must lie on the sphere's surface. First, let me set up some coordinates to model this. Maybe if I place the sphere with its center at the origin (0,0,0) and radius R. Then, the equation of the sphere is x² + y² + z² = R². Now, the cone. Let's assume the cone is symmetric along one axis, say the z-axis. That would make the problem easier. So the apex of the cone (the tip) would be at the north pole of the sphere, which is (0,0,R), and the base of the cone would be a circle in some plane parallel to the xy-plane at a lower z-coordinate. Wait, but if the apex is at the north pole, then the cone is pointing downward. Alternatively, maybe the cone is oriented with its apex at a lower point and the base somewhere else. Hmm, actually, I need to clarify.Wait, inscribing a cone in a sphere usually means that all points of the cone (the apex and all points on the base) lie on the sphere. So, the apex is a single point on the sphere, and the base is a circle also lying on the sphere. Is that right? Or does only the apex and the circumference of the base lie on the sphere? Because if the entire base is on the sphere, then the base would be a flat circular region on the sphere, which is only possible if the sphere's radius is the same as the cone's height, but that might not make sense. Wait, maybe only the apex and the edge of the base are on the sphere. Let me check.Actually, in a standard inscribed cone in a sphere, the apex is at one end (like the north pole) and the base is a circle on the sphere's surface. Wait, no, if the base is on the sphere's surface, then the base would be part of the sphere, but a flat circle can't lie on a sphere unless it's a great circle, which would make the cone's height equal to the sphere's diameter. But that's a cone with maximum height, but maybe not maximum volume. Wait, volume of a cone is (1/3)πr²h, so to maximize that, we need to balance r and h. But let's think again. If the cone is inscribed in the sphere, all points of the cone must be inside or on the sphere, and the cone touches the sphere at least at the apex and the base's circumference. So the apex is a point on the sphere, and the base is a circle whose circumference is on the sphere. So the apex is at, say, (0,0,a) and the base is a circle in the plane z = b, with all points (x,y,b) such that x² + y² + b² = R². Then the radius of the base would be sqrt(R² - b²). The height of the cone would be |a - b|. But since the apex is on the sphere, a must satisfy a² = R²? Wait, no, if the apex is at (0,0,a), then since it's on the sphere, (0)^2 + (0)^2 + a² = R², so a = ±R. Let's assume the apex is at (0,0,R), the north pole, and the base is somewhere below it on the sphere. Then the base circle is in the plane z = c, where c < R, and the radius of the base is sqrt(R² - c²). The height of the cone would be R - c. So in this case, the volume of the cone is (1/3)πr²h = (1/3)π(R² - c²)(R - c). So we need to maximize this volume with respect to c. Let's denote c as the z-coordinate of the base plane. Since the apex is at z = R, and the base is at z = c, then c must be between -R and R, but since we want the cone to be below the apex, c should be less than R. But if c is negative, the base is below the center of the sphere. But maybe that's allowed. Wait, but if c is negative, the height R - c becomes larger, but the radius sqrt(R² - c²) becomes smaller. So maybe the maximum volume occurs somewhere in between.Wait, let's set up the problem with variables. Let's let the sphere have radius R. Let’s assume the apex is at the top of the sphere, (0,0,R). The base of the cone is a circle in the plane z = h, where h is between -R and R. The radius of the base of the cone, let's call it r, satisfies r² + h² = R², so r = sqrt(R² - h²). The height of the cone is not h, but rather the distance from the apex at z = R to the base at z = h, so the height H is R - h. Therefore, the volume V of the cone is (1/3)πr²H = (1/3)π(R² - h²)(R - h). Now, we need to find the value of h that maximizes V. So we can write V(h) = (1/3)π(R - h)(R² - h²). Let me expand this to make differentiation easier. First, note that (R² - h²) = (R - h)(R + h). So V(h) = (1/3)π(R - h)^2(R + h). Alternatively, multiply out: (R - h)(R² - h²) = (R - h)(R - h)(R + h) = (R - h)^2(R + h). So, V(h) = (1/3)π(R - h)^2(R + h). Alternatively, expanding this expression:First, (R - h)^2 = R² - 2Rh + h², so V(h) = (1/3)π(R² - 2Rh + h²)(R + h). Let me multiply this out:(R² - 2Rh + h²)(R + h) = R³ + R²h - 2R²h - 2Rh² + R h² + h³Simplify term by term:R³ + R²h - 2R²h - 2Rh² + R h² + h³= R³ - R²h - Rh² + h³So V(h) = (1/3)π(R³ - R²h - Rh² + h³)To find the maximum, take derivative of V with respect to h, set to zero.dV/dh = (1/3)π[-R² - 2Rh + 3h²]Set derivative equal to zero:-R² - 2Rh + 3h² = 0Multiply both sides by -1:R² + 2Rh - 3h² = 0This is a quadratic in h: -3h² + 2Rh + R² = 0Multiply both sides by -1: 3h² - 2Rh - R² = 0Use quadratic formula: h = [2R ± sqrt(4R² + 12R²)] / 6Simplify discriminant: sqrt(16R²) = 4RSo h = [2R ± 4R]/6Thus, two solutions:h = (2R + 4R)/6 = 6R/6 = Rh = (2R - 4R)/6 = (-2R)/6 = -R/3But h = R would mean the base is at the same point as the apex, which gives a cone with zero volume. So the meaningful solution is h = -R/3.Therefore, the base of the cone is at z = -R/3. Therefore, the height H = R - h = R - (-R/3) = (4R)/3.The radius r = sqrt(R² - h²) = sqrt(R² - (R²/9)) = sqrt(8R²/9) = (2√2 R)/3.Therefore, the volume is (1/3)πr²H = (1/3)π*(8R²/9)*(4R/3) = (1/3)*(32R³/27)π = (32/81)πR³.Wait, but is this correct? Let me verify.So if h = -R/3, then H = 4R/3, r = 2√2 R /3.So r² = 8R²/9, H = 4R/3, so Volume = (1/3)*π*(8R²/9)*(4R/3) = (1/3)*(32 R³/27)π = 32πR³/81.But is this indeed the maximum? Let me check the second derivative to ensure it's a maximum.Compute the second derivative of V with respect to h:First, dV/dh = (1/3)π(-R² - 2Rh + 3h²)Then d²V/dh² = (1/3)π(-2R + 6h)At h = -R/3,d²V/dh² = (1/3)π(-2R + 6*(-R/3)) = (1/3)π(-2R - 2R) = (1/3)π(-4R) = -4πR/3 < 0Since the second derivative is negative, this critical point is indeed a maximum.Therefore, the maximum volume occurs when the base of the cone is at z = -R/3, giving the cone a height of 4R/3 and base radius 2√2 R /3, with volume 32πR³/81.But wait, is there a different configuration where the apex isn't at the north pole? The problem says "inscribe a cone in a sphere", but didn't specify that the apex has to be at the top. Maybe the cone can be placed anywhere inside the sphere, with both the apex and the base's edge on the sphere. But in that case, would that give a larger volume?Wait, perhaps my initial assumption that the apex is at the north pole is restrictive. Maybe positioning the cone such that both the apex and the base are on the sphere but not necessarily aligned with the poles could give a larger volume.Let me consider that scenario. Let's suppose the apex is at some arbitrary point on the sphere, and the base is another circle on the sphere. To simplify, let's still use a coordinate system where the apex is along the z-axis at (0,0,a), and the base is a circle in the plane z = b, with radius r, such that all points on the circumference of the base are on the sphere. So, for the apex (0,0,a), we have a² = R² (since it's on the sphere). Similarly, for the base, any point (x,y,b) on the base must satisfy x² + y² + b² = R², so the radius r of the base satisfies r² + b² = R², so r = sqrt(R² - b²). The height of the cone is the distance between the apex and the base. Since the apex is at z = a and the base is at z = b, the height H is |a - b|. However, in this case, if the apex is at (0,0,a), then a can be anywhere on the sphere, so a = ±R. Wait, but if the apex is not at the north pole, but somewhere else, how would that affect the height and radius?Wait, actually, if the apex is at an arbitrary point on the sphere, not necessarily at (0,0,R), then the axis of the cone can be along any direction. But due to symmetry, the maximum volume should occur when the cone is aligned along the diameter, so we can assume without loss of generality that the apex is at (0,0,R) and the base is a circle in the plane z = c, as before. Because rotating the sphere wouldn't change the volume, so the maximum should be achieved in this symmetric configuration.Therefore, my initial approach was correct. So the maximum volume is 32πR³/81 when the cone has height 4R/3 and base radius 2√2 R/3, with the base located at z = -R/3.But just to make sure, let me think again. Suppose the apex is not at the top, but somewhere else. Let's say the apex is at (0,0,d), where d is between -R and R. Then, the base is at some plane z = e, and the radius of the base is sqrt(R² - e²). The height of the cone would be |d - e|. However, since the apex is also on the sphere, d² = R², so d = ±R. If d = R, same as before. If d = -R, then the apex is at the south pole, and the base is above it. The height would be | -R - e |. If e is less than R, but the maximum volume in that case would be similar, just mirrored. It should give the same maximum volume. So regardless of whether the apex is at the north or south pole, the maximum volume would be the same. So the initial calculation holds.Alternatively, if the apex is somewhere else on the sphere, not at the poles. Let me suppose the apex is at (0,0,a) where a ≠ ±R. Wait, no, since the apex is on the sphere, a² must equal R². Therefore, a must be ±R. So the apex can only be at the north or south pole if we're keeping the cone's axis along the z-axis. So that's the only possibility. Therefore, my initial model is correct.Therefore, the maximum volume is indeed 32πR³/81. So the answer should be boxed{dfrac{32}{81} pi R^3}Final AnswerThe maximum volume of the cone inscribed in a sphere is boxed{dfrac{32}{81} pi R^3}.</think>To inscribe a cone in a sphere such that the volume of the cone is maximized, we start by considering a sphere with radius ( R ) centered at the origin. The cone is assumed to be symmetric along the z-axis with its apex at the north pole ((0, 0, R)) and its base on the sphere at some plane ( z = h ).The radius ( r ) of the base of the cone is given by the equation of the sphere:[ r^2 + h^2 = R^2 implies r = sqrt{R^2 - h^2} ]The height ( H ) of the cone is the distance from the apex at ( z = R ) to the base at ( z = h ):[ H = R - h ]The volume ( V ) of the cone is:[ V = frac{1}{3} pi r^2 H = frac{1}{3} pi (R^2 - h^2)(R - h) ]To maximize this volume, we express ( V ) as a function of ( h ):[ V(h) = frac{1}{3} pi (R - h)(R^2 - h^2) ]Expanding and simplifying:[ V(h) = frac{1}{3} pi (R^3 - R^2h - Rh^2 + h^3) ]Taking the derivative with respect to ( h ) and setting it to zero:[ frac{dV}{dh} = frac{1}{3} pi (-R^2 - 2Rh + 3h^2) = 0 ][ -R^2 - 2Rh + 3h^2 = 0 ]Solving the quadratic equation:[ 3h^2 - 2Rh - R^2 = 0 ]Using the quadratic formula ( h = frac{2R pm sqrt{4R^2 + 12R^2}}{6} ):[ h = frac{2R pm 4R}{6} ][ h = R text{ (trivial solution)} text{ or } h = -frac{R}{3} ]The critical point ( h = -frac{R}{3} ) is verified by the second derivative test:[ frac{d^2V}{dh^2} = frac{1}{3} pi (-2R + 6h) ]At ( h = -frac{R}{3} ):[ frac{d^2V}{dh^2} = frac{1}{3} pi (-2R - 2R) = -frac{4Rpi}{3} < 0 ]Thus, the maximum volume occurs when ( h = -frac{R}{3} ). The height ( H ) and radius ( r ) of the cone are:[ H = R - left(-frac{R}{3}right) = frac{4R}{3} ][ r = sqrt{R^2 - left(-frac{R}{3}right)^2} = sqrt{frac{8R^2}{9}} = frac{2sqrt{2}R}{3} ]The maximum volume is:[ V = frac{1}{3} pi left(frac{8R^2}{9}right) left(frac{4R}{3}right) = frac{32}{81} pi R^3 ]The maximum volume of the cone inscribed in a sphere is (boxed{dfrac{32}{81} pi R^3}).,<think>Okay, so I need to figure out how to inscribe a cone in a sphere such that the cone has the maximum possible volume. Hmm, let me start by visualizing this. A sphere with a cone inside it, right? The cone has to touch the sphere at certain points. Since it's inscribed, I guess the base of the cone must lie on the surface of the sphere, and the tip of the cone should also touch the sphere. Wait, or maybe the entire cone is inside the sphere, with its base on a great circle? Hmm, I need to clarify that.First, let's recall some basics. The volume of a cone is given by (1/3)πr²h, where r is the radius of the base and h is the height. The sphere has a fixed radius, let's say R. So the challenge is to express the cone's volume in terms of variables related to the sphere and then maximize it.But how exactly is the cone positioned inside the sphere? Let me think. If the cone is inscribed, that probably means that the apex of the cone is at the top of the sphere, and the base is somewhere on the sphere's surface. Alternatively, maybe the cone is symmetric with its axis passing through the center of the sphere. That makes sense for maximizing volume, as symmetry often gives extremal values.Let me sketch this mentally. Imagine a sphere with radius R. The cone is inside the sphere, with its apex at the north pole and its base on the sphere's surface somewhere in the southern hemisphere. Wait, but actually, since the sphere is symmetrical, the apex could be anywhere, but placing it at the north pole might simplify calculations.Alternatively, maybe the cone's base is a circle on the sphere's surface, and the height of the cone is the distance from the apex to the base. If the apex is on the sphere, then the height of the cone would be the distance from the apex to the base along the central axis.Wait, maybe I should model this using coordinates. Let's place the sphere with its center at the origin (0,0,0) and radius R. Let’s suppose the cone's apex is at the top of the sphere, at (0,0,R), and the base is a circle lying on the sphere's surface. The height of the cone would then be R + z, where z is the z-coordinate of the base's center? Wait, no. If the apex is at (0,0,R), and the base is a circle lying on the sphere, then the height h of the cone would be the distance from (0,0,R) to the base's plane. Let's say the base is located at z = a, where a is between -R and R. Then the height h would be R - a.But the base is a circle on the sphere at z = a. The radius of the base of the cone can be found using the sphere's equation. The sphere's equation is x² + y² + z² = R². At height z = a, the radius r of the circular section is sqrt(R² - a²). So the base radius of the cone is r = sqrt(R² - a²).Therefore, the volume V of the cone would be (1/3)πr²h = (1/3)π(R² - a²)(R - a). Now, we need to maximize this volume with respect to a.So the problem reduces to finding the value of a (between -R and R) that maximizes V(a) = (1/3)π(R² - a²)(R - a). Since (1/3)π is a constant, we can focus on maximizing the function f(a) = (R² - a²)(R - a).Let me expand this function to make differentiation easier. Let's compute f(a):f(a) = (R² - a²)(R - a) = (R - a)(R + a)(R - a) = (R - a)²(R + a)Alternatively, expand it as:First, multiply (R² - a²)(R - a):= R²(R - a) - a²(R - a)= R³ - R²a - a²R + a³So f(a) = R³ - R²a - a²R + a³To find the maximum, take the derivative f'(a) with respect to a:f'(a) = -R² - 2aR + 3a²Set the derivative equal to zero:- R² - 2aR + 3a² = 0Multiply both sides by -1:R² + 2aR - 3a² = 0This is a quadratic equation in terms of a:-3a² + 2Ra + R² = 0Let me write it as:3a² - 2Ra - R² = 0Solving for a using quadratic formula:a = [2R ± sqrt( (2R)^2 + 12R² ) ] / 6Compute discriminant:(2R)^2 + 12R² = 4R² + 12R² = 16R²Thus,a = [2R ± 4R] / 6So two solutions:a = (2R + 4R)/6 = 6R/6 = Ra = (2R - 4R)/6 = (-2R)/6 = -R/3But a = R would give us a cone with height h = R - R = 0, which is degenerate. So the meaningful solution is a = -R/3.Therefore, the base of the cone is at z = -R/3. Let's verify this.So the height h of the cone is R - a = R - (-R/3) = R + R/3 = (4/3)RThe radius r of the base is sqrt(R² - a²) = sqrt(R² - (R²/9)) = sqrt(8R²/9) = (2√2 R)/3Therefore, the volume is (1/3)πr²h = (1/3)π*(8R²/9)*(4R/3) = (1/3)*(8/9)*(4/3)*πR³ = (32/81)πR³Wait, but let me check that. Let me compute r² = (8R²/9), h = (4R/3), so (1/3)*π*(8R²/9)*(4R/3) = (1/3)*(8*4)/(9*3) π R³ = (32/81)πR³. Yes.But is this the maximum volume? Let me verify by checking the second derivative.We had f(a) = (R² - a²)(R - a) = R³ - R²a - Ra² + a³First derivative f'(a) = -R² - 2Ra + 3a²Second derivative f''(a) = -2R + 6aAt a = -R/3,f''(-R/3) = -2R + 6*(-R/3) = -2R - 2R = -4RSince R is positive, f''(-R/3) = -4R < 0, which means the function has a local maximum at a = -R/3. Therefore, this is indeed the maximum volume.So, the cone with maximum volume inscribed in a sphere of radius R has height h = 4R/3 and base radius r = 2√2 R/3, giving a volume of 32πR³/81.But wait, let me make sure that the apex of the cone is at (0,0,R) and the base is at z = -R/3. The height of the cone would be from z = R to z = -R/3, which is indeed a distance of (4/3)R, correct. The base radius is sqrt(R² - ( (-R/3)^2 )) = sqrt(R² - R²/9) = sqrt(8R²/9) = 2√2 R/3, which is correct.Alternatively, is there another way to parameterize the cone? Maybe using the slant height or something else? But I think this approach is correct.Wait, another way to check is to recall that in some references, the maximum volume of a cone inscribed in a sphere is known. Let me recall from memory. I think the maximum occurs when the height of the cone is 4R/3 and the radius is 2√2 R/3, so the volume is 32πR³/81, which matches our result. So that seems correct.Therefore, the answer should be that the maximum volume is 32πR³/81, achieved when the cone has height 4R/3 and base radius 2√2 R/3.But the problem just says "inscribe a cone in a sphere, such that the volume of the cone is maximum." So the answer is the maximum volume, which is 32πR³/81. However, the problem might expect the answer in terms of the sphere's radius. Since the sphere is given but its radius isn't specified, we need to present the answer with R.Wait, but in the problem statement, they didn't give a specific radius, so the answer should be in terms of the sphere's radius R. So, yes, boxed answer would be boxed{dfrac{32}{81} pi R^3}.But just to make sure, let's recheck the steps:1. Position the sphere at center (0,0,0) with radius R.2. Place the cone's apex at (0,0,R) and base at z = a, with base radius sqrt(R² - a²).3. Express the volume in terms of a: V(a) = (1/3)π(R² - a²)(R - a).4. Take derivative, set to zero, solve for a, get a = -R/3.5. Compute the volume at a = -R/3, getting 32πR³/81.All steps check out. The maximum volume is indeed 32/81 π R³.Final AnswerThe maximum volume is boxed{dfrac{32}{81} pi R^3}.</think>To find the maximum volume of a cone inscribed in a sphere, we start by considering a sphere with radius ( R ) and a cone inscribed within it. The volume of a cone is given by ( frac{1}{3} pi r^2 h ), where ( r ) is the radius of the base and ( h ) is the height of the cone.We position the sphere with its center at the origin ((0,0,0)). The apex of the cone is at the top of the sphere ((0,0,R)), and the base of the cone lies on a plane ( z = a ) within the sphere. The height ( h ) of the cone is then ( R - a ), and the radius ( r ) of the base of the cone is ( sqrt{R^2 - a^2} ).The volume ( V ) of the cone can be expressed as:[V = frac{1}{3} pi (R^2 - a^2)(R - a)]To maximize this volume, we define the function ( f(a) = (R^2 - a^2)(R - a) ) and take its derivative with respect to ( a ):[f(a) = R^3 - R^2a - Ra^2 + a^3][f'(a) = -R^2 - 2Ra + 3a^2]Setting the derivative equal to zero to find critical points:[-R^2 - 2Ra + 3a^2 = 0][3a^2 - 2Ra - R^2 = 0]Solving this quadratic equation using the quadratic formula:[a = frac{2R pm sqrt{(2R)^2 + 12R^2}}{6} = frac{2R pm 4R}{6}]This gives two solutions, ( a = R ) (degenerate cone) and ( a = -frac{R}{3} ). The meaningful solution is ( a = -frac{R}{3} ).Substituting ( a = -frac{R}{3} ) back into the expressions for height and radius:[h = R - left(-frac{R}{3}right) = frac{4R}{3}][r = sqrt{R^2 - left(-frac{R}{3}right)^2} = sqrt{frac{8R^2}{9}} = frac{2sqrt{2}R}{3}]The maximum volume is then:[V = frac{1}{3} pi left(frac{8R^2}{9}right) left(frac{4R}{3}right) = frac{32}{81} pi R^3]Thus, the maximum volume of a cone inscribed in a sphere is (boxed{dfrac{32}{81} pi R^3}).]

🔑:Let the radius of the sphere be R, the radius of the base of the cone be x, and its height be y.The volume of the cone is:K=frac{x^{2} cdot y pi}{3}but sincex^{2}=y(2 R-y)thereforeK=frac{pi}{3} cdot y^{2}(2 R-y)The volume of the cone will be maximum when the functionz=y^{2}(2 R-y)takes its maximum value.Multiplying both sides of the equation by 2:2 z=y cdot y(4 R-2 y)However, the sum of the factors on the right side of the equation (4 R) is a constant number, which is why the volume of the cone will be the largest possible when the factors are equal (K.M.L.V.36.1.), that is, when4 R-2 y=yfrom which the height of the cone is:y=frac{4}{3} Rin which case the radius of the base of the cone isx=frac{2}{3} R sqrt{2}(Krisztián György, Pécs.)The problem was also solved by: Krausz B., Lukhaub Gy., Sasvári G.

🧠:[<think>Okay, so I need to figure out the digits A, B, and C such that when I take the number AB.C (which is a decimal number, right?), multiply it by C, round it to the nearest integer, and get BC as a two-digit number. All digits A, B, C are distinct. Hmm, let's break this down step by step.First, let's understand the notation. The number "AB whole, C tenth" is written as AB.C in decimal form. So, AB is the whole number part, and C is the tenths place. That means the value of the number is 10*A + B + C/10. Then, we multiply this by C, round to the nearest integer, and the result should be the two-digit number BC, which is 10*B + C.So, the equation we have is: (10*A + B + C/10) * C ≈ 10*B + C, where ≈ means rounded to the nearest integer. Since we need to round, the actual product before rounding must be within 0.5 of the target integer BC. Therefore, the product (10*A + B + C/10) * C should be in the interval [10*B + C - 0.5, 10*B + C + 0.5).Let me write that as an inequality:10*B + C - 0.5 ≤ (10*A + B + C/10) * C < 10*B + C + 0.5Our goal is to find integers A, B, C (digits from 0 to 9, distinct) that satisfy this inequality.Given that A, B, C are digits, A can't be 0 because AB is a two-digit number, so A is from 1 to 9. B and C can be from 0 to 9, but all three digits must be distinct.Let me consider possible values for C first, since C appears in multiple places. Since we're multiplying by C and rounding to get a two-digit number BC, C must be a digit such that when multiplied by the decimal number AB.C, the result is around BC. Also, BC is a two-digit number, so B can't be 0 if C is 0, but wait, C can't be 0 because otherwise, the product would be AB.0 * 0 = 0, which is 00, but digits must be distinct. Wait, if C is 0, the product is 0, which would imply BC is 00, but B and C can't both be 0 (since digits must be distinct). So C cannot be 0. Therefore, C is from 1 to 9.Similarly, B can be from 0 to 9, but since BC is a two-digit number, B can be 0 only if C is non-zero. But since digits must be distinct, B and C can't be the same. So possible values for B and C are 0-9, distinct.So, C ∈ {1,2,...,9}, B ∈ {0,1,...,9}, B ≠ C, A ∈ {1,...,9}, A ≠ B, A ≠ C.Let me try to rephrase the inequality:Let’s denote N = (10*A + B + C/10) * CWe need N ∈ [10*B + C - 0.5, 10*B + C + 0.5)So, N must be very close to 10*B + C. Let’s write:(10*A + B + 0.1*C) * C ≈ 10*B + CMultiply out the left side:10*A*C + B*C + 0.1*C² ≈ 10*B + CSo, 10AC + BC + 0.1C² ≈ 10B + CLet me rearrange terms:10AC + BC + 0.1C² - 10B - C ≈ 0Factor terms where possible:10AC + BC - 10B - C + 0.1C² ≈ 0Looking at the terms 10AC - C, which can be factored as C(10A - 1), and terms BC -10B, which is B(C - 10). So:C(10A - 1) + B(C - 10) + 0.1C² ≈ 0Hmm, not sure if that helps. Maybe it's better to express it as:10AC + BC + 0.1C² = 10B + C + delta, where delta is in (-0.5, 0.5]But maybe trying specific values for C would be more straightforward.Since C is a digit from 1 to 9, let's iterate through possible C values and see what A and B could be.Starting with C=1:Then, N = (10A + B + 0.1) * 1 = 10A + B + 0.1We need this to round to 10B + 1. So,10B + 1 - 0.5 ≤ 10A + B + 0.1 < 10B + 1 + 0.5Simplify:10B + 0.5 ≤ 10A + B + 0.1 < 10B + 1.5Subtract B from all parts:9B + 0.5 ≤ 10A + 0.1 < 9B + 1.5Subtract 0.1:9B + 0.4 ≤ 10A < 9B + 1.4Divide by 10:0.9B + 0.04 ≤ A < 0.9B + 0.14But A is an integer from 1 to 9. So, 0.9B + 0.04 ≤ A < 0.9B + 0.14Since A must be an integer, let's check possible B:For B=0:0.04 ≤ A < 0.14 → A must be 0, but A can't be 0. Not possible.For B=1:0.94 ≤ A < 1.04 → A=1Check if A=1, B=1, but digits must be distinct. B=1, C=1, but C=1 here. Wait, but B=1 and C=1 would be same digit, which is not allowed. So invalid.For B=2:0.9*2 +0.04=1.84 ≤ A <0.9*2 +0.14=1.94→ A=2But A=2, B=2, same digit. Not allowed.Similarly, B=3:0.9*3 +0.04=2.74 ≤ A <2.84→ A=3. But same digit. Not allowed.Continuing, for each B, A would need to be the integer part, but since 0.9B +0.04 is approximately 0.9B, and A must be the next integer, but then A would equal B, which is invalid. So for C=1, there are no solutions.Moving on to C=2:N = (10A + B + 0.2)*2 = 20A + 2B + 0.4Rounded to nearest integer should be 10B + 2.So,10B + 2 - 0.5 ≤ 20A + 2B + 0.4 <10B + 2 + 0.5Simplify:10B + 1.5 ≤20A + 2B + 0.4 <10B + 2.5Subtract 2B:8B + 1.5 ≤20A +0.4 <8B +2.5Subtract 0.4:8B +1.1 ≤20A <8B +2.1Divide by 20:0.4B +0.055 ≤A <0.4B +0.105Again, A must be integer. Let's check possible B:B=0: 0.055 ≤A <0.105→A=0. Not allowed.B=1: 0.455 ≤A <0.505→A=0. Not possible.B=2: 0.8 +0.055=0.855≤A<0.8+0.105=0.905→A=0. Not possible.Wait, B ranges from 0 to 9. Let's compute 0.4B +0.055 and 0.4B +0.105 for each B:For B=0: [0.055, 0.105) → A=0 invalid.B=1: [0.455, 0.505) → A=0 invalid.B=2: [0.855, 0.905) → A=0. Still no.B=3: 1.255 ≤A <1.305→A=1Check if digits distinct: A=1, B=3, C=2. All distinct. Possible.Check if 20A + 2B +0.4 =20*1 +2*3 +0.4=20 +6 +0.4=26.4. Rounded to nearest integer is 26. Which is 10*3 +2=32? Wait, 26 is not 32. So this doesn't work. Hmm.Wait, maybe I miscalculated. Let's see:If A=1, B=3, C=2:Number is 13.2. Multiply by 2: 13.2*2=26.4. Rounded is 26. But 10B + C =10*3 +2=32. 26 ≠32. So this is invalid.Wait, so even if A is 1 here, the result is 26.4, which would round to 26, not 32. So that's not matching. So perhaps even if A fits the inequality, the actual calculation doesn't give the desired result. Therefore, my approach might be missing something.Wait, but the inequality is supposed to bound N between 10B + C -0.5 and 10B + C +0.5. Let's check for B=3, C=2:10B + C =32. So N must be between 31.5 and 32.5.But N in this case is 26.4, which is way lower. So even though A=1 satisfies the earlier inequality (which I derived incorrectly?), the actual value is not in the required range.Therefore, my method must have an error. Let me check the steps again.Original inequality for C=2:10B + 2 -0.5 ≤ (10A + B +0.2)*2 <10B +2 +0.5Which is:10B +1.5 ≤20A +2B +0.4 <10B +2.5Subtract 2B:8B +1.5 ≤20A +0.4 <8B +2.5Subtract 0.4:8B +1.1 ≤20A <8B +2.1Divide by 20:0.4B +0.055 ≤A <0.4B +0.105So for B=3, 0.4*3=1.2, so 1.255 ≤A <1.305→A=1. So A=1. Then, plug back into N=20A +2B +0.4=20+6+0.4=26.4. So 26.4 is supposed to be between 31.5 and 32.5? That's not true. So why does the inequality say that?Wait, this seems contradictory. It looks like my algebra is correct, but the result doesn't match. Therefore, perhaps I made a mistake in setting up the inequality.Wait, the product before rounding is (10A + B +0.2)*2. This should be in [10B +2 -0.5,10B +2 +0.5). So:10B +2 -0.5 ≤ (10A + B +0.2)*2 <10B +2 +0.5Which is:10B +1.5 ≤20A +2B +0.4 <10B +2.5Then, subtract 2B:8B +1.5 ≤20A +0.4 <8B +2.5Subtract 0.4:8B +1.1 ≤20A <8B +2.1Divide by 20:0.4B +0.055 ≤A <0.4B +0.105Yes, that's correct. Then, for B=3, A≈1.255 to 1.305, so A=1. But as we saw, with A=1, B=3, we get N=26.4, which is not in [31.5,32.5). So, something is wrong here.Ah! Wait, perhaps my initial setup is wrong. Let me verify:The original problem states: "the product of the number 'AB whole, C tenth' and C, rounded to the nearest integer, results in the number BC".So, the product is (AB.C) * C, which is (10A + B + 0.1C) * C.But when I calculated for C=2, B=3, A=1, the product was 13.2 *2 =26.4, which rounds to 26. But BC is 32. So clearly not matching.Therefore, my inequality is correct, but the conclusion is that for B=3, C=2, there is no solution. So the fact that A=1 fits into the inequality doesn't mean that the product is in the required range. Wait, but why?Wait, maybe my algebra is correct, but when I converted 8B +1.1 ≤20A <8B +2.1, and divided by 20, perhaps I need to take into account that A is an integer. So even if 0.4B +0.055 ≤A <0.4B +0.105, but since A must be an integer, this interval might not contain any integer A. For example, for B=3, 0.4*3 +0.055=1.255 and 0.4*3 +0.105=1.305, so A must be 1.255 to 1.305. But A must be integer, so there is no integer A in this interval. Wait, but I thought A=1. But 1 is less than 1.255. So actually, there is no solution for B=3. Wait, but then why did I think A=1?Wait, I think I made a mistake here. If the inequality is 0.4B +0.055 ≤A <0.4B +0.105, and A must be an integer, then we need to check if there's an integer A in that interval. For B=3:0.4*3 +0.055=1.2 +0.055=1.2550.4*3 +0.105=1.2 +0.105=1.305So A must be between 1.255 and 1.305. But A must be integer. There is no integer in this interval. Therefore, no solution for B=3. So my previous conclusion that A=1 was incorrect because 1 is not within [1.255,1.305). Therefore, there is actually no solution for C=2 either.Wait, then how does this work? Let me check B=5:For B=5:0.4*5 +0.055=2.0550.4*5 +0.105=2.105So A must be between 2.055 and 2.105. So A=2. So check if A=2, B=5, C=2. But C=2, so digits must be distinct. A=2, B=5, C=2. But A and C are both 2. Not allowed.Similarly, for B=6:0.4*6 +0.055=2.4550.4*6 +0.105=2.505So A=2.455 to 2.505, so A=2. But check:A=2, B=6, C=2. Again, duplicate digits. Not allowed.Wait, for B=7:0.4*7 +0.055=2.8550.4*7 +0.105=2.905So A=2.855 to 2.905. So A=3. Check:A=3, B=7, C=2. All distinct. Let's compute N:(10*3 +7 +0.2)*2 = (37.2)*2=74.4. Rounded is 74. But BC is 72. So 74≠72. Not matching.Wait, but according to the inequality, A=3 would be in the interval for B=7:Original inequality:10*7 +2 -0.5 ≤ (37.2)*2 <10*7 +2 +0.572 -0.5=71.5 ≤74.4 <72 +0.5=72.5But 74.4 is not in [71.5,72.5). Therefore, even though A=3 is in the computed interval, the actual product is way higher. Hence, my approach is missing something.Wait, but how come? Let's check:If I have B=7, then 10B + C =72. Then, N must be in [71.5,72.5). But the product is 74.4, which is outside. So clearly, even though A=3 satisfies the inequality derived from the rounded equation, the actual product is not in the required range. Therefore, my initial approach is flawed.I think the problem is that when I set up the inequality, I rearranged terms and divided by something, which might have introduced an error. Alternatively, perhaps the initial setup is wrong. Let me start over.Given (10A + B + 0.1C) * C ≈ 10B + C (rounded to nearest integer). Therefore, the exact value must be within 0.5 of 10B + C. So:10B + C - 0.5 ≤ (10A + B + 0.1C) * C <10B + C +0.5Let me define this as:Lower bound: 10B + C - 0.5Upper bound: 10B + C + 0.5Therefore, for each C from 1 to 9, we can try to find integers A and B (distinct digits, A ≠0) such that (10A + B +0.1C)*C is within [10B + C -0.5, 10B + C +0.5)Alternatively, rearranged:(10A + B +0.1C)*C -10B -C ∈ [-0.5,0.5)Let me compute this expression:=10AC + BC +0.1C² -10B -C=10AC -10B + BC -C +0.1C²Factor terms:=10B(A -1) + C(B -1 +0.1C) +10AC -10B ?Wait, perhaps not helpful. Alternatively, group terms with A and B:=10AC + (BC -10B) +0.1C² -C=10AC + B(C -10) +0.1C² -CHmm. Maybe not helpful. Alternatively, solve for A:From the inequality:10B + C -0.5 ≤ (10A + B +0.1C)*C <10B + C +0.5Divide all parts by C (since C ≥1):(10B + C -0.5)/C ≤10A + B +0.1C < (10B + C +0.5)/CThen subtract B +0.1C:[(10B + C -0.5)/C - B -0.1C] ≤10A <[(10B + C +0.5)/C - B -0.1C]Simplify left side:(10B + C -0.5)/C - B -0.1C = (10B/C +1 -0.5/C) - B -0.1C=10B/C - B +1 -0.5/C -0.1C= B(10/C -1) +1 -0.5/C -0.1CSimilarly, the right side:(10B + C +0.5)/C - B -0.1C =10B/C +1 +0.5/C - B -0.1C= B(10/C -1) +1 +0.5/C -0.1CTherefore, the inequality becomes:B(10/C -1) +1 -0.5/C -0.1C ≤10A <B(10/C -1) +1 +0.5/C -0.1CThis is quite complex. Maybe plugging in values for C and solving for B and A is better.Let me try C=5.C=5:Then, the inequality:10B +5 -0.5 ≤(10A + B +0.5)*5 <10B +5 +0.5Simplify:10B +4.5 ≤50A +5B +2.5 <10B +5.5Subtract 5B:5B +4.5 ≤50A +2.5 <5B +5.5Subtract 2.5:5B +2 ≤50A <5B +3Divide by 50:0.1B +0.04 ≤A <0.1B +0.06Since A must be an integer between 1 and 9, 0.1B +0.04 ≤A <0.1B +0.06For example:B=0: 0.04 ≤A <0.06 → No solution.B=1: 0.14 ≤A <0.16 → No.B=2: 0.24 ≤A <0.26 → No....Up to B=9:0.94 ≤A <0.96 → No.No solutions for C=5.C=9:Let's try C=9.Inequality:10B +9 -0.5 ≤(10A + B +0.9)*9 <10B +9 +0.5Which is:10B +8.5 ≤90A +9B +8.1 <10B +9.5Subtract 9B:B +8.5 ≤90A +8.1 <B +9.5Subtract 8.1:B +0.4 ≤90A <B +1.4Divide by 90:(B +0.4)/90 ≤A <(B +1.4)/90Since A is at least 1:(B +0.4)/90 ≤1 ⇒ B +0.4 ≤90 ⇒ B ≤89.6. Which is always true since B is a digit (0-9). So for B from 0 to9:For each B:(B +0.4)/90 ≤A <(B +1.4)/90But (B +1.4)/90 is less than (9 +1.4)/90=10.4/90≈0.115. But A must be at least 1. So no solution for C=9.Hmm, this approach isn't working. Maybe trying specific C values where the multiplication might result in something feasible.Let’s think of possible C where multiplying a two-digit decimal by C could result in a two-digit integer when rounded. For example, if C is small, say 1 or 2, the product might be too small. If C is large, like 9, the product could be large. Maybe try C=3,4,5,6,7,8.Let’s try C=6:Inequality:10B +6 -0.5 ≤(10A + B +0.6)*6 <10B +6 +0.5Simplify:10B +5.5 ≤60A +6B +3.6 <10B +6.5Subtract 6B:4B +5.5 ≤60A +3.6 <4B +6.5Subtract 3.6:4B +1.9 ≤60A <4B +2.9Divide by 60:(4B +1.9)/60 ≤A <(4B +2.9)/60Compute for B=0:(0 +1.9)/60≈0.0317 ≤A <≈0.0483→No.B=1:(4 +1.9)/60≈5.9/60≈0.0983 ≤A <(4 +2.9)/60≈6.9/60≈0.115→No.B=2:(8 +1.9)/60≈9.9/60≈0.165 ≤A <(8 +2.9)/60≈10.9/60≈0.181→No.B=3:12 +1.9=13.9→13.9/60≈0.231 ≤A <14.9/60≈0.248→No.B=4:16 +1.9=17.9→17.9/60≈0.298 ≤A <18.9/60≈0.315→No.B=5:20 +1.9=21.9→21.9/60≈0.365 ≤A <22.9/60≈0.381→No.B=6:24 +1.9=25.9→25.9/60≈0.431 ≤A <26.9/60≈0.448→No.B=7:28 +1.9=29.9→29.9/60≈0.498 ≤A <30.9/60≈0.515→A=0.5 is not integer. So no.B=8:32 +1.9=33.9→33.9/60≈0.565 ≤A <34.9/60≈0.581→No.B=9:36 +1.9=37.9→37.9/60≈0.631 ≤A <38.9/60≈0.648→No.No solutions for C=6.C=4:Inequality:10B +4 -0.5 ≤(10A + B +0.4)*4 <10B +4 +0.5Simplify:10B +3.5 ≤40A +4B +1.6 <10B +4.5Subtract 4B:6B +3.5 ≤40A +1.6 <6B +4.5Subtract 1.6:6B +1.9 ≤40A <6B +2.9Divide by 40:(6B +1.9)/40 ≤A <(6B +2.9)/40Compute for B=0:1.9/40≈0.0475 ≤A <2.9/40≈0.0725→No.B=1:7.9/40≈0.1975 ≤A <8.9/40≈0.2225→No.B=2:13.9/40≈0.3475 ≤A <14.9/40≈0.3725→No.B=3:19.9/40≈0.4975 ≤A <20.9/40≈0.5225→No.B=4:25.9/40≈0.6475 ≤A <26.9/40≈0.6725→No.B=5:31.9/40≈0.7975 ≤A <32.9/40≈0.8225→No.B=6:37.9/40≈0.9475 ≤A <38.9/40≈0.9725→A=1.Check if A=1, B=6, C=4. All distinct. Then compute N=(10*1 +6 +0.4)*4=16.4*4=65.6. Rounded is 66. Which should be BC=64. 66≠64. Not a match.B=7:43.9/40≈1.0975 ≤A <44.9/40≈1.1225→A=1.Check A=1, B=7, C=4: (10+7+0.4)*4=17.4*4=69.6. Rounded to 70. BC=74. Doesn't match.B=8:49.9/40≈1.2475 ≤A <50.9/40≈1.2725→A=1. Not valid since A=1.2475 is higher than 1. So no integer A here.Wait, if B=8:6*8 +1.9=49.9→49.9/40≈1.2475≤A <(48 +2.9)/40=50.9/40≈1.2725. So A=1.2475 to 1.2725. So no integer A.Similarly, B=9:55.9/40≈1.3975 ≤A <56.9/40≈1.4225→No integer A.No solutions for C=4.C=7:Inequality:10B +7 -0.5 ≤(10A + B +0.7)*7 <10B +7 +0.5Which is:10B +6.5 ≤70A +7B +4.9 <10B +7.5Subtract 7B:3B +6.5 ≤70A +4.9 <3B +7.5Subtract 4.9:3B +1.6 ≤70A <3B +2.6Divide by 70:(3B +1.6)/70 ≤A <(3B +2.6)/70For B=0:1.6/70≈0.023 ≤A <2.6/70≈0.037→No.B=1:4.6/70≈0.066 ≤A <5.6/70≈0.08→No....B=5:15 +1.6=16.6/70≈0.237 ≤A <17.6/70≈0.251→No.B=6:18 +1.6=19.6/70≈0.28 ≤A <20.6/70≈0.294→No.B=7:21 +1.6=22.6/70≈0.323 ≤A <23.6/70≈0.337→No.B=8:24 +1.6=25.6/70≈0.366 ≤A <26.6/70≈0.38→No.B=9:27 +1.6=28.6/70≈0.409 ≤A <29.6/70≈0.423→No.No solutions for C=7.C=8:Inequality:10B +8 -0.5 ≤(10A + B +0.8)*8 <10B +8 +0.5Simplify:10B +7.5 ≤80A +8B +6.4 <10B +8.5Subtract 8B:2B +7.5 ≤80A +6.4 <2B +8.5Subtract 6.4:2B +1.1 ≤80A <2B +2.1Divide by 80:(2B +1.1)/80 ≤A <(2B +2.1)/80Which is approximately:For B=0:1.1/80≈0.01375 ≤A <2.1/80≈0.02625→No....B=4:8 +1.1=9.1/80≈0.11375 ≤A <10.1/80≈0.12625→No.B=5:10 +1.1=11.1/80≈0.13875 ≤A <12.1/80≈0.15125→No.B=6:12 +1.1=13.1/80≈0.16375 ≤A <14.1/80≈0.17625→No.B=7:14 +1.1=15.1/80≈0.18875 ≤A <16.1/80≈0.20125→No.B=8:16 +1.1=17.1/80≈0.21375 ≤A <18.1/80≈0.22625→No.B=9:18 +1.1=19.1/80≈0.23875 ≤A <20.1/80≈0.25125→No.No solutions for C=8.C=3:Inequality:10B +3 -0.5 ≤(10A + B +0.3)*3 <10B +3 +0.5Simplify:10B +2.5 ≤30A +3B +0.9 <10B +3.5Subtract 3B:7B +2.5 ≤30A +0.9 <7B +3.5Subtract 0.9:7B +1.6 ≤30A <7B +2.6Divide by 30:(7B +1.6)/30 ≤A <(7B +2.6)/30For B=0:1.6/30≈0.053 ≤A <2.6/30≈0.086→No.B=1:7 +1.6=8.6/30≈0.287 ≤A <9.6/30≈0.32→A=0. Not allowed.B=2:14 +1.6=15.6/30≈0.52 ≤A <16.6/30≈0.553→No.B=3:21 +1.6=22.6/30≈0.753 ≤A <23.6/30≈0.786→No.B=4:28 +1.6=29.6/30≈0.987 ≤A <30.6/30≈1.02→A=1.Check A=1, B=4, C=3. All distinct. Compute N=(10 +4 +0.3)*3=14.3*3=42.9. Rounded to 43. Which should be BC=43. Yes! 43 is equal to BC. So digits A=1, B=4, C=3.Check if distinct: 1,4,3 are distinct. Correct. So this is a possible solution.But wait, let's verify:AB.C is 14.3. Multiply by C=3: 14.3*3=42.9. Rounded to the nearest integer is 43, which is BC=43. Perfect. So this works.Are there other possible solutions?Let me check B=5:7*5 +1.6=36.6/30=1.22 ≤A <37.6/30≈1.253→A=1.22 to 1.253→A=1. So A=1, B=5, C=3.Check N=(10 +5 +0.3)*3=15.3*3=45.9. Rounded is 46. BC=53. Not equal.B=6:7*6 +1.6=43.6/30≈1.453 ≤A <44.6/30≈1.486→A=1. Not in interval.B=7:7*7 +1.6=50.6/30≈1.687 ≤A <51.6/30≈1.72→A=1.687 to 1.72. A must be 2.Check A=2, B=7, C=3. Digits distinct. Compute N=(20 +7 +0.3)*3=27.3*3=81.9. Rounded to 82. BC=73. Not equal.B=8:7*8 +1.6=57.6 +1.6=59.2? Wait, no:Wait, original calculation: For B=8:(7*8 +1.6)/30=(56 +1.6)/30=57.6/30=1.92 ≤A <(56 +2.6)/30=58.6/30≈1.953→A=1.92 to1.953→A=1. Not possible. So no solution.B=9:7*9 +1.6=63 +1.6=64.6/30≈2.153 ≤A <65.6/30≈2.186→A=2. Check:A=2, B=9, C=3. Compute N=(20 +9 +0.3)*3=29.3*3=87.9→88. BC=93. Not equal.So only B=4 gives a valid solution.Therefore, the solution is A=1, B=4, C=3.But wait, let's check C=3 for other B values just in case.Wait, for C=3, B=4, A=1 is the only solution. Let me check if there are other C values where this could happen.Wait, we tried C=3,4,5,6,7,8,9. C=3 was the only one with a solution. But let me check C=5 again in case I missed something.C=5:As before, we saw no solution, but let's check manually. Suppose A=1, B=2, C=5.N=(10 +2 +0.5)*5=12.5*5=62.5. Rounded is 63. BC=25. Not equal.Alternatively, A=2, B=5, C=5: but duplicate digits.Alternatively, A=3, B=5, C=5: duplicates.Not helpful.Alternatively, maybe C=9:Wait, let's try A=1, B=8, C=9:N=(10 +8 +0.9)*9=18.9*9=170.1. Rounded is 170. BC=89. No.Alternatively, maybe C=6:If A=2, B=1, C=6:N=(20 +1 +0.6)*6=21.6*6=129.6→130. BC=16. No.This seems not working.Alternatively, check C=2 again with different B.Wait, earlier I saw for C=2, B=3, but the product was 26.4, which didn't match. But suppose there is a rounding up.Wait, maybe I need to check if there's a case where the decimal part is exactly 0.5. For example, if the product is xx.x5, which rounds to xx.x+1. But since we are rounding to integer, 0.5 rounds up.But in our inequality, we included up to but not including 10B + C +0.5. So if the product is exactly 10B + C +0.5, it would round to 10B + C +1, which is outside the interval. Therefore, the upper bound is exclusive.But let's see, if there's a case where the product is exactly halfway, but it's not included. So only products less than 10B + C +0.5 are considered.Therefore, the solution for C=3, B=4, A=1 seems to be the only one.Let me verify once again:A=1, B=4, C=3.AB.C =14.3Multiply by C=3:14.3*3=42.9Rounded to nearest integer:43Which is BC=43. Yes, all distinct digits. Perfect.Are there any others?Suppose A=2, B=8, C=4:AB.C=28.4Multiply by 4:28.4*4=113.6→114. BC=84. Not equal.Another try: A=1, B=5, C=2:AB.C=15.2Multiply by 2:30.4→30. BC=52. Not equal.A=1, B=3, C=9:13.9*9=125.1→125. BC=39. No.A=2, B=5, C=7:25.7*7=179.9→180. BC=57. No.A=1, B=9, C=5:19.5*5=97.5→98. BC=95. No.Another possible check: A=2, B=3, C=4:23.4*4=93.6→94. BC=34. Not equal.Wait, maybe C=3, B=4, A=1 is the only solution.Yes, I think that's the case. Therefore, the distinct digits are A=1, B=4, C=3.Final AnswerThe distinct digits are (boxed{A=1}), (boxed{B=4}), and (boxed{C=3}).</think>Given the problem, we need to determine the distinct digits (A), (B), and (C) such that the product of the number (overline{AB.C}) (where (overline{AB}) is the whole number part and (C) is the tenths digit) and (C), rounded to the nearest integer, results in the number (overline{BC}).The key steps are as follows:1. Formulate the equation: [ (10A + B + 0.1C) cdot C approx 10B + C ] This product must be within the interval ([10B + C - 0.5, 10B + C + 0.5)).2. Check possible values for (C): - For (C = 3), the inequality becomes: [ 10B + 3 - 0.5 leq (10A + B + 0.3) cdot 3 < 10B + 3 + 0.5 ] Simplifying this, we get: [ 7B + 1.6 leq 30A < 7B + 2.6 ] Dividing by 30: [ frac{7B + 1.6}{30} leq A < frac{7B + 2.6}{30} ] For (B = 4): [ frac{7 cdot 4 + 1.6}{30} = frac{29.6}{30} approx 0.987 leq A < frac{30.6}{30} approx 1.02 ] This gives (A = 1).3. Verify the solution: - Check if (A = 1), (B = 4), (C = 3) are distinct and satisfy the equation: [ overline{AB.C} = 14.3 ] Multiplying by (C = 3): [ 14.3 cdot 3 = 42.9 ] Rounded to the nearest integer, this is 43, which matches (overline{BC} = 43).Thus, the distinct digits are (boxed{A=1}), (boxed{B=4}), and (boxed{C=3}).,<think>Okay, let's try to solve this problem. So we have to find distinct digits A, B, C such that when we take the number "AB whole, C tenth" (which I think is the number AB.C) and multiply it by C, then round it to the nearest integer, we get the number BC. The digits A, B, C all have to be different from each other. Alright, let's break this down step by step.First, let's understand the notation. The number "AB whole, C tenth" is written as overline{A B, C}, which in decimal form is 10*A + B + C/10. So, for example, if A=1, B=2, C=3, then the number would be 12.3. Then, we multiply this number by C, so 12.3 * 3 in this example. After multiplying, we round the result to the nearest integer, and this should equal the number BC, which is 10*B + C. So in the example, BC would be 23, and 12.3*3 is 36.9, which rounds to 37. But 37 is not 23, so that example doesn't work.Our job is to find A, B, C (distinct digits from 0 to 9) such that this condition holds. Let's formalize this with equations. Let me write it out:Let N = (10*A + B + C/10) * CThen, rounding N to the nearest integer should equal 10*B + C.So, N ≈ 10*B + C, where ≈ means rounded to the nearest integer.Therefore, N must be in the interval [ (10*B + C) - 0.5, (10*B + C) + 0.5 )That is, (10*B + C) - 0.5 ≤ N < (10*B + C) + 0.5So substituting N:(10*B + C) - 0.5 ≤ (10*A + B + C/10) * C < (10*B + C) + 0.5That's the inequality we need to satisfy. Also, A, B, C are distinct digits, so they must be between 0-9, all different. Let me note that A cannot be zero because "AB" is a two-digit number, so A must be from 1-9. B and C can be from 0-9, but all three digits must be distinct.So, our variables:A: 1-9B: 0-9, B ≠ AC: 0-9, C ≠ A, C ≠ BSo, our approach can be to iterate over possible values of A, B, C (with A ≥1, all distinct) and check if the inequality holds. However, since this is a math problem, maybe we can find a smarter way.Alternatively, perhaps we can manipulate the inequality.Let me expand the expression for N:N = (10*A + B + C/10) * C = (10*A + B)*C + (C^2)/10So N = (10*A + B)*C + (C^2)/10Therefore, the inequality becomes:(10*B + C) - 0.5 ≤ (10*A + B)*C + (C^2)/10 < (10*B + C) + 0.5Let me rearrange this inequality:Let's subtract (10*B + C) from all parts:-0.5 ≤ (10*A + B)*C + (C^2)/10 - (10*B + C) < 0.5So:| (10*A + B)*C + (C^2)/10 - (10*B + C) | < 0.5That's the absolute value of that expression must be less than 0.5.Let me compute the expression inside the absolute value:E = (10*A + B)*C + (C^2)/10 - 10*B - CLet me expand (10*A + B)*C:= 10*A*C + B*CSo, E = 10*A*C + B*C + (C^2)/10 - 10*B - CNow, combine like terms:= 10*A*C + B*C - 10*B - C + (C^2)/10Factor terms where possible:Let's look at B*C - 10*B = B*(C - 10)Similarly, 10*A*C - C = C*(10*A -1)So:E = C*(10*A -1) + B*(C -10) + (C^2)/10Alternatively, maybe that's not helpful. Let's see.Alternatively, group terms differently:E = 10*A*C + (B*C - 10*B) + (-C) + (C^2)/10= 10*A*C + B*(C -10) - C + (C^2)/10Hmm, perhaps that's not particularly helpful. Let's consider that the entire expression must be between -0.5 and 0.5. So maybe we can model E ≈ 0, with error less than 0.5.Alternatively, perhaps approximate E ≈ 0, so:(10*A + B)*C + (C^2)/10 ≈ 10*B + CBut perhaps this approach is better. Let me denote:(10*A + B)*C ≈ 10*B + C - (C^2)/10But perhaps even that is not helpful. Alternatively, since (C^2)/10 is a small term (since C is a digit from 0-9, so C^2 is up to 81, so (C^2)/10 is up to 8.1). But 10*B + C is a two-digit number, so up to 99. So subtracting up to 8.1 from it gives down to 90.9. So (10*A + B)*C must be close to 10*B + C - (C^2)/10.Alternatively, perhaps approximate (10*A + B)*C ≈ 10*B + C, ignoring the (C^2)/10 term. Then, we can write:(10*A + B)*C ≈ 10*B + CWhich would lead to:10*A*C + B*C ≈ 10*B + CBut maybe not. Let's see. Let's bring all terms to one side:10*A*C + B*C - 10*B - C ≈ 0Which is similar to the expression E we had before. So:10*A*C + B*C - 10*B - C ≈ 0Factor terms:Let's factor B from the second and third terms:10*A*C + B*(C - 10) - C ≈ 0Alternatively, factor C from the first and fourth terms:C*(10*A -1) + B*(C -10) ≈ 0So:C*(10*A -1) ≈ - B*(C -10)So:C*(10*A -1) ≈ B*(10 - C)This is an equation we can perhaps work with.Since all variables are integers from 0-9 (except A, which is 1-9), and all distinct, we can try to find integers A, B, C such that C*(10*A -1) is approximately equal to B*(10 - C). The approximation here is because we have to account for the (C^2)/10 term, but maybe this can guide us.Alternatively, perhaps look for values where C*(10*A -1) = B*(10 - C). If this equation holds exactly, then:(10*A -1)/B = (10 - C)/CBut since all variables are integers, this ratio must be rational. Let's see. Alternatively, cross-multiplying:C*(10*A -1) = B*(10 - C)So:10*A*C - C = 10*B - B*CBring all terms to one side:10*A*C - C -10*B + B*C = 0Factor:C*(10*A -1 + B) -10*B =0Hmm. Let's write this as:C*(10*A -1 + B) =10*BTherefore:C = (10*B)/(10*A -1 + B)But since C must be an integer between 0 and 9, and A and B are integers with A ≥1, this equation can help us find possible A, B, C.But this equation is derived under the assumption that (C^2)/10 is negligible, which might not be the case, but perhaps this can give us a starting point.So, let's try to use this equation:C = (10*B) / (10*A -1 + B)We can attempt to find integers A, B, C (with A ≥1, all distinct) such that this equation holds. Since C must be an integer, (10*A -1 + B) must divide 10*B.Let's let’s denote D = 10*A -1 + B. Then D divides 10*B. So D | 10*B.But D =10*A -1 + B. So, D =10*A + B -1.Therefore, 10*A + B -1 divides 10*B.Given that A and B are digits with A ≥1, B from 0-9, D can range from 10*1 + 0 -1=9 up to 10*9 +9 -1=98.So, D is between 9 and 98, and D divides 10*B. Let's note that 10*B is between 0 and 90 (since B is 0-9). So D must be a divisor of 10*B, and D ≥9. Also, since D =10*A + B -1, D is at least 10*1 +0 -1=9, as mentioned.But let's note that if B=0, then 10*B=0, so D divides 0, which is possible for any D, but in that case, C=0/Bla=0, but C would have to be 0. But since B=0 and C=0, they are not distinct. Therefore, B cannot be 0. So B must be from 1-9.Similarly, if B=1, then 10*B=10. So D must divide 10. D=10*A +1 -1=10*A. So 10*A divides 10, so 10*A |10. Therefore, A must be 1. Then D=10*1=10. 10 divides 10, so C=10*1/10=1. But then C=1, B=1, which are not distinct. So B=1 is invalid.Similarly, B=2, 10*B=20. So D divides 20. D=10*A +2 -1=10*A +1. So 10*A +1 divides 20. So possible divisors of 20 greater than or equal to 9 (since A ≥1, D=10*1 +1=11, which is a divisor of 20? 20 divided by 11 is not an integer. Next A=2: D=21, which doesn't divide 20. A=3: D=31, etc. So no solution for B=2.Wait, maybe this approach is too tedious. Let's see.Alternatively, perhaps iterate over possible values of C. Since C is a digit from 0-9, and C cannot be equal to A or B.Also, since in the original equation, when you multiply AB.C by C and round, you get BC. Let's think about the magnitude.The number AB.C is between 10.0 (if A=1, B=0, C=0, but digits must be distinct) up to 99.9. So multiplying by C (digits 0-9) gives a number between 0 (if C=0) up to 899.1 (99.9 *9). However, BC is a two-digit number (10*B + C), so between 10 and 99. Therefore, the product AB.C * C must be between 9.5 and 99.5 (since it rounds to BC, which is 10 to 99). Therefore, AB.C * C must be in [9.5, 99.5). Therefore, AB.C is in [9.5/C, 99.5/C). Since AB.C is at least 10.0 (since A is at least 1, and if B and C are 0, but digits must be distinct. Wait, if A=1, B=0, then C cannot be 0, so the minimum AB.C would be 10.1 (if B=0, C=1). So AB.C is between 10.1 and 99.9. Then, when multiplied by C, which is between 1 and 9 (since if C=0, then the product is 0, which would round to 0, but BC would be 0C which is 00, but digits must be distinct. So C can't be 0, because BC would have B and C both 0, which is invalid. So C must be from 1-9.Therefore, C ∈ {1,2,...,9}, and AB.C * C ≈ BC (two-digit number). So let's note that BC is 10*B + C, so B can be from 0-9, but since BC is a two-digit number, B can be 0, but then BC would be 0C, which is a single-digit number if C is non-zero. Wait, but BC is two digits, so even if B=0, it's written as 0C, but 0C is actually just C, a single-digit number. But in the problem statement, it's written as overline{B C}, which is a two-digit number. Therefore, B cannot be 0 because leading zeros are not allowed in two-digit numbers. Therefore, B must be from 1-9 as well. Therefore, B and C are from 1-9, distinct, and A is from 1-9, distinct from B and C.So all three digits A, B, C are distinct digits from 1-9. Wait, but B can be 0? Wait, the problem says "the number overline{B C}", which is a two-digit number. Therefore, B can't be 0 because overline{B C} would be 0C, which is a single-digit number. Therefore, B must be from 1-9 as well. Therefore, A, B, C are distinct digits from 1-9.Therefore, A, B, C ∈ {1,2,...,9}, all distinct.So, now we can narrow down the possibilities.Let me consider possible values for C. Since the product AB.C * C is approximately BC, which is 10*B + C. Let's note that AB.C is a number between 10.1 and 99.9. Let's see:If C=1, then AB.1 * 1 = AB.1, which rounds to AB.1 ≈ AB. So BC would be AB, but BC is 10*B + C =10*B +1. Therefore, 10*B +1 ≈ AB. But AB is 10*A + B. So 10*A + B ≈10*B +1. Then, 10*A ≈9*B +1. Therefore, 10*A -9*B ≈1. Since A and B are integers from 1-9, let's see if this is possible.Looking for integers A, B where 10*A -9*B =1 (since the difference must be approximately 1, but considering the rounding, perhaps up to 0.5). Let's solve 10*A -9*B =1:10*A =9*B +1Looking for B such that 9*B +1 is divisible by 10. So 9*B ≡ -1 mod10 => 9*B ≡9 mod10 => B ≡1 mod10. So B=1. Then, 10*A=9*1 +1=10 => A=1. But then A=B=1, which is invalid because they must be distinct. So no solution for C=1.Alternatively, if the difference is 0.5, but since 10*A and 9*B are integers, their difference must be an integer. Therefore, the equation 10*A -9*B =1 is the closest we can get, but that gives no valid solution. So C=1 is not possible.Next, C=2.Then, AB.2 *2 ≈ BC=10*B +2. So compute AB.2 *2 =2*(10*A + B +0.2)=20*A +2*B +0.4. This must round to 10*B +2. Therefore, 20*A +2*B +0.4 must be within [10*B +2 -0.5, 10*B +2 +0.5).Therefore:10*B +2 -0.5 ≤20*A +2*B +0.4 <10*B +2 +0.5Simplify:10*B +1.5 ≤20*A +2*B +0.4 <10*B +2.5Subtract 2*B from all parts:10*B -2*B +1.5 ≤20*A +0.4 <10*B -2*B +2.5Which is:8*B +1.5 ≤20*A +0.4 <8*B +2.5Subtract 0.4:8*B +1.1 ≤20*A <8*B +2.1Divide by 20:(8*B +1.1)/20 ≤A <(8*B +2.1)/20Since A is an integer between 1-9, let's compute (8*B +1.1)/20 and (8*B +2.1)/20 for B=1 to 9.For example, B=1:Left side: (8 +1.1)/20=9.1/20≈0.455Right side: (8 +2.1)/20=10.1/20≈0.505So A must be between ~0.455 and ~0.505, but A must be integer ≥1. So no solution for B=1.B=2:Left: (16 +1.1)/20=17.1/20=0.855Right: (16 +2.1)/20=18.1/20=0.905Again, A must be between ~0.855 and ~0.905. No integer A here.B=3:Left: (24 +1.1)/20=25.1/20≈1.255Right: (24 +2.1)/20=26.1/20≈1.305So A must be ≈1.255 to1.305, so A=1. But 1 is in the lower end. But check if A=1 satisfies:Original inequality:8*3 +1.1=25.1 ≤20*1=20 <26.1? 20 is not ≥25.1. So no.B=4:Left: (32 +1.1)/20=33.1/20≈1.655Right: (32 +2.1)/20=34.1/20≈1.705A must be ≈1.655 to1.705, so A=2. Check if A=2:20*2=40. So check:33.1 ≤40 <34.1? No, 40 is not less than34.1. Not valid.B=5:Left: (40 +1.1)/20=41.1/20≈2.055Right: (40 +2.1)/20=42.1/20≈2.105Thus, A must be ≈2.055 to2.105, so A=2. Check:Left:41.1 ≤20*2=40 <42.1? 40 <42.1, but 40 is not ≥41.1. No.B=6:Left: (48 +1.1)/20=49.1/20≈2.455Right: (48 +2.1)/20=50.1/20≈2.505Thus, A≈2.455 to2.505, so A=2. Check:49.1 ≤40 <50.1? 40 <50.1 but 40 is not ≥49.1. No.B=7:Left: (56 +1.1)/20=57.1/20≈2.855Right: (56 +2.1)/20=58.1/20≈2.905Thus, A=3. Check:57.1 ≤60 <58.1? 60 is not <58.1. No.B=8:Left: (64 +1.1)/20=65.1/20≈3.255Right: (64 +2.1)/20=66.1/20≈3.305Thus, A=3. Check:65.1 ≤60 <66.1? 60 is not ≥65.1. No.B=9:Left: (72 +1.1)/20=73.1/20≈3.655Right: (72 +2.1)/20=74.1/20≈3.705Thus, A=4. Check:73.1 ≤80 <74.1? 80 is not <74.1. No.So, C=2 yields no solutions. Moving on to C=3.AB.3 *3 ≈ BC=10*B +3. So compute AB.3 *3 =3*(10*A + B +0.3)=30*A +3*B +0.9. This must round to 10*B +3. Therefore:10*B +3 -0.5 ≤30*A +3*B +0.9 <10*B +3 +0.5Simplify:10*B +2.5 ≤30*A +3*B +0.9 <10*B +3.5Subtract 3*B:7*B +2.5 ≤30*A +0.9 <7*B +3.5Subtract 0.9:7*B +1.6 ≤30*A <7*B +2.6Divide by 30:(7*B +1.6)/30 ≤A <(7*B +2.6)/30Again, iterate over B=1-9.B=1:(7 +1.6)/30=8.6/30≈0.287 ≤A <(7 +2.6)/30=9.6/30≈0.32. No integer A.B=2:(14 +1.6)/30=15.6/30≈0.52 ≤A <16.6/30≈0.553. No.B=3:21 +1.6=22.6/30≈0.753 ≤A <23.6/30≈0.787. No.B=4:28 +1.6=29.6/30≈0.987 ≤A <30.6/30≈1.02. So A=1. Check if A=1:7*4 +1.6=29.6 ≤30*1=30 <7*4 +2.6=30.6. 30 is within [29.6,30.6). So yes. So possible A=1, B=4, C=3. Check if this works.Compute AB.C =14.3. Multiply by 3:14.3*3=42.9. Round to nearest integer:43. BC is 43. So yes! 43=43. So digits A=1, B=4, C=3. All distinct. This seems to be a solution.But let's check if there are other possible B for C=3.B=5:35 +1.6=36.6/30≈1.22 ≤A <37.6/30≈1.253. No integer A.B=6:42 +1.6=43.6/30≈1.453 ≤A <44.6/30≈1.486. No.B=7:49 +1.6=50.6/30≈1.687 ≤A <51.6/30≈1.72. A=2? Check:7*7 +1.6=50.6 ≤30*2=60 <51.6. 60 is not <51.6. No.B=8:56 +1.6=57.6/30≈1.92 ≤A <58.6/30≈1.953. A=2. Check:57.6 ≤60 <58.6. 60 is not <58.6. No.B=9:63 +1.6=64.6/30≈2.153 ≤A <65.6/30≈2.186. A=2. Check:64.6 ≤60 <65.6. 60 is not ≥64.6. No.Thus, only possible solution for C=3 is A=1, B=4, C=3. Let's check the distinctness: A=1, B=4, C=3, all distinct. The product 14.3 *3=42.9, which rounds to 43, which is BC=43. Perfect. This seems valid.But we need to check other values of C to see if there are other solutions. Let's check C=4.AB.4 *4 ≈ BC=10*B +4. Compute AB.4 *4=4*(10*A + B +0.4)=40*A +4*B +1.6. This must round to 10*B +4.Thus:10*B +4 -0.5 ≤40*A +4*B +1.6 <10*B +4 +0.5Simplify:10*B +3.5 ≤40*A +4*B +1.6 <10*B +4.5Subtract 4*B:6*B +3.5 ≤40*A +1.6 <6*B +4.5Subtract 1.6:6*B +1.9 ≤40*A <6*B +2.9Divide by 40:(6*B +1.9)/40 ≤A <(6*B +2.9)/40For B=1 to9:B=1:(6 +1.9)/40≈7.9/40≈0.1975 ≤A <8.9/40≈0.2225. No.B=2:12 +1.9=13.9/40≈0.3475 ≤A <14.9/40≈0.3725. No.B=3:18 +1.9=19.9/40≈0.4975 ≤A <20.9/40≈0.5225. No.B=4:24 +1.9=25.9/40≈0.6475 ≤A <26.9/40≈0.6725. No.B=5:30 +1.9=31.9/40≈0.7975 ≤A <32.9/40≈0.8225. No.B=6:36 +1.9=37.9/40≈0.9475 ≤A <38.9/40≈0.9725. So A=1.Check if A=1:6*6 +1.9=37.9 ≤40*1=40 <38.9? 40 is not <38.9. No.B=7:42 +1.9=43.9/40≈1.0975 ≤A <44.9/40≈1.1225. So A=1.Check:43.9 ≤40 <44.9. 40 is not ≥43.9. No.B=8:48 +1.9=49.9/40≈1.2475 ≤A <50.9/40≈1.2725. A=1. No.B=9:54 +1.9=55.9/40≈1.3975 ≤A <56.9/40≈1.4225. A=1. No.No solutions for C=4.C=5:AB.5 *5 ≈ BC=10*B +5. Compute AB.5 *5=5*(10*A + B +0.5)=50*A +5*B +2.5. This must round to 10*B +5.Therefore:10*B +5 -0.5 ≤50*A +5*B +2.5 <10*B +5 +0.5Simplify:10*B +4.5 ≤50*A +5*B +2.5 <10*B +5.5Subtract 5*B:5*B +4.5 ≤50*A +2.5 <5*B +5.5Subtract 2.5:5*B +2 ≤50*A <5*B +3Divide by50:(5*B +2)/50 ≤A <(5*B +3)/50But since A is integer ≥1, let's check for B=1-9:For any B, (5*B +2)/50 is less than1 (for B=9: 47/50=0.94). So no A≥1 satisfies this. Therefore, no solution for C=5.C=6:AB.6 *6 ≈ BC=10*B +6. Compute AB.6 *6=6*(10*A + B +0.6)=60*A +6*B +3.6. Must round to 10*B +6.Therefore:10*B +6 -0.5 ≤60*A +6*B +3.6 <10*B +6 +0.5Simplify:10*B +5.5 ≤60*A +6*B +3.6 <10*B +6.5Subtract 6*B:4*B +5.5 ≤60*A +3.6 <4*B +6.5Subtract 3.6:4*B +1.9 ≤60*A <4*B +2.9Divide by60:(4*B +1.9)/60 ≤A <(4*B +2.9)/60For B=1-9:B=1:(4 +1.9)/60≈5.9/60≈0.098 ≤A <6.9/60≈0.115. No.B=2:8 +1.9=9.9/60≈0.165 ≤A <10.9/60≈0.181. No.B=3:12 +1.9=13.9/60≈0.231 ≤A <14.9/60≈0.248. No.B=4:16 +1.9=17.9/60≈0.298 ≤A <18.9/60≈0.315. No.B=5:20 +1.9=21.9/60≈0.365 ≤A <22.9/60≈0.381. No.B=6:24 +1.9=25.9/60≈0.431 ≤A <26.9/60≈0.448. No.B=7:28 +1.9=29.9/60≈0.498 ≤A <30.9/60≈0.515. So A≈0.498 to0.515. No.B=8:32 +1.9=33.9/60≈0.565 ≤A <34.9/60≈0.581. No.B=9:36 +1.9=37.9/60≈0.631 ≤A <38.9/60≈0.648. No.Thus, no solutions for C=6.C=7:AB.7 *7 ≈ BC=10*B +7. Compute AB.7 *7=7*(10*A + B +0.7)=70*A +7*B +4.9. Rounded to 10*B +7.Therefore:10*B +7 -0.5 ≤70*A +7*B +4.9 <10*B +7 +0.5Simplify:10*B +6.5 ≤70*A +7*B +4.9 <10*B +7.5Subtract 7*B:3*B +6.5 ≤70*A +4.9 <3*B +7.5Subtract 4.9:3*B +1.6 ≤70*A <3*B +2.6Divide by70:(3*B +1.6)/70 ≤A <(3*B +2.6)/70For B=1-9:B=1:(3 +1.6)/70≈4.6/70≈0.066 ≤A <5.6/70≈0.08. No.B=2:6 +1.6=7.6/70≈0.108 ≤A <8.6/70≈0.123. No.B=3:9 +1.6=10.6/70≈0.151 ≤A <11.6/70≈0.166. No.B=4:12 +1.6=13.6/70≈0.194 ≤A <14.6/70≈0.209. No.B=5:15 +1.6=16.6/70≈0.237 ≤A <17.6/70≈0.251. No.B=6:18 +1.6=19.6/70≈0.28 ≤A <20.6/70≈0.294. No.B=7:21 +1.6=22.6/70≈0.323 ≤A <23.6/70≈0.337. No.B=8:24 +1.6=25.6/70≈0.366 ≤A <26.6/70≈0.38. No.B=9:27 +1.6=28.6/70≈0.409 ≤A <29.6/70≈0.423. No.No solutions for C=7.C=8:AB.8 *8 ≈ BC=10*B +8. Compute AB.8 *8=8*(10*A + B +0.8)=80*A +8*B +6.4. Rounds to 10*B +8.Thus:10*B +8 -0.5 ≤80*A +8*B +6.4 <10*B +8 +0.5Simplify:10*B +7.5 ≤80*A +8*B +6.4 <10*B +8.5Subtract 8*B:2*B +7.5 ≤80*A +6.4 <2*B +8.5Subtract 6.4:2*B +1.1 ≤80*A <2*B +2.1Divide by80:(2*B +1.1)/80 ≤A <(2*B +2.1)/80For B=1-9:B=1:2 +1.1=3.1/80≈0.03875 ≤A <4.1/80≈0.05125. No.B=2:4 +1.1=5.1/80≈0.06375 ≤A <6.1/80≈0.07625. No.B=3:6 +1.1=7.1/80≈0.08875 ≤A <8.1/80≈0.10125. No.B=4:8 +1.1=9.1/80≈0.11375 ≤A <10.1/80≈0.12625. No.B=5:10 +1.1=11.1/80≈0.13875 ≤A <12.1/80≈0.15125. No.B=6:12 +1.1=13.1/80≈0.16375 ≤A <14.1/80≈0.17625. No.B=7:14 +1.1=15.1/80≈0.18875 ≤A <16.1/80≈0.20125. No.B=8:16 +1.1=17.1/80≈0.21375 ≤A <18.1/80≈0.22625. No.B=9:18 +1.1=19.1/80≈0.23875 ≤A <20.1/80≈0.25125. No.No solutions for C=8.C=9:AB.9 *9 ≈ BC=10*B +9. Compute AB.9 *9=9*(10*A + B +0.9)=90*A +9*B +8.1. Rounded to 10*B +9.Therefore:10*B +9 -0.5 ≤90*A +9*B +8.1 <10*B +9 +0.5Simplify:10*B +8.5 ≤90*A +9*B +8.1 <10*B +9.5Subtract 9*B:1*B +8.5 ≤90*A +8.1 <1*B +9.5Subtract 8.1:B +0.4 ≤90*A <B +1.4Divide by90:(B +0.4)/90 ≤A <(B +1.4)/90But since B is from1-9, (B +0.4)/90 ranges from (1.4)/90≈0.015 to(9.4)/90≈0.104. Similarly, upper bound (B +1.4)/90≈0.026 to0.115. Therefore, A must be less than0.115, which is impossible since A≥1. Thus, no solutions for C=9.So, the only solution we found is when C=3, A=1, B=4. Let's double-check that:AB.C =14.3Multiply by C=3:14.3*3=42.9Round to nearest integer:43Which is BC=43. Indeed, B=4, C=3. All digits A=1, B=4, C=3 are distinct. So this satisfies all conditions.Therefore, the distinct digits are A=1, B=4, C=3.Final AnswerThe distinct digits are boxed{A=1}, boxed{B=4}, and boxed{C=3}.</think>Given the problem, we need to determine the distinct digits ( A, B, C ) such that the product of the number ( overline{A B, C} ) (which is ( 10A + B + frac{C}{10} )) and ( C ), rounded to the nearest integer, results in the number ( overline{B C} ) (which is ( 10B + C )).We start by formalizing the problem with the equation:[(10A + B + frac{C}{10}) cdot C approx 10B + C]This implies that the product must be within the interval ([10B + C - 0.5, 10B + C + 0.5)).After analyzing the equation and testing various values for ( C ), we find that the valid solution occurs when ( C = 3 ). For ( C = 3 ), we derive the equation:[(10A + B + 0.3) cdot 3 approx 10B + 3]Testing possible values for ( A ) and ( B ), we find that ( A = 1 ) and ( B = 4 ) satisfy the condition. The calculation is as follows:[overline{A B, C} = 14.3]Multiplying by ( C = 3 ):[14.3 cdot 3 = 42.9]Rounding 42.9 to the nearest integer gives 43, which matches ( overline{B C} = 43 ).Thus, the distinct digits are ( A = 1 ), ( B = 4 ), and ( C = 3 ).[boxed{A=1}, boxed{B=4}, boxed{C=3}]]

🔑:If A, B, C are distinct, then it is obvious that C cannot be 1. B must be at least A cdot C, because there must be at least this many tens in the product, so A and C must be larger than A. If C geq 4, then A can only be 1, because otherwise the product would be greater than 25 cdot 4=100. It cannot be C geq 6, because then the product would be greater than 17 cdot 6>100. Therefore, we need to examine the cases C=2,3,4, and 5.We can write our approximate equality in the form of the following double inequality (allowing both rounding up and down of the tens digit):overline{B C}-0.5 leq overline{A B, C} cdot C leq overline{B C}+0.5Multiplying by 10 and subtracting overline{B C} cdot C:(10-C) cdot overline{B C}-5 leq 100 cdot A cdot C leq (10-C) cdot overline{B C}+5Adding 5 to the first inequality and subtracting 5 from the second, we can then recombine the resulting two inequalities in reverse order:100 cdot A cdot C-5 leq (10-C) cdot overline{B C} leq 100 cdot A cdot C+5Thus, (10-C) cdot overline{B C}, when rounded up or down to the nearest ten, must result in a number divisible by 100, where the number of hundreds is divisible by C. This is impossible for C=5, because then the number of tens in the middle product would be 2 or 7. If C=4, then on one hand A=1, and on the other hand, the middle 6 cdot overline{B 4} must be rounded down, so 6 cdot B must end in 8, meaning B=3 or 8. The former is smaller for smaller C, and with the latter, rounding gives 500 instead of 400.If C=3, the middle 7 cdot overline{B 3} will be rounded to a multiple of 100 when B=4, and indeed 300, which is of the form 100 cdot A cdot C for A=1. Indeed, 14.3 cdot 3=42.9 approx 43, meeting the requirements of the problem.If C=2, 8 cdot overline{B 2}, when rounded to the nearest ten (upward), only gives round hundreds for B=1 or 6. The former is smaller for smaller C, and with the latter, the rounded product is 500, and the number of hundreds is not divisible by C. Thus, the only solution to the problem is A=1, B=4, C=3.