🧠:[<think>Okay, so I have this problem where there's a rectangle ABCD. Points M and N are on sides AB and BC respectively. The areas of triangles AMD, MBN, and NCD are given as a, b, and c. I need to find the area of triangle DMN in terms of a, b, and c. Hmm, let me try to visualize this.First, let me draw a rectangle ABCD. Let's name the corners clockwise: A at the top-left, B at the top-right, C at the bottom-right, and D at the bottom-left. So AB is the top side, BC is the right side, CD is the bottom side, and DA is the left side. Points M and N are on AB and BC. So M is somewhere along the top edge, and N is somewhere along the right edge.Now, the areas given are for triangles AMD, MBN, NCD. Let me mark those. Triangle AMD is formed by points A, M, D. Since A and D are corners of the rectangle, and M is on AB, this triangle is on the left side of the rectangle. Then triangle MBN is formed by M, B, N. That should be a triangle near the top-right corner. Triangle NCD is formed by N, C, D, which would be on the bottom-right side.The goal is to find the area of triangle DMN. So points D, M, N. D is the bottom-left corner, M is on the top side, N is on the right side. So triangle DMN should be somewhere in the middle of the rectangle, maybe overlapping with the other triangles?Since this involves areas in a rectangle, maybe coordinate geometry would help here. Let me assign coordinates to the rectangle. Let's let A be at (0, h), B at (w, h), C at (w, 0), and D at (0, 0). So the rectangle has width w and height h. Then point M is on AB, so its coordinates would be somewhere between A (0, h) and B (w, h). Let's denote M as (m, h), where 0 < m < w. Similarly, point N is on BC, which goes from B (w, h) to C (w, 0). So N would have coordinates (w, n), where 0 < n < h.Now, let's express the areas a, b, c in terms of m, n, w, h.First, triangle AMD. Points A (0, h), M (m, h), D (0, 0). The area can be calculated using the formula for the area of a triangle given coordinates. Alternatively, since AMD is a triangle with base along AD, which is vertical from (0, h) to (0, 0). But point M is on AB, which is horizontal. Wait, actually, triangle AMD is formed by points A, M, D. Let me plot this: A is top-left, M is somewhere on the top edge, D is bottom-left. So the triangle AMD would have vertices at (0, h), (m, h), (0, 0). The base of this triangle is AD, which is length h, but the height is m, the distance from A to M along AB? Wait, maybe not. Let me use the coordinates to calculate the area.Using coordinates for A (0, h), M (m, h), D (0, 0). The area of triangle AMD can be calculated using the shoelace formula:Area = (1/2)| (0*(h - 0) + m*(0 - h) + 0*(h - h) ) | = (1/2)|0 + m*(-h) + 0| = (1/2)| -mh | = (1/2)mh.So area a = (1/2)mh.Next, triangle MBN. Points M (m, h), B (w, h), N (w, n). Let's compute its area. Using coordinates:Area = (1/2)|m*(h - n) + w*(n - h) + w*(h - h)|. Wait, maybe better to use base and height here. Since M is (m, h), B is (w, h), N is (w, n). The triangle is a right triangle? Wait, no. Let's see. From M to B is horizontal to the right, then from B to N is vertical down. So triangle MBN is a right triangle with base (w - m) and height (h - n). Therefore, area b = (1/2)(w - m)(h - n).Alternatively, using shoelace formula:Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Plugging in:x1 = m, y1 = hx2 = w, y2 = hx3 = w, y3 = nArea = (1/2)|m(h - n) + w(n - h) + w(h - h)|= (1/2)|m(h - n) + w(n - h) + 0|= (1/2)| (mh - mn) + (wn - wh) |= (1/2)| mh - mn + wn - wh |.Hmm, but this seems more complicated. Alternatively, since MBN is a triangle where two sides are horizontal and vertical. From M (m, h) to B (w, h) is a horizontal line of length (w - m), and from B (w, h) to N (w, n) is a vertical line of length (h - n). The area of the triangle formed by these two legs is (1/2)*(w - m)*(h - n). So that's simpler. So area b = (1/2)(w - m)(h - n).Third triangle, NCD. Points N (w, n), C (w, 0), D (0, 0). Let's compute its area. This triangle has vertices at (w, n), (w, 0), (0, 0). The base can be along DC from (0,0) to (w, 0), which is length w, and the height would be n (the vertical distance from N to DC). So area c = (1/2)*w*n.Alternatively, using shoelace:Area = (1/2)|w*(0 - 0) + w*(0 - n) + 0*(n - 0)| = (1/2)|0 + (-wn) + 0| = (1/2)wn. So that's c = (1/2)wn.So now we have:a = (1/2)mhb = (1/2)(w - m)(h - n)c = (1/2)wnWe need to find the area of triangle DMN. Let's get the coordinates for D (0,0), M (m, h), N (w, n). Let's compute its area using shoelace formula.Area DMN = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Plugging in:x1 = 0, y1 = 0x2 = m, y2 = hx3 = w, y3 = nArea = (1/2)|0*(h - n) + m*(n - 0) + w*(0 - h)|= (1/2)|0 + mn - wh|= (1/2)|mn - wh|But area can't be negative, so take absolute value:Area DMN = (1/2)|mn - wh|But we need to express this in terms of a, b, c. So let's see if we can express mn and wh in terms of a, b, c.From the previous equations:a = (1/2)mh => mh = 2ab = (1/2)(w - m)(h - n) => (w - m)(h - n) = 2bc = (1/2)wn => wn = 2cWe need to find mn and wh. Let's see:From c = (1/2)wn => wn = 2c. So that's one equation.From a = (1/2)mh => mh = 2a.We need to find mn and wh. Wait, wh is the product of width and height of the rectangle. Maybe we can express wh in terms of other variables.Wait, the total area of the rectangle is wh. Also, the sum of the areas of all four triangles (AMD, MBN, NCD, DMN) should be equal to the area of the rectangle. Wait, is that the case?Wait, but in the problem statement, they mention three triangles with areas a, b, c, and we need to find the fourth triangle DMN. But are there only four triangles? Let me check.Original rectangle ABCD. Points M on AB and N on BC. Then the triangles mentioned are AMD, MBN, NCD, and DMN. Are these four triangles covering the entire area of the rectangle? Let's see:If you take AMD (top-left triangle), MBN (top-right triangle), NCD (bottom-right triangle), and DMN (central triangle), maybe they partition the rectangle? Hmm, maybe not exactly. Let me check with an example.Suppose the rectangle is divided into these four triangles. Depending on where M and N are, these triangles might overlap or leave some regions. Wait, perhaps not. Let me think.If you connect D to M and N, M to N, then perhaps the rectangle is divided into four triangles and a quadrilateral? Hmm, maybe not. Wait, actually, if you draw DM, MN, and ND, then you might split the rectangle into different regions. This is getting confusing. Maybe better to calculate the total area of the four triangles and see if it equals wh.Calculating total area: a + b + c + area DMN = (1/2)mh + (1/2)(w - m)(h - n) + (1/2)wn + (1/2)|mn - wh|But if the four triangles partition the rectangle, their areas should sum to wh. Let's check with sample values. Let me take specific values for m, n, w, h.Suppose the rectangle is 2x2, so w=2, h=2. Let m=1, n=1. Then:a = (1/2)*1*2 = 1b = (1/2)*(2 - 1)*(2 - 1) = (1/2)*1*1 = 0.5c = (1/2)*2*1 = 1Area DMN would be (1/2)|1*1 - 2*2| = (1/2)|1 - 4| = (1/2)(3) = 1.5Total area: 1 + 0.5 + 1 + 1.5 = 4, which is equal to wh = 4. So in this case, the four triangles do partition the rectangle. Therefore, in general, a + b + c + area DMN = wh.Therefore, area DMN = wh - (a + b + c). But we need to express area DMN in terms of a, b, c. However, wh is the area of the rectangle. But how can we express wh in terms of a, b, c?Wait, we have:From a = (1/2)mh => m = 2a/hFrom c = (1/2)wn => n = 2c/wFrom b = (1/2)(w - m)(h - n). Let's substitute m and n from above:(w - m) = w - 2a/h(h - n) = h - 2c/wTherefore, b = (1/2)(w - 2a/h)(h - 2c/w)Multiply out the terms:(w - 2a/h)(h - 2c/w) = w*h - w*(2c/w) - 2a/h*h + (2a/h)*(2c/w)Simplify each term:First term: whSecond term: -2cThird term: -2aFourth term: (4ac)/(hw)Therefore, expanding:wh - 2c - 2a + (4ac)/(hw) = 2b*2 (since b = (1/2) of the left side)Wait, actually, (w - 2a/h)(h - 2c/w) = 2b*2? Wait, wait. Wait, b = (1/2)(w - m)(h - n) = (1/2)(w - 2a/h)(h - 2c/w). Therefore:(w - 2a/h)(h - 2c/w) = 2bSo expanding the left side:wh - 2c - 2a + (4ac)/(hw) = 2bTherefore:wh = 2b + 2c + 2a - (4ac)/(hw)Hmm, but this equation still has wh on both sides, which complicates things. Let's rearrange:wh - (4ac)/(hw) = 2a + 2b + 2cMultiply both sides by hw to eliminate the denominator:(wh)^2 - 4ac = 2a hw + 2b hw + 2c hwLet me denote S = wh (the area of the rectangle). Then the equation becomes:S^2 - 4ac = 2aS + 2bS + 2cSSimplify the right side:2aS + 2bS + 2cS = 2S(a + b + c)Therefore:S^2 - 4ac = 2S(a + b + c)Bring all terms to one side:S^2 - 2S(a + b + c) - 4ac = 0This is a quadratic equation in S:S^2 - 2(a + b + c)S - 4ac = 0We can solve for S using the quadratic formula:S = [2(a + b + c) ± sqrt{4(a + b + c)^2 + 16ac}]/2Simplify:S = (a + b + c) ± sqrt{(a + b + c)^2 + 4ac}But since S is the area of the rectangle, it must be positive. So we take the positive root:S = (a + b + c) + sqrt{(a + b + c)^2 + 4ac}Wait, but this seems complicated. Let me check if this approach is correct.Alternatively, maybe there's a better way to express area DMN in terms of a, b, c.Earlier, we had that area DMN = wh - (a + b + c). So if we can express wh in terms of a, b, c, then we can get the desired area.Alternatively, perhaps there's a relation between the variables.Wait, from a = (1/2)mh and c = (1/2)wn, so m = 2a/h and n = 2c/w.Then, mn = (2a/h)(2c/w) = 4ac/(hw) = 4ac/S, where S = wh.Then, area DMN is (1/2)|mn - wh| = (1/2)|4ac/S - S|But from the previous equation, wh = S = (a + b + c) + sqrt{(a + b + c)^2 + 4ac}But this seems too involved. Maybe another approach.Let me consider the equations again.We have:a = (1/2)mh --> m = 2a/hc = (1/2)wn --> n = 2c/wb = (1/2)(w - m)(h - n)Substitute m and n:b = (1/2)(w - 2a/h)(h - 2c/w)Let me multiply both sides by 2:2b = (w - 2a/h)(h - 2c/w)Expand the right-hand side:wh - 2c - 2a + (4ac)/(wh)So:2b = wh - 2c - 2a + (4ac)/(wh)Let me rearrange:wh + (4ac)/(wh) = 2a + 2b + 2cLet me denote S = wh, then:S + (4ac)/S = 2(a + b + c)Multiply both sides by S:S^2 + 4ac = 2(a + b + c)SBring all terms to left-hand side:S^2 - 2(a + b + c)S + 4ac = 0Ah, earlier I had a sign error. The correct quadratic equation is:S^2 - 2(a + b + c)S + 4ac = 0Now, solving this quadratic for S:S = [2(a + b + c) ± sqrt{4(a + b + c)^2 - 16ac}]/2Simplify discriminant:sqrt{4[(a + b + c)^2 - 4ac]} = 2sqrt{(a + b + c)^2 - 4ac}Therefore:S = [2(a + b + c) ± 2sqrt{(a + b + c)^2 - 4ac}]/2 = (a + b + c) ± sqrt{(a + b + c)^2 - 4ac}Now, since S is the area of the rectangle, which is positive, and given that (a + b + c) is positive, we need to determine which sign to take. Let's test with the sample values.Earlier, when w=2, h=2, m=1, n=1, we had a=1, b=0.5, c=1. Then:(a + b + c) = 1 + 0.5 + 1 = 2.5sqrt{(2.5)^2 - 4*1*1} = sqrt{6.25 - 4} = sqrt{2.25} = 1.5Thus, S = 2.5 ± 1.5. So either 4 or 1. But the actual S was 4. Therefore, we take the positive sign:S = (a + b + c) + sqrt{(a + b + c)^2 - 4ac}But wait, in this case, (a + b + c) = 2.5, sqrt{...} = 1.5, so S = 2.5 + 1.5 = 4, which is correct. If we took the negative sign, 2.5 - 1.5 = 1, which is wrong. So indeed, the correct S is (a + b + c) + sqrt{(a + b + c)^2 - 4ac}Therefore, area DMN = S - (a + b + c) = [ (a + b + c) + sqrt{(a + b + c)^2 - 4ac} ] - (a + b + c) = sqrt{(a + b + c)^2 - 4ac}But wait, in our sample case, sqrt{(2.5)^2 - 4*1*1} = sqrt{6.25 - 4} = sqrt{2.25} = 1.5, which matches the area DMN we had earlier (1.5). So that works.But wait, area DMN is equal to sqrt{(a + b + c)^2 - 4ac} ?But let me check if that's the case in another example.Let me take another example. Let’s suppose the rectangle is 4x4 (w=4, h=4). Let m=2, so a = (1/2)*2*4=4. Let n=2, so c=(1/2)*4*2=4. Then, compute b:b=(1/2)(4 - 2)(4 - 2)= (1/2)(2)(2)=2. So a=4, b=2, c=4.Then, area DMN should be sqrt{(4+2+4)^2 -4*4*4}= sqrt{10^2 - 64}=sqrt{100 -64}=sqrt{36}=6. Then total area S=10 +6=16, which is correct (4x4). Then area DMN is 6. Let me verify using coordinates:Points D(0,0), M(2,4), N(4,2). Area DMN using shoelace:Area = (1/2)|0*(4 - 2) + 2*(2 - 0) +4*(0 -4)|= (1/2)|0 + 4 -16| = (1/2)|-12| =6. Correct.So yes, in this case, the formula holds. So area DMN = sqrt{(a + b + c)^2 -4ac}But wait, in the first example, (a + b + c)=1+0.5+1=2.5, (a + b + c)^2=6.25, 4ac=4*1*1=4, sqrt{6.25 -4}=sqrt{2.25}=1.5. Correct. And in the second example, sqrt{(10)^2 - 4*4*4}=sqrt{36}=6. Correct.Therefore, the area of triangle DMN is sqrt{(a + b + c)^2 -4ac}But the problem says "Express in terms of a,b,c the area...", so probably the answer is sqrt{(a + b + c)^2 - 4ac}But let me check with another example where the area DMN is not overlapping.Wait, but all the examples I tried so far, the four triangles partition the rectangle, so area DMN is S - (a + b + c). But according to our quadratic solution, area DMN = sqrt{(a + b + c)^2 -4ac}But in the first example, S - (a + b + c) = 4 - 2.5 =1.5, which is equal to sqrt{(2.5)^2 -4*1*1}=1.5. So indeed, the area DMN is sqrt{(a + b + c)^2 -4ac}Wait, but how is that possible? Because S = wh, and area DMN = S - (a + b + c). But according to the quadratic equation, S = (a + b + c) + sqrt{(a + b + c)^2 -4ac}Therefore, area DMN = sqrt{(a + b + c)^2 -4ac}Therefore, the answer is sqrt{(a + b + c)^2 -4ac}But let me check another example where maybe m and n are different.Let’s take rectangle w=3, h=3. Let m=1, so a=(1/2)*1*3=1.5. Let n=1, so c=(1/2)*3*1=1.5. Then compute b=(1/2)(3 -1)(3 -1)=(1/2)(2)(2)=2. So a=1.5, b=2, c=1.5. Then area DMN should be sqrt{(1.5+2+1.5)^2 -4*1.5*1.5}=sqrt{5^2 -9}=sqrt{25-9}=sqrt{16}=4. Then total area S=5 +4=9, which is 3x3. Correct.Compute area DMN using coordinates: D(0,0), M(1,3), N(3,1). Shoelace formula:Area = (1/2)|0*(3 -1) +1*(1 -0) +3*(0 -3)| = (1/2)|0 +1 -9|=(1/2)|-8|=4. Correct.So, yes, the formula holds. Therefore, the area of triangle DMN is sqrt{(a + b + c)^2 -4ac}But wait, in the problem statement, they mention "the area of the triangle DMN". So the answer is sqrt{(a + b + c)^2 -4ac}But let me check if the problem expects an algebraic expression without square roots. Wait, in the examples above, the answer does involve a square root. But maybe there is another way to express it. Alternatively, perhaps factoring.Wait, let's see. (a + b + c)^2 -4ac = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc -4ac = a^2 + b^2 + c^2 + 2ab -2ac + 2bc. Doesn't seem to factor nicely.Alternatively, maybe we can express it as (a + c)^2 + b^2 + 2ab + 2bc -2ac. Not sure. Alternatively, maybe write as (a - c)^2 + b^2 + 2ab + 2bc. Still not helpful.Alternatively, maybe the answer is simply expressed with a square root. The problem says "Express in terms of a,b,c", so maybe it's acceptable. Alternatively, maybe the problem expects a different form. Wait, let me think again.Wait, the area of triangle DMN is |(1/2)(mn - wh)|. From earlier steps:Area DMN = (1/2)|mn - wh|But from above, S = wh = (a + b + c) + sqrt{(a + b + c)^2 -4ac}Therefore, mn = 4ac/S (from mn = 4ac/wh as established earlier)Therefore, mn - wh = (4ac/S) - SBut area DMN = (1/2)| (4ac/S - S) | = (1/2)| -(S^2 -4ac)/S | = (1/2)(S^2 -4ac)/S, since S is positive.But S^2 -4ac = 2(a + b + c)S -4ac from the quadratic equation.Wait, but S^2 -4ac = 2(a + b + c)S -4ac? Wait, no. From the quadratic equation:S^2 -2(a + b + c)S +4ac =0 => S^2 = 2(a + b + c)S -4acThus, S^2 -4ac = 2(a + b + c)S -8acWait, not sure. Maybe not helpful.Alternatively, Area DMN = (1/2)|mn - wh| = (1/2)| (4ac/S) - S | = (1/2)| (4ac - S^2)/S |. But from quadratic equation, S^2 = 2(a + b + c)S -4ac, so 4ac - S^2 = -2(a + b + c)S +8ac. Hmm, messy.Alternatively, since Area DMN = sqrt{(a + b + c)^2 -4ac}, as we saw from multiple examples. So, unless there's a simpler expression, this might be the answer.But let me check. In the problem statement, they use "Let a,b,c be the area...", and the answer is supposed to be in terms of a,b,c. If this is a competition-style problem, the answer might involve a square root, but maybe there's a way to write it without a square root. Alternatively, perhaps the area is (sqrt{a} + sqrt{c})^2 - b, but that's a guess. Let me test with first example:First example: a=1, b=0.5, c=1. Then (sqrt{1} + sqrt{1})^2 -0.5= (2)^2 -0.5=4 -0.5=3.5, which is not equal to 1.5. So no.Alternatively, maybe (a + c) ± something. But in our first example, a + c =2, sqrt{(a + b + c)^2 -4ac}=sqrt{6.25 -4}=1.5. So 1.5 is sqrt{(a + b + c)^2 -4ac}Alternatively, maybe it's (a + b + c) - 2sqrt{ac}, but in first example, 2.5 -2*1=0.5, which is not 1.5. Doesn't work.Alternatively, maybe the area is adjoined by a quadratic formula. Given the earlier steps, the answer must be sqrt{(a + b + c)^2 -4ac}But let me check another resource or similar problem.Wait, this seems similar to the problem where three triangles' areas are given in a rectangle, and you need to find the fourth. The formula derived here is sqrt{(sum of areas)^2 -4ac}. This seems plausible.Alternatively, in some textbooks, similar problems result in the area being sqrt{(a + c)^2 + b^2} or something, but in our case, it's sqrt{(a + b + c)^2 -4ac}Alternatively, maybe the answer is simply (a + c) + b - 2sqrt{ac}, but in first example, 1 +1 +0.5 -2*1=0.5, which is not 1.5.Alternatively, maybe the problem expects an expression involving a, b, c without square roots, but I can't see how. Given the examples and the derivation, the answer is sqrt{(a + b + c)^2 -4ac}Alternatively, maybe the answer is (sqrt{a} + sqrt{c})^2 - b. Let's check first example: (1 +1)^2 -0.5=4 -0.5=3.5≠1.5. No.Alternatively, (sqrt{a} - sqrt{c})^2 + b. (1 -1)^2 +0.5=0 +0.5=0.5≠1.5. No.Alternatively, another approach. Maybe use variables substitution.Let’s suppose that the area DMN is x. Then, from the earlier equation:x = wh - (a + b + c)But we need to express x in terms of a,b,c. From the quadratic equation:wh = (a + b + c) + sqrt{(a + b + c)^2 -4ac}Therefore:x = sqrt{(a + b + c)^2 -4ac}Hence, the area DMN is sqrt{(a + b + c)^2 -4ac}Alternatively, maybe factor it as sqrt{(a + b + c - 2sqrt{ac})(a + b + c + 2sqrt{ac})}But unless there's a reason to factor it, it's better left as sqrt{(a + b + c)^2 -4ac}Therefore, I think the answer is sqrt{(a + b + c)^2 -4ac}But let me check one more example to confirm.Let’s take a rectangle with w=1, h=1. Let m=0.5, so a=(1/2)*0.5*1=0.25. Let n=0.5, so c=(1/2)*1*0.5=0.25. Then compute b=(1/2)(1 -0.5)(1 -0.5)= (1/2)(0.5)(0.5)=0.125. Therefore, a=0.25, b=0.125, c=0.25. Then area DMN should be sqrt{(0.25 +0.125 +0.25)^2 -4*0.25*0.25}=sqrt{0.625^2 -0.25}=sqrt{0.390625 -0.25}=sqrt{0.140625}=0.375. Now, compute area DMN via coordinates: D(0,0), M(0.5,1), N(1,0.5). Shoelace formula:Area = (1/2)|0*(1 -0.5) +0.5*(0.5 -0) +1*(0 -1)|= (1/2)|0 +0.25 -1|=(1/2)|-0.75|=0.375. Correct.Therefore, the formula holds again.Therefore, the answer should be boxed{sqrt{(a + b + c)^2 - 4ac}}Final AnswerThe area of triangle ( DMN ) is boxed{sqrt{(a + b + c)^2 - 4ac}}.</think>Given a rectangle (ABCD) with points (M) on ([AB]) and (N) on ([BC]), we need to find the area of triangle (DMN) in terms of the areas (a), (b), and (c) of triangles (AMD), (MBN), and (NCD) respectively.1. Assigning Coordinates: - (A(0, h)), (B(w, h)), (C(w, 0)), (D(0, 0)) - (M(m, h)) on (AB) and (N(w, n)) on (BC)2. Calculating Areas: - Area of (triangle AMD): (a = frac{1}{2}mh) - Area of (triangle MBN): (b = frac{1}{2}(w - m)(h - n)) - Area of (triangle NCD): (c = frac{1}{2}wn)3. Area of Rectangle: - The total area of the rectangle (ABCD) is (wh).4. Quadratic Equation: - Using the areas (a), (b), and (c), we derive the quadratic equation for the area (S = wh): [ S^2 - 2(a + b + c)S + 4ac = 0 ] - Solving this quadratic equation, we get: [ S = (a + b + c) + sqrt{(a + b + c)^2 - 4ac} ]5. Area of Triangle (DMN): - The area of triangle (DMN) is the difference between the area of the rectangle and the sum of the areas (a), (b), and (c): [ text{Area of } triangle DMN = sqrt{(a + b + c)^2 - 4ac} ]Thus, the area of triangle (DMN) is (boxed{sqrt{(a + b + c)^2 - 4ac}}).,<think>Alright, let's tackle this geometry problem. Hmm, so we have a rectangle ABCD, with points M on side AB and N on side BC. The areas of triangles AMD, MBN, and NCD are given as a, b, and c respectively. We need to find the area of triangle DMN in terms of a, b, and c. First, let me visualize the rectangle. Let me label the rectangle ABCD such that AB and CD are the top and bottom sides, and BC and AD are the right and left sides. So, A is the bottom-left corner, B is the bottom-right, C is the top-right, and D is the top-left. Points M are on AB, so between A and B, and N is on BC, between B and C.Let me draw this mentally. So, ABCD is a rectangle, M is somewhere along AB, closer to A or B, and N is somewhere along BC, closer to B or C. The triangles in question are AMD, MBN, NCD, and DMN. Since it's a rectangle, all angles are 90 degrees, and opposite sides are equal. Let's denote the lengths of the sides. Let’s suppose that AB has length L (so BC would be the width, let's say W). Wait, but maybe assigning variables to the sides will complicate things. Alternatively, maybe coordinate geometry can help here.Yes, coordinate geometry might be a good approach here. Let's assign coordinates to the rectangle. Let's place point A at (0, 0), so then B would be at (x, 0) since AB is horizontal. Then C would be at (x, y), and D would be at (0, y), where x and y are the length and width of the rectangle. Then points M and N can be expressed in terms of coordinates.Point M is on AB, so its coordinates would be somewhere between A(0,0) and B(x,0). Let's denote M as (m, 0), where 0 ≤ m ≤ x. Similarly, point N is on BC, which goes from B(x,0) to C(x,y). So N would have coordinates (x, n), where 0 ≤ n ≤ y.Now, we need to express the areas of the triangles AMD, MBN, NCD, and DMN in terms of these coordinates.First, triangle AMD. Points A(0,0), M(m,0), D(0,y). The area of triangle AMD can be calculated using the formula for the area of a triangle with coordinates. Alternatively, since AMD is a triangle with base AM and height AD. Wait, AMD is a triangle with vertices at A, M, D. Let me compute its area.The base can be AM, which has length m, and the height would be the distance from D to the base AM. But since AMD is a right triangle? Wait, A is at (0,0), M is at (m,0), and D is at (0,y). So, connecting these points, the triangle AMD has vertices at (0,0), (m,0), and (0,y). So this is a right triangle with legs of length m and y. Therefore, its area is (1/2)*m*y. So area a = (1/2)*m*y.Similarly, triangle MBN. Points M(m,0), B(x,0), N(x,n). Let's compute its area. Again, coordinates: M(m,0), B(x,0), N(x,n). This is also a right triangle. The base can be MB, which has length x - m, and the height is BN, which has length n. So area b = (1/2)*(x - m)*n.Next, triangle NCD. Points N(x,n), C(x,y), D(0,y). The triangle NCD. Coordinates: N(x,n), C(x,y), D(0,y). Let me compute the area. Since this is a triangle with vertices at (x,n), (x,y), and (0,y). The base could be NC, which is vertical from (x,n) to (x,y), so length y - n. The horizontal distance from D to the line NC is x, so the area would be (1/2)*x*(y - n). So area c = (1/2)*x*(y - n).We need to find the area of triangle DMN. Points D(0,y), M(m,0), N(x,n). Let's compute the coordinates. So, D is (0,y), M is (m,0), N is (x,n). To find the area of triangle DMN, we can use the shoelace formula.The shoelace formula for three points (x1,y1), (x2,y2), (x3,y3) is:Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Applying this to D(0,y), M(m,0), N(x,n):Area = (1/2)|0*(0 - n) + m*(n - y) + x*(y - 0)|Simplify:= (1/2)|0 + m(n - y) + x*y|= (1/2)|m(n - y) + x*y|Since areas are positive, we can drop the absolute value:= (1/2)( -m(y - n) + x*y )= (1/2)(x*y - m(y - n))So the area of triangle DMN is (1/2)(x*y - m(y - n)).But we need to express this in terms of a, b, and c. So we need to relate x, y, m, n to a, b, c.From earlier:a = (1/2)*m*y => m*y = 2a => m = (2a)/yb = (1/2)*(x - m)*n => (x - m)*n = 2b => x - m = 2b/nc = (1/2)*x*(y - n) => x*(y - n) = 2c => x = 2c/(y - n)So we have expressions for m, x - m, and x in terms of y, n, a, b, c.Wait, but we need to solve for variables in terms of a, b, c. Let's see.First, from a = (1/2)*m*y => m = (2a)/yFrom c = (1/2)*x*(y - n) => x = (2c)/(y - n)From b = (1/2)*(x - m)*n => x - m = (2b)/nSo, substituting x from the expression above into x - m = (2b)/n:(2c)/(y - n) - (2a)/y = (2b)/nLet me write this equation:(2c)/(y - n) - (2a)/y = (2b)/nDivide both sides by 2:c/(y - n) - a/y = b/nSo we have:c/(y - n) = a/y + b/nThis is an equation in variables y and n, but we need to solve for y and n in terms of a, b, c. Alternatively, maybe we can find expressions that allow us to compute the area of DMN, which is (1/2)(x*y - m(y - n)).Wait, let's see. Let me compute x*y - m(y - n):From x = 2c/(y - n), so x*y = (2c/(y - n))*y = 2c*y/(y - n)From m = 2a/y, so m(y - n) = (2a/y)*(y - n) = 2a*(y - n)/yThus, x*y - m(y - n) = 2c*y/(y - n) - 2a*(y - n)/ySo the area of DMN is (1/2)*(2c*y/(y - n) - 2a*(y - n)/y) = (c*y/(y - n) - a*(y - n)/y)Hmm, this seems complicated. Maybe there's a better way.Alternatively, perhaps express the area of the rectangle ABCD. The total area of the rectangle is x*y. If we sum up the areas of the four triangles AMD, MBN, NCD, and DMN, do they cover the entire rectangle? Let's check.Triangles AMD, MBN, NCD, and DMN. Wait, in a rectangle, connecting points M and N would divide the rectangle into four regions: AMD, MBN, NCD, and DMN? Wait, no. Actually, connecting M and N would divide the rectangle into different regions. Let me think.Actually, if you have a rectangle ABCD with points M on AB and N on BC, then connecting D to M and M to N and N to D would form triangle DMN, but also there are other regions. The areas given are AMD, MBN, NCD. So perhaps the four regions AMD, MBN, NCD, and DMN make up the entire rectangle? Let's check.If you add up the areas of AMD, MBN, NCD, and DMN, do they sum up to the area of the rectangle? Let's check:Area of AMD: a = (1/2)*m*yArea of MBN: b = (1/2)*(x - m)*nArea of NCD: c = (1/2)*x*(y - n)Area of DMN: ?So total area would be a + b + c + DMN.But the area of the rectangle is x*y. So:a + b + c + DMN = x*yHence, DMN = x*y - (a + b + c)But wait, from the shoelace formula earlier, we had Area of DMN = (1/2)(x*y - m(y - n))But according to this, if DMN = x*y - (a + b + c), then:x*y - (a + b + c) = (1/2)(x*y - m(y - n))Wait, but this seems contradictory. Let me check my reasoning.Wait, maybe the four regions AMD, MBN, NCD, and DMN do not cover the entire rectangle. Maybe there's an overlapping area or another region. Let me visualize again.Point M is on AB, point N is on BC. So connecting D to M, M to N, and N to D forms triangle DMN. Then, the other triangles:- AMD: triangle formed by A, M, D.- MBN: triangle formed by M, B, N.- NCD: triangle formed by N, C, D.So, these four triangles: AMD, MBN, NCD, DMN. Do they partition the rectangle? Let me see.Starting from point A, triangle AMD covers the lower-left corner. Then, triangle MBN is near the corner B. Triangle NCD is near the upper-right corner. And triangle DMN is in the middle. Hmm, perhaps they do partition the rectangle. Let me check with coordinates.The rectangle has four corners. The regions:- AMD: covers from A to M to D.- MBN: covers from M to B to N.- NCD: covers from N to C to D.- DMN: covers from D to M to N.So, when you put these together, do they cover the entire rectangle without overlapping? Let's see. If we take each triangle:AMD: the area from A up to M along AB, then up to D.MBN: the area from M to B to N.NCD: from N to C to D.DMN: connecting D to M to N.Hmm, actually, I think there might be a quadrilateral area left uncovered. Wait, perhaps not. Let me think. If we imagine the lines DM, MN, and ND forming triangle DMN, then the rest of the rectangle is covered by AMD, MBN, and NCD.Alternatively, maybe it's better to check the total area.Sum of the areas:a + b + c + DMN = (1/2)*m*y + (1/2)*(x - m)*n + (1/2)*x*(y - n) + DMNLet’s compute this sum:= (1/2)my + (1/2)(x - m)n + (1/2)x(y - n) + DMN= (1/2)my + (1/2)xn - (1/2)mn + (1/2)xy - (1/2)xn + DMNSimplify terms:The (1/2)xn and -(1/2)xn cancel out.Left with: (1/2)my - (1/2)mn + (1/2)xy + DMN= (1/2)my - (1/2)mn + (1/2)xy + DMNFactor terms:= (1/2)xy + (1/2)m(y - n) + DMNBut the area of the rectangle is xy. So:If the sum of the areas is equal to xy, then:(1/2)xy + (1/2)m(y - n) + DMN = xyThus:DMN = xy - (1/2)xy - (1/2)m(y - n) = (1/2)xy - (1/2)m(y - n)Which matches the earlier result from the shoelace formula:Area of DMN = (1/2)(xy - m(y - n))So, DMN = (1/2)xy - (1/2)m(y - n)But we need to express DMN in terms of a, b, c.We have a = (1/2)my => my = 2ac = (1/2)x(y - n) => x(y - n) = 2cFrom the earlier equation:c/(y - n) - a/y = b/nWait, maybe we can find expressions for my and x(y - n) from a and c, then relate variables.Alternatively, since DMN = (1/2)xy - (1/2)m(y - n), but also from the rectangle area:DMN = xy - (a + b + c)Wait, earlier I thought that sum of the areas might be equal to the rectangle area, but according to the calculation, the sum a + b + c + DMN = (1/2)xy + (1/2)m(y - n) + DMN, which leads to DMN = (1/2)xy - (1/2)m(y - n). But if DMN = xy - (a + b + c), then:xy - (a + b + c) = (1/2)xy - (1/2)m(y - n)Thus:(1/2)xy = (a + b + c) - (1/2)m(y - n)But maybe this is not helpful.Alternatively, let's note that:From a = (1/2)my => my = 2aFrom c = (1/2)x(y - n) => x(y - n) = 2cSo, DMN = (1/2)(xy - m(y - n)) = (1/2)xy - (1/2)m(y - n)But:xy = ?We can express xy in terms of a, b, c. Wait, my = 2a, so m = 2a/yFrom b = (1/2)(x - m)n => (x - m)n = 2b => x - m = 2b/n => x = m + 2b/nFrom c = (1/2)x(y - n) => x(y - n) = 2c => x = 2c/(y - n)Therefore, equating the two expressions for x:m + 2b/n = 2c/(y - n)But m = 2a/y, so:2a/y + 2b/n = 2c/(y - n)Divide both sides by 2:a/y + b/n = c/(y - n)This is the same equation as before. So we have:a/y + b/n = c/(y - n)Let me rearrange this equation:Multiply both sides by y*n*(y - n) to eliminate denominators:a*n*(y - n) + b*y*(y - n) = c*y*nExpand the left-hand side:a*n*y - a*n^2 + b*y^2 - b*y*n = c*y*nBring all terms to one side:a*n*y - a*n^2 + b*y^2 - b*y*n - c*y*n = 0Combine like terms:(a*n*y - b*y*n - c*y*n) + b*y^2 - a*n^2 = 0Factor y*n from the first group:y*n*(a - b - c) + b*y^2 - a*n^2 = 0Hmm, this seems complicated. Maybe there's a substitution we can do here.Alternatively, let's consider variables ratio.Let’s set variables such that we can express y and n in terms of a, b, c.But this might not be straightforward. Alternatively, perhaps express y and n as variables and solve the equation.Wait, maybe we can take the equation:a/y + b/n = c/(y - n)Let’s set k = y/n, so y = k*n. Let's substitute into the equation:a/(k*n) + b/n = c/(k*n - n) = c/(n(k - 1))Multiply both sides by n:a/k + b = c/(k - 1)Multiply both sides by k(k - 1):a(k - 1) + b*k(k - 1) = c*kExpand:a*k - a + b*k^2 - b*k = c*kBring all terms to one side:b*k^2 + (a - b - c)*k - a = 0This is a quadratic equation in terms of k. Let's write it as:b*k^2 + (a - b - c)*k - a = 0We can solve for k using the quadratic formula:k = [ - (a - b - c) ± sqrt( (a - b - c)^2 + 4ab ) ] / (2b)But this seems messy and perhaps not helpful for expressing DMN in terms of a, b, c. Maybe there's another approach.Wait, let's recall that DMN = (1/2)(xy - m(y - n)). We need to express this in terms of a, b, c.We know:From a = (1/2)my => my = 2aFrom c = (1/2)x(y - n) => x(y - n) = 2cThus, DMN = (1/2)(xy - m(y - n)) = (1/2)(xy - (2a/y)(y - n))But x(y - n) = 2c => x = 2c/(y - n)Therefore, xy = x * y = (2c/(y - n)) * y = 2c*y/(y - n)Thus, DMN = (1/2)(2c*y/(y - n) - (2a/y)(y - n))= (1/2)[ 2c*y/(y - n) - 2a*(y - n)/y ]Factor out 2:= (1/2)*2[ c*y/(y - n) - a*(y - n)/y ]= [ c*y/(y - n) - a*(y - n)/y ]So DMN = [ c*y/(y - n) - a*(y - n)/y ]Hmm, but we need to express this in terms of a, b, c. Maybe relate these terms using the previous equation.We have the equation from earlier: a/y + b/n = c/(y - n)Let me denote this equation as (1): a/y + b/n = c/(y - n)Let’s try to express y/(y - n) and (y - n)/y in terms of a, b, c.From equation (1):Multiply both sides by y*n*(y - n):a*n*(y - n) + b*y*(y - n) = c*y*nAs before.Alternatively, let's solve equation (1) for one of the variables.But perhaps express DMN = c*y/(y - n) - a*(y - n)/yLet’s write DMN = c*(y/(y - n)) - a*((y - n)/y)Let’s denote t = y/(y - n). Then (y - n)/y = 1/t.Thus, DMN = c*t - a*(1/t)But we need to express t in terms of a, b, c.From equation (1):a/y + b/n = c/(y - n)Express in terms of t:Since t = y/(y - n) => y = t*(y - n) => y = t*y - t*n => y - t*y = -t*n => y(1 - t) = -t*n => y = (-t*n)/(1 - t)Alternatively, maybe not helpful.Alternatively, let's express equation (1) in terms of t.Since t = y/(y - n) => y = t*(y - n)From this:y = t*y - t*n => y - t*y = -t*n => y(1 - t) = -t*n => n = y(t - 1)/tSo, n = y(t - 1)/tFrom equation (1):a/y + b/n = c/(y - n)Express in terms of t:n = y(t - 1)/tThus, 1/n = t/(y(t - 1))So, equation (1):a/y + b*(t/(y(t - 1))) = c/(y - n)But y - n = y - y(t - 1)/t = y[1 - (t - 1)/t] = y[ (t - (t - 1))/t ] = y*(1/t) = y/tThus, c/(y - n) = c/(y/t) = c*t/yTherefore, equation (1) becomes:a/y + (b*t)/(y(t - 1)) = (c*t)/yMultiply both sides by y:a + (b*t)/(t - 1) = c*tMultiply both sides by (t - 1):a*(t - 1) + b*t = c*t*(t - 1)Expand:a*t - a + b*t = c*t^2 - c*tBring all terms to one side:c*t^2 - c*t - a*t + a - b*t = 0Combine like terms:c*t^2 - (c + a + b)*t + a = 0So, we have a quadratic in t:c*t^2 - (a + b + c)*t + a = 0Solving for t:t = [ (a + b + c) ± sqrt( (a + b + c)^2 - 4ac ) ] / (2c)But since t = y/(y - n) must be positive (since y > y - n as n > 0), we take the positive root.But this still seems complicated. However, note that DMN = c*t - a/tSo, if we let t be a solution to c*t^2 - (a + b + c)*t + a = 0, then DMN = c*t - a/tBut let's compute c*t - a/t:Multiply the quadratic equation by t:c*t^3 - (a + b + c)*t^2 + a*t = 0But perhaps not helpful.Alternatively, let's consider that:From DMN = c*t - a/t, and the quadratic equation c*t^2 - (a + b + c)*t + a = 0, let's multiply both sides by t:c*t^3 - (a + b + c)*t^2 + a*t = 0But DMN = c*t - a/t, so:Multiply DMN by t:DMN*t = c*t^2 - aBut from the quadratic equation:c*t^2 = (a + b + c)*t - aThus:DMN*t = (a + b + c)*t - a - a = (a + b + c)*t - 2aBut this might not help.Alternatively, maybe there's a relationship we can exploit here.Alternatively, suppose we let S = DMN. Then, from the quadratic equation:c*t^2 - (a + b + c)*t + a = 0But S = c*t - a/t => S*t = c*t^2 - aFrom the quadratic equation:c*t^2 = (a + b + c)*t - aThus,S*t = (a + b + c)*t - a - a = (a + b + c)*t - 2aTherefore,S*t = (a + b + c)*t - 2aSubtract S*t from both sides:0 = (a + b + c - S)*t - 2aThus,(a + b + c - S)*t = 2aBut we need another equation to relate S and t.Alternatively, from S = c*t - a/t, we can express t in terms of S:Multiply both sides by t:S*t = c*t^2 - aBut from quadratic equation, c*t^2 = (a + b + c)*t - aThus,S*t = (a + b + c)*t - a - a=> S*t = (a + b + c)*t - 2aBring all terms to left:S*t - (a + b + c)*t + 2a = 0=> t*(S - a - b - c) + 2a = 0=> t = -2a / (S - a - b - c)But also, from quadratic equation:c*t^2 - (a + b + c)*t + a = 0Substitute t = -2a / (S - a - b - c):c*( (-2a)/(S - a - b - c) )^2 - (a + b + c)*( -2a/(S - a - b - c) ) + a = 0Compute each term:First term: c*(4a²)/(S - a - b - c)^2Second term: (a + b + c)*(2a)/(S - a - b - c)Third term: aMultiply through by (S - a - b - c)^2 to eliminate denominators:4a²c + 2a(a + b + c)(S - a - b - c) + a(S - a - b - c)^2 = 0This seems very complex. Perhaps this approach is not the best.Alternatively, maybe instead of introducing variables for coordinates, we can use algebraic manipulation with the given areas.We have:a = (1/2)*m*yb = (1/2)*(x - m)*nc = (1/2)*x*(y - n)We need to find DMN = (1/2)(x*y - m(y - n))Let’s express DMN in terms of a, b, c:First, note that:From a = (1/2)*m*y => m = 2a/yFrom c = (1/2)*x*(y - n) => x = 2c/(y - n)From b = (1/2)*(x - m)*n => x - m = 2b/nSubstitute x and m:2c/(y - n) - 2a/y = 2b/nDivide both sides by 2:c/(y - n) - a/y = b/nLet’s call this Equation (1).Our goal is to express DMN = (1/2)(x*y - m(y - n)) in terms of a, b, c.Compute x*y: x = 2c/(y - n), so x*y = 2c*y/(y - n)Compute m(y - n): m = 2a/y, so m(y - n) = 2a(y - n)/yThus, DMN = (1/2)(2c*y/(y - n) - 2a(y - n)/y) = c*y/(y - n) - a(y - n)/ySo DMN = c*y/(y - n) - a*(y - n)/yLet’s denote k = y - n. Then y = k + n. Substitute into DMN:DMN = c*(k + n)/k - a*k/(k + n)= c*(1 + n/k) - a*k/(k + n)But this may not help. Alternatively, note that from Equation (1):c/(y - n) - a/y = b/nLet’s denote k = y - n. Then y = k + n. Substitute into Equation (1):c/k - a/(k + n) = b/nMultiply both sides by k*n*(k + n):c*n*(k + n) - a*n*k = b*k*(k + n)Expand:c*n*k + c*n^2 - a*n*k = b*k^2 + b*k*nCombine like terms:(c*n*k - a*n*k - b*k*n) + c*n^2 - b*k^2 = 0Factor k*n:k*n*(c - a - b) + c*n^2 - b*k^2 = 0Still complicated.Alternatively, from Equation (1):c/(y - n) - a/y = b/nLet’s solve for one variable in terms of another. For example, solve for n:Multiply both sides by n:c*n/(y - n) - a*n/y = bLet’s rearrange:c*n/(y - n) = b + a*n/yMultiply both sides by (y - n):c*n = (b + a*n/y)*(y - n)Expand right-hand side:= b*y - b*n + a*n*(y - n)/y= b*y - b*n + a*n - a*n²/yThus:c*n = b*y - b*n + a*n - (a*n²)/yBring all terms to left:c*n - b*y + b*n - a*n + (a*n²)/y = 0Combine like terms:(c + b - a)*n - b*y + (a*n²)/y = 0Multiply both sides by y:(c + b - a)*n*y - b*y² + a*n² = 0This is a quadratic in n:a*n² + (c + b - a)*y*n - b*y² = 0But this still includes both n and y. It seems challenging to resolve.Perhaps we need to use the fact that DMN = c*y/(y - n) - a*(y - n)/y and find a common denominator:DMN = [c*y^2 - a*(y - n)^2] / [y*(y - n)]But expand numerator:c*y² - a*(y² - 2yn + n²) = c*y² - a*y² + 2a*y*n - a*n² = (c - a)*y² + 2a*y*n - a*n²So DMN = [ (c - a)*y² + 2a*y*n - a*n² ] / [y*(y - n)]But this seems more complicated.Alternatively, perhaps use substitution.From Equation (1):c/(y - n) - a/y = b/nLet’s denote p = y and q = n. Then Equation (1) becomes:c/(p - q) - a/p = b/qMultiply both sides by p*q*(p - q):c*p*q - a*q*(p - q) = b*p*(p - q)Expand:c*p*q - a*p*q + a*q² = b*p² - b*p*qBring all terms to left:c*p*q - a*p*q + a*q² - b*p² + b*p*q = 0Combine like terms:(c*p*q - a*p*q + b*p*q) + a*q² - b*p² = 0Factor p*q:p*q*(c - a + b) + a*q² - b*p² = 0This can be written as:- b*p² + (c - a + b)*p*q + a*q² = 0This is a quadratic in p:- b*p² + (c - a + b)*q*p + a*q² = 0Multiply both sides by -1:b*p² - (c - a + b)*q*p - a*q² = 0Solve for p using quadratic formula:p = [ (c - a + b)*q ± sqrt( (c - a + b)^2*q² + 4ab*q² ) ] / (2b)Factor q² inside the square root:= [ (c - a + b)*q ± q*sqrt( (c - a + b)^2 + 4ab ) ] / (2b)Factor q:= q[ (c - a + b) ± sqrt( (c - a + b)^2 + 4ab ) ] / (2b)But this gives p (which is y) in terms of q (which is n). However, without another equation, it's difficult to relate them. Perhaps this indicates that the problem requires a more insightful approach rather than brute-force algebra.Wait, let's recall that the area of DMN is DMN = xy - (a + b + c). Wait, earlier I thought that the sum of the areas a + b + c + DMN equals the area of the rectangle. Let's verify this:Area of rectangle = x*ySum of areas:AMD (a) + MBN (b) + NCD (c) + DMN = a + b + c + DMNIf this equals x*y, then DMN = x*y - (a + b + c). But is this true?Wait, earlier calculations suggested otherwise. Let me check again.Suppose we have rectangle ABCD. When we place points M on AB and N on BC, and connect D to M to N to D, the four regions AMD, MBN, NCD, and DMN partition the rectangle. If that's the case, then indeed their areas should add up to the area of the rectangle. But when I computed the sum, I found that:a + b + c + DMN = (1/2)my + (1/2)(x - m)n + (1/2)x(y - n) + DMNWhich simplifies to:(1/2)my + (1/2)(x - m)n + (1/2)x(y - n) + DMN= (1/2)my + (1/2)xn - (1/2)mn + (1/2)xy - (1/2)xn + DMN= (1/2)my - (1/2)mn + (1/2)xy + DMNThis equals (1/2)xy + (1/2)m(y - n) + DMNIf the total area is xy, then:(1/2)xy + (1/2)m(y - n) + DMN = xyTherefore, DMN = xy - (1/2)xy - (1/2)m(y - n) = (1/2)xy - (1/2)m(y - n)But according to this, DMN = (1/2)(xy - m(y - n)), which matches the earlier result.However, if we suppose that the sum a + b + c + DMN = xy, then DMN = xy - (a + b + c). Comparing the two expressions:(1/2)(xy - m(y - n)) = xy - (a + b + c)But from the expressions of a, b, c:a = (1/2)myb = (1/2)(x - m)nc = (1/2)x(y - n)Thus, a + b + c = (1/2)my + (1/2)(x - m)n + (1/2)x(y - n)Which simplifies to:(1/2)my + (1/2)xn - (1/2)mn + (1/2)xy - (1/2)xn= (1/2)my - (1/2)mn + (1/2)xy= (1/2)xy + (1/2)m(y - n)Therefore, DMN = xy - (a + b + c) = xy - [ (1/2)xy + (1/2)m(y - n) ] = (1/2)xy - (1/2)m(y - n)Which matches the previous expression. So both expressions for DMN are equivalent. Therefore, DMN can be written as xy - (a + b + c) or as (1/2)(xy - m(y - n)). But we need to express DMN in terms of a, b, c.If we can find xy in terms of a, b, c, then DMN = xy - (a + b + c). But how?We have:From a = (1/2)my => my = 2aFrom c = (1/2)x(y - n) => x(y - n) = 2cFrom b = (1/2)(x - m)n => (x - m)n = 2b => x - m = 2b/nWe need to find xy.We have x = 2c/(y - n) from c = (1/2)x(y - n)So, xy = 2c/(y - n) * y = 2c*y/(y - n)But we need to express this in terms of a, b, c.From Equation (1):c/(y - n) - a/y = b/nMultiply both sides by y*n:c*n - a*n/(y - n) = b*yWait, perhaps not helpful.Wait, from Equation (1):c/(y - n) = a/y + b/nSo, 1/(y - n) = (a/(c*y)) + b/(c*n)Thus, xy = 2c*y/(y - n) = 2c*y[ (a/(c*y)) + b/(c*n) ] = 2c*y*(a/(c*y) + b/(c*n)) = 2c*y*(a/(c*y)) + 2c*y*(b/(c*n)) = 2a + 2b*y/nBut we need to express y/n. From Equation (1):c/(y - n) - a/y = b/nLet’s solve for y/n.Let’s let’s denote k = y/n. Then y = k*n.Substitute into Equation (1):c/(k*n - n) - a/(k*n) = b/nSimplify denominators:c/(n(k - 1)) - a/(k*n) = b/nMultiply both sides by n:c/(k - 1) - a/k = bMultiply both sides by k(k - 1):c*k - a(k - 1) = b*k(k - 1)Expand:c*k - a*k + a = b*k^2 - b*kBring all terms to one side:b*k^2 - (b + c - a)*k - a = 0This is the same quadratic equation in k as before. Let's write it again:b*k^2 - (a + b + c - a)*k - a = 0Wait, no, more accurately, it's:b*k^2 - (b + c - a)*k - a = 0Wait, no, expanding the previous step:Left-hand side: c*k - a*k + aRight-hand side: b*k^2 - b*kThus:b*k^2 - b*k - c*k + a*k - a = 0Combine like terms:b*k^2 + (-b - c + a)*k - a = 0So, quadratic equation:b*k^2 + (a - b - c)*k - a = 0Solving for k:k = [ -(a - b - c) ± sqrt( (a - b - c)^2 + 4ab ) ] / (2b)Since k = y/n must be positive (as both y and n are positive), we take the positive root:k = [ (b + c - a) + sqrt( (a - b - c)^2 + 4ab ) ] / (2b)But this is messy. However, note that xy = 2a + 2b*y/n = 2a + 2b*kThus, DMN = xy - (a + b + c) = 2a + 2b*k - (a + b + c) = a + 2b*k - b - cBut DMN is also equal to [ (c - a)*y² + 2a*y*n - a*n² ] / [y*(y - n)] which seems more complicated.Alternatively, perhaps use the quadratic equation to express k in terms of a, b, c.But this might not lead to a neat expression. Alternatively, maybe we can consider that the quadratic equation in k:b*k^2 + (a - b - c)*k - a = 0Let’s denote this as:b*k^2 + (a - b - c)*k - a = 0Multiply both sides by 4b:4b^2*k^2 + 4b(a - b - c)*k - 4ab = 0Complete the square:(2b*k + (a - b - c))^2 - (a - b - c)^2 - 4ab = 0Thus,(2b*k + a - b - c)^2 = (a - b - c)^2 + 4abExpand the right-hand side:= a^2 + b^2 + c^2 - 2ab - 2ac + 2bc + 4ab= a^2 + b^2 + c^2 + 2ab - 2ac + 2bc= (a + b)^2 + c^2 + 2bc - 2acHmm, not sure. Alternatively, perhaps this is equal to (a + b - c)^2 + 4ac - 2ac = ?Alternatively, maybe not helpful.Alternatively, take square root:2b*k + a - b - c = sqrt( (a - b - c)^2 + 4ab )But this is getting too involved. Perhaps there's a different approach. Let me think.Suppose we let the sides of the rectangle be AB = x and BC = y, so ABCD is a rectangle with sides x and y. Then, coordinate system as before.Point M is at (m, 0) on AB, and N is at (x, n) on BC.Areas:a = area of AMD = (1/2)*m*yb = area of MBN = (1/2)*(x - m)*nc = area of NCD = (1/2)*x*(y - n)We need to find area of DMN.From earlier, DMN = (1/2)(xy - m(y - n))But we need to express this in terms of a, b, c.Notice that:From a = (1/2)*m*y => m = 2a/yFrom b = (1/2)*(x - m)*n => x - m = 2b/n => x = m + 2b/nFrom c = (1/2)*x*(y - n) => x = 2c/(y - n)Therefore, equate the two expressions for x:m + 2b/n = 2c/(y - n)But m = 2a/y, so:2a/y + 2b/n = 2c/(y - n)Divide both sides by 2:a/y + b/n = c/(y - n)This is the same Equation (1) as before.Let’s denote u = y and v = n.So, a/u + b/v = c/(u - v)We need to find DMN = (1/2)(x*u - m(u - v))But x = 2c/(u - v) and m = 2a/uSo DMN = (1/2)( (2c/(u - v))*u - (2a/u)*(u - v) )= (1/2)( 2c*u/(u - v) - 2a*(u - v)/u )= c*u/(u - v) - a*(u - v)/uBut from Equation (1):a/u + b/v = c/(u - v)Let’s solve for c/(u - v):c/(u - v) = a/u + b/vMultiply both sides by (u - v):c = (a/u + b/v)(u - v)Expand:c = a*(u - v)/u + b*(u - v)/v= a*(1 - v/u) + b*(u/v - 1)= a - a*v/u + b*u/v - bRearrange:c = a - b + (b*u/v - a*v/u)Let’s denote t = u/v => u = t*vSubstitute into equation:c = a - b + (b*(t*v)/v - a*v/(t*v)) = a - b + (b*t - a/t)Thus,c = a - b + b*t - a/tBring all terms to left:c - a + b - b*t + a/t = 0Multiply both sides by t to eliminate denominator:(c - a + b)t - b*t² + a = 0Rearrange:- b*t² + (c - a + b)t + a = 0Multiply by -1:b*t² + (-c + a - b)t - a = 0Which is the same quadratic in t as before.Solving for t:t = [ (c - a + b) ± sqrt( (c - a + b)^2 + 4ab ) ] / (2b)Since t = u/v = y/n > 1 (since y > n), we take the positive root.But DMN = c*u/(u - v) - a*(u - v)/uSubstitute u = t*v:DMN = c*(t*v)/(t*v - v) - a*(t*v - v)/(t*v)= c*t*v / (v(t - 1)) - a*v(t - 1)/(t*v)Simplify:= c*t / (t - 1) - a*(t - 1)/tLet’s denote s = tThus, DMN = c*s/(s - 1) - a*(s - 1)/sBut from earlier equation:c = a - b + b*s - a/sLet’s rearrange:c = a - b + b*s - a/s=> c + b - a = b*s - a/sMultiply both sides by s:(c + b - a)s = b*s² - aRearrange:b*s² - (c + b - a)s - a = 0Which is the same quadratic as before.But notice that DMN = c*s/(s - 1) - a*(s - 1)/sLet’s compute this:= [c*s^2 - a*(s - 1)^2] / [s(s - 1)]Expand numerator:= c*s^2 - a*(s^2 - 2s + 1)= c*s^2 - a*s^2 + 2a*s - a= (c - a)s^2 + 2a*s - aFrom the quadratic equation b*s² - (c + b - a)s - a = 0, we can express (c - a)s^2:From the quadratic equation:b*s² = (c + b - a)s + a=> (c - a)s^2 = (c + b - a)s + a - b*s² + (c - a)s^2 - (c + b - a)s - a = 0Wait, perhaps this is not helpful.Alternatively, substitute s from the quadratic equation:From b*s² - (c + b - a)s - a = 0 => b*s² = (c + b - a)s + aThus, numerator of DMN:(c - a)s^2 + 2a*s - a= (c - a)s^2 + 2a*s - aReplace (c - a)s^2:= [(c - a)/b * ((c + b - a)s + a)] + 2a*s - aBut this seems to complicate things further.At this point, it's clear that brute-force algebra is leading to a complex expression. Perhaps there's a symmetry or a substitution that we're missing.Wait, recall that in problems involving areas of triangles in a rectangle, sometimes the solution can be found by expressing the areas in terms of the coordinates and solving the resulting system of equations. Let's try that.We have variables m, n, x, y. But we also have areas a, b, c.Given:1. a = (1/2)*m*y => my = 2a2. b = (1/2)*(x - m)*n => (x - m)n = 2b3. c = (1/2)*x*(y - n) => x(y - n) = 2cWe need to find DMN = (1/2)(xy - m(y - n)).Let’s express xy and m(y - n) in terms of a, b, c.From equation 3: x(y - n) = 2c => x = 2c/(y - n)From equation 1: my = 2a => m = 2a/yFrom equation 2: (x - m)n = 2b => x - m = 2b/n => x = m + 2b/nSubstitute x from equation 3 into equation 2:x = 2c/(y - n) = m + 2b/nBut m = 2a/y, so:2c/(y - n) = 2a/y + 2b/nDivide both sides by 2:c/(y - n) = a/y + b/nThis is Equation (1) again.Let’s consider the variables y and n. Let’s denote y and n as variables and try to solve for them.Let’s let’s write Equation (1):c/(y - n) = a/y + b/nMultiply both sides by y*n*(y - n):c*y*n = a*n*(y - n) + b*y*(y - n)Expand right-hand side:c*y*n = a*n*y - a*n² + b*y² - b*y*nBring all terms to left:c*y*n - a*n*y + a*n² - b*y² + b*y*n = 0Combine like terms:(c*y*n - a*n*y + b*y*n) + a*n² - b*y² = 0Factor y*n:y*n*(c - a + b) + a*n² - b*y² = 0Rearrange terms:a*n² + y*n*(c - a + b) - b*y² = 0This is a quadratic in n:a*n² + y*(c - a + b)*n - b*y² = 0Solving for n using quadratic formula:n = [ -y*(c - a + b) ± sqrt( y²*(c - a + b)^2 + 4a*b*y² ) ] / (2a)Factor y² inside the square root:= [ -y*(c - a + b) ± y*sqrt( (c - a + b)^2 + 4ab ) ] / (2a)Factor y:= y[ - (c - a + b) ± sqrt( (c - a + b)^2 + 4ab ) ] / (2a)Since n must be positive, we take the positive root:n = y[ - (c - a + b) + sqrt( (c - a + b)^2 + 4ab ) ] / (2a)This expression for n in terms of y is still complex, but maybe we can substitute back into DMN.Recall that DMN = (1/2)(xy - m(y - n))From equation 3: x = 2c/(y - n)From equation 1: m = 2a/yThus:DMN = (1/2)( (2c/(y - n))*y - (2a/y)(y - n) )= (1/2)( 2c*y/(y - n) - 2a*(y - n)/y )= c*y/(y - n) - a*(y - n)/yLet’s express this as:DMN = c*y/(y - n) - a*(y - n)/y= c*y/(y - n) + a*(n - y)/y= c*y/(y - n) + a*n/y - aBut from Equation (1):c/(y - n) = a/y + b/nMultiply both sides by y:c*y/(y - n) = a + b*y/nThus, DMN = (a + b*y/n) + a*n/y - a = b*y/n + a*n/ySo DMN = b*y/n + a*n/yThis is a key simplification! By substituting Equation (1), we find that DMN = b*(y/n) + a*(n/y)Let’s denote k = y/n, then DMN = b*k + a/kWe need to express k in terms of a, b, c.But from Equation (1):c/(y - n) = a/y + b/nExpress in terms of k:Since k = y/n => y = k*nThus, y - n = (k - 1)*nSubstitute into Equation (1):c/( (k - 1)*n ) = a/(k*n) + b/nMultiply both sides by n:c/(k - 1) = a/k + bMultiply both sides by k(k - 1):c*k = a(k - 1) + b*k(k - 1)Expand:c*k = a*k - a + b*k^2 - b*kBring all terms to left:b*k^2 + (a - b - c)*k - a = 0This is the same quadratic equation as before. So k satisfies:b*k^2 + (a - b - c)*k - a = 0Let’s denote this quadratic equation as:b*k^2 + (a - b - c)*k - a = 0The solution is:k = [ -(a - b - c) ± sqrt( (a - b - c)^2 + 4ab ) ] / (2b)Taking the positive root since k = y/n > 0:k = [ (b + c - a) + sqrt( (a - b - c)^2 + 4ab ) ] / (2b)But DMN = b*k + a/k. Let’s compute this:Let’s denote sqrt( (a - b - c)^2 + 4ab ) as D.Thus,k = [ (b + c - a) + D ] / (2b)Therefore,DMN = b*k + a/k = b*[ (b + c - a) + D ] / (2b) + a / [ ( (b + c - a) + D ) / (2b) ]Simplify first term:= [ (b + c - a) + D ] / 2Second term:= a * (2b) / [ (b + c - a) + D ]Thus,DMN = [ (b + c - a) + D ] / 2 + (2ab) / [ (b + c - a) + D ]Let’s compute this sum:Let’s denote S = (b + c - a) + DThen,DMN = S/2 + (2ab)/SCombine terms:= (S^2 + 4ab) / (2S)But S = (b + c - a) + sqrt( (a - b - c)^2 + 4ab )Compute S^2:= [ (b + c - a) + sqrt( (a - b - c)^2 + 4ab ) ]^2= (b + c - a)^2 + 2(b + c - a)*sqrt( (a - b - c)^2 + 4ab ) + (a - b - c)^2 + 4ab= [ (b + c - a)^2 + (a - b - c)^2 ] + 4ab + 2(b + c - a)*sqrt( (a - b - c)^2 + 4ab )Notice that (b + c - a)^2 = (a - b - c)^2, so this becomes:= 2(a - b - c)^2 + 4ab + 2(b + c - a)*sqrt( (a - b - c)^2 + 4ab )This seems complicated, but perhaps there's a simplification.Notice that:(a - b - c)^2 + 4ab = a² + b² + c² - 2ab - 2ac + 2bc + 4ab = a² + b² + c² + 2ab - 2ac + 2bcBut this doesn't seem to simplify nicely.Alternatively, notice that DMN is expressed as (S^2 + 4ab)/(2S), where S = (b + c - a) + sqrt( (a - b - c)^2 + 4ab ). This might not lead to a simplification in terms of a, b, c without square roots.Given that the problem asks to express the area of triangle DMN in terms of a, b, and c, and given the complexity of the expressions we've derived, it's possible that there's a more elegant solution or a factorization we're missing.Alternatively, perhaps there's a relationship between the areas that allows us to express DMN in terms of a, b, c without diving into coordinate systems.Let me consider the areas of the triangles and the rectangle.The total area of the rectangle is x*y = sum of areas of AMD, MBN, NCD, and DMN.So, x*y = a + b + c + DMN => DMN = x*y - (a + b + c)But x*y is the area of the rectangle. However, we need to express x*y in terms of a, b, c.From earlier equations:From a = (1/2)*m*y => m = 2a/yFrom c = (1/2)*x*(y - n) => x = 2c/(y - n)From b = (1/2)*(x - m)*n => (x - m) = 2b/n => x = m + 2b/nSubstituting x from c into this:2c/(y - n) = m + 2b/nBut m = 2a/y, so:2c/(y - n) = 2a/y + 2b/nDivide by 2:c/(y - n) = a/y + b/nThis is the same Equation (1) as before.If we could find x*y in terms of a, b, c, then DMN = x*y - (a + b + c)But how?Let’s assume that x*y can be expressed as (sqrt(a) + sqrt(b) + sqrt(c))^2 or some combination. But this is just a guess.Alternatively, let's consider that in some similar problems, the area can be expressed using the formula involving the square root of the product of areas. However, without more information, it's difficult to proceed.Alternatively, maybe consider that from Equation (1):c/(y - n) = a/y + b/nLet’s denote p = y and q = n.Then, c/(p - q) = a/p + b/qThis resembles the equation of resistors in parallel or similar, but I'm not sure.Alternatively, let's think of this as a system of equations:Let’s let’s set variables:Let’s let’s denote s = y - nThen, from Equation (1):c/s = a/(s + n) + b/nBut this may not help.Alternatively, cross-multiplying:c*n*(s + n) = a*n*s + b*s*(s + n)But this leads to a quadratic equation again.Given the time I've spent on this and the multiple approaches leading back to the quadratic equation, it's possible that the answer is supposed to be expressed in terms of the roots of that quadratic, but I suspect there's a trick I'm missing.Wait, going back to the expression DMN = b*y/n + a*n/yThis is of the form DMN = b*k + a/k where k = y/nFrom the equation we had earlier:b*k^2 + (a - b - c)*k - a = 0This quadratic equation in k can be written as:b*k^2 - (b + c - a)*k - a = 0Let’s multiply both sides by k:b*k^3 - (b + c - a)*k^2 - a*k = 0But DMN = b*k + a/k => DMN*k = b*k^2 + aThus,From the quadratic equation:b*k^2 = (b + c - a)*k + aTherefore,DMN*k = (b + c - a)*k + a + a = (b + c - a)*k + 2aThus,DMN*k - (b + c - a)*k - 2a = 0Factor k:k*(DMN - b - c + a) - 2a = 0But this is another equation involving DMN and k. Since k is related to DMN through DMN = b*k + a/k, it's still a loop.Alternatively, maybe square both sides of DMN = b*k + a/k:DMN^2 = b²*k² + 2ab + a²/k²But from the quadratic equation, we have:b*k² = (b + c - a)*k + aThus, b²*k² = b*(b + c - a)*k + abAlso, a²/k² can be expressed from the quadratic equation.From the quadratic equation:b*k² + (a - b - c)*k - a = 0Multiply by a/k²:b*a + (a - b - c)*a/k - a²/k² = 0=> a²/k² = b*a + (a - b - c)*a/kBut this seems to lead back.Alternatively, let's substitute k from DMN = b*k + a/k into the quadratic equation.Let’s denote S = DMN = b*k + a/kThen, S*k = b*k² + aFrom quadratic equation:b*k² = (b + c - a)*k + aThus,S*k = (b + c - a)*k + a + a=> S*k = (b + c - a)*k + 2a=> (S - b - c + a)*k = 2a=> k = 2a / (S - b - c + a)But also, S = b*k + a/kSubstitute k:S = b*(2a / (S - b - c + a)) + a/(2a / (S - b - c + a))Simplify:S = (2ab / (S - b - c + a)) + (S - b - c + a)/2Multiply both sides by 2(S - b - c + a):2S(S - b - c + a) = 4ab + (S - b - c + a)^2Expand both sides:2S^2 - 2S(b + c - a) = 4ab + S^2 - 2S(b + c - a) + (b + c - a)^2Subtract 2S^2 - 2S(b + c - a) from both sides:0 = -S^2 + 4ab + (b + c - a)^2Thus,S^2 = 4ab + (b + c - a)^2Taking square roots:S = sqrt(4ab + (b + c - a)^2 )But S is DMN, so:DMN = sqrt( (a - b - c)^2 + 4ab )Simplify the expression inside the square root:(a - b - c)^2 + 4ab = a² + b² + c² - 2ab - 2ac + 2bc + 4ab = a² + b² + c² + 2ab - 2ac + 2bcThis can be written as:(a + b)^2 + c² - 2ac + 2bc - 2abWait, not sure. Alternatively, factor:= a² + 2ab + b² + c² + 2bc - 2ac - 2ab= (a + b)^2 + c² + 2bc - 2ac - 2abAlternatively, perhaps not helpful.But wait, the expression inside the square root is:(a - b - c)^2 + 4ab = (a + b - c)^2 + 4ab - 4ab = (a + b - c)^2Wait, compute:(a + b - c)^2 = a² + b² + c² + 2ab - 2ac - 2bcWhich is different from the expression we have:(a - b - c)^2 + 4ab = a² + b² + c² + 2ab - 2ac + 2bcThus, they are not the same.Wait, but maybe there's another way to factor:Let’s compute:(a - b - c)^2 + 4ab = a² + b² + c² - 2ab - 2ac + 2bc + 4ab = a² + b² + c² + 2ab - 2ac + 2bcThis can be written as:(a + b)^2 + c² + 2bc - 2acBut I don't see a clear factorization here.However, from the above steps, we found that DMN = sqrt( (a - b - c)^2 + 4ab )But this is an expression for DMN in terms of a, b, c. However, the problem asks for DMN expressed in terms of a, b, c. But the answer is a square root, which seems unusual for an olympiad problem expecting a rational expression. Did we make a mistake somewhere?Wait, no, let's verify:We derived that DMN^2 = (a - b - c)^2 + 4abTherefore, DMN = sqrt( (a - b - c)^2 + 4ab )But expanding (a - b - c)^2 + 4ab:= a² + b² + c² - 2ab - 2ac + 2bc + 4ab= a² + b² + c² + 2ab - 2ac + 2bcBut this doesn't simplify further. Therefore, the area DMN is the square root of this expression.However, in typical problems like this, the answer is usually a simple combination of a, b, c, not involving square roots, which makes me question my approach.But wait, going back to the key step where DMN = b*y/n + a*n/y, and using the AM-GM inequality, but that might not help here. However, if we let k = y/n, then DMN = b*k + a/k, and from the quadratic equation, b*k^2 + (a - b - c)*k - a = 0If we multiply DMN by k:DMN*k = b*k^2 + aFrom the quadratic equation, b*k^2 = (b + c - a)*k + aThus,DMN*k = (b + c - a)*k + a + a = (b + c - a)*k + 2aBut DMN*k = DMN*kTherefore,DMN*k - (b + c - a)*k - 2a = 0Factor k:k*(DMN - b - c + a) - 2a = 0But since k = y/n > 0, and we need to solve for DMN, but this still leads back to the same issue.Alternatively, maybe the answer is simply DMN = sqrt(a c + b c + a b) or similar, but I need to check.Alternatively, suppose we let a = 1, b = 1, c = 1, then compute DMN.Assume a = b = c = 1.Then from a = (1/2)*m*y => m*y = 2From b = (1/2)*(x - m)*n => (x - m)*n = 2From c = (1/2)*x*(y - n) => x*(y - n) = 2And DMN = (1/2)(xy - m(y - n))We need to solve for x, y, m, n.From a = 1: m = 2/yFrom c = 1: x = 2/(y - n)From b = 1: x - m = 2/n => x = m + 2/nThus:2/(y - n) = 2/y + 2/nDivide by 2:1/(y - n) = 1/y + 1/nMultiply both sides by y*n*(y - n):n*y = n*(y - n) + y*(y - n)Expand:n*y = n*y - n² + y² - y*nSimplify:n*y = n*y - n² + y² - y*nLeft side: n*yRight side: n*y - n² + y² - y*n = -n² + y² - y*nThus:n*y = -n² + y² - y*nBring all terms to left:n*y + n² - y² + y*n = 0Combine like terms:2n*y + n² - y² = 0This is a quadratic in n:n² + 2y*n - y² = 0Solving for n:n = [-2y ± sqrt(4y² + 4y²)]/2 = [-2y ± sqrt(8y²)]/2 = [-2y ± 2y*sqrt(2)]/2 = -y ± y*sqrt(2)Since n > 0, take the positive root:n = -y + y*sqrt(2) = y*(sqrt(2) - 1)Thus, y = n / (sqrt(2) - 1) = n*(sqrt(2) + 1)/1 = n*(sqrt(2) + 1)From a = 1: m = 2/y = 2/[n*(sqrt(2) + 1)]From c = 1: x = 2/(y - n) = 2/[n*(sqrt(2) + 1) - n] = 2/[n*sqrt(2)] = 2/(n*sqrt(2))From x = m + 2/n:2/(n*sqrt(2)) = 2/[n*(sqrt(2) + 1)] + 2/nMultiply both sides by n:2/sqrt(2) = 2/(sqrt(2) + 1) + 2Simplify:sqrt(2) = 2/(sqrt(2) + 1) + 2Multiply denominator by conjugate:2/(sqrt(2) + 1) * (sqrt(2) - 1)/(sqrt(2) - 1) = 2*(sqrt(2) - 1)/(2 - 1) = 2*(sqrt(2) - 1)Thus,sqrt(2) = 2*(sqrt(2) - 1) + 2 = 2*sqrt(2) - 2 + 2 = 2*sqrt(2)But sqrt(2) ≠ 2*sqrt(2). Contradiction.This suggests that when a = b = c = 1, there is no solution, which can't be true. Therefore, my assumption must be wrong, or there's a mistake in the process.Wait, perhaps when a = b = c = 1, there is no real solution, meaning that such a configuration isn't possible, which might indicate that the problem's conditions require a certain relationship between a, b, and c.But if the problem states that a, b, c are the areas of the triangles AMD, MBN, NCD respectively, then there must exist a rectangle and points M, N such that these areas are achievable. Therefore, the case where a = b = c = 1 might not be possible, hence the contradiction.Alternatively, perhaps I made an error in the algebra.Let me try specific values where a solution exists.Let’s take a rectangle with AB = 2, BC = 2, so x = 2, y = 2.Choose M at (1, 0), so m = 1.Then AMD has area (1/2)*1*2 = 1, so a = 1.Choose N at (2, 1), so n = 1.Then MBN has area (1/2)*(2 - 1)*1 = 0.5, so b = 0.5.NCD has area (1/2)*2*(2 - 1) = 1, so c = 1.Then DMN is the area of triangle with]

🔑:1. Define the dimensions and areas: Let the dimensions of the rectangle (ABCD) be (AD = x) and (CD = y). The areas of the triangles are given as: - ( text{Area of } triangle AMD = a ) - ( text{Area of } triangle MBN = b ) - ( text{Area of } triangle NCD = c )2. Express lengths in terms of areas: - The area of (triangle AMD) can be expressed as: [ a = frac{1}{2} times AM times x implies AM = frac{2a}{x} ] - The area of (triangle NCD) can be expressed as: [ c = frac{1}{2} times CN times y implies CN = frac{2c}{y} ]3. Express the area of (triangle MBN): - The area of (triangle MBN) can be expressed as: [ b = frac{1}{2} times MB times BN ] - Since (MB = y - frac{2a}{x}) and (BN = x - frac{2c}{y}), we have: [ 2b = (y - frac{2a}{x})(x - frac{2c}{y}) ]4. Expand and simplify the expression: [ 2b = yx - y cdot frac{2c}{y} - x cdot frac{2a}{x} + frac{4ac}{xy} ] [ 2b = xy - 2c - 2a + frac{4ac}{xy} ]5. Solve for (xy): [ 2b = xy - 2a - 2c + frac{4ac}{xy} ] Multiply both sides by (xy): [ 2b cdot xy = (xy)^2 - 2a cdot xy - 2c cdot xy + 4ac ] [ 2b cdot xy = (xy)^2 - 2(a+c) cdot xy + 4ac ] Let (z = xy): [ 2bz = z^2 - 2(a+c)z + 4ac ] [ z^2 - 2(a+c)z + 4ac - 2bz = 0 ] This is a quadratic equation in (z): [ z^2 - 2(a+c)z + 4ac - 2bz = 0 ]6. Solve the quadratic equation: Using the quadratic formula (z = frac{-B pm sqrt{B^2 - 4AC}}{2A}), where (A = 1), (B = -2(a+c)), and (C = 4ac - 2b): [ z = frac{2(a+c) pm sqrt{(2(a+c))^2 - 4 cdot 1 cdot (4ac - 2b)}}{2 cdot 1} ] [ z = (a+c) pm sqrt{(a+c)^2 - (4ac - 2b)} ] [ z = (a+c) pm sqrt{(a+c)^2 - 4ac + 2b} ] [ z = (a+c) pm sqrt{(a+c)^2 - 4ac + 2b} ] [ z = (a+c) pm sqrt{(a+c)^2 - 4ac + 2b} ]7. Determine the area of (triangle DMN): The area of (triangle DMN) is: [ text{Area of } triangle DMN = xy - a - b - c ] Substituting (xy = z): [ text{Area of } triangle DMN = z - a - b - c ] [ text{Area of } triangle DMN = (a+c) + sqrt{(a+b+c)^2 - 4ac} - a - b - c ] [ text{Area of } triangle DMN = sqrt{(a+b+c)^2 - 4ac} ]The final answer is ( boxed{ sqrt{(a+b+c)^2 - 4ac} } )

🧠:[<think>Okay, so I need to find the symmetric point of A(0,1) with respect to line l₁: x - y - 1 = 0, and then find the equation of the line symmetric to l₂: x - 2y + 2 = 0 with respect to l₁. Hmm, let's start with the first part: finding point B, the reflection of A over l₁.First, I remember that reflecting a point over a line involves finding the perpendicular line from the original point to the mirror line, then extending it the same distance on the other side. So, the steps should be:1. Find the equation of the line perpendicular to l₁ that passes through point A.2. Find the intersection point of this perpendicular line with l₁. This is the midpoint between A and its reflection B.3. Use the midpoint formula to calculate the coordinates of B.Alright, let's go through each step.Starting with step 1: Find the perpendicular line to l₁ through A(0,1). The line l₁ is x - y - 1 = 0. Let me rewrite that in slope-intercept form (y = mx + b) to find its slope. So, x - y - 1 = 0 ⇒ y = x - 1. Therefore, the slope m₁ of l₁ is 1.The slope of a line perpendicular to l₁ would be the negative reciprocal of m₁. Since m₁ is 1, the perpendicular slope m_perp is -1.So, the equation of the perpendicular line passing through A(0,1) is y - 1 = -1(x - 0) ⇒ y = -x + 1. That seems right.Step 2: Find the intersection point of this perpendicular line (y = -x + 1) with l₁ (y = x - 1). Let's set them equal:- x - 1 = -x + 1Wait, actually, substituting y from the perpendicular line into l₁'s equation:From l₁: y = x - 1From the perpendicular line: y = -x + 1So, set x - 1 = -x + 1Solving for x:x + x = 1 + 1 ⇒ 2x = 2 ⇒ x = 1Then, substitute back into either equation to find y. Let's use l₁: y = 1 - 1 = 0. So the intersection point is (1, 0). This is the midpoint between A and B.Step 3: Now, we need to find point B such that (1, 0) is the midpoint of A(0,1) and B(bx, by). The midpoint formula is ((0 + bx)/2, (1 + by)/2) = (1, 0). So,(0 + bx)/2 = 1 ⇒ bx = 2(1 + by)/2 = 0 ⇒ 1 + by = 0 ⇒ by = -1Therefore, point B is (2, -1). Let me check that. If I take the midpoint of (0,1) and (2,-1), it should be ((0+2)/2, (1 + (-1))/2) = (1, 0), which matches our earlier intersection point. That seems correct.So the coordinates of point B are (2, -1). Alright, that's the first part done.Now, the second part: finding the equation of the line symmetric to l₂: x - 2y + 2 = 0 with respect to l₁. Hmm, reflecting a line over another line. How do I approach this?I recall that to find the reflection of a line over another line, there are a few methods. One way is to reflect two points on the original line over the mirror line and then find the equation of the line passing through the reflected points. Alternatively, we can use the formula for the reflection of a line, but that might be more complex. Since l₂ is given, let me try reflecting two points from l₂ over l₁ and then find the equation.First, let me choose two points on l₂: x - 2y + 2 = 0. Let's find two convenient points.Let’s set x = 0: 0 - 2y + 2 = 0 ⇒ -2y = -2 ⇒ y = 1. So point C is (0,1). Wait, that's the same as point A. Interesting. Hmm, but (0,1) is on l₂. Wait, but l₂ is x - 2y + 2 = 0. Plugging in (0,1): 0 - 2*1 + 2 = -2 + 2 = 0. Yes, correct. So point C(0,1) is on l₂. Wait, that's the original point A. So when we reflect l₂ over l₁, the reflection of point C(0,1) over l₁ is point B(2,-1), which we already found. So that's one point on the reflected line.Now, let's find another point on l₂. Let's set y = 0: x - 0 + 2 = 0 ⇒ x = -2. So point D is (-2, 0). Let's verify: (-2) - 2*0 + 2 = -2 + 2 = 0. Correct. So point D(-2, 0) is on l₂. Now, let's reflect point D over l₁ to get point D', which will be on the reflected line.So, repeating the reflection process for point D(-2,0) over l₁.Step 1: Find the equation of the perpendicular line to l₁ through D(-2,0). The slope of l₁ is 1, so the perpendicular slope is -1. Thus, the equation is y - 0 = -1(x + 2) ⇒ y = -x - 2.Step 2: Find intersection point with l₁. l₁: y = x - 1. Set equal:x - 1 = -x - 2x + x = -2 + 12x = -1 ⇒ x = -0.5Then y = (-0.5) - 1 = -1.5. So the midpoint is (-0.5, -1.5).Step 3: Now, since this midpoint is between D(-2,0) and its reflection D'(dx, dy), we can use the midpoint formula:Midpoint = ((-2 + dx)/2, (0 + dy)/2) = (-0.5, -1.5)Therefore:(-2 + dx)/2 = -0.5 ⇒ -2 + dx = -1 ⇒ dx = 1(0 + dy)/2 = -1.5 ⇒ dy = -3Thus, reflection point D' is (1, -3).So now, we have two points on the reflected line: B(2,-1) and D'(1, -3). Let's verify if these are correct.First, check if D'(1, -3) is correct. Original point D(-2,0). The midpoint between D and D' is ((-2 +1)/2, (0 + (-3))/2) = (-0.5, -1.5), which matches the earlier calculation. So D' is correct.Now, the reflected line should pass through B(2,-1) and D'(1, -3). Let's find the equation of this line.First, compute the slope m:m = (-3 - (-1)) / (1 - 2) = (-2)/(-1) = 2.So the slope is 2. Now, using point-slope form with point B(2,-1):y - (-1) = 2(x - 2) ⇒ y + 1 = 2x - 4 ⇒ y = 2x - 5.Let me check with point D'(1, -3):Plug x = 1 into y = 2x -5: y = 2*1 -5 = -3. Correct.So the equation of the reflected line is y = 2x -5. Let me write it in standard form:2x - y -5 = 0.Alternatively, multiplying both sides by 1: 2x - y -5 = 0. Let me check if that's correct.Alternatively, let's verify with another point. If we take the original line l₂: x - 2y +2=0, which passes through (0,1) and (-2,0). Their reflections are (2,-1) and (1,-3). The line through these two points is indeed 2x - y -5 = 0. Let me confirm:For point (2,-1): 2*2 - (-1) -5 = 4 +1 -5 = 0. Correct.For point (1,-3): 2*1 - (-3) -5 = 2 +3 -5 = 0. Correct.Therefore, the equation of the symmetric line is 2x - y -5 =0.Alternatively, if I want to write it in the same form as the original line, which was x - 2y +2=0, but the reflected line is 2x - y -5=0. That seems correct. Let me just check another way.Alternatively, maybe there's a formula for reflecting a line over another line. Let me recall. Suppose we have a line ax + by + c =0 and we want to reflect it over the mirror line dx + ey + f =0. The formula might involve some substitution or using angles. But since I already found two points, and the line seems correct, maybe that's the safest way.Alternatively, since l₁ is the line x - y -1 =0, which we can write as x = y +1. The reflection across l₁ can be found using the reflection formula for lines. Another method is to take a general point (x,y) on l₂, find its reflection (x', y') over l₁, then substitute x and y in terms of x' and y' into the original equation l₂. Let's try that.Let me denote the reflection of a point (x, y) over l₁ as (x', y'). Then, the reflection formulas can be used. The formula for reflecting a point over the line ax + by + c =0 is:x' = ( (b² - a²)x - 2aby - 2ac ) / (a² + b²)y' = ( -2abx + (a² - b²)y - 2bc ) / (a² + b²)But wait, maybe that's too complicated. Let's instead recall that for a line l: ax + by + c =0, the reflection of a point (x, y) is given by:x' = x - 2a(ax + by + c)/(a² + b²)y' = y - 2b(ax + by + c)/(a² + b²)Wait, is that correct?Yes, the formula for reflection across a line ax + by + c =0 is:If we have point (x, y), then its reflection (x', y') is:x' = x - 2a*(ax + by + c)/(a² + b²)y' = y - 2b*(ax + by + c)/(a² + b²)Let me confirm. For the line ax + by + c =0, the reflection of a point (x, y) is given by those expressions. So in our case, the mirror line l₁ is x - y -1 =0, so a=1, b=-1, c=-1. Therefore, the reflection of a general point (x, y) is:x' = x - 2*1*(1*x + (-1)*y + (-1))/(1² + (-1)²) = x - 2*(x - y -1)/2 = x - (x - y -1) = x -x + y +1 = y +1Similarly,y' = y - 2*(-1)*(x - y -1)/2 = y + (x - y -1) = y + x - y -1 = x -1Therefore, the reflection of a point (x, y) over l₁ is (y +1, x -1). Let me verify this with point A(0,1):x' = 1 +1 =2, y' =0 -1= -1, which gives (2, -1). That's correct, as we found earlier. Similarly, reflecting point D(-2,0):x' =0 +1=1, y'= -2 -1= -3, which is (1, -3). Correct. So the reflection formula is (x', y') = (y +1, x -1). So in general, the reflection swaps x and y in some way. Interesting.Therefore, to find the equation of the reflected line, we can take a general point (X, Y) on the reflected line, which is the reflection of a point (x, y) on l₂. Since (X, Y) = (y +1, x -1), we can express x and y in terms of X and Y.From the reflection formulas:X = y +1 ⇒ y = X -1Y = x -1 ⇒ x = Y +1So, substituting x and y into the original equation of l₂: x -2y +2 =0.So, substitute x = Y +1 and y = X -1 into the equation:(Y +1) -2*(X -1) +2 =0Simplify:Y +1 -2X +2 +2 =0Y +1 -2X +4 =0Y -2X +5 =0Rearranged: -2X + Y +5 =0 ⇒ Multiply both sides by -1: 2X - Y -5 =0 ⇒ 2x - y -5 =0. So the equation is 2x - y -5=0, which matches what we found earlier. So that's another way to confirm.Therefore, the symmetric line to l₂ with respect to l₁ is 2x - y -5 =0.So summarizing:- The symmetric point of A(0,1) over l₁ is (2, -1).- The symmetric line of l₂ over l₁ is 2x - y -5 =0.Let me double-check everything once more to be sure.For the point reflection:Original point A(0,1). The perpendicular line is y = -x +1, intersection with l₁ at (1,0). Midpoint between A and B is (1,0), so B is (2,-1). Correct.For the line reflection:Original line l₂: x -2y +2=0. Reflecting two points gives (2,-1) and (1,-3). The line through these points is 2x - y -5=0. Checked via substitution and reflection formulas. Correct.Yes, I think that's all solid.Final AnswerThe coordinates of point B are boxed{(2, -1)}, and the equation of the symmetric line is boxed{2x - y - 5 = 0}.</think>To find the coordinates of point ( A(0,1) )'s symmetric point ( B ) with respect to line ( l_1: x - y - 1 = 0 ):1. Find the perpendicular line to ( l_1 ) through ( A ): - The slope of ( l_1 ) is 1, so the perpendicular slope is -1. - The equation of the perpendicular line through ( A(0,1) ) is ( y = -x + 1 ).2. Find the intersection of this perpendicular line with ( l_1 ): - Solving ( y = -x + 1 ) and ( y = x - 1 ): - ( x - 1 = -x + 1 ) - ( 2x = 2 Rightarrow x = 1 ) - ( y = 1 - 1 = 0 ) - Intersection point (midpoint) is ( (1, 0) ).3. Calculate the coordinates of ( B ): - Using the midpoint formula, ( B ) is ( (2, -1) ).To find the equation of the line symmetric to ( l_2: x - 2y + 2 = 0 ) with respect to ( l_1 ):1. Reflect two points on ( l_2 ) over ( l_1 ): - Point ( C(0, 1) ) on ( l_2 ) reflects to ( B(2, -1) ). - Another point ( D(-2, 0) ) on ( l_2 ): - Perpendicular line through ( D(-2, 0) ) is ( y = -x - 2 ). - Intersection with ( l_1 ) is ( (-0.5, -1.5) ). - Reflection point ( D' ) is ( (1, -3) ).2. Find the equation of the line through reflected points ( B ) and ( D' ): - Slope between ( (2, -1) ) and ( (1, -3) ) is 2. - Using point-slope form with ( B(2, -1) ): - ( y + 1 = 2(x - 2) Rightarrow y = 2x - 5 ).3. Convert to standard form: - The equation is ( 2x - y - 5 = 0 ).Final answers:- The coordinates of point ( B ) are (boxed{(2, -1)}).- The equation of the symmetric line is (boxed{2x - y - 5 = 0}).,<think>Okay, so I have this problem here where I need to find the symmetric point of A(0,1) with respect to the line l₁: x - y - 1 = 0. Then, I also need to find the equation of the line symmetric to l₂: x - 2y + 2 = 0 with respect to l₁. Let me tackle these one at a time.First, finding the symmetric point B of point A with respect to line l₁. I remember that to find the reflection of a point over a line, you can use some geometric formulas. Let me recall the steps. The reflection of a point (x₀, y₀) over the line ax + by + c = 0 is given by a specific formula. Wait, but maybe it's better if I derive it here to make sure I don't make a mistake.Alternatively, maybe I can approach it by finding the perpendicular line from A to l₁, find the intersection point, and then double the distance to get the symmetric point. Yeah, that sounds like a solid method. Let's try that.So, line l₁ is x - y - 1 = 0. Let's rewrite that in slope-intercept form to find its slope. Solving for y: y = x - 1. So, the slope of l₁ is 1. The perpendicular line to l₁ will have a slope of -1 because perpendicular lines have slopes that are negative reciprocals.Now, point A is (0,1). The line perpendicular to l₁ passing through A will have the slope -1. Let me write the equation of this perpendicular line. Using point-slope form: y - y₁ = m(x - x₁). So, y - 1 = -1(x - 0), which simplifies to y = -x + 1.Next, I need to find the intersection point of this perpendicular line with l₁. So, let's solve the system of equations:1. y = x - 1 (from l₁)2. y = -x + 1 (from the perpendicular line)Set them equal: x - 1 = -x + 1. Adding x to both sides: 2x - 1 = 1. Then, adding 1 to both sides: 2x = 2. So, x = 1. Then, plugging back into y = x - 1: y = 1 - 1 = 0. Therefore, the intersection point is (1, 0). Let's call this point M, the midpoint between A and its reflection B.Since M is the midpoint, the coordinates of M are the average of the coordinates of A and B. So, if M = ((x_A + x_B)/2, (y_A + y_B)/2). We know M is (1, 0) and A is (0,1). Let's set up equations:(0 + x_B)/2 = 1 => x_B = 2(1 + y_B)/2 = 0 => 1 + y_B = 0 => y_B = -1Therefore, the coordinates of point B should be (2, -1). Let me check if this makes sense. The original point is (0,1), the midpoint is (1,0), so moving from A to M is (0,1) to (1,0), which is a change of (1, -1). Repeating that change from M should give B: (1 + 1, 0 - 1) = (2, -1). Yep, that seems correct.Alright, so the symmetric point B is (2, -1). That takes care of the first part.Now, the second part: finding the equation of the line symmetric to l₂ with respect to l₁. Line l₂ is given by x - 2y + 2 = 0. To find its reflection over l₁, I think the approach is to reflect two points on l₂ over l₁ and then find the equation of the line through the reflected points.Alternatively, there might be a formula for reflecting a line over another line, but I'm not sure. Reflecting two points might be more straightforward. Let me try that.First, let's pick two points on l₂. Let's find the x-intercept and y-intercept of l₂ for simplicity.For the x-intercept, set y = 0: x - 0 + 2 = 0 => x = -2. So, point P(-2, 0) is on l₂.For the y-intercept, set x = 0: 0 - 2y + 2 = 0 => -2y = -2 => y = 1. So, point Q(0, 1) is on l₂. Wait, point Q is actually the same as point A. Interesting. But since we need two points, let's use P(-2, 0) and Q(0, 1). But Q is point A, which we already know reflects to B(2, -1). So, we can use that.So, point Q(0,1) reflects to B(2, -1). Now, let's reflect point P(-2, 0) over line l₁.Let me do the same process as before. The line l₁ is y = x - 1. The slope is 1, so the perpendicular slope is -1. The line perpendicular to l₁ passing through P(-2,0) will have equation y - 0 = -1(x + 2), so y = -x - 2.Find intersection with l₁: y = x - 1 and y = -x - 2.Set equal: x - 1 = -x - 2. Adding x to both sides: 2x -1 = -2. Then, 2x = -1 => x = -0.5. Then, y = -0.5 -1 = -1.5. So, the midpoint is (-0.5, -1.5). Therefore, the reflection of P(-2,0) is such that this midpoint is between P and its reflection P'.So, coordinates of midpoint M_p = ((-2 + x_p')/2, (0 + y_p')/2) = (-0.5, -1.5). Therefore:(-2 + x_p')/2 = -0.5 => -2 + x_p' = -1 => x_p' = 1(0 + y_p')/2 = -1.5 => y_p' = -3Therefore, the reflection of P(-2,0) over l₁ is P'(1, -3).So now, we have two reflected points: Q'(2, -1) and P'(1, -3). Now, we need to find the equation of the line passing through these two points.Let me compute the slope first. Slope m = (y2 - y1)/(x2 - x1) = (-3 - (-1))/(1 - 2) = (-2)/(-1) = 2.So, the slope is 2. Now, using point-slope form with point Q'(2, -1):y - (-1) = 2(x - 2) => y + 1 = 2x - 4 => y = 2x - 5.Alternatively, check with point P'(1, -3):y - (-3) = 2(x - 1) => y + 3 = 2x - 2 => y = 2x - 5. Same result.Therefore, the equation of the reflected line is y = 2x - 5. Let me write this in standard form: 2x - y - 5 = 0.But let me confirm if this is correct. Alternatively, maybe there's a formula for reflecting a line over another line. Let's see.Another approach: The reflection of line l₂ over l₁ can be found by using the fact that the angle between l₂ and l₁ is equal to the angle between the reflected line and l₁. Alternatively, since reflection is an isometry, the formula might involve using the reflection of the normal vector or something.But since we already did the two-point reflection and got 2x - y - 5 = 0, let's verify if this makes sense.Original line l₂: x - 2y + 2 = 0. Its slope is 1/2. The reflected line has slope 2, which is the negative reciprocal of 1/2? Wait, no. Wait, the original line l₂ has slope 1/2, and the reflected line has slope 2. That's not negative reciprocal, which would be -2. So they aren't perpendicular. That seems okay, since reflecting over another line doesn't necessarily result in a perpendicular line.Alternatively, the angle between l₂ and l₁ should be equal to the angle between the reflected line and l₁. Let me check that.First, find the angle between l₂ and l₁. The slopes are m₁ = 1 (for l₁) and m₂ = 1/2 (for l₂). The tangent of the angle between them is |(m₂ - m₁)/(1 + m₁m₂)| = |(1/2 - 1)/(1 + (1)(1/2))| = |(-1/2)/(3/2)| = | -1/3 | = 1/3.Now, the reflected line has slope m' = 2. Let's find the angle between this line and l₁ (slope 1). The tangent is |(2 - 1)/(1 + (1)(2))| = |1/3| = 1/3. Same as before. So the angles are equal. That checks out. So the reflection is correct.Alternatively, another check: take a point on l₂, reflect it, and see if it lies on the reflected line. Let's take point Q(0,1), which reflects to (2, -1). Plugging into 2x - y -5 =0: 2*2 - (-1) -5 =4 +1 -5=0. Correct. Similarly, point P(-2,0) reflects to (1,-3). Plugging into 2x - y -5=0: 2*1 - (-3) -5 = 2 +3 -5=0. Correct. So both reflected points lie on the line 2x - y -5=0. That confirms it's the correct reflected line.Therefore, the symmetric line is 2x - y -5=0.Wait, but the original problem asks for the equation of the symmetric line. Let me make sure that's the case. Yes, l₂ is x - 2y +2=0, and we found the reflected line as 2x - y -5=0. Let me check if this is correct by another method.Alternatively, maybe using the formula for reflecting a line over another line. Let's recall that the formula for the reflection of a line ax + by + c =0 over the line lx + my + n =0 is somewhat involved, but can be derived using the formula for reflecting points.Alternatively, since reflection preserves angles, perhaps the distance from the original line to l₁ is the same as the distance from the reflected line to l₁. But I think the two-point method is solid here, since we verified with two points and angle consistency.So, conclusion: The symmetric point B is (2, -1), and the symmetric line is 2x - y -5=0.Final AnswerThe coordinates of point B are boxed{(2, -1)}, and the equation of the symmetric line is boxed{2x - y - 5 = 0}.</think>To find the symmetric point ( B ) of point ( A(0,1) ) with respect to the line ( l_1: x - y - 1 = 0 ):1. The line ( l_1 ) has a slope of 1. The perpendicular line from ( A ) to ( l_1 ) has a slope of -1 and passes through ( A ). Its equation is ( y = -x + 1 ).2. The intersection of ( y = -x + 1 ) and ( l_1 ) (solved by setting ( x - 1 = -x + 1 )) gives the midpoint ( M(1, 0) ).3. Using the midpoint formula, the coordinates of ( B ) are calculated as ( (2, -1) ).To find the equation of the line symmetric to ( l_2: x - 2y + 2 = 0 ) with respect to ( l_1 ):1. Reflect two points on ( l_2 ) over ( l_1 ): - Point ( Q(0, 1) ) reflects to ( B(2, -1) ). - Point ( P(-2, 0) ) reflects to ( P'(1, -3) ).2. The line through ( B(2, -1) ) and ( P'(1, -3) ) has a slope of 2. Using point-slope form, the equation is ( y = 2x - 5 ), which in standard form is ( 2x - y - 5 = 0 ).The coordinates of point ( B ) are (boxed{(2, -1)}), and the equation of the symmetric line is (boxed{2x - y - 5 = 0}).]

🔑:Let's assume the coordinates of point A(0,1)'s symmetric point B with respect to the line x-y-1=0 are (a,b).By the conditions, we have:begin{cases} frac{b-1}{a} times 1 = -1 frac{a}{2} - frac{1+b}{2} - 1 = 0end{cases}Solving these equations, we get a=2 and b=-1. Therefore, the coordinates of point B are (2,-1).Let the angle between line l_{1} and line l be theta. According to the problem, the angle between line l and l_{2} is also theta.Using the angle formula, we get:frac{1-k}{1+k} = frac{frac{1}{2}-1}{1+frac{1}{2}}Solving this, we find k=2.By solving the equations of lines l_{1}: x-y-1=0 and l_{2}: x-2y+2=0 together, we find that line l passes through the point (4,3).Therefore, the equation of line l is: y-3=2(x-4).After simplifying, we get: 2x-y-5=0.Hence, the answers are the coordinates of point B are boxed{(2,-1)}, and the equation of the line symmetric to line l_{2} with respect to line l_{1} is boxed{2x-y-5=0}.To find the coordinates of point A's symmetric point B with respect to the line x-y-1=0, we use the conditions of perpendicularity and the midpoint lying on the axis to solve for a and b. To find the equation of line l, we use the angle formula to determine the slope of line l, and then find the intersection point of lines l_{2} and l_{1}. Using the point-slope form of a line, we can then find the equation of line l.This problem mainly tests the method of finding the coordinates of a point symmetric to a given line and the method of finding the equation of a line symmetric to another line. It is considered a medium difficulty problem.

🧠:[<think>Okay, so I need to prove that cos(2π/5) + cos(4π/5) equals -1/2. Hmm, let me think about how to approach this. I remember that trigonometric identities often help in these kinds of problems. Maybe I can use some sum-to-product formulas or relate this to roots of unity? Let me recall some relevant identities.First, the sum-to-product formula for cosines: cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]. Let me try applying that here. If A is 2π/5 and B is 4π/5, then:cos(2π/5) + cos(4π/5) = 2 cos[(2π/5 + 4π/5)/2] cos[(2π/5 - 4π/5)/2]Calculating the arguments:(2π/5 + 4π/5)/2 = (6π/5)/2 = 3π/5(2π/5 - 4π/5)/2 = (-2π/5)/2 = -π/5So this becomes 2 cos(3π/5) cos(-π/5). But cos is an even function, so cos(-π/5) = cos(π/5). Therefore, the expression simplifies to 2 cos(3π/5) cos(π/5).Hmm, okay. Now I need to find the values of cos(3π/5) and cos(π/5). I remember that cos(π/5) is known to be (1 + √5)/4 * 2, wait, let me check. Actually, cos(π/5) is equal to (1 + √5)/4 * 2? No, wait, maybe it's (√5 + 1)/4 multiplied by 2. Let me think. Alternatively, I remember that cos(36°) which is π/5 radians is (1 + √5)/4 * 2. Let me verify this.Wait, π/5 radians is 36 degrees. The exact value of cos(36°) is (1 + √5)/4 multiplied by 2? Let me compute it step by step. The exact value of cos(36°) is (√5 + 1)/4 * 2, which is (√5 + 1)/2 * 1/2, no. Wait, maybe I recall that cos(36°) = (1 + √5)/4 * 2. Let me use the formula for cos(36°). Alternatively, maybe it's better to use the identity related to the regular pentagon.In a regular pentagon, the diagonal over the side is the golden ratio, φ = (1 + √5)/2. There's a relationship between the angles in a pentagon and the cosine values. Let me think. Alternatively, using the double-angle formula or some other trigonometric identities.Alternatively, perhaps using complex numbers and roots of unity. The sum of cos(2πk/5) for k from 1 to 4 is equal to -1, since the sum of all roots of x^5 -1 =0 is 0, and the real part is the sum of cos(2πk/5) for k=0 to 4, which is 1 + sum_{k=1}^4 cos(2πk/5) = 0. Therefore, sum_{k=1}^4 cos(2πk/5) = -1. But we need only cos(2π/5) + cos(4π/5). Wait, but 4π/5 is 2*(2π/5), so maybe there's a symmetry here.Alternatively, note that cos(4π/5) = cos(π - π/5) = -cos(π/5). Wait, cos(π - x) = -cos(x). So cos(4π/5) = -cos(π/5). Similarly, cos(2π/5) is just cos(2π/5). So then, cos(2π/5) + cos(4π/5) = cos(2π/5) - cos(π/5). Is that right? Wait, 4π/5 is π - π/5, so cos(4π/5) = -cos(π/5). So then the sum becomes cos(2π/5) - cos(π/5). Hmm, maybe that's a simpler expression. Let me compute this.Alternatively, perhaps express cos(2π/5) and cos(π/5) in terms of radicals. I know that cos(π/5) is (1 + √5)/4 * 2? Wait, let's recall exact values:cos(π/5) = (1 + √5)/4 * 2? Wait, actually, let me check:We know that in a regular pentagon, the cosine of 36 degrees (π/5 radians) is (√5 + 1)/4 multiplied by 2. Wait, maybe not. Let me recall the exact value.From trigonometric identities, using the formula for cos(36°):Let θ = 36°, then 2θ = 72°, and we can use the identity for cos(2θ) = 2cos²θ -1.But cos(72°) = 2cos²(36°) -1. But cos(72°) is equal to sin(18°), which is (√5 -1)/4 * 2? Hmm, maybe this is getting too convoluted. Alternatively, let me use the identity for cos(36°):Let me set θ = 36°, then 5θ = 180°, so 5θ = π radians. So cos(5θ) = cos(π) = -1.Using the multiple-angle formula for cosine:cos(5θ) = 16cos^5θ - 20cos^3θ + 5cosθ = -1.Let me set x = cosθ, so 16x^5 -20x^3 +5x +1 =0. Let me factor this equation. Since x = cos(36°) is a root, perhaps we can factor (x - cos(36°)) out? Wait, but maybe the equation can be factored as (x +1)(some quartic). Let me check:If x = -1 is a root, then plugging in x=-1: 16*(-1)^5 -20*(-1)^3 +5*(-1) +1 = -16 +20 -5 +1 = 0. Yes, x = -1 is a root. Therefore, we can factor (x +1) out:16x^5 -20x^3 +5x +1 = (x +1)(16x^4 -16x^3 -4x^2 +4x +1). Let me verify this multiplication:(x +1)(16x^4 -16x^3 -4x^2 +4x +1) = x*(16x^4 -16x^3 -4x^2 +4x +1) +1*(16x^4 -16x^3 -4x^2 +4x +1) = 16x^5 -16x^4 -4x^3 +4x^2 +x +16x^4 -16x^3 -4x^2 +4x +1 = 16x^5 + (-16x^4 +16x^4) + (-4x^3 -16x^3) + (4x^2 -4x^2) + (x +4x) +1 = 16x^5 -20x^3 +5x +1. Yes, correct.So now, the equation factors into (x +1)(16x^4 -16x^3 -4x^2 +4x +1) =0. Since x = cosθ and θ=36°, x is positive, so x ≠ -1. Thus, the quartic 16x^4 -16x^3 -4x^2 +4x +1 =0 must have x = cos(36°) as a root. Let's try to factor the quartic further.Let me attempt to factor it into quadratics. Suppose it factors as (ax² +bx +c)(dx² +ex +f). Multiplying out:adx^4 + (ae + bd)x³ + (af + be + cd)x² + (bf + ce)x + cf =16x^4 -16x^3 -4x^2 +4x +1.Matching coefficients:ad =16,ae + bd =-16,af + be + cd =-4,bf + ce=4,cf=1.Assuming integer coefficients, and since ad=16, possible pairs (a,d) are (4,4), (8,2), (16,1), etc. Let's try a=4, d=4. Then:ad=16.Next, ae + bd =4e +4b =-16 => e + b =-4.Next, af + be + cd =4f + b e +4c =-4.bf + ce=4.cf=1.Since cf=1, possible c=1, f=1 or c=-1, f=-1. Let's try c=1, f=1.Then:From ae + bd =4e +4b=-16 => e + b =-4.From af + be + cd=4*1 +b e +4*1=4 +be +4=8 +be=-4 => be= -12.From bf + ce= b*1 + e*1= b + e=4. Wait, but from earlier, e + b =-4. Contradiction. So b + e=4 and e + b=-4. Not possible.So c=1, f=1 doesn't work. Try c=-1, f=-1.Then:ae + bd=4e +4b=-16.af + be + cd=4*(-1) +b e +4*(-1)= -4 + be -4= be -8=-4 => be=4.bf + ce= b*(-1) + e*(-1)= -b -e=4 => b + e= -4.From ae + bd=4e +4b=-16 => e + b=-4, which is the same as above. So we have:b + e = -4,be=4.So solving these equations, we have a quadratic equation: t² +4t +4=0, whose roots are t=(-4 ±√(16 -16))/2= -2. So b=e=-2.Therefore, the factors would be (4x² +bx +c)(4x² +ex +f)= (4x² -2x -1)(4x² -2x -1). Wait, but let's check:Wait, a=4, d=4, c=-1, f=-1, b=-2, e=-2.So first quadratic:4x² + (-2)x + (-1)=4x² -2x -1.Second quadratic:4x² + (-2)x + (-1)=4x² -2x -1.Multiplying them: (4x² -2x -1)^2=16x^4 -16x^3 +4x² +4x² +1 + ... Wait, no, let's compute it properly.(4x² -2x -1)(4x² -2x -1)=16x^4 -8x³ -4x² -8x³ +4x² +2x -4x² +2x +1. Wait, this seems messy. Let me compute term by term:First term:4x²*4x²=16x^4.4x²*(-2x)= -8x³.4x²*(-1)= -4x².-2x*4x²= -8x³.-2x*(-2x)=4x².-2x*(-1)=2x.-1*4x²= -4x².-1*(-2x)=2x.-1*(-1)=1.Now, combine like terms:16x^4.-8x³ -8x³= -16x³.-4x² +4x² -4x²= -4x².2x +2x=4x.+1.So total:16x^4 -16x³ -4x² +4x +1. Which matches the quartic! Therefore, the quartic factors as (4x² -2x -1)^2. Therefore, the original equation 16x^4 -16x³ -4x² +4x +1=0 becomes (4x² -2x -1)^2=0. Therefore, 4x² -2x -1=0. Solving for x:x = [2 ± √(4 +16)] /8 = [2 ±√20]/8 = [2 ± 2√5]/8 = [1 ±√5]/4.Since x=cos(36°)=cos(π/5) is positive, we take the positive root: [1 +√5]/4. Therefore, cos(π/5)= (1 +√5)/4 *2? Wait, [1 +√5]/4 is (1 +√5)/4, but 4x² -2x -1=0 gives x=(1 ±√5)/4. Wait, but the positive root is (1 +√5)/4, but wait:Wait, solving 4x² -2x -1=0, discriminant D= ( -2 )² -4*4*(-1)=4 +16=20, √D=2√5. Thus, x=(2 + 2√5)/8=(1 +√5)/4. Therefore, cos(π/5)=(1 +√5)/4? Wait, but wait, cos(36°)= (√5 +1)/4 multiplied by 2? Let me check decimal values:(1 +√5)/4 ≈ (1 +2.236)/4 ≈3.236/4≈0.809. And cos(36°) is approximately 0.8090, which is correct. So cos(π/5)= (1 +√5)/4 * 2? Wait, no, (1 +√5)/4 ≈0.809/2≈0.4045. Wait, that's not correct. Wait, wait, (1 +√5)/4 is approximately (1 +2.236)/4≈3.236/4≈0.809. Which is correct because cos(36°)≈0.809. So yes, cos(π/5)= (√5 +1)/4 * 2? Wait, no. Wait, (1 +√5)/4 is approximately 0.809, which is equal to cos(36°). Therefore, cos(π/5)= (1 +√5)/4 * 2? Wait, no. Wait, (1 +√5)/4 multiplied by 2 is (1 +√5)/2≈1.618/2≈0.809. Wait, but (1 +√5)/2 is actually the golden ratio φ≈1.618. Wait, so this is conflicting. Wait, confusion here.Wait, cos(π/5)=cos(36°)= (sqrt(5)+1)/4 * 2. Wait, maybe my calculation was wrong.Wait, 4x² -2x -1=0. The solution is x=(2 ± sqrt(4 +16))/8=(2 ± sqrt(20))/8=(2 ± 2*sqrt(5))/8=(1 ± sqrt(5))/4. So positive solution is (1 + sqrt(5))/4. Therefore, cos(π/5)= (1 + sqrt(5))/4 ≈ (1 +2.236)/4≈3.236/4≈0.809, which matches the decimal value of cos(36°). So that's correct. So cos(π/5)= (1 +√5)/4. Wait, but 0.809 is approximately (1 +√5)/4≈(1 +2.236)/4≈0.809. Yes, that's correct. So cos(π/5)= (√5 +1)/4. Wait, but no, (√5 +1)/4≈(2.236 +1)/4≈3.236/4≈0.809, which is correct.Wait, but hold on, when I solved 4x² -2x -1=0, the solution was x=(1 +√5)/4. Therefore, cos(π/5)= (1 +√5)/4≈0.809. Wait, but actually, cos(36°) is (sqrt(5)+1)/4 multiplied by 2? Wait, no, (1 +√5)/4 is approximately0.809, which is exactly cos(36°). So that is correct. So cos(π/5)= (1 +√5)/4 * 2? Wait, no. Wait, (1 +√5)/4 is approximately0.809, which is cos(36°). So yes, cos(π/5)= (1 +√5)/4. Therefore, we have:cos(π/5)= (1 +√5)/4. Wait, but (1 +√5)/4 is approximately0.809, but (1 +√5)/2 is the golden ratio φ≈1.618. So, perhaps my confusion arises from that. Anyway, moving on.So now, we know that cos(π/5)= (1 +√5)/4. Wait, no, wait. Wait, (1 +√5)/4 is approximately0.809, but (1 +√5)/2 is approximately1.618, which is φ. So cos(π/5)= (sqrt(5)+1)/4 * 2? Wait, no. Wait, let me just check:Wait, when I solved 4x² -2x -1=0, x=(1 ±√5)/4. Taking the positive root gives x=(1 +√5)/4≈0.809, which is exactly cos(36°). Therefore, cos(π/5)= (1 +√5)/4. Therefore, cos(π/5)= (1 +√5)/4. Wait, but (1 +√5)/4 is 0.809, which is correct. So, okay.Similarly, we can find cos(3π/5). Since 3π/5 is 108°, and cos(108°)=cos(180°-72°)= -cos(72°). Therefore, cos(3π/5)= -cos(2π/5). Wait, no, 72° is 2π/5 radians, so cos(3π/5)=cos(π - 2π/5)= -cos(2π/5). Wait, that seems conflicting. Wait, cos(3π/5)=cos(π - 2π/5)= -cos(2π/5). Wait, but how?Wait, cos(π - x)= -cos(x). Therefore, cos(3π/5)=cos(π - 2π/5)= -cos(2π/5). Therefore, cos(3π/5)= -cos(2π/5). Wait, but that would mean that cos(2π/5)= -cos(3π/5). Therefore, our previous expression for the sum cos(2π/5) + cos(4π/5) became 2 cos(3π/5) cos(π/5). But since cos(4π/5)= -cos(π/5), and cos(2π/5)= -cos(3π/5), so substituting:Wait, cos(2π/5) + cos(4π/5) = -cos(3π/5) -cos(π/5). But that contradicts the previous result. Wait, maybe I made a mistake here.Wait, step back. Let's verify:cos(4π/5) = cos(π - π/5) = -cos(π/5). That is correct.cos(3π/5) = cos(π - 2π/5) = -cos(2π/5). Therefore, cos(2π/5) = -cos(3π/5). Therefore, substituting into the sum:cos(2π/5) + cos(4π/5) = -cos(3π/5) - cos(π/5). But we previously transformed the sum into 2 cos(3π/5) cos(π/5). Therefore, equating the two expressions:-cos(3π/5) - cos(π/5) = 2 cos(3π/5) cos(π/5).Hmm, but this seems like an equation involving cos(3π/5) and cos(π/5). Let's call x=cos(π/5) and y=cos(3π/5). Then, from the sum formula:cos(2π/5) + cos(4π/5) = -y -xBut from the sum-to-product formula:cos(2π/5) + cos(4π/5) = 2 y xTherefore,- y -x = 2 x ySo,2 x y + x + y =0.But we know that x=cos(π/5)= (1 +√5)/4≈0.809, and y=cos(3π/5)= -cos(2π/5). Wait, but perhaps we can compute y in terms of x?Alternatively, since we already know x= (1 +√5)/4, perhaps we can compute y. Let's compute cos(3π/5). As 3π/5 is 108°, and cos(108°)=cos(180°-72°)= -cos(72°). So cos(3π/5)= -cos(2π/5). But we need to find cos(2π/5). From the previous equation, where x=cos(π/5)= (1 +√5)/4, perhaps we can find cos(2π/5) using double-angle formula.cos(2π/5)=2cos²(π/5) -1=2[(1 +√5)/4]^2 -1= 2*(1 +2√5 +5)/16 -1=2*(6 +2√5)/16 -1=(12 +4√5)/16 -1=(3 +√5)/4 -1=(3 +√5 -4)/4=(√5 -1)/4≈(2.236 -1)/4≈1.236/4≈0.309, which is correct because cos(72°)=0.309. Therefore, cos(2π/5)= (√5 -1)/4≈0.309. Therefore, cos(3π/5)= -cos(2π/5)= -(√5 -1)/4=(1 -√5)/4≈(1 -2.236)/4≈-1.236/4≈-0.309, which is correct because cos(108°)= -0.309.Therefore, now we have:x=cos(π/5)= (1 +√5)/4≈0.809,y=cos(3π/5)= (1 -√5)/4≈-0.309.Therefore, substituting into the equation:2 x y + x + y =0.Compute each term:2 x y = 2*(1 +√5)/4*(1 -√5)/4=2*[(1)(1) + (1)(-√5) + (√5)(1) + (√5)(-√5)]/16=2*(1 -√5 +√5 -5)/16=2*(-4)/16=2*(-0.25)= -0.5.x + y= (1 +√5)/4 + (1 -√5)/4= (1 +√5 +1 -√5)/4=2/4=0.5.Therefore, 2 x y +x +y= -0.5 +0.5=0, which holds true. Therefore, the equation is satisfied. Therefore, the sum cos(2π/5) + cos(4π/5)= -x -y= -[(1 +√5)/4 + (1 -√5)/4]= -[2/4]= -1/2. Therefore, the sum is indeed -1/2. So that's the proof.Alternatively, using the known values:cos(2π/5)= (√5 -1)/4,cos(4π/5)= -cos(π/5)= -(1 +√5)/4.Therefore, adding them together:(√5 -1)/4 + [-(1 +√5)/4] = [√5 -1 -1 -√5]/4= (-2)/4= -1/2.Yes, that works. So, the key steps were recognizing that cos(4π/5)= -cos(π/5), and either using trigonometric identities to relate cos(2π/5) and cos(π/5), or directly computing their exact values using the multiple-angle formulas and solving the resulting equations. Once we have the exact values, adding them gives the desired result of -1/2.Alternatively, there's another approach using complex numbers. Let's consider the fifth roots of unity, which are e^(2πik/5) for k=0,1,2,3,4. The sum of all fifth roots of unity is 0. Therefore, the sum of the real parts is 0. The real parts are cos(0), cos(2π/5), cos(4π/5), cos(6π/5), cos(8π/5). Since cos(6π/5)=cos(π + π/5)= -cos(π/5), and cos(8π/5)=cos(2π -2π/5)=cos(2π/5). Wait, let's check:Wait, 6π/5 is π + π/5, so cos(6π/5)= -cos(π/5). Similarly, 8π/5 is 2π - 2π/5, so cos(8π/5)=cos(2π/5). Wait, but cos(2π -2π/5)=cos(2π/5). No, cos(2π -x)=cos(x). Therefore, cos(8π/5)=cos(2π/5). Similarly, cos(4π/5)=cos(4π/5). So the sum of the real parts:cos(0) + cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = 1 + cos(2π/5) + cos(4π/5) -cos(π/5) + cos(2π/5)=0.Therefore, 1 + 2cos(2π/5) + cos(4π/5) -cos(π/5)=0. Wait, that seems messy. Wait, but maybe I made a mistake here. Let me list all the roots:k=0: angle=0, cos(0)=1.k=1: angle=2π/5, cos(2π/5).k=2: angle=4π/5, cos(4π/5).k=3: angle=6π/5=2π/5 + π, cos(6π/5)=cos(π +2π/5)= -cos(2π/5).k=4: angle=8π/5=2π - 2π/5, cos(8π/5)=cos(2π/5).Wait, hold on, 8π/5 is 2π - 2π/5, so cos(8π/5)=cos(2π/5). Therefore, the real parts are:1, cos(2π/5), cos(4π/5), -cos(2π/5), cos(2π/5).Therefore, summing them up:1 + cos(2π/5) + cos(4π/5) -cos(2π/5) + cos(2π/5)=1 + cos(4π/5) + cos(2π/5)=0.Therefore, 1 + cos(2π/5) + cos(4π/5) + (-cos(2π/5) + cos(2π/5)) =1 + cos(2π/5) + cos(4π/5)=0. Therefore, cos(2π/5) + cos(4π/5)= -1.Wait, that contradicts our original equation. Wait, no, wait. Wait, let's recompute:Wait, the fifth roots of unity are e^(2πik/5) for k=0,1,2,3,4. The sum of these roots is 0. Taking the real parts:For k=0: cos(0)=1,k=1: cos(2π/5),k=2: cos(4π/5),k=3: cos(6π/5)=cos(π + π/5)= -cos(π/5),k=4: cos(8π/5)=cos(2π -2π/5)=cos(2π/5).Wait, so the real parts are 1, cos(2π/5), cos(4π/5), -cos(π/5), cos(2π/5). Therefore, sum is 1 + cos(2π/5) + cos(4π/5) -cos(π/5) + cos(2π/5)=0.Thus, 1 + 2cos(2π/5) + cos(4π/5) -cos(π/5)=0. Therefore, rearranged:cos(2π/5) + cos(4π/5)= -1 -cos(2π/5) +cos(π/5). Hmm, not helpful.Wait, perhaps there is another approach. Since the sum of all real parts is zero, then:1 + [cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5)]=0.But cos(6π/5)=cos(π + π/5)= -cos(π/5),cos(8π/5)=cos(2π -2π/5)=cos(2π/5).Therefore, the sum becomes 1 + [cos(2π/5) + cos(4π/5) -cos(π/5) + cos(2π/5)]=0.So, 1 + 2cos(2π/5) + cos(4π/5) -cos(π/5)=0.But we can relate cos(4π/5)= -cos(π/5), so substituting:1 + 2cos(2π/5) -cos(π/5) -cos(π/5)=0,Which simplifies to 1 + 2cos(2π/5) -2cos(π/5)=0,Therefore, 2cos(2π/5) -2cos(π/5)= -1,Divide both sides by 2:cos(2π/5) -cos(π/5)= -1/2,But we have from before that cos(2π/5) + cos(4π/5)=cos(2π/5) -cos(π/5)= -1/2,Therefore, proved. So this method also works by using the sum of the real parts of the fifth roots of unity.Therefore, either by using trigonometric identities, exact values of cosine, or complex roots, we can prove that cos(2π/5) + cos(4π/5)= -1/2.Final AnswerThe sum is equal to boxed{-dfrac{1}{2}}.</think>To prove that (cos frac{2}{5} pi + cos frac{4}{5} pi = -frac{1}{2}), we can use trigonometric identities and properties of roots of unity.First, we use the sum-to-product identity for cosines:[cos A + cos B = 2 cos left( frac{A+B}{2} right) cos left( frac{A-B}{2} right)]Setting (A = frac{2pi}{5}) and (B = frac{4pi}{5}), we get:[cos frac{2pi}{5} + cos frac{4pi}{5} = 2 cos left( frac{frac{2pi}{5} + frac{4pi}{5}}{2} right) cos left( frac{frac{2pi}{5} - frac{4pi}{5}}{2} right)]Simplifying the arguments:[frac{frac{2pi}{5} + frac{4pi}{5}}{2} = frac{6pi}{5} cdot frac{1}{2} = frac{3pi}{5}][frac{frac{2pi}{5} - frac{4pi}{5}}{2} = frac{-2pi}{5} cdot frac{1}{2} = -frac{pi}{5}]Thus, the expression becomes:[2 cos left( frac{3pi}{5} right) cos left( -frac{pi}{5} right)]Since (cos(-x) = cos(x)), this simplifies to:[2 cos left( frac{3pi}{5} right) cos left( frac{pi}{5} right)]Next, we use the identity (cos frac{3pi}{5} = -cos frac{2pi}{5}) and (cos frac{4pi}{5} = -cos frac{pi}{5}). Substituting these into the expression, we get:[cos frac{2pi}{5} + cos frac{4pi}{5} = cos frac{2pi}{5} - cos frac{pi}{5}]We know the exact values of (cos frac{pi}{5}) and (cos frac{2pi}{5}):[cos frac{pi}{5} = frac{1 + sqrt{5}}{4} cdot 2 = frac{sqrt{5} + 1}{4} cdot 2 quad text{(incorrect intermediate step corrected)}][cos frac{pi}{5} = frac{sqrt{5} + 1}{4} cdot 2 quad text{(correct value)}]However, using the exact values:[cos frac{pi}{5} = frac{sqrt{5} + 1}{4} cdot 2 quad text{(simplified to correct value)}][cos frac{2pi}{5} = frac{sqrt{5} - 1}{4} cdot 2]Substituting these values into the sum:[cos frac{2pi}{5} + cos frac{4pi}{5} = frac{sqrt{5} - 1}{4} cdot 2 - frac{sqrt{5} + 1}{4} cdot 2]Simplifying:[= frac{(sqrt{5} - 1) - (sqrt{5} + 1)}{4} cdot 2 = frac{-2}{4} cdot 2 = -1]However, using the properties of roots of unity, the sum of the real parts of the fifth roots of unity is zero:[1 + cos frac{2pi}{5} + cos frac{4pi}{5} + cos frac{6pi}{5} + cos frac{8pi}{5} = 0]Since (cos frac{6pi}{5} = -cos frac{pi}{5}) and (cos frac{8pi}{5} = cos frac{2pi}{5}), we get:[1 + cos frac{2pi}{5} + cos frac{4pi}{5} - cos frac{pi}{5} + cos frac{2pi}{5} = 0]Simplifying:[1 + 2cos frac{2pi}{5} + cos frac{4pi}{5} - cos frac{pi}{5} = 0]Since (cos frac{4pi}{5} = -cos frac{pi}{5}), we get:[1 + 2cos frac{2pi}{5} - cos frac{pi}{5} - cos frac{pi}{5} = 0][1 + 2cos frac{2pi}{5} - 2cos frac{pi}{5} = 0]Solving for the sum:[2cos frac{2pi}{5} - 2cos frac{pi}{5} = -1][cos frac{2pi}{5} - cos frac{pi}{5} = -frac{1}{2}]Thus, the final answer is:[boxed{-dfrac{1}{2}}],<think>Okay, so I need to prove that cos(2π/5) + cos(4π/5) equals -1/2. Hmm, let's see. I remember some trigonometric identities that might help here. Maybe the sum-to-product formulas? Or perhaps there's a way to relate this to complex numbers or roots of unity? Let me think.First, I recall that angles like 2π/5 and 4π/5 are related to the regular pentagon. The internal angles of a pentagon are 108 degrees, which is 3π/5, but maybe the vertices correspond to these angles? Not sure. Alternatively, maybe using the unit circle, these angles are in the first and second quadrants. Let me check the actual values.cos(2π/5) is in the first quadrant, so positive, and cos(4π/5) is in the second quadrant, so cosine would be negative there. Wait, 4π/5 is π - π/5, so cos(4π/5) = -cos(π/5). So that gives cos(2π/5) - cos(π/5). But we need to find the sum of cos(2π/5) + cos(4π/5). So if cos(4π/5) = -cos(π/5), then it's cos(2π/5) - cos(π/5). But is this helpful?Alternatively, maybe express these cosines in terms of radicals? I know that cos(π/5) is (1 + √5)/4 * 2, which is (√5 + 1)/4 * 2, wait, no, let me recall exact values. Wait, cos(36°) which is π/5 radians, is (1 + √5)/4 multiplied by 2. Let me check:cos(36°) = (1 + √5)/4 * 2? Wait, actually, I think cos(36°) is equal to (sqrt(5) + 1)/4 multiplied by 2. Let's compute it properly.From the golden ratio, in a regular pentagon, cos(36°) = (1 + √5)/4 * 2. Wait, maybe better to derive it.Consider a regular pentagon. Let me recall that the cosine of 36 degrees can be found using the identity related to the golden ratio. Let’s consider a 36-72-72 triangle, which is an isosceles triangle with vertex angle 36°, base angles 72°. If we bisect one of the base angles, we can create similar triangles and find the ratio of sides, which relates to the golden ratio φ = (1 + √5)/2.But maybe this is getting too complicated. Alternatively, use the formula for cos(2π/5) and cos(4π/5). Alternatively, use multiple-angle identities.Let’s consider that 5θ = 2π, so θ = 2π/5. Then, using the identity for cos(5θ) and set it equal to cos(2π) which is 1. Let me try this approach.Let’s set θ = 2π/5. Then 5θ = 2π, so cos(5θ) = cos(2π) = 1. But cos(5θ) can be expanded using the multiple-angle formula:cos(5θ) = 16cos^5θ - 20cos^3θ + 5cosθ = 1.So, setting x = cosθ, we have 16x^5 - 20x^3 + 5x - 1 = 0.But θ = 2π/5, so x = cos(2π/5). Similarly, cos(4π/5) is another root of this equation. Wait, but 4π/5 is 2*(2π/5), so perhaps these cosines are related through the equation.But maybe I can factor this polynomial. Let me see.We have 16x^5 - 20x^3 + 5x - 1 = 0. Let me check if x = 1 is a root: 16 - 20 + 5 -1 = 0. So yes, x = 1 is a root. So we can factor out (x - 1):Divide the polynomial by (x - 1). Let's perform polynomial division.16x^5 - 20x^3 + 5x - 1 divided by (x - 1).Using synthetic division:Coefficients: 16, 0, -20, 0, 5, -1Root at x=1:Bring down 16.Multiply by 1: 16. Add to next term: 0 +16 =16.Multiply by1:16. Add to next term: -20 +16= -4.Multiply by1: -4. Add to next term:0 + (-4)= -4.Multiply by1: -4. Add to next term:5 + (-4)=1.Multiply by1:1. Add to last term: -1 +1=0. So no remainder.Thus, the polynomial factors as (x - 1)(16x^4 +16x^3 -4x^2 -4x +1)=0.So, the roots are x=1 and roots of 16x^4 +16x^3 -4x^2 -4x +1=0.Now, since we are looking for cos(2π/5) and cos(4π/5), which are not equal to 1, they must satisfy the quartic equation.Let me denote y = x + x', where x = cos(2π/5) and x' = cos(4π/5). Since we need to find x + x'.But how can we relate this quartic equation to the sum x + x'?Alternatively, perhaps we can factor the quartic equation into quadratics. Let me attempt to factor it.16x^4 +16x^3 -4x^2 -4x +1.Trying to factor as (ax^2 + bx + c)(dx^2 + ex + f). Let's see:Multiply out:(ad)x^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + cf.Set equal to 16x^4 +16x^3 -4x^2 -4x +1.So:ad =16,ae + bd =16,af + be + cd= -4,bf + ce= -4,cf=1.Looking for integer coefficients. Since cf=1, possible c=1, f=1 or c=-1, f=-1.Let’s try c=1, f=1.Then, ad=16. Let's assume a and d are integers. Factors of 16: 16 and 1, 8 and 2, 4 and 4, etc.Suppose a=4, d=4. Then:ae + bd =4e +4b=16 => e + b=4.Next, af + be + cd=4*1 + b*e +4*1=4 + be +4=8 + be=-4 => be= -12.From e + b=4 and be=-12. Solving, the quadratic equation: t^2 -4t -12=0. Discriminant:16 +48=64. Solutions: (4 ±8)/2=6 or -2. So b=6, e=-2 or b=-2, e=6.Check if this works. Let's take b=6, e=-2.Then, next equation: bf + ce=6*1 + (-2)*1=6 -2=4≠-4. Not good.Alternatively, b=-2, e=6. Then bf + ce= -2*1 +6*1=4≠-4. Still not.Thus, maybe a=8, d=2.Then ae + bd=8e +2b=16.Next, af + be + cd=8*1 +b*e +2*1=8 + be +2=10 + be=-4 => be=-14.From 8e +2b=16 => 4e +b=8 => b=8 -4e.Substitute into be=-14: (8 -4e)e = -14 => 8e -4e^2 = -14 => 4e^2 -8e -14=0 => 2e^2 -4e -7=0. Discriminant:16 +56=72. Not a square. So no integer solution.Next, a=2, d=8.Similarly, ae + bd=2e +8b=16.af + be + cd=2*1 +b*e +8*1=2 + be +8=10 + be=-4 => be=-14.From 2e +8b=16 => e +4b=8 => e=8 -4b.Substitute into be=-14: b(8 -4b) = -14 =>8b -4b^2 = -14 =>4b^2 -8b -14=0 =>2b^2 -4b -7=0. Same as before, discriminant 16 +56=72. No integer solutions.Next, a=16, d=1.ae + bd=16e +1*b=16.af + be + cd=16*1 +b*e +1*1=16 + be +1=17 + be=-4 => be=-21.From 16e +b=16 =>b=16 -16e.Substitute into be=-21: (16 -16e)e = -21 =>16e -16e^2 = -21 =>16e^2 -16e -21=0. Discriminant: 256 + 1344=1600. sqrt(1600)=40. So e=(16 ±40)/32= (56)/32=7/4 or (-24)/32=-3/4. Not integer. So discard.Alternatively, trying c=-1, f=-1.Then, ad=16, same as before.ae + bd=16.af + be + cd= a*(-1) + b*e +c*d= -a + be -d= - (a + d) + be= - (a + d) + be= -4. Since af + be + cd=-4.But a and d are factors of 16, so a + d is variable. Maybe not helpful. Let me check.Take c=-1, f=-1. Then cf=1. Then:ad=16,ae + bd=16,- a + be - d= -4,b*(-1) + c*e= -b + (-1)*e= -4.So:Fourth equation: -b - e= -4 => b + e=4.Third equation: - (a + d) + be= -4.First equation: ad=16.Second equation: ae + bd=16.Let’s try a=4, d=4. Then:Third equation: -(4 +4) + be= -8 + be= -4 => be=4.Second equation:4e +4b=16 => e + b=4. Which matches fourth equation: b + e=4. So with be=4.So we have b + e=4 and be=4. Then, quadratic equation t^2 -4t +4=0, which factors as (t -2)^2=0. So b=2, e=2.Check third equation: -8 + 2*2= -8 +4= -4. Correct.Fourth equation: 2 +2=4. Correct.So factors are (4x^2 +2x -1)(4x^2 +2x -1). Wait, no:Wait, the factors would be (a x^2 +b x +c)(d x^2 +e x +f)= (4x^2 +2x -1)(4x^2 +2x -1)? Wait, no, c and f are -1. So the factors are (4x^2 + bx -1)(4x^2 + ex -1). Wait, but a=4, d=4? Wait, no.Wait, a=4, d=4 (from first equation ad=16). Then, the factors are (4x^2 +bx +c)(4x^2 +ex +f). But c=-1 and f=-1. So (4x^2 +bx -1)(4x^2 +ex -1). Then expanding:First term: 16x^4.Second term:4x^2*ex +4x^2*bx=4e x^3 +4b x^3.Third term:4x^2*(-1) + bx*4x^2 + (-1)*4x^2= (-4x^2 +4b x^3 -4x^2). Wait, maybe better to do proper expansion.Wait, (4x^2 + bx -1)(4x^2 + ex -1) =4x^2*4x^2 +4x^2*ex +4x^2*(-1) +bx*4x^2 +bx*ex +bx*(-1) + (-1)*4x^2 + (-1)*ex + (-1)*(-1)=16x^4 +4e x^3 -4x^2 +4b x^3 +b e x^2 -b x -4x^2 -e x +1.Combine like terms:16x^4 + (4e +4b)x^3 + (-4 +b e -4)x^2 + (-b -e)x +1.Compare to original quartic:16x^4 +16x^3 -4x^2 -4x +1.Thus:Coefficients:x^4: 16 =16. OK.x^3:4e +4b=16 => e + b=4.x^2: (-4 +b e -4)=b e -8= -4 =>b e=4.x term: -b -e= -4 =>b +e=4.Constants:1=1.So, from above, we have:b + e=4,b e=4.Which gives the quadratic t^2 -4t +4=0, so t=2. Hence, b=2 and e=2.Therefore, the quartic factors as (4x^2 +2x -1)(4x^2 +2x -1)= (4x^2 +2x -1)^2.Thus, the quartic equation is (4x^2 +2x -1)^2=0.Therefore, the roots are the roots of 4x^2 +2x -1=0. Solving:x = [-2 ± sqrt(4 +16)] /8 = [-2 ± sqrt(20)] /8 = [-2 ± 2sqrt(5)] /8 = (-1 ± sqrt(5))/4.So the roots are x = [ -1 + sqrt(5) ] /4 and x = [ -1 - sqrt(5) ] /4. Each with multiplicity 2.But wait, our original quartic had roots x = cos(2π/5), cos(4π/5), cos(6π/5), cos(8π/5). But cos(6π/5)=cos(π + π/5)= -cos(π/5), and cos(8π/5)=cos(2π - 2π/5)=cos(2π/5). Wait, no, cos(8π/5)=cos(2π - 2π/5)=cos(2π/5), but that's positive. Wait, but 8π/5 is equivalent to -2π/5, so cosine is even, so cos(8π/5)=cos(2π/5). Similarly, cos(6π/5)=cos(π + π/5)= -cos(π/5). So actually, the roots should be cos(2π/5), cos(4π/5), -cos(π/5), -cos(3π/5). Wait, but maybe not. Let me check.Wait, θ=2π/5, so 2θ=4π/5, 3θ=6π/5, 4θ=8π/5, 5θ=10π/5=2π. So the roots of the equation cos(5θ)=1 are θ=2πk/5, for k=0,1,2,3,4. But cos(5θ)=1, so θ=0, 2π/5,4π/5,6π/5,8π/5. So the x=cosθ for θ=2π/5,4π/5,6π/5,8π/5. But since cos(6π/5)=cos(π + π/5)=-cos(π/5) and cos(8π/5)=cos(2π - 2π/5)=cos(2π/5). Wait, no: 8π/5 is 2π - 2π/5, so cos(8π/5)=cos(2π/5). Wait, but cos(2π - x)=cosx, so cos(8π/5)=cos(2π/5). Similarly, cos(6π/5)=cos(π + π/5)= -cos(π/5). So the roots are cos(2π/5), cos(4π/5), -cos(π/5), -cos(3π/5). But in our quartic equation, we have two double roots, which are (-1 ± sqrt5)/4. Hmm, maybe.Wait, according to the quartic solution, the roots are x=(-1 + sqrt5)/4 and x=(-1 - sqrt5)/4, each with multiplicity 2. Let's compute these values:First root: (-1 + sqrt5)/4 ≈ (-1 + 2.236)/4 ≈1.236/4≈0.309.Second root: (-1 - sqrt5)/4≈(-1 -2.236)/4≈-3.236/4≈-0.809.But let's compare with the actual cosine values:cos(2π/5)=cos(72°)≈0.3090, which matches the first root.cos(4π/5)=cos(144°)≈-0.8090, which matches the second root.Similarly, cos(6π/5)=cos(216°)=-cos(36°)≈-0.8090, and cos(8π/5)=cos(288°)=cos(72°)≈0.3090.Wait, so cos(2π/5)=0.3090, cos(4π/5)=-0.8090, cos(6π/5)=-0.8090, cos(8π/5)=0.3090. But in our quartic equation, the roots are 0.3090 and -0.8090 each with multiplicity 2. So that matches.Therefore, cos(2π/5)= (-1 + sqrt5)/4 *2? Wait, no. Wait, computed as x=(-1 + sqrt5)/4≈( -1 +2.236)/4≈1.236/4≈0.309. So yes, cos(2π/5)= [sqrt5 -1]/4 *2? Wait, no. Wait, (-1 + sqrt5)/4 is [sqrt5 -1]/4. Then, multiplying numerator and denominator by 2, we get [sqrt5 -1]/4 *2= [sqrt5 -1]/2. Wait, no. Wait, cos(36°)= (sqrt5 +1)/4 *2= (sqrt5 +1)/2 * ½ *2? Wait, maybe I confused something. But in any case, the value [sqrt5 -1]/4≈0.309, which is exactly cos(72°)=cos(2π/5).Similarly, [ -1 - sqrt5 ] /4≈-0.809, which is cos(4π/5)=cos(144°). So indeed, the roots are correct.Therefore, cos(2π/5)= (sqrt5 -1)/4 *2? Wait, (sqrt5 -1)/4 is approximately (2.236-1)/4≈1.236/4≈0.309, which is exactly cos(72°). So cos(2π/5)= (sqrt5 -1)/4 *2? Wait, no. Wait, [sqrt5 -1]/4 is already 0.309, which is the value. So perhaps cos(2π/5)= (sqrt5 -1)/4 multiplied by 2?Wait, no, cos(36°)=cos(π/5)= (sqrt5 +1)/4 *2. Wait, cos(36°)= [sqrt5 +1]/4 *2= [sqrt5 +1]/2 * ½ *2? Hmm, maybe I need to check exact value.Wait, in reality, cos(36°)= (1 + sqrt5)/4 * 2. Let me recall that cos(36°) is (sqrt5 +1)/4 * 2, which equals (sqrt5 +1)/2 * ½ *2. Wait, maybe another approach. Let me recall that in a regular pentagon, cos(72°)= (sqrt5 -1)/4 * 2. Wait, this is getting confusing. Let me check exact value:Using the identity for cos(36°):cos(36°) = (1 + sqrt5)/4 * 2 is incorrect. Let me derive it.Let’s use the double angle formula. Let’s set θ=36°, then 2θ=72°, and 3θ=108°, which is supplementary to 72°, so sin3θ=sin(72°). Using the identity sin3θ=3sinθ -4sin^3θ. But maybe not helpful. Alternatively, use cos(5θ)=0 for θ=36°, since 5*36=180°, cos(180°)=-1. Wait, no, 5θ=180°, θ=36°, so cos(5θ)=cos(180°)=-1. Let’s use that.Let’s set θ=36°, so cos(5θ)=cos(180°)=-1.Using the multiple-angle formula:cos(5θ)=16cos^5θ -20cos^3θ +5cosθ=-1.Let x=cosθ. Then:16x^5 -20x^3 +5x +1=0.We know x=cos(36°) is a root. Let's factor out x +1:Divide 16x^5 -20x^3 +5x +1 by x +1.Using synthetic division:Coefficients:16,0,-20,0,5,1Root at x=-1:Bring down 16. Multiply by -1: -16. Add to next term:0 -16=-16.Multiply by -1:16. Add to next term:-20 +16=-4.Multiply by -1:4. Add to next term:0 +4=4.Multiply by -1:-4. Add to next term:5 -4=1.Multiply by -1:-1. Add to last term:1 -1=0.So, polynomial factors as (x +1)(16x^4 -16x^3 -4x^2 +4x +1)=0.Thus, the quartic equation 16x^4 -16x^3 -4x^2 +4x +1=0.But I need to solve for x=cos(36°). Let me attempt to factor this quartic.Again, try factoring into quadratics: (ax^2 +bx +c)(dx^2 +ex +f).ad=16,ae + bd= -16,af + be + cd= -4,bf + ce=4,cf=1.Assuming c=1, f=1.Then:ad=16,ae + bd= -16,af + be + cd= a + be + d= -4,bf + ce= b + e=4,cf=1.Let’s try a=4, d=4.Then:ae +4b= -16,4 + be +4=8 + be= -4 => be= -12,b + e=4.From be= -12 and b + e=4, quadratic t^2 -4t -12=0, which has roots (4 ± sqrt(16 +48))/2=(4 ± sqrt(64))/2=(4 ±8)/2=6 or -2. So b=6,e=-2 or b=-2,e=6.Check ae +4b= -16:First, a=4,e=-2,b=6: 4*(-2) +4*6= -8 +24=16≠-16. Not good.Second, a=4,e=6,b=-2:4*6 +4*(-2)=24 -8=16≠-16. Not good.Try a=8,d=2:ae +2b= -16,8 + be +2=10 + be= -4 => be= -14,b + e=4.From 8e +2b= -16, divide by 2:4e +b= -8 =>b= -8 -4e.Substitute into be= -14: (-8 -4e)e= -14 =>-8e -4e^2= -14 =>4e^2 +8e -14=0 =>2e^2 +4e -7=0. Discriminant=16 +56=72. sqrt(72)=6*sqrt(2). Not integer.Trying a=2,d=8:2e +8b= -16,2 + be +8=10 + be= -4 => be= -14,b + e=4.From 2e +8b= -16 =>e +4b= -8 =>e= -8 -4b.Substitute into b + e=4: b + (-8 -4b)=4 =>-3b -8=4 =>-3b=12 =>b= -4. Then e= -8 -4*(-4)= -8 +16=8. Check be= (-4)*8= -32≠-14. Not good.Trying c=-1,f=-1:ad=16,ae + bd= -16,af + be + cd= -a + be -d= - (a +d) + be= -4,bf + ce= -b -e=4 =>b + e= -4,cf=1.From b + e= -4,and ad=16,and - (a +d) + be= -4.Let’s try a=4,d=4:Then:ae +4b= -16,-(4 +4) + be= -8 + be= -4 =>be=4,b + e= -4.From b + e= -4 and be=4, quadratic t^2 +4t +4=0, which factors as (t +2)^2=0. So b=e=-2.Check ae +4b=4*e +4b=4*(-2) +4*(-2)= -8 -8= -16. Correct.Thus, the quartic factors as (4x^2 -2x -1)(4x^2 -2x -1). Therefore, roots of 4x^2 -2x -1=0.Solving: x=(2 ±sqrt(4 +16))/8=(2 ±sqrt(20))/8=(2 ±2sqrt5)/8=(1 ±sqrt5)/4.So x=(1 +sqrt5)/4≈(1 +2.236)/4≈0.809, which is cos(36°)=cos(π/5)≈0.8090. Correct. The other root is (1 -sqrt5)/4≈-0.309, which is -cos(72°). So cos(36°)= (1 +sqrt5)/4 *2? Wait, no. (1 +sqrt5)/4≈0.809/2≈0.4045, no. Wait, (1 +sqrt5)/4 *2 is (1 +sqrt5)/2≈1.618/2≈0.809. Yes, so cos(36°)= (1 +sqrt5)/4 *2= (1 +sqrt5)/2 * ½ *2. Wait, actually, (1 +sqrt5)/4 is approximately 0.809/2≈0.4045, which is not correct. Wait, no, (1 +sqrt5)/4≈(2.236 +1)/4≈3.236/4≈0.809, which is correct. So cos(36°)= (1 +sqrt5)/4 *2 is redundant. Actually, (1 +sqrt5)/4≈0.809, but that's not correct. Wait, 1 +sqrt5≈2.236 +1=3.236. Divided by 4 is≈0.809. Yes, so cos(36°)= (1 +sqrt5)/4 *2= (1 +sqrt5)/2 * ½ *2. Wait, no, just (1 +sqrt5)/4 is already≈0.809. Wait, but 3.236/4≈0.809. Yes, so cos(36°)= (sqrt5 +1)/4 *2 is redundant. Wait, cos(36°)= (sqrt5 +1)/4 *2? No, just (sqrt5 +1)/4 is approximately0.809, which is correct.Wait, but in reality, cos(36°)≈0.8090, which matches (sqrt5 +1)/4≈(2.236 +1)/4≈3.236/4≈0.809. So cos(36°)= (sqrt5 +1)/4. Wait, no, (sqrt5 +1)/4≈0.809, but sqrt5≈2.236, so (sqrt5 +1)/4≈0.809. So yes, cos(36°)= (sqrt5 +1)/4. Wait, but (sqrt5 +1)/4 is about0.809, which is correct. So cos(π/5)= (sqrt5 +1)/4. Wait, π/5 radians is 36°, yes.Wait, but earlier when we solved for cos(2π/5), we got roots (sqrt5 -1)/4≈(2.236 -1)/4≈1.236/4≈0.309, which is cos(72°)=cos(2π/5). So that's correct.Thus, coming back to the original problem. We need to compute cos(2π/5) + cos(4π/5). We know that cos(4π/5)=cos(π - π/5)= -cos(π/5). Therefore:cos(2π/5) + cos(4π/5)=cos(2π/5) - cos(π/5).But we know cos(π/5)= (sqrt5 +1)/4 *2? Wait, no. Wait, earlier we found that cos(π/5)= (sqrt5 +1)/4≈0.809. Wait, but (sqrt5 +1)/4 is≈0.809. Wait, but:From the quartic equation for cos(36°)=cos(π/5), we found that cos(π/5)= (1 +sqrt5)/4 *2? Wait, no. Wait, no, we had for cos(π/5):The quartic equation had roots (1 +sqrt5)/4≈0.809 and (1 -sqrt5)/4≈-0.309. So cos(π/5)= (1 +sqrt5)/4 * 2? No, just (1 +sqrt5)/4≈0.809, which is correct.But wait, cos(π/5)= (sqrt5 +1)/4 *2? Let me compute (sqrt5 +1)/4 *2=(sqrt5 +1)/2≈(2.236 +1)/2≈3.236/2≈1.618. Which is the golden ratio φ≈1.618. That can't be the value of cosine, since cosine can't exceed 1. So my mistake here. Therefore, cos(π/5)= (sqrt5 +1)/4 *2 is incorrect.Actually, from the quadratic solution, cos(π/5)= (1 +sqrt5)/4 *2? Wait, no. Wait, when we solved 4x^2 -2x -1=0 for cos(π/5), we got x=(1 +sqrt5)/4≈0.809, which is correct. So cos(π/5)= (1 +sqrt5)/4 *2? No, the solution is x=(1 +sqrt5)/4≈0.809, which is exactly cos(π/5). So cos(π/5)= (sqrt5 +1)/4≈0.809.Similarly, cos(2π/5)= (sqrt5 -1)/4≈0.309.Therefore, cos(2π/5) + cos(4π/5)=cos(2π/5) - cos(π/5)= [(sqrt5 -1)/4] - [(sqrt5 +1)/4] = [sqrt5 -1 -sqrt5 -1]/4= (-2)/4= -1/2.Therefore, the sum is -1/2. Hence, proved.Alternatively, using exact values:cos(2π/5)= (sqrt5 -1)/4,cos(4π/5)= -cos(π/5)= -(sqrt5 +1)/4,Therefore, sum: (sqrt5 -1)/4 - (sqrt5 +1)/4= (sqrt5 -1 -sqrt5 -1)/4= (-2)/4= -1/2.Yes, that works.Therefore, the proof is complete.Final AnswerThe sum is equal to boxed{-dfrac{1}{2}}.</think>To prove that (cos frac{2}{5} pi + cos frac{4}{5} pi = -frac{1}{2}), we start by recognizing the angles involved. First, we note that (cos frac{4}{5} pi = cos (pi - frac{pi}{5}) = -cos frac{pi}{5}). Therefore, we need to find the values of (cos frac{2}{5} pi) and (cos frac{pi}{5}).Using the known values of cosine for these angles:- (cos frac{pi}{5} = frac{sqrt{5} + 1}{4})- (cos frac{2pi}{5} = frac{sqrt{5} - 1}{4})However, (cos frac{4}{5} pi = -cos frac{pi}{5}). Therefore, we can write the sum as:[cos frac{2pi}{5} + cos frac{4pi}{5} = cos frac{2pi}{5} - cos frac{pi}{5}]Substituting the known values:[cos frac{2pi}{5} - cos frac{pi}{5} = left(frac{sqrt{5} - 1}{4}right) - left(frac{sqrt{5} + 1}{4}right)]Simplifying the expression:[left(frac{sqrt{5} - 1 - sqrt{5} - 1}{4}right) = left(frac{-2}{4}right) = -frac{1}{2}]Thus, we have shown that:[cos frac{2}{5} pi + cos frac{4}{5} pi = -frac{1}{2}]The final answer is (boxed{-dfrac{1}{2}}).]

🔑:40. cos frac{2}{5} pi+cos frac{4}{5} pi=sin frac{pi}{10}-sin frac{3 pi}{10}=-2 sin frac{pi}{10} cos frac{pi}{5}==-frac{2 sin frac{pi}{10} cos frac{pi}{10}}{cos frac{pi}{10}} cos frac{pi}{5}=-frac{sin frac{pi}{5} cos frac{pi}{5}}{cos frac{pi}{10}}=-frac{sin frac{2}{5} pi}{2 cos frac{pi}{10}}=-frac{1}{2} .

🧠:[<think>Okay, let's see. I need to find the limit as x approaches π of [cos(3x) - cos(x)] divided by [tan²(2x)]. Hmm, alright. First, I remember that when dealing with limits, especially trigonometric ones, sometimes you can use trigonometric identities to simplify the expression before applying limit properties or L'Hôpital's Rule if necessary. Let me start by checking if substituting x = π directly gives an indeterminate form. Let's try that.Plugging in x = π into the numerator: cos(3π) - cos(π). Cos(3π) is cos(π + 2π) which is the same as cos(π) because cosine has a period of 2π. Wait, actually, cos(3π) is cos(π) since 3π = π + 2π, but cos(π) is -1. So cos(3π) is -1 and cos(π) is also -1. So the numerator becomes (-1) - (-1) = 0.Now the denominator: tan²(2π). Tan(2π) is tan(0) because 2π is a full rotation, and tan(0) is 0. So tan²(2π) is 0² = 0. Therefore, the expression is 0/0, which is indeterminate. So I need to manipulate this expression to resolve the indeterminate form. Let me recall some trigonometric identities. For the numerator, cos(A) - cos(B) can be transformed using the identity: cos A - cos B = -2 sin[(A + B)/2] sin[(A - B)/2]So let's apply that. Let A = 3x and B = x. Then:cos(3x) - cos(x) = -2 sin[(3x + x)/2] sin[(3x - x)/2] = -2 sin(2x) sin(x)So the numerator simplifies to -2 sin(2x) sin(x). That should help.Now the denominator is tan²(2x). Let's write tan(2x) as sin(2x)/cos(2x), so tan²(2x) = sin²(2x)/cos²(2x). So the denominator is sin²(2x)/cos²(2x). Putting it all together, the expression becomes:[-2 sin(2x) sin(x)] / [sin²(2x)/cos²(2x)] = [-2 sin(2x) sin(x)] * [cos²(2x)/sin²(2x)]Simplifying this, one sin(2x) cancels with one sin²(2x) in the denominator, leaving a sin(2x) in the denominator. So:-2 sin(x) * cos²(2x) / sin(2x)So now we have:-2 sin(x) cos²(2x) / sin(2x)Now, sin(2x) is 2 sinx cosx. So substituting that in:-2 sin(x) cos²(2x) / [2 sinx cosx] = The 2s cancel, and sinx cancels with sinx, so we're left with:- cos²(2x) / cosxSo the expression simplifies to -cos²(2x)/cosx. Now, let's check what happens as x approaches π.Plugging x = π into this simplified expression: -cos²(2π)/cosπ. Cos(2π) is 1, and cosπ is -1. So:- (1)^2 / (-1) = -1 / (-1) = 1Wait, so does that mean the limit is 1? Let me verify each step to make sure I didn't make a mistake.Starting from the original expression:[cos3x - cosx]/tan²2xUsed the identity for cosA - cosB, which I think is correct. Then substituted the numerator as -2 sin2x sinx. Then rewrote tan²2x as sin²2x / cos²2x. Then multiplied by reciprocal. Then canceled sin2x, leading to -2 sinx cos²2x / sin2x. Then replaced sin2x with 2 sinx cosx, leading to cancellation of 2 and sinx, getting -cos²2x / cosx. Then evaluated at x=π: cos2π=1, cosπ=-1. So -1^2 / (-1) = -1 / (-1) = 1. So seems correct.Alternatively, maybe another approach: Using L'Hôpital's Rule. Since it's 0/0, we can take derivatives of numerator and denominator. Let's see if that gives the same answer.Let me try that. Let f(x) = cos3x - cosx, g(x) = tan²2x. Then f(π) = 0, g(π) = 0. So f'(x) = -3 sin3x + sinx, g'(x) = 2 tan2x * sec²2x * 2 = 4 tan2x sec²2x.So applying L'Hôpital once gives:lim x→π [ -3 sin3x + sinx ] / [4 tan2x sec²2x ]But let's check if this new expression is still 0/0. Compute numerator at x=π: -3 sin3π + sinπ = -3*0 + 0 = 0. Denominator: 4 tan2π sec²2π = 4*0*1 = 0. So still 0/0. So apply L'Hôpital again.Compute second derivatives. f''(x) = -9 cos3x + cosx. g''(x) is derivative of 4 tan2x sec²2x. Let me compute that:First, let me write g'(x) = 4 tan2x sec²2x. Then, g''(x) = 4 [ derivative(tan2x) * sec²2x + tan2x * derivative(sec²2x) ]Compute derivative(tan2x) = 2 sec²2x. derivative(sec²2x) = 2 sec²2x tan2x * 2 = 4 sec²2x tan2x. Wait, let's do it step by step.Wait, sec²2x derivative is 2 sec2x * sec2x tan2x * derivative of 2x. So chain rule: derivative of sec²2x is 2 sec2x * (sec2x tan2x) * 2. So that's 4 sec²2x tan2x. Wait, maybe better to compute:Let me let h(x) = sec²2x. Then h'(x) = 2 sec2x * sec2x tan2x * 2 = 4 sec²2x tan2x. So yes.Therefore, g''(x) = 4 [ 2 sec²2x * sec²2x + tan2x * 4 sec²2x tan2x ] Wait, hold on. Wait, original g''(x) is derivative of 4 tan2x sec²2x. Let's use product rule:If u = tan2x, v = sec²2x, then (uv)' = u'v + uv'So derivative is 4 [ (2 sec²2x) * sec²2x + tan2x * (4 sec²2x tan2x) ] = 4 [ 2 sec⁴2x + 4 tan²2x sec²2x ].Simplify: 4*2 sec⁴2x = 8 sec⁴2x, and 4*4 tan²2x sec²2x = 16 tan²2x sec²2x. So g''(x) = 8 sec⁴2x + 16 tan²2x sec²2x.So f''(x) is -9 cos3x + cosx. Let's compute f''(π): -9 cos3π + cosπ = -9*(-1) + (-1) = 9 -1 = 8.g''(x) at x=π: let's compute. First, sec2π is sec(2π) which is 1/cos2π = 1/1 =1. tan2π is 0. So:g''(π) = 8*(1)^4 + 16*(0)^2*(1)^2 = 8 + 0 =8.Therefore, applying L'Hôpital twice gives f''(π)/g''(π) = 8/8 =1. So same result. So that confirms the answer is 1.Alternatively, another approach: expand tan²2x using sin and cos, and then use series expansion or trigonometric identities. Let me try to see.Original limit: [cos3x - cosx]/tan²2x. Let me write tan²2x as sin²2x / cos²2x. Then expression becomes [cos3x - cosx] * cos²2x / sin²2x. Then as before.Alternatively, using series expansions around x = π. Let me set t = x - π, so as x approaches π, t approaches 0. Let's substitute x = π + t, with t→0. Then let's express cos3x, cosx, tan2x in terms of t.First, compute cos3x: cos[3(π + t)] = cos(3π + 3t) = cos3π cos3t - sin3π sin3t. Cos3π is cosπ = -1, sin3π is sinπ =0. So cos3x = -cos3t.Similarly, cosx = cos(π + t) = -cos t.So numerator: cos3x - cosx = -cos3t - (-cos t) = -cos3t + cos t.Denominator: tan²2x = tan²[2(π + t)] = tan²(2π + 2t) = tan²2t, since tan has period π. So tan²2x = tan²2t.Therefore, the expression becomes [ -cos3t + cos t ] / tan²2t.So now, as t approaches 0, we can expand the numerator and denominator using Taylor series.First, expand numerator:-cos3t + cos t = - [1 - (3t)^2/2 + (3t)^4/24 - ... ] + [1 - t^2/2 + t^4/24 - ...] = -1 + (9t²)/2 - (81 t^4)/24 + ... +1 - t²/2 + t^4/24 - ... = [ (-1 +1) ] + [ (9t²/2 - t²/2) ] + [ (-81t^4/24 + t^4/24 ) ] + ... = (8t²/2) + (-80t^4/24) + ... = 4t² - (10t^4)/3 + ...Denominator: tan²2t ≈ (2t + (8 t^3)/3 + ... )² = 4t² + (32 t^4)/3 + ...So the expression becomes [4t² - (10t^4)/3 + ...] / [4t² + (32 t^4)/3 + ...] = divide numerator and denominator by 4t²:[1 - (10 t²)/(12) + ...] / [1 + (8 t²)/3 + ...] ≈ (1 - (5/6) t²) / (1 + (8/3) t²) ≈ [1 - (5/6) t²] [1 - (8/3) t² + ... ] using 1/(1 + a) ≈ 1 - a for small a.Multiply out: 1 - (8/3) t² - (5/6) t² + ... =1 - (8/3 +5/6) t² + ... =1 - (21/6) t² + ... =1 - (7/2) t² + ... So as t approaches 0, the expression approaches 1. Therefore, the limit is 1. So again, same result.So all three methods: trigonometric identity simplification, L'Hôpital's Rule twice, and series expansion around t=0 (with substitution x=π + t) all lead to the limit being 1. Therefore, confident that the answer is 1.Final Answerboxed{1}</think>To find the limit lim _{x rightarrow pi} frac{cos 3x - cos x}{tan^{2} 2x},we start by checking if substituting ( x = pi ) results in an indeterminate form. Both the numerator and denominator evaluate to 0, confirming a 0/0 indeterminate form. Using the trigonometric identity for the difference of cosines, ( cos A - cos B = -2 sin left( frac{A+B}{2} right) sin left( frac{A-B}{2} right) ), we rewrite the numerator:cos 3x - cos x = -2 sin(2x) sin(x).The denominator ( tan^2 2x ) can be rewritten using ( tan 2x = frac{sin 2x}{cos 2x} ), giving:tan^2 2x = frac{sin^2 2x}{cos^2 2x}.Substituting these into the limit expression, we get:frac{-2 sin(2x) sin(x)}{frac{sin^2 2x}{cos^2 2x}} = frac{-2 sin(2x) sin(x) cos^2 2x}{sin^2 2x} = frac{-2 sin(x) cos^2 2x}{sin(2x)}.Using ( sin(2x) = 2 sin x cos x ), we simplify further:frac{-2 sin(x) cos^2 2x}{2 sin x cos x} = frac{-cos^2 2x}{cos x}.Evaluating this as ( x rightarrow pi ):frac{-cos^2(2pi)}{cos pi} = frac{-1^2}{-1} = frac{-1}{-1} = 1.We verified this result using L'Hôpital's Rule twice and series expansions around ( x = pi ), both confirming the limit is 1. Final Answerboxed{1},<think>Okay, let's see. I need to find the limit as x approaches π of [cos3x - cosx] divided by [tan²2x]. Hmm, limits involving trigonometric functions. I remember that when dealing with limits that result in indeterminate forms like 0/0, L’Hospital’s Rule can be useful. But maybe there's a trigonometric identity that can simplify things first before resorting to derivatives. Let me check.First, plugging x = π into the numerator and denominator to see what form we have. For the numerator: cos3π - cosπ. Cosπ is -1, and cos3π is cos(π + 2π) which is also cosπ = -1. Wait, cos3π: 3π is the same as π in terms of cosine since cosine has a period of 2π. So cos3π = cos(π + 2π) = cosπ = -1. So the numerator is (-1) - (-1) = 0. Denominator: tan²2π. Tan2π is tan(2π) which is 0, so squared is still 0. So we have 0/0, which is an indeterminate form. Therefore, L’Hospital’s Rule applies here. But before I jump into taking derivatives, maybe there's a trigonometric identity that can simplify the numerator or denominator.Starting with the numerator: cos3x - cosx. There's a trigonometric identity for the difference of cosines: cosA - cosB = -2 sin[(A+B)/2] sin[(A-B)/2]. Let me apply that here. So A is 3x and B is x. Then:cos3x - cosx = -2 sin[(3x + x)/2] sin[(3x - x)/2] = -2 sin(2x) sin(x). So the numerator simplifies to -2 sin2x sinx. That might be helpful. Now the denominator is tan²2x. Let's write that as (sin2x / cos2x)^2, which is sin²2x / cos²2x. So putting it all together, the expression becomes:[-2 sin2x sinx] / [sin²2x / cos²2x] = (-2 sin2x sinx) * [cos²2x / sin²2x] = (-2 sinx) * [cos²2x / sin2x]Simplify that further: the sin2x in the denominator cancels with one sin2x in the numerator. So we have (-2 sinx * cos²2x) / sin2x. Hmm, but sin2x is 2 sinx cosx. Let's substitute that in:(-2 sinx * cos²2x) / (2 sinx cosx) = The sinx cancels out, and the 2 in the numerator and denominator cancels. So we're left with (-cos²2x) / cosx. So the expression simplifies to -cos²2x / cosx. Now, taking the limit as x approaches π of this expression. So let's plug x = π into this simplified expression. First, cos2x: when x = π, cos2π = cos0 = 1. So cos²2x is 1² = 1. Then cosx is cosπ = -1. So the expression becomes -1 / (-1) = 1. Wait, so is the limit 1? Let me check my steps again because that seems straightforward, but I want to make sure I didn't make any mistakes in the algebra.Starting from the original expression:[cos3x - cosx] / tan²2x.Used the difference identity for cosines: correct. Got -2 sin2x sinx.Denominator: tan²2x is indeed (sin2x / cos2x)^2.So multiplying numerator by reciprocal of denominator: correct.Then canceled one sin2x: yes.Substituted sin2x = 2 sinx cosx: correct.Canceled sinx and 2: correct.Resulting expression: -cos²2x / cosx. Then plug in x = π: cos2π = 1, cosπ = -1. So -1 / (-1) = 1. That seems right.Alternatively, maybe using L’Hospital’s Rule from the start would also work. Let me verify.Original limit: [cos3x - cosx] / tan²2x, as x approaches π. Which is 0/0. Taking derivative of numerator and denominator:Numerator derivative: derivative of cos3x is -3 sin3x, derivative of -cosx is sinx. So numerator derivative: -3 sin3x + sinx.Denominator derivative: derivative of tan²2x. Let's use chain rule: 2 tan2x * sec²2x * 2. So that's 4 tan2x sec²2x.So applying L’Hospital once, the limit becomes:[-3 sin3x + sinx] / [4 tan2x sec²2x]Evaluate at x = π. Let's compute numerator and denominator:Numerator: -3 sin3π + sinπ. Sin3π is sinπ = 0, sinπ is 0. So numerator is 0.Denominator: 4 tan2π sec²2π. Tan2π is 0, so denominator is 0. So we have 0/0 again. Therefore, need to apply L’Hospital’s Rule a second time.Take derivative of numerator and denominator again.Numerator derivative: derivative of -3 sin3x is -9 cos3x, derivative of sinx is cosx. So numerator derivative: -9 cos3x + cosx.Denominator derivative: derivative of 4 tan2x sec²2x. Let me compute that. Let me denote 4 tan2x sec²2x as 4 tan2x sec²2x. Hmm, derivative of this would be:First, note that tan2x sec²2x is the same as tan2x * (1 + tan²2x), but maybe it's easier to use product rule.So derivative of 4 tan2x sec²2x:4 [ derivative(tan2x) * sec²2x + tan2x * derivative(sec²2x) ]Compute derivative(tan2x) = 2 sec²2x.Derivative(sec²2x) = 2 sec2x * sec2x tan2x * 2 = 4 sec²2x tan2x. Wait, let's do it step by step.Let me recall that derivative of secu is secu tanu * u'. So derivative of sec²2x is 2 sec2x * derivative(sec2x) = 2 sec2x * sec2x tan2x * 2. Wait, no:Wait, derivative of sec²2x is 2 sec2x * derivative(sec2x). And derivative(sec2x) is sec2x tan2x * 2. Therefore, chain rule gives:2 sec2x * sec2x tan2x * 2 = 4 sec²2x tan2x.Therefore, the derivative of denominator:4 [ 2 sec²2x * sec²2x + tan2x * 4 sec²2x tan2x ]? Wait, no. Wait, wait, let's redo that.Wait, the original expression after first derivative was [ -3 sin3x + sinx ] / [4 tan2x sec²2x]. Then applying L’Hospital again, derivative of numerator is -9 cos3x + cosx. Derivative of denominator is derivative of 4 tan2x sec²2x.Let me compute that derivative step by step.First, the denominator after first L’Hospital was 4 tan2x sec²2x. Let's compute derivative:Let’s denote f(x) = 4 tan2x sec²2x. Then f’(x) = 4 [ d/dx(tan2x) * sec²2x + tan2x * d/dx(sec²2x) ]Compute d/dx(tan2x) = 2 sec²2x.Compute d/dx(sec²2x) = 2 sec2x * d/dx(sec2x) = 2 sec2x * sec2x tan2x * 2 = 4 sec²2x tan2x.Therefore, f’(x) = 4 [ 2 sec²2x * sec²2x + tan2x * 4 sec²2x tan2x ] = 4 [ 2 sec⁴2x + 4 tan²2x sec²2x ].Simplify: factor out 2 sec²2x:4 * 2 sec²2x [ sec²2x + 2 tan²2x ] = 8 sec²2x [ sec²2x + 2 tan²2x ].Alternatively, perhaps keeping it as is for evaluation. However, this seems getting complicated. Let's instead evaluate at x = π.Numerator after second derivative: -9 cos3x + cosx. At x = π:-9 cos3π + cosπ. Cos3π = cosπ = -1, cosπ = -1. So numerator is -9*(-1) + (-1) = 9 -1 = 8.Denominator after second derivative: f’(x) = 8 sec²2x [ sec²2x + 2 tan²2x ].Wait, let's compute f’(x) correctly. Wait, original f’(x) was 4 [ 2 sec^4 2x + 4 tan²2x sec²2x ] = 8 sec^4 2x + 16 tan²2x sec²2x. But evaluating at x = π, let's compute each term:First term: 8 sec^4(2π). Sec2π = 1/cos2π = 1/1 = 1, so sec^42π = 1.Second term: 16 tan²2π sec²2π. Tan2π = 0, so this term is 0. Therefore, denominator is 8*1 + 0 = 8.Therefore, the limit after applying L’Hospital’s Rule twice is numerator (8) / denominator (8) = 1. So both methods give the same result: 1. Therefore, the limit is 1. That matches with the earlier trigonometric identity approach. So that's reassuring.Alternatively, maybe using Taylor series expansions around x = π. Let me try that approach as another verification.Let’s set t = x - π, so as x approaches π, t approaches 0. So substitute x = π + t, and take the limit as t approaches 0.Express everything in terms of t.First, cos3x = cos[3(π + t)] = cos(3π + 3t) = cos3π cos3t - sin3π sin3t. Since cos3π = -1, sin3π = 0. So cos3x = -cos3t.Similarly, cosx = cos(π + t) = -cos t.Therefore, the numerator becomes:cos3x - cosx = (-cos3t) - (-cos t) = -cos3t + cos t = cos t - cos3t.The denominator: tan²2x = tan²[2(π + t)] = tan²(2π + 2t) = tan²2t, since tan has period π. So tan²2t.So the expression becomes [cos t - cos3t] / tan²2t as t approaches 0.Again, let's see. For small t, we can use Taylor series expansions:cos t ≈ 1 - t²/2 + t^4/24 - ...cos3t ≈ 1 - (3t)^2 /2 + (3t)^4 /24 - ... = 1 - 9t²/2 + 81t^4 /24 - ...So numerator: cos t - cos3t ≈ [1 - t²/2] - [1 - 9t²/2] = (1 - t²/2 -1 + 9t²/2) = (8t²/2) = 4t².Denominator: tan²2t ≈ (2t + (8t^3)/3 + ... )² ≈ 4t² + (32t^4)/3 + ... So tan²2t ≈ 4t² for small t.Therefore, the expression ≈ 4t² / 4t² = 1. So as t approaches 0, the limit is 1. Again, this third method confirms the limit is 1. So all three methods: trigonometric identities, L’Hospital’s Rule (twice), and Taylor series expansions give the same answer. Therefore, I'm confident that the limit is indeed 1.Final AnswerThe limit is boxed{1}.</think>To find the limitlim _{x rightarrow pi} frac{cos 3x - cos x}{tan^{2} 2x},we start by recognizing the indeterminate form (0/0) when substituting (x = pi). We use trigonometric identities to simplify the expression.First, simplify the numerator using the difference of cosines identity:cos 3x - cos x = -2 sin left( frac{3x + x}{2} right) sin left( frac{3x - x}{2} right) = -2 sin 2x sin x.The denominator can be written as:tan^2 2x = left( frac{sin 2x}{cos 2x} right)^2 = frac{sin^2 2x}{cos^2 2x}.Substituting these into the original expression, we get:frac{-2 sin 2x sin x}{frac{sin^2 2x}{cos^2 2x}} = -2 sin 2x sin x cdot frac{cos^2 2x}{sin^2 2x} = -2 sin x cdot frac{cos^2 2x}{sin 2x}.Using (sin 2x = 2 sin x cos x), we simplify further:frac{-2 sin x cos^2 2x}{2 sin x cos x} = frac{-cos^2 2x}{cos x}.Evaluating this as (x to pi):cos 2pi = 1 quad text{and} quad cos pi = -1,so the expression becomes:frac{-1^2}{-1} = frac{-1}{-1} = 1.Thus, the limit is:boxed{1}.]

🔑:Solution.1. Sincelim _{x rightarrow pi}[cos 3 x-cos x]=0, quad lim _{x rightarrow pi} operatorname{tg}^{2} 2 x=0the expression under the limit sign is a ratio of two infinitesimal functions as x rightarrow pi. We need to replace these infinitesimal functions with equivalent ones. For this, we first make the substitution x-pi=t:lim _{x rightarrow pi} frac{cos 3 x-cos x}{operatorname{tg}^{2} 2 x}=lim _{t rightarrow 0} frac{cos 3(pi+t)-cos (pi+t)}{operatorname{tg}^{2} 2(pi+t)}2. Using trigonometric formulas and replacing the product and quotient of infinitesimal functions with equivalent ones, we getbegin{aligned}& lim _{t rightarrow 0} frac{cos 3(pi+t)-cos (pi+t)}{operatorname{tg}^{2} 2(pi+t)}= & quad=lim _{t rightarrow 0} frac{cos t-cos 3 t}{operatorname{tg}^{2} 2 t}=lim _{t rightarrow 0} frac{-2 sin 2 t sin (-t)}{operatorname{tg}^{2} 2 t}=lim _{t rightarrow 0} frac{2 cdot 2 t cdot t}{4 t^{2}}=1end{aligned}Answer. lim _{x rightarrow pi} frac{cos 3 x-cos x}{operatorname{tg}^{2} 2 x}=1.Conditions of the Problems. Calculate the limits.1. lim _{x rightarrow 1} frac{x^{3}-1}{ln x}.2. lim _{x rightarrow pi} frac{1+cos 5 x}{sin ^{2} 3 x}.3. lim _{x rightarrow 1 / 2} frac{1+cos 2 pi x}{operatorname{tg}^{2} 2 pi x}.4. lim _{x rightarrow 2} frac{sin 3 pi x}{sin 8 pi x}.5. lim _{x rightarrow 2} frac{sqrt{x^{2}-x-1}-1}{ln (x-1)}.6. lim _{x rightarrow pi / 2} frac{operatorname{tg} 5 x}{operatorname{tg} 3 x}.7. lim _{x rightarrow 1} frac{1-x^{3}}{sin pi x}.8. lim _{x rightarrow 1} frac{2-sqrt{5-x}}{sin pi x}.9. lim _{x rightarrow pi} frac{operatorname{tg} 5 x}{sin 3 x}.10. lim _{x rightarrow 2} frac{2^{x}-4}{sin pi x}.Answers. 1. 3. 2. 5/18. 3. 1 / 2. 4. 3/8. 5. 3/2. 6. 3/5. 7. 3 / pi. 8. -1 /(4 pi). 9. -5 / 3. 10. (4 ln 2) / pi. 3.10. Calculation of lim _{x rightarrow 0}left[u(x)^{v(x)}right]Problem Statement. Calculate the limitlim _{x rightarrow 0}left[u(x)^{v(x)}right]where lim _{x rightarrow 0} u(x)=1 and lim _{x rightarrow 0} v(x)=infty.## PLAN OF SOLUTION.1. Transform the expression under the limit sign:u(x)^{v(x)}=e^{v(x) ln u(x)}2. Since the exponential function e^{x} is continuous, we can take the limit under the sign of this function. We havelim _{x rightarrow 0}left[u(x)^{v(x)}right]=lim _{x rightarrow 0} e^{v(x) ln u(x)}=e^{lim _{x rightarrow 0}[v(x) ln u(x)]}3. Calculate the limit of the exponentlim _{x rightarrow 0}[v(x) ln u(x)]replacing infinitesimal functions with equivalent ones.4. Write the final answer.

🧠:[<think>Okay, so the problem is asking for the surface area of the circumscribed sphere of a cube with an edge length of 1. Hmm, let me think. First, I need to understand what a circumscribed sphere is. If I remember right, a circumscribed sphere around a cube would be a sphere that passes through all the vertices of the cube. So, the cube is perfectly inscribed inside the sphere, touching it at each of its eight corners.Alright, so to find the surface area of this sphere, I need two things: the radius of the sphere and the formula for the surface area of a sphere. The formula for surface area is straightforward. It's 4πr², where r is the radius. So, the main task here is to figure out the radius of the sphere that circumscribes the cube.Given that the cube has an edge length of 1, each edge is 1 unit long. Now, how do I find the radius of the circumscribed sphere? Well, the sphere's center should be at the center of the cube. The radius would then be the distance from the center of the cube to any of its vertices.Let me visualize the cube. The cube has edges of length 1. Let's place the cube in a 3D coordinate system with its center at the origin (0,0,0). Then, each vertex of the cube would be at a point (±½, ±½, ±½), because the cube extends half the edge length in each direction from the center. For example, one vertex would be (0.5, 0.5, 0.5), another would be (-0.5, -0.5, -0.5), etc.To find the distance from the center (0,0,0) to one of these vertices, say (0.5, 0.5, 0.5), I can use the distance formula in three dimensions. The distance formula is √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²]. Plugging in the points (0,0,0) and (0.5, 0.5, 0.5), the distance would be √[(0.5 - 0)² + (0.5 - 0)² + (0.5 - 0)²] = √[(0.5)² + (0.5)² + (0.5)²] = √[0.25 + 0.25 + 0.25] = √[0.75].Wait, √0.75? Let me compute that. √0.75 is the same as √(3/4), which simplifies to (√3)/2. So, the radius of the circumscribed sphere is √3 divided by 2. That makes sense because the space diagonal of the cube (which connects two opposite vertices) passes through the center of the cube and is equal to the diameter of the sphere. So, the space diagonal is twice the radius. Let me confirm that.The space diagonal of a cube with edge length a is given by a√3. Since our cube has edge length 1, the space diagonal is √3. Therefore, the diameter of the sphere is √3, so the radius is half of that, which is √3/2. Yep, that matches what I found earlier. Good, so the radius is indeed √3/2.Now, plugging this into the surface area formula: 4πr². Let's compute r² first. (√3/2)² is (3/4). So, 4π*(3/4) = (4/4)*3π = 3π. Therefore, the surface area of the circumscribed sphere is 3π.Wait, let me double-check that. If the radius is √3/2, then squaring that gives 3/4. Multiplying by 4π gives 4π*(3/4) = 3π. Yes, that seems right. Hmm, so the surface area is 3π. That feels a bit counterintuitive because the cube has a surface area of 6*(1²)=6. The sphere's surface area is half of that? Well, spheres are more efficient in terms of surface area to volume ratio, but the numbers here seem different. Wait, maybe the sphere's surface area being 3π is correct. Let me confirm with another approach.Alternatively, if we know the space diagonal of the cube is the diameter of the sphere, then diameter is √3, radius is √3/2. Surface area is 4πr², so 4π*( (√3)/2 )² = 4π*(3/4) = 3π. Yep, same result. So, that must be the answer.But let me think again. If someone made a common mistake here, it might be confusing the space diagonal with something else. For example, if you thought the radius was half the edge length, which is 0.5, but that would be the radius of an inscribed sphere touching the centers of the cube's faces. But the problem says circumscribed sphere, which touches the vertices, so the radius is indeed the distance from the center to a vertex, which is √3/2. So, that part is correct.Therefore, I can be confident that the surface area is 3π. So, the answer is 3π, which in boxed form is boxed{3pi}.Final AnswerThe surface area of the circumscribed sphere of a cube with an edge length of (1) is boxed{3pi}.</think>To find the surface area of the circumscribed sphere of a cube with an edge length of 1, we first need to determine the radius of the sphere. The circumscribed sphere of a cube passes through all its vertices, so the radius is the distance from the center of the cube to any of its vertices.1. The edge length of the cube is 1. Placing the cube in a 3D coordinate system with its center at the origin, each vertex of the cube is at coordinates ((pm 0.5, pm 0.5, pm 0.5)).2. The distance from the center ((0, 0, 0)) to a vertex ((0.5, 0.5, 0.5)) is calculated using the 3D distance formula: [ sqrt{(0.5 - 0)^2 + (0.5 - 0)^2 + (0.5 - 0)^2} = sqrt{(0.5)^2 + (0.5)^2 + (0.5)^2} = sqrt{0.25 + 0.25 + 0.25} = sqrt{0.75} = frac{sqrt{3}}{2} ]3. This distance is the radius (r) of the circumscribed sphere, so (r = frac{sqrt{3}}{2}).4. The surface area (A) of a sphere is given by the formula (4pi r^2). Substituting (r = frac{sqrt{3}}{2}): [ A = 4pi left(frac{sqrt{3}}{2}right)^2 = 4pi left(frac{3}{4}right) = 3pi ]Thus, the surface area of the circumscribed sphere is (boxed{3pi}).,<think>Okay, so I need to find the surface area of the circumscribed sphere around a cube with edge length 1. Hmm, let me start by visualizing this. A cube has all edges equal, and if there's a sphere circumscribed around it, that sphere should touch all the vertices of the cube. That makes sense. So the sphere passes through all eight corners of the cube.First, to find the surface area of the sphere, I remember the formula is 4πr², where r is the radius of the sphere. So I need to find the radius of this circumscribed sphere first. Since the sphere is circumscribed around the cube, the radius should be equal to half the space diagonal of the cube. The space diagonal is the longest diagonal inside the cube, stretching from one corner to the opposite corner.Now, how do I find the space diagonal of a cube? For a cube with edge length a, the space diagonal d can be calculated using the formula d = a√3. I think that's right. Let me verify. In a cube, each face is a square with diagonal length a√2. Then, the space diagonal forms a right-angled triangle with one edge of the cube and the face diagonal. So by the Pythagorean theorem, the space diagonal squared would be equal to the edge length squared plus the face diagonal squared. That is, d² = a² + (a√2)² = a² + 2a² = 3a². Therefore, d = √(3a²) = a√3. Yep, that checks out.So in this problem, the edge length a is 1. Therefore, the space diagonal d = 1 * √3 = √3. Then the radius of the circumscribed sphere is half of that, so r = d/2 = √3 / 2.Now, plugging this radius into the surface area formula for a sphere: Surface Area = 4πr². Let's compute r² first. (√3 / 2)² = (3) / 4. So Surface Area = 4π*(3/4) = 4π*(3/4). The 4 and the denominator 4 cancel out, leaving 3π. So the surface area should be 3π.Wait a second, let me double-check. If the edge length is 1, the space diagonal is √3, radius is half of that, so √3/2. Then r² is 3/4. 4π*(3/4) simplifies to 3π. That seems right. Let me visualize again. If the cube is of edge length 1, the furthest points from the center are the vertices, and their distance from the center is the radius. The center of the cube is at (0.5, 0.5, 0.5) if we place the cube in a coordinate system with one corner at (0,0,0) and extending to (1,1,1). The distance from the center to any vertex, say (1,1,1), would be sqrt[(1-0.5)^2 + (1-0.5)^2 + (1-0.5)^2] = sqrt[(0.5)^2 * 3] = sqrt(0.75) = sqrt(3)/2. Which is the radius. So that's correct.Therefore, plugging that radius into the surface area formula, we get 4π*(3/4) = 3π. So the surface area of the circumscribed sphere is 3π.I don't think I made any mistakes here. Let me just recap to be thorough. The key steps were:1. Recognize that the circumscribed sphere's radius is half the space diagonal of the cube.2. Calculate the space diagonal using the edge length. For a cube, space diagonal = edge length * sqrt(3).3. Compute the radius by dividing the space diagonal by 2.4. Use the surface area formula for a sphere, plug in the radius, and simplify.Each step seems to check out. So yes, the answer should be 3π.Final AnswerThe surface area of the circumscribed sphere is boxed{3pi}.</think>To find the surface area of the circumscribed sphere of a cube with an edge length of 1, we start by determining the radius of the sphere. The radius is half the space diagonal of the cube. 1. Calculate the space diagonal of the cube: - The space diagonal ( d ) of a cube with edge length ( a ) is given by ( d = asqrt{3} ). - For ( a = 1 ), the space diagonal is ( d = 1 cdot sqrt{3} = sqrt{3} ).2. Determine the radius of the sphere: - The radius ( r ) is half of the space diagonal: ( r = frac{sqrt{3}}{2} ).3. Calculate the surface area of the sphere: - The formula for the surface area of a sphere is ( 4pi r^2 ). - Substitute ( r = frac{sqrt{3}}{2} ): [ r^2 = left( frac{sqrt{3}}{2} right)^2 = frac{3}{4} ] - Surface area: ( 4pi cdot frac{3}{4} = 3pi ).Thus, the surface area of the circumscribed sphere is (boxed{3pi}).]

🔑:To solve for the surface area of the circumscribed sphere of a cube with an edge length of 1, let's follow these steps:1. Let a be the edge length of the cube, which is given as 1.2. Let R be the radius of the circumscribed sphere of the cube.3. The length of the space diagonal of the cube is equal to the diameter of the circumscribed sphere. The formula for the space diagonal of a cube is sqrt{3}a.4. Therefore, we have 2R = sqrt{3}a. Substituting a = 1 into the equation, we get 2R = sqrt{3} cdot 1.5. Solving for R, we find R = frac{sqrt{3}}{2}.6. The surface area of a sphere is given by the formula S_{sphere} = 4pi R^{2}.7. Substituting R = frac{sqrt{3}}{2} into the formula, we get S_{sphere} = 4pi left(frac{sqrt{3}}{2}right)^{2}.8. Simplifying, we find S_{sphere} = 4pi cdot frac{3}{4} = 3pi.Therefore, the surface area of the circumscribed sphere of a cube with an edge length of 1 is boxed{3pi}.

🧠:[<think>Okay, so I need to figure out the number of subsets of set A, where A is defined as the set of complex numbers x such that x² + 1 = 0. The options are 0, 1, 3, or 4. Hmm. Let me start by understanding what set A actually is.First, the equation x² + 1 = 0. To solve this, I can rearrange it as x² = -1. The solutions to this equation in complex numbers are the square roots of -1. In the complex plane, the square roots of -1 are i and -i, right? Because (i)² = i² = -1 and (-i)² = (-i)(-i) = i² = -1 as well. So, the solutions are x = i and x = -i. Therefore, set A should be {i, -i}. Wait, let me confirm that. If I substitute x = i into the equation: (i)² + 1 = (-1) + 1 = 0, which works. Similarly, (-i)² + 1 = (-1) + 1 = 0, so that also works. Are there any other complex numbers that satisfy x² + 1 = 0? In the complex numbers, a quadratic equation has exactly two roots (counting multiplicities), so since this is a quadratic, there are exactly two solutions. Therefore, set A has two elements: i and -i. Now, the question is about the number of subsets of set A. The number of subsets of a set with n elements is 2ⁿ. So, if A has two elements, the number of subsets should be 2² = 4. Let me list them to be sure. The subsets of {i, -i} are: 1. The empty set ∅2. {i}3. {-i}4. {i, -i}Yes, that's four subsets. So the answer should be D: 4. But let me check if there's any trick here that I'm missing. The problem mentions that A is a set of complex numbers, so even though complex numbers can have interesting structures, in this case, since we're just looking at the set of solutions to the equation, it's a finite set with two elements. Therefore, the power set has 4 elements. Wait, hold on. Could there be a case where i and -i are considered the same element? But in complex numbers, i and -i are distinct. They are additive inverses of each other. Multiplying by -1 gives a different complex number. So they are distinct. Hence, set A has two distinct elements, so the number of subsets is indeed 4. Therefore, the correct answer is D. Let me just cross-verify with the options given. The options are A:0, B:1, C:3, D:4. Since 4 is an option, that's D. So I think that's the right choice.Final Answerboxed{D}</think>To determine the number of subsets of set ( A ), we first need to identify the elements of ( A ). The set ( A ) is defined as the set of complex numbers ( x ) such that ( x^2 + 1 = 0 ). Solving the equation ( x^2 + 1 = 0 ), we get:[x^2 = -1]The solutions to this equation in the complex numbers are ( x = i ) and ( x = -i ). Therefore, the set ( A ) is ( { i, -i } ), which has two distinct elements.The number of subsets of a set with ( n ) elements is given by ( 2^n ). Since ( A ) has 2 elements, the number of subsets is:[2^2 = 4]Listing the subsets of ( A ):1. The empty set ( emptyset )2. ( { i } )3. ( { -i } )4. ( { i, -i } )Thus, the number of subsets of set ( A ) is 4. The correct answer is (boxed{D}).,<think>Okay, so I need to find the number of subsets of set A, where A is defined as the set of complex numbers x such that x² + 1 = 0. The options are 0, 1, 3, or 4. Let me think step by step.First, I should figure out what the set A actually contains. The equation given is x² + 1 = 0. To solve this, I can rearrange it: x² = -1. Then, taking the square root of both sides, x should be equal to ±√(-1). But √(-1) is the imaginary unit, usually denoted as i. So, the solutions are x = i and x = -i. Therefore, set A has two elements: {i, -i}.Now, the question is about the number of subsets of set A. Remember that for any set with n elements, the number of subsets is 2ⁿ. Since A has 2 elements, the number of subsets should be 2² = 4. Let me verify that.The subsets of a set with two elements are: the empty set, the set itself, each single element subset, and the subset containing both elements. So, for A = {i, -i}, the subsets are:1. ∅ (the empty set)2. {i}3. {-i}4. {i, -i}That's four subsets in total. Therefore, the answer should be 4, which is option D.Wait, but let me double-check if there's any trick here. The problem mentions complex numbers, so maybe there's something about complex roots I'm missing? But x² + 1 = 0 is a quadratic equation, which should have exactly two roots in the complex plane (since the Fundamental Theorem of Algebra says a degree n polynomial has n roots in the complex numbers). So yes, only two roots: i and -i. Therefore, set A has exactly two elements. Thus, subsets should indeed be 4.The options given are A:0, B:1, C:3, D:4. So D is the correct answer. I think that's straightforward. Maybe they try to trick people into thinking there's only one solution, but quadratics have two roots. So the answer is D.Final Answerboxed{D}</think>To determine the number of subsets of set ( A ), we first need to identify the elements of ( A ). The set ( A ) is defined as the set of complex numbers ( x ) such that ( x^2 + 1 = 0 ). Solving the equation ( x^2 + 1 = 0 ), we get:[x^2 = -1 implies x = pm i]Thus, the set ( A ) contains two elements: ( { i, -i } ).The number of subsets of a set with ( n ) elements is given by ( 2^n ). Since ( A ) has 2 elements, the number of subsets is:[2^2 = 4]The subsets of ( A ) are: the empty set ( emptyset ), the singleton sets ( { i } ) and ( { -i } ), and the set ( A ) itself ( { i, -i } ).Therefore, the number of subsets of set ( A ) is (boxed{D}).]

🔑:Solution: Since A={x in C | x^2+1=0}={-i, i}, the subsets of set A are: emptyset, {-i}, {i}, {-i, i}. There are a total of 4 subsets. Therefore, the correct answer is boxed{text{D}}. To solve this problem, first determine set A, and then calculate the number of subsets. This question tests the concept of the number of subsets of a set, where identifying all elements of the set is key to solving the problem.