🧠:Okay, so I need to solve the equation (16 sin pi x cos pi x = 16 x + frac{1}{x}). Let me start by understanding what this equation is asking. First, I notice that on the left side, there's a trigonometric expression involving sine and cosine of πx. Maybe I can simplify that using a trigonometric identity. I remember that ( sin 2theta = 2 sin theta cos theta ), so maybe that applies here. Let me check:If I have (16 sin pi x cos pi x), that can be written as (8 times 2 sin pi x cos pi x), which simplifies to (8 sin 2pi x). So the left side becomes (8 sin 2pi x). That simplifies the equation to:(8 sin 2pi x = 16x + frac{1}{x})Hmm, so that's better. Let me write that down:(8 sin 2pi x = 16x + frac{1}{x})Now, the equation is (8 sin(2pi x) = 16x + frac{1}{x}). Let me divide both sides by 8 to simplify further:(sin(2pi x) = 2x + frac{1}{8x})Okay, so the equation is now (sin(2pi x) = 2x + frac{1}{8x}). My goal is to find all real solutions x that satisfy this equation.Let me think about the domains here. The right side of the equation, (2x + frac{1}{8x}), is defined for all x ≠ 0. So x cannot be zero. The left side, (sin(2pi x)), is defined for all real numbers. So the domain is all real numbers except x = 0.Now, I need to find x ≠ 0 such that (sin(2pi x) = 2x + frac{1}{8x}). Let me analyze both sides of the equation. The left side, (sin(2pi x)), is a sine function with amplitude 1 and period 1. The right side is a function that involves a linear term and a reciprocal term. Let's denote the right side as f(x) = 2x + 1/(8x). I need to find the points where the sine curve intersects with f(x). Since the sine function oscillates between -1 and 1, any solutions must satisfy that f(x) is also between -1 and 1. Therefore, we can first find the regions where ( -1 leq 2x + frac{1}{8x} leq 1 ).So, let's first find the values of x for which (2x + frac{1}{8x} leq 1) and (2x + frac{1}{8x} geq -1).This seems a bit complicated. Maybe I can consider the function f(x) = 2x + 1/(8x) and analyze its behavior.First, let's note that as x approaches positive infinity, 2x dominates and f(x) tends to positive infinity. Similarly, as x approaches negative infinity, 2x tends to negative infinity, so f(x) tends to negative infinity. As x approaches 0 from the positive side, 1/(8x) tends to positive infinity, so f(x) tends to positive infinity. As x approaches 0 from the negative side, 1/(8x) tends to negative infinity, so f(x) tends to negative infinity.Therefore, f(x) has vertical asymptotes at x = 0 and behaves linearly at infinity. The function f(x) is continuous on its domain (all real numbers except x=0). Let's see if f(x) has any critical points where it could attain local maxima or minima. To find that, we can take the derivative of f(x):f'(x) = 2 - 1/(8x²)Set the derivative equal to zero to find critical points:2 - 1/(8x²) = 01/(8x²) = 21/(16x²) = 1Wait, let me solve that again. 2 - 1/(8x²) = 0 => 1/(8x²) = 2 => 8x² = 1/2 => x² = 1/(16) => x = ±1/4So f(x) has critical points at x = 1/4 and x = -1/4. Let's compute the value of f(x) at these points.For x = 1/4:f(1/4) = 2*(1/4) + 1/(8*(1/4)) = 1/2 + 1/(2) = 1/2 + 1/2 = 1For x = -1/4:f(-1/4) = 2*(-1/4) + 1/(8*(-1/4)) = -1/2 + 1/(-2) = -1/2 - 1/2 = -1Interesting! So at x = 1/4, f(x) reaches a local maximum of 1, and at x = -1/4, f(x) reaches a local minimum of -1. Therefore, the range of f(x) is (-∞, -1] ∪ [1, ∞) for x < 0 and x > 0 respectively? Wait, let's check:Wait, for x > 0, f(x) has a minimum at x = 1/4, where f(x) = 1. As x increases beyond 1/4, f(x) increases to infinity. As x approaches 0 from the right, f(x) approaches infinity. So for x > 0, f(x) ≥ 1.Similarly, for x < 0, f(x) has a maximum at x = -1/4, where f(x) = -1. As x decreases past -1/4, f(x) decreases to negative infinity. As x approaches 0 from the left, f(x) approaches negative infinity. So for x < 0, f(x) ≤ -1.But the left side of our equation is sin(2πx), which oscillates between -1 and 1. Therefore, the equation (sin(2πx) = f(x)) can only have solutions when f(x) is between -1 and 1. However, from our analysis:- For x > 0, f(x) ≥ 1, with equality only at x = 1/4.- For x < 0, f(x) ≤ -1, with equality only at x = -1/4.Therefore, the only possible points where f(x) can intersect the sine function (which is between -1 and 1) are exactly at x = 1/4 and x = -1/4, where f(x) is exactly 1 and -1, respectively.But wait, let's verify if these points actually satisfy the original equation.First, check x = 1/4:Left side: sin(2π*(1/4)) = sin(π/2) = 1Right side: 2*(1/4) + 1/(8*(1/4)) = 1/2 + 1/(2) = 1/2 + 1/2 = 1So equality holds. Therefore, x = 1/4 is a solution.Check x = -1/4:Left side: sin(2π*(-1/4)) = sin(-π/2) = -1Right side: 2*(-1/4) + 1/(8*(-1/4)) = -1/2 + 1/(-2) = -1/2 - 1/2 = -1Equality holds here as well. Therefore, x = -1/4 is also a solution.Now, are there any other solutions? Let's think.Since for x > 0, f(x) ≥ 1, and the sine function is ≤ 1. Therefore, the only possible intersection point when x > 0 is at x = 1/4 where both sides equal 1. Similarly, for x < 0, f(x) ≤ -1, and the sine function is ≥ -1. Therefore, the only possible intersection point when x < 0 is at x = -1/4 where both sides equal -1.But wait, let me think again. The sine function is periodic. For example, sin(2πx) = 1 occurs when 2πx = π/2 + 2πk, which implies x = 1/4 + k, where k is an integer. Similarly, sin(2πx) = -1 occurs when 2πx = -π/2 + 2πk, which implies x = -1/4 + k.So, if there are other solutions, they would have to occur at these points where sin(2πx) = ±1, but in such cases, the right-hand side would also have to equal ±1. However, as we saw earlier, except at x = 1/4 and x = -1/4, the right-hand side f(x) is either greater than 1 (for x >0) or less than -1 (for x <0). So, if we take x = 1/4 + k, where k is an integer other than 0, then x would be 1/4 + k. Let's check if for such x, f(x) equals 1.Take k = 1: x = 1/4 + 1 = 5/4. Then f(x) = 2*(5/4) + 1/(8*(5/4)) = 5/2 + 1/(10) = 2.5 + 0.1 = 2.6, which is greater than 1. So sin(2πx) = 1, but f(x) = 2.6 ≠ 1. So no solution here.Similarly, x = 1/4 -1 = -3/4. Let's compute f(-3/4): 2*(-3/4) + 1/(8*(-3/4)) = -3/2 + 1/(-6) ≈ -1.5 - 0.1667 ≈ -1.6667 < -1. But sin(2πx) at x = -3/4 is sin(2π*(-3/4)) = sin(-3π/2) = 1. Wait, sin(-3π/2) = sin(π/2) because sine is periodic with period 2π. Wait, no:Wait, 2πx when x = -3/4 is 2π*(-3/4) = -3π/2. The sine of -3π/2 is sin(-3π/2) = sin(π/2) = 1. Wait, no. sin(-3π/2) = -sin(3π/2) = -(-1) = 1. Wait, sin(3π/2) = -1, so sin(-3π/2) = 1. So sin(2πx) = 1 here, but f(x) ≈ -1.6667 ≠ 1. So no solution here.Similarly, for other integers k, x = 1/4 +k would lead to f(x) being either greater than 1 or less than -1. Therefore, the only solutions are x = 1/4 and x = -1/4.But wait, let me check for x values where sin(2πx) might cross f(x) even if f(x) is outside the range. Wait, no. Because for x >0, f(x) is always ≥1, and sin(2πx) is ≤1. So the only time they can meet is when both are equal to 1. Similarly, for x <0, f(x) ≤ -1, and sin(2πx) ≥ -1, so the only time they can meet is when both are equal to -1. Therefore, the only solutions are x = 1/4 and x = -1/4.But let me test another point just to be sure. For example, take x = 1/2. Then sin(2π*(1/2)) = sin(π) = 0. The right side is 2*(1/2) + 1/(8*(1/2)) = 1 + 1/(4) = 1.25. So 0 ≠ 1.25. So no solution there.Take x = 1/8. Then sin(2π*(1/8)) = sin(π/4) = √2/2 ≈0.707. The right side: 2*(1/8) + 1/(8*(1/8)) = 1/4 + 1/(1) = 1.25. So 0.707 ≈ 1.25? No, not equal.Take x = 1/4. As before, both sides equal 1.Take x = 3/4. sin(2π*(3/4)) = sin(3π/2) = -1. The right side: 2*(3/4) + 1/(8*(3/4)) = 3/2 + 1/(6) ≈1.5 + 0.1667 ≈1.6667. So -1 ≠1.6667.For x negative, let's check x = -1/2. sin(2π*(-1/2))= sin(-π) = 0. The right side: 2*(-1/2) + 1/(8*(-1/2)) = -1 + 1/(-4) = -1 -0.25 = -1.25. 0 ≠ -1.25.x = -1/8. sin(2π*(-1/8)) = sin(-π/4) = -√2/2 ≈-0.707. Right side: 2*(-1/8) + 1/(8*(-1/8)) = -0.25 + 1/(-1) = -0.25 -1 = -1.25. So -0.707 ≈-1.25? Not equal.x = -3/4. sin(2π*(-3/4)) = sin(-3π/2) =1. Right side: 2*(-3/4) + 1/(8*(-3/4)) = -3/2 + 1/(-6) ≈-1.5 -0.1667 ≈-1.6667. So 1 ≠ -1.6667.Thus, all other x's tested do not satisfy the equation. Therefore, the only solutions are x = 1/4 and x = -1/4.But let me check one more thing. Since the equation involves both x and 1/x, maybe there's a symmetry here. Let me substitute x = -y, where y is positive. Then, the equation becomes:sin(2π*(-y)) = 2*(-y) + 1/(8*(-y)) => -sin(2πy) = -2y -1/(8y)Multiply both sides by -1: sin(2πy) = 2y + 1/(8y)Which is the same equation as the original but for y positive. So if x is a negative solution, then y = -x is a positive solution. But we already found that the only positive solution is y = 1/4, so x = -1/4. Therefore, this symmetry confirms that the solutions are symmetric around zero.Another thought: perhaps x = 1/4 and x = -1/4 are the only solutions. But wait, what if there are other points where f(x) equals sin(2πx) even if f(x) is outside [-1,1]? But since sin(2πx) is bounded between -1 and 1, and for x >0, f(x) ≥1, equality can only occur at f(x)=1. Similarly, for x <0, f(x) ≤ -1, equality can only occur at f(x) = -1. Therefore, the only solutions are at the points where f(x) reaches exactly 1 and -1, which are x =1/4 and x=-1/4.Therefore, the solution set is { -1/4, 1/4 }.Wait, but let me check x=0. But x cannot be 0, as division by zero occurs. So those are the only two solutions.But just to be thorough, let's consider the original equation again: (16 sin pi x cos pi x = 16x + frac{1}{x}). We converted it to (8 sin 2πx = 16x + 1/x), then divided by 8 to get ( sin 2πx = 2x + 1/(8x) ). So, the steps are correct. Then, analyzing f(x) and its critical points, leading to the conclusion that only x=1/4 and x=-1/4 satisfy the equation. Therefore, the solution set is { -1/4, 1/4 }.Final AnswerThe solution set is boxed{left{ -dfrac{1}{4}, dfrac{1}{4} right}}.

✅:To solve the equation (16 sin pi x cos pi x = 16 x + frac{1}{x}), we start by simplifying the trigonometric expression on the left side. Using the double-angle identity, we rewrite (16 sin pi x cos pi x) as (8 sin 2pi x). This transforms the equation into:[8 sin 2pi x = 16x + frac{1}{x}]Dividing both sides by 8, we get:[sin 2pi x = 2x + frac{1}{8x}]Next, we analyze the function (f(x) = 2x + frac{1}{8x}). The sine function (sin 2pi x) oscillates between -1 and 1, so we need (f(x)) to be within this range. We find the critical points of (f(x)) by taking its derivative:[f'(x) = 2 - frac{1}{8x^2}]Setting the derivative to zero, we solve:[2 - frac{1}{8x^2} = 0 implies x^2 = frac{1}{16} implies x = pm frac{1}{4}]Evaluating (f(x)) at these critical points:- For (x = frac{1}{4}), (fleft(frac{1}{4}right) = 2 left(frac{1}{4}right) + frac{1}{8 left(frac{1}{4}right)} = 1)- For (x = -frac{1}{4}), (fleft(-frac{1}{4}right) = 2 left(-frac{1}{4}right) + frac{1}{8 left(-frac{1}{4}right)} = -1)These values correspond to the maximum and minimum values of (f(x)). Since (sin 2pi x) must equal these values at these points, we check:- For (x = frac{1}{4}), (sin 2pi left(frac{1}{4}right) = sin frac{pi}{2} = 1)- For (x = -frac{1}{4}), (sin 2pi left(-frac{1}{4}right) = sin left(-frac{pi}{2}right) = -1)Both points satisfy the equation. Further analysis shows that (f(x)) is outside the range ([-1, 1]) for all other (x), confirming no additional solutions.Thus, the solution set is (boxed{left{ -dfrac{1}{4}, dfrac{1}{4} right}}).

🔑:We start with the given equation:[ 16 sin pi x cos pi x = 16 x + frac{1}{x} ]1. Simplify the Left-Hand Side: [ 16 sin pi x cos pi x ] Using the double-angle identity for sine, we know that: [ sin 2theta = 2 sin theta cos theta ] Therefore, let ( theta = pi x ), [ 2 sin pi x cos pi x = sin (2 pi x) ] Thus, [ 16 sin pi x cos pi x = 8 cdot 2 sin pi x cos pi x = 8 sin (2 pi x) ]2. Substitute the simplified expression into the original equation: [ 8 sin (2 pi x) = 16 x + frac{1}{x} ]3. Analyze the equation ( 8 sin (2 pi x) = 16 x + frac{1}{x} ): Considering the range of ( sin (2 pi x) ): [ -1 leq sin (2 pi x) leq 1 ] Therefore, the range of ( 8 sin (2 pi x) ) will be: [ -8 leq 8 sin (2 pi x) leq 8 ]4. At the boundaries, evaluate ( 16 x + frac{1}{x} ): - When ( x > 0 ), [ 16 x + frac{1}{x} geq 8 ] - When ( x < 0 ), [ 16 x + frac{1}{x} leq -8 ] Hence, for these inequalities to hold, the conditions are: [ 16 x + frac{1}{x} = 8 quad text{or} quad 16 x + frac{1}{x} = -8 ]5. Solve for (x): - For ( 16 x + frac{1}{x} = 8 ): [ 16 x^2 + 1 = 8x ] [ 16 x^2 - 8 x + 1 = 0 ] Solving this quadratic equation using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ x = frac{8 pm sqrt{(-8)^2 - 4 cdot 16 cdot 1}}{2 cdot 16} ] [ x = frac{8 pm sqrt{64 - 64}}{32} ] [ x = frac{8 pm 0}{32} ] [ x = frac{1}{4} ] - For ( 16 x + frac{1}{x} = -8 ): [ 16 x^2 + 1 = -8x ] [ 16 x^2 + 8 x + 1 = 0 ] Solving this quadratic equation using the quadratic formula: [ x = frac{-8 pm sqrt{(8)^2 - 4 cdot 16 cdot 1}}{2 cdot 16} ] [ x = frac{-8 pm sqrt{64 - 64}}{32} ] [ x = frac{-8 pm 0}{32} ] [ x = -frac{1}{4} ]6. Conclusion:Therefore, the solution set for the original equation is:[ left{ frac{1}{4}, -frac{1}{4} right} ](boxed{left{ frac{1}{4}, -frac{1}{4} right}})

🧠:Okay, let's tackle this trapezoid problem. So, we have a trapezoid ABCD with AB parallel to CD. The diagonals intersect at point M, dividing the trapezoid into four parts. We know the areas of triangles AMD and DCM are 8 cm² and 4 cm², respectively. We need to find the area of the entire trapezoid.First, I need to visualize the trapezoid. Let me sketch it mentally: AB is the top base, CD is the bottom base, and the legs are AD and BC. The diagonals AC and BD intersect at point M. This intersection creates four triangles: AMD, DCM, and the other two triangles, which must be BMC and AMB. The areas of AMD and DCM are given, so we need to find the areas of the other two triangles and sum them all up.I remember that in a trapezoid, the diagonals divide each other proportionally. That is, the ratio of the segments of one diagonal is equal to the ratio of the segments of the other diagonal. Also, the areas of the triangles formed by the diagonals are related to the lengths of the parallel sides.Let me recall the theorem: In a trapezoid, the ratio of the areas of triangles formed by the intersection of the diagonals is equal to the square of the ratio of the lengths of the two bases. Wait, is that correct? Or is it the ratio of the areas related to the ratio of the bases?Alternatively, maybe the ratio of the areas of triangles that share the same vertex is equal to the ratio of the bases. For example, triangles AMD and DCM share the base MD, but their heights from A and C to MD would be proportional to the lengths of AB and CD since AB is parallel to CD. Hmm, maybe I need to think in terms of similar triangles.Wait, triangles AMD and DCM: are they similar? Let's see. Since AB is parallel to CD, angles at M might be equal. Let me check. Angle AMD is vertically opposite to angle BMC, but maybe that's not directly helpful. Alternatively, angles at M for triangles AMD and CMB? Not sure.Alternatively, since AB is parallel to CD, the triangles AMB and CMD are similar. Wait, is that the case? Let's think. If AB is parallel to CD, then angles at A and C are equal when considering the transversal AC. Hmm, maybe triangles AMB and CMD are similar by AA similarity. Let's verify.In triangle AMB and triangle CMD: angle at M is common? Wait, no. The diagonals intersect at M, so angle AMB is opposite to angle CMD. Are they vertical angles? Wait, when diagonals intersect, the vertical angles are equal. So angle AMB is equal to angle CMD. Also, because AB is parallel to CD, the alternate interior angles would be equal. So angle BAM is equal to angle DCM. Therefore, triangles AMB and CMD are similar by AA similarity.If that's the case, then the ratio of their areas is the square of the ratio of their corresponding sides. Let me denote the ratio of similarity as k. Let’s suppose that the ratio of AB to CD is k. Then the ratio of areas of triangles AMB to CMD would be k². But wait, the areas given are AMD = 8 and DCM = 4. Wait, triangle DCM is triangle DCM, which is part of triangle CMD? Wait, maybe my initial similarity is not correct.Wait, let me clarify the labels. The trapezoid is ABCD, with AB parallel to CD. Diagonals AC and BD intersect at M. The four triangles formed are:1. Triangle AMD (area 8)2. Triangle DCM (area 4)3. Triangle CMB4. Triangle AMBSo, triangles AMD and DCM are adjacent along diagonal AC. Then triangles CMB and AMB are adjacent along diagonal BD.Wait, maybe another approach. Since the diagonals intersect at M, the ratio of the areas of triangles on either side of a diagonal relates to the ratio of the bases. For example, the area of triangle AMD over the area of triangle AMB should be equal to the ratio of MD over MB, because they share the same base AM and have heights proportional to MD and MB. Similarly, since AB is parallel to CD, the ratio MD/MB is equal to the ratio of the lengths of CD/AB.Wait, let me think again. If two triangles share the same base and their heights are in the ratio h1/h2, then their areas are in the ratio h1/h2. Alternatively, if two triangles share the same height, then their areas are in the ratio of their bases.In this case, triangles AMD and AMB share the same vertex A and their bases MD and MB lie on the same line BD. Therefore, the ratio of their areas is equal to the ratio MD/MB. Similarly, triangles DCM and CMB share the same vertex C and their bases MD and MB lie on BD. So their areas are also in the ratio MD/MB.Given that, the area of AMD is 8, and the area of DCM is 4. Let's denote MD/MB = r. Then area AMD / area AMB = r, and area DCM / area CMB = r. Therefore, we can write:8 / area AMB = r, and 4 / area CMB = r. Hence, area AMB = 8 / r, and area CMB = 4 / r.But we also know that triangles AMD and DCM are on the other diagonal AC. Similarly, maybe we can relate the ratio of areas along diagonal AC.Alternatively, since the trapezoid is divided by the diagonals into four triangles, and the areas of two are given, perhaps we can find the ratio of the two bases AB and CD, then express the areas of the other two triangles in terms of that ratio.Let’s denote AB = a and CD = b. Then the ratio of AB to CD is a/b. Because of the properties of the trapezoid, the ratio in which point M divides the diagonals is equal to a/b. That is, AM/MC = BM/MD = a/b.Wait, that seems familiar. In a trapezoid, the diagonals intersect each other in the same ratio. Specifically, the ratio of the segments of each diagonal is equal to the ratio of the lengths of the two bases. So, AM/MC = BM/MD = AB/CD.Let me confirm this. If AB is parallel to CD, then the diagonals intersect at M such that AM/MC = BM/MD = AB/CD. So if AB = a and CD = b, then AM/MC = a/b and BM/MD = a/b.Given that, we can relate the areas of the triangles. Let's first note that triangles that are on the same base and between the same parallels have areas proportional to their heights. But here, since the heights from M to the bases would be in proportion.Alternatively, since the ratio of the segments of the diagonals is a/b, the areas of the triangles can be related through this ratio.Let’s consider triangles AMD and DCM. Wait, triangle AMD has area 8 and triangle DCM has area 4. These two triangles share the same base MD (if we consider them as part of diagonal BD) but have different heights from A and C to BD. Alternatively, they might share a common height if considered from another perspective.Wait, perhaps triangles AMD and CMD share the same base MD and their heights are in the ratio of the distances from A and C to the line BD. However, since AB is parallel to CD, the distance from A to BD is proportional to the distance from C to BD? Not sure. Maybe another approach.Alternatively, since we know the ratio of AM/MC = a/b, and BM/MD = a/b, perhaps we can use this ratio to find the areas of the other triangles.Let’s denote the ratio AB/CD = k. So AM/MC = BM/MD = k.Given that, let's analyze the areas. Let's consider triangles along diagonal AC. Triangles AMD and CMD share the same base MD (if we consider diagonal BD), but maybe another way. Wait, maybe triangles AMD and AMB are along diagonal BD. Let me think.Alternatively, let's consider triangles AMD and AMB. They share the vertex A and their bases are MD and MB on diagonal BD. The ratio of their areas is equal to MD/MB = (MD)/(MB) = 1/k (since BM/MD = k). So area AMD / area AMB = MD/MB = 1/k. But we know area AMD is 8, so 8 / area AMB = 1/k, which gives area AMB = 8k.Similarly, triangles DCM and CMB share vertex C and their bases are MD and MB on diagonal BD. The ratio of their areas is MD/MB = 1/k. Given that area DCM is 4, then 4 / area CMB = 1/k, so area CMB = 4k.Now, the total area of the trapezoid is the sum of the areas of AMD (8), DCM (4), AMB (8k), and CMB (4k). So total area = 8 + 4 + 8k + 4k = 12 + 12k.But we need to find k. How?We can use another relation. Let's look at triangles along diagonal AC. For instance, triangles AMD and AMC. Wait, triangle AMD is part of triangle AMC. Wait, no. Diagonal AC divides the trapezoid into triangles ABC and ADC. The areas of these triangles are equal because in a trapezoid, the areas of the triangles formed by a diagonal are equal. Wait, is that true?Wait, in a trapezoid, the two triangles formed by one of the diagonals (say AC) are triangle ABC and triangle ADC. These two triangles have the same base AC and their heights are the distances from B and D to AC. Since AB is parallel to CD, the heights from B and D to AC should be equal. Therefore, the areas of triangles ABC and ADC should be equal.But triangle ABC is composed of triangles AMB and BMC, and triangle ADC is composed of triangles AMD and DCM. Therefore, area ABC = area AMB + area BMC = area ADC = area AMD + area DCM. So area AMB + area BMC = 8 + 4 = 12.But from earlier, we have area AMB = 8k and area BMC = 4k. Therefore, 8k + 4k = 12k = 12. Therefore, 12k = 12 ⇒ k = 1. Wait, that can’t be right. If k = 1, then AB/CD = 1, meaning AB = CD, which would make the trapezoid a parallelogram. But a trapezoid with both bases equal is a parallelogram, but the problem states it's a trapezoid, so maybe it's allowed. However, in that case, the areas might not make sense. Wait, let's check.If k = 1, then area AMB = 8*1 = 8, and area BMC = 4*1 = 4. Then the total area would be 8 + 4 + 8 + 4 = 24 cm². But if the trapezoid is a parallelogram, then all four triangles formed by the diagonals would have equal areas. But here, AMD is 8 and DCM is 4, which contradicts that. Therefore, there must be a mistake in my reasoning.Wait, so if k = 1, then the areas of AMB and BMC would be 8 and 4, same as AMD and DCM. But in a parallelogram, the diagonals bisect each other, so all four triangles should have equal area. But in this problem, AMD is 8 and DCM is 4, so it's not a parallelogram. Therefore, k cannot be 1. Hence, my previous step where I set 12k = 12 must be wrong.Wait, where did I go wrong? Let's retrace.I said that area ABC = area ADC. But is this true? In a trapezoid, the areas of triangles ABC and ADC are actually equal. Let me confirm. Yes, because both triangles share the base AC, and the heights from B and D to AC are equal (since AB is parallel to CD). Therefore, their areas must be equal.Therefore, area ABC (AMB + BMC) = area ADC (AMD + DCM) = 8 + 4 = 12. Therefore, AMB + BMC = 12. But from earlier, AMB = 8k and BMC = 4k. Therefore, 8k + 4k = 12k = 12 ⇒ k = 1. But this contradicts the given areas unless the trapezoid is a parallelogram. Hmm, this suggests a problem in my approach.Wait, maybe the error is in assuming that area ABC = area ADC. Let me double-check. Take a trapezoid with AB parallel to CD. The area of triangle ABC is (1/2)*AB*h1, where h1 is the height from C to AB. The area of triangle ADC is (1/2)*CD*h2, where h2 is the height from A to CD. But in a trapezoid, the heights from A and C to the opposite bases are the same as the height of the trapezoid. Wait, actually, in a trapezoid, the distance between the two bases (AB and CD) is constant, which is the height of the trapezoid. Therefore, the heights h1 and h2 for triangles ABC and ADC would both be equal to the height of the trapezoid. Therefore, the areas of triangles ABC and ADC would be (1/2)*AB*h and (1/2)*CD*h, respectively. Therefore, unless AB = CD, these areas are not equal. Therefore, my earlier assertion that area ABC = area ADC is incorrect unless it's a parallelogram. Therefore, that was a mistake.So, scratch that approach. Let me find another way.Given that the diagonals intersect at M, and AM/MC = BM/MD = AB/CD = k (let's denote k as the ratio of AB to CD). Let’s set AB = k*CD.We know the areas of triangles AMD and DCM: 8 and 4. Let's see if we can relate these areas to k.Triangles AMD and DCM lie on the same base MD and between the parallels AB and CD. Wait, no. Alternatively, since AB is parallel to CD, the heights of triangles AMD and DCM from A and C to the line BD (which contains MD) would be proportional to AB and CD? Hmm, perhaps.Alternatively, since AM/MC = k, the ratio of areas of triangles AMD and DCM would be equal to k, because they share the base MD and their heights from A and C are proportional to AM and MC. Wait, triangles AMD and CMD share the base MD, and their heights are proportional to AM and MC. Therefore, area AMD / area CMD = AM / MC = k. But the area of AMD is 8, and the area of DCM is 4. Wait, but is triangle CMD the same as DCM?Yes, triangle DCM is the same as triangle CMD. So area AMD / area CMD = 8 / 4 = 2 = k. Therefore, k = 2. Therefore, AB / CD = 2. So AB is twice CD.Therefore, the ratio of the segments of the diagonals is 2:1. So AM/MC = 2/1, and BM/MD = 2/1.Now, knowing that, we can find the areas of the remaining triangles.First, let's consider triangles AMB and BMC. These two triangles share the same height from B to diagonal AC. The ratio of their areas is equal to the ratio of their bases AM and MC, which is 2:1. So area AMB / area BMC = 2 / 1.Similarly, triangles AMB and AMD share the same vertex A and their bases BM and MD on diagonal BD. The ratio BM/MD = 2/1. Since they share the same height from A to BD, the ratio of their areas is 2:1. So area AMB / area AMD = 2 / 1. Therefore, area AMB = 2 * area AMD = 2*8 = 16 cm².Similarly, triangles BMC and DCM share the same vertex C and their bases BM and MD on diagonal BD. The ratio BM/MD = 2/1. Therefore, area BMC / area DCM = 2 / 1 ⇒ area BMC = 2*4 = 8 cm².Alternatively, since area AMB is 16 and area BMC is 8, we can confirm that area AMB / area BMC = 16 / 8 = 2 / 1, which matches the ratio of AM/MC.Now, adding all the areas together: AMD (8) + DCM (4) + AMB (16) + BMC (8) = 8 + 4 + 16 + 8 = 36 cm².Wait, but let's verify this another way. Since AB is twice CD, let's denote CD = x, so AB = 2x. The height of the trapezoid can be found using the areas of the triangles. But maybe not necessary.Alternatively, using the formula for the area of a trapezoid: (AB + CD)/2 * height. We need to find AB and CD in terms of x, but we might need more information.Alternatively, in a trapezoid with diagonals intersecting in the ratio k:1, the areas of the four triangles are proportional to k², k, k, and 1. Wait, but in our case, k = 2. So the areas would be proportional to 4, 2, 2, 1. But in our problem, the given areas are 8 and 4, which would correspond to 4 and 2 in the proportion, scaled by a factor. Wait, maybe scaling factor is 2, so the areas would be 8, 4, 4, 2. But that doesn't match our case. Hmm.Alternatively, if the ratio is 2:1, then the areas of the triangles could be calculated based on the ratios.Alternatively, since AM/MC = 2/1, the area of triangle ABD (which is composed of AMD and AMB) compared to the area of triangle CBD (composed of DCM and BMC). Wait, triangle ABD and triangle CBD share the base BD and their heights are in the ratio of AM/MC = 2/1. Therefore, area ABD / area CBD = 2/1. Area ABD = AMD + AMB = 8 + 16 = 24. Area CBD = DCM + BMC = 4 + 8 = 12. So 24 / 12 = 2/1, which checks out.Similarly, the entire trapezoid area is 24 + 12 = 36 cm². Therefore, the answer is 36 cm².But wait, let me confirm once more. Let's check the areas step by step.1. Given areas: AMD = 8, DCM = 4.2. From the ratio of areas AMD/DCM = 8/4 = 2, which is equal to the ratio AM/MC = AB/CD = 2. So AB = 2*CD.3. The diagonals intersect at M, dividing each diagonal into segments with ratio 2:1.4. Therefore, BM/MD = 2:1 as well.5. Triangles AMB and AMD share the same vertex A and their bases are BM and MD on diagonal BD. The ratio BM/MD = 2/1, so their areas are in the same ratio. Thus, area AMB = 2 * area AMD = 2*8 = 16.6. Similarly, triangles BMC and DCM share vertex C and their bases are BM and MD. Ratio BM/MD = 2/1, so area BMC = 2 * area DCM = 2*4 = 8.7. Sum all areas: 8 (AMD) + 4 (DCM) + 16 (AMB) + 8 (BMC) = 36.Yes, this seems consistent. Let me also think if there's an alternative formula or method to cross-verify.Another approach: The area of the trapezoid can also be calculated if we know the areas of the triangles formed by the diagonals. There's a formula that relates the areas of these triangles to the area of the trapezoid. Specifically, if the areas of triangles AMD and DCM are known, we can find the ratio k, then use that ratio to find the areas of the other two triangles, and sum them all up.Given that the ratio k = AB/CD = 2, we can use the formula for the area of a trapezoid in terms of the areas of the triangles formed by the diagonals. The formula is:Area = (sqrt(area AMD) + sqrt(area BMC))^2 + (sqrt(area DCM) + sqrt(area AMB))^2But I don't recall such a formula. Alternatively, the areas of the triangles can be used to find the ratio k, which in turn helps in calculating the total area.Alternatively, since the ratio of AB to CD is 2:1, let’s denote CD = x, so AB = 2x. The height of the trapezoid can be related to the areas of the triangles. Let’s denote h as the height of the trapezoid.The area of triangle AMD is 8. Triangle AMD has base MD and height from A to MD. Similarly, triangle DCM has base MD and height from C to MD. The ratio of the heights from A and C to MD should be the same as the ratio of AB to CD, which is 2:1.Let’s denote the height from A to MD as h1 and from C to MD as h2. Then h1/h2 = 2/1. The area of AMD is (1/2)*MD*h1 = 8, and the area of DCM is (1/2)*MD*h2 = 4.Let’s write the equations:(1/2)*MD*h1 = 8 => MD*h1 = 16(1/2)*MD*h2 = 4 => MD*h2 = 8So MD*h1 = 16 and MD*h2 = 8. Then dividing these equations: (MD*h1)/(MD*h2) = 16/8 => h1/h2 = 2, which confirms our ratio.Therefore, h1 = 2*h2.Also, the total height of the trapezoid h = h1 + h2 = 2h2 + h2 = 3h2.From MD*h2 = 8, we can write h2 = 8 / MD. Similarly, h1 = 16 / MD.But we might need another relation to find MD. Alternatively, we can relate MD and MB.Since BM/MD = 2/1 (from the ratio AB/CD = 2), let BM = 2*MD. Let’s denote MD = y, then BM = 2y. Therefore, the entire diagonal BD = BM + MD = 3y.But how does this help us? Maybe we can relate MD with the heights h1 and h2.Alternatively, consider triangles AMD and AMB. They share the same height from A to BD, which is h1. The bases are MD and BM. Since BM = 2y and MD = y, the ratio of their areas is BM/MD = 2/1. So area AMB = 2*area AMD = 16, which matches our previous calculation.Similarly, triangles DCM and BMC share the same height from C to BD, which is h2. The bases are MD = y and BM = 2y. Therefore, area BMC = 2*area DCM = 8, which also matches.Now, to find the height of the trapezoid h, we have h = h1 + h2 = 2h2 + h2 = 3h2. From triangle DCM: (1/2)*MD*h2 = 4. We know MD = y, so (1/2)*y*h2 = 4 ⇒ y*h2 = 8. But we also have from triangle AMD: (1/2)*MD*h1 = 8 ⇒ (1/2)*y*(2h2) = 8 ⇒ y*h2 = 8. Which is consistent. So y*h2 = 8.But we need another equation to find h. However, maybe it's not necessary since we already found the total area by summing the areas of the four triangles.Alternatively, the area of the trapezoid can be found using the formula:Area = (AB + CD)/2 * hWe know AB = 2x, CD = x, so Area = (2x + x)/2 * h = (3x/2)*h.But we need to relate x and h. From the triangles, we have h = 3h2, and y*h2 = 8. But we also might relate y to the bases AB and CD.Since the diagonals intersect in the ratio 2:1, we can use coordinate geometry to model the trapezoid and find relations.Let’s place the trapezoid in coordinate system. Let’s set point D at (0, 0), C at (c, 0), since CD is the lower base. Point A is somewhere up, since AB is parallel to CD. Let’s set AB such that point A is at (a, h) and point B is at (b, h). Then AB has length |b - a|, and CD has length c - 0 = c. Since AB is parallel to CD, both are horizontal lines.The diagonals AC and BD intersect at point M. Let's find the coordinates of M.Diagonal AC connects (a, h) to (c, 0). Diagonal BD connects (b, h) to (0, 0).The intersection point M can be found by solving the equations of the two diagonals.Equation of AC: parametric form. Starting at A(a, h), moving towards C(c, 0). The parametric equations are x = a + t(c - a), y = h - t*h, for t in [0, 1].Equation of BD: parametric form. Starting at B(b, h), moving towards D(0, 0). The parametric equations are x = b - s*b, y = h - s*h, for s in [0, 1].To find intersection M, set the coordinates equal:a + t(c - a) = b - s*bh - t*h = h - s*hFrom the y-coordinate equation: h - t*h = h - s*h ⇒ t = s.Substitute t = s into the x-coordinate equation:a + t(c - a) = b - t*ba + t(c - a + b) = bt(c - a + b) = b - at = (b - a)/(c - a + b)Now, the coordinates of M are:x = a + t(c - a) = a + [(b - a)/(c - a + b)]*(c - a)y = h - t*h = h[1 - (b - a)/(c - a + b)] = h*(c - a + b - b + a)/(c - a + b) = h*(c)/(c - a + b)But we know that the ratio AM/MC = 2/1. In terms of the coordinates, since AC is from (a, h) to (c, 0), the point M divides AC in the ratio AM:MC = 2:1. Therefore, the coordinates of M can also be expressed using the section formula:x = (2*c + 1*a)/(2 + 1) = (2c + a)/3y = (2*0 + 1*h)/3 = h/3Similarly, from the parametric equations, we have:x = a + t(c - a) = (2c + a)/3y = h - t*h = h/3 ⇒ t = 2/3From the parametric equations, t = 2/3, so plugging into x-coordinate:x = a + (2/3)(c - a) = (3a + 2c - 2a)/3 = (a + 2c)/3Which matches the section formula.Similarly, from the BD diagonal, point M divides BD in the ratio BM:MD = 2:1. The coordinates from BD should also satisfy this.Coordinates of M from BD: Since BD is from B(b, h) to D(0,0), the coordinates of M should be:x = (2*0 + 1*b)/(2 + 1) = b/3y = (2*0 + 1*h)/3 = h/3But from the previous calculation, x = (a + 2c)/3 and y = h/3. Therefore, equating x-coordinates:(a + 2c)/3 = b/3 ⇒ a + 2c = b.Therefore, b = a + 2c.Now, we have AB = |b - a| = |a + 2c - a| = 2c. CD = c. Therefore, AB = 2*CD, confirming our ratio k = 2.Now, let's relate this to the areas of the triangles.The area of triangle AMD is 8. Let's compute it using coordinates.Points A(a, h), M(b/3, h/3), D(0, 0).Using the formula for the area of a triangle given coordinates:Area = (1/2)| (a*(h/3 - 0) + (b/3)*(0 - h) + 0*(h - h/3) ) |= (1/2)| a*(h/3) + (b/3)*(-h) + 0 |= (1/2)| (ah/3 - bh/3) | = (1/2)| h(a - b)/3 | = (1/6)|h(a - b)|Given that area AMD = 8:(1/6)|h(a - b)| = 8 ⇒ |h(a - b)| = 48But we know that AB = 2c = b - a (since AB is from a to b), so b - a = 2c ⇒ a - b = -2c. Therefore, |h(a - b)| = |h*(-2c)| = 2hc. So:2hc = 48 ⇒ hc = 24Similarly, the area of triangle DCM is 4. Let's compute that.Points D(0,0), C(c,0), M(b/3, h/3).Area = (1/2)|0*(0 - h/3) + c*(h/3 - 0) + (b/3)*(0 - 0)|= (1/2)|0 + c*h/3 + 0| = (1/2)*(ch/3) = ch/6Given that area DCM = 4:ch/6 = 4 ⇒ ch = 24Which matches our previous result hc =24. Therefore, ch =24.So from this, we know that ch =24. Also, since AB =2c, CD =c.The area of the trapezoid is (AB + CD)/2 * h = (2c + c)/2 * h = (3c/2)*h = (3/2)*c*h. Since c*h =24, the area is (3/2)*24 = 36 cm².Therefore, confirming the earlier result. The area of the trapezoid is 36 cm².Final AnswerThe area of the trapezoid is boxed{36} mathrm{cm}^{2}.

✅:Given a trapezoid (ABCD) with (AB parallel CD), the diagonals intersect at point (M), dividing the trapezoid into four parts. The areas of triangles (AMD) and (DCM) are given as (8 mathrm{cm}^2) and (4 mathrm{cm}^2) respectively. We need to determine the area of the entire trapezoid.1. Ratio of Diagonals: - The ratio of the areas of triangles (AMD) and (DCM) is (8:4 = 2:1). - This ratio is equal to the ratio of the segments of the diagonals divided by their intersection point (M). Therefore, (AM/MC = BM/MD = 2:1).2. Areas of Other Triangles: - Triangles (AMB) and (AMD) share the same vertex (A) and their bases (BM) and (MD) are in the ratio (2:1). Thus, the area of (AMB) is twice the area of (AMD), giving (8 times 2 = 16 mathrm{cm}^2). - Similarly, triangles (BMC) and (DCM) share the same vertex (C) and their bases (BM) and (MD) are in the ratio (2:1). Thus, the area of (BMC) is twice the area of (DCM), giving (4 times 2 = 8 mathrm{cm}^2).3. Total Area of the Trapezoid: - Summing the areas of all four triangles: [ text{Total Area} = text{Area of } AMD + text{Area of } DCM + text{Area of } AMB + text{Area of } BMC ] [ = 8 + 4 + 16 + 8 = 36 mathrm{cm}^2 ]Thus, the area of the trapezoid (ABCD) is (boxed{36} mathrm{cm}^{2}).

🔑:Trapezoid ABCD, where side AB is parallel to side CD, is divided by its diagonals intersecting at point M into four parts. We are given that the area of triangle AMD is 8 , mathrm{cm}^2 and the area of triangle DCM is 4 , mathrm{cm}^2. Let’s find the area of the trapezoid ABCD.# Step 1: Establish Angle RelationshipsAngles angle BMA and angle DMC are equal as they are vertically opposite angles. Similarly, angles angle ABM and angle CDM are equal because they are alternate interior angles. Therefore, triangles ABM and CDM are similar by the AA criterion (Angle-Angle similarity).# Step 2: Find Relationship of Heights and AreasLet's denote the height of trapezoid ABCD as v and the height of triangle CDM perpendicular to side CD as x. Given the areas, we write:[S_{CDM} = frac{1}{2} |CD| cdot x = 4 , mathrm{cm}^2,]and[S_{CDA} = frac{1}{2} |CD| cdot v = 12 , mathrm{cm}^2.]From the above, the ratio of the areas gives us:[frac{v}{x} = frac{12}{4} = 3 quad Rightarrow quad v = 3x.]# Step 3: Determine Height of Triangle ABMNext, examine the height of triangle ABM perpendicular to side AB, which is given by v - x. Substituting v:[v - x = 3x - x = 2x.]Since triangles CDM and ABM are similar with a height ratio of 1:2, their area ratio is the square of this ratio:[text{Area ratio} = left(frac{1}{2}right)^2 = frac{1}{4}.]Thus, the area S_{ABM} is:[S_{ABM} = 4 cdot S_{CDM} = 4 cdot 4, mathrm{cm}^2 = 16 , mathrm{cm}^2.]# Step 4: Find Area of Triangle MBCFrom the diagram, we can establish that:[S_{AMD} = S_{ABD} - S_{ABM},]and[S_{MBC} = S_{ABC} - S_{ABM}.]Since S_{ABC} = S_{ABD}, it follows that:[S_{MBC} = S_{AMD} = 8 , mathrm{cm}^2.]# Step 5: Calculate Total Area of Trapezoid ABCDSumming the areas of the triangles:[S_{ABCD} = S_{CDM} + S_{AMD} + S_{ABM} + S_{MBC} = 4 + 8 + 16 + 8 = 36 , mathrm{cm}^2.] ConclusionThe total area of trapezoid ABCD is:[boxed{36, mathrm{cm}^2}]

🧠:Okay, let's try to solve this problem. Hmm, Ellis's Eel Emporium has three types of eels: electric, moray, and freshwater. The notice gives three statements:1. All the eels are electric eels except 12.2. All the eels are moray eels except 14.3. All the eels are freshwater eels except 16.We need to find the total number of eels in the tank. The options are A)42, B)33, C)24, D)21, E)20.Let me think. The statements are a bit confusing at first glance. Let's parse them one by one. First statement: All the eels are electric eels except 12. That means if I subtract the number of electric eels from the total, I get 12. So, total eels minus electric eels equals 12. Similarly, the second statement says all are moray except 14, so total minus moray equals 14. Third statement: total minus freshwater equals 16.Wait, let's write equations for each. Let me denote:Let E = number of electric eels,M = number of moray eels,F = number of freshwater eels,T = total number of eels.From the first statement: All eels except 12 are electric. So, the non-electric eels are 12. So E = T - 12.Wait, wait, no. If all are electric except 12, that means that the number of non-electric eels is 12. So, M + F = 12.Similarly, all are moray except 14. So, non-moray eels are E + F = 14.All are freshwater except 16. So, non-freshwater eels are E + M = 16.So, we have three equations:1. M + F = 122. E + F = 143. E + M = 16And the total T = E + M + F.So, we have three equations with three variables. Let me write them again:Equation 1: M + F = 12Equation 2: E + F = 14Equation 3: E + M = 16We need to solve for E, M, F, and then sum them to get T.Let me see. Let's try to solve these equations step by step. First, from equation 1: M = 12 - FFrom equation 2: E = 14 - FNow substitute M and E into equation 3:E + M = (14 - F) + (12 - F) = 26 - 2F = 16So, 26 - 2F = 16Subtract 26 from both sides: -2F = 16 -26 = -10Divide by -2: F = 5Then, from equation 2: E = 14 - F = 14 -5 = 9From equation 1: M = 12 - F = 12 -5 =7So, E=9, M=7, F=5. Total T = 9+7+5=21.Wait, 21 is option D. But let me check again. Let's verify the original statements with these numbers.First statement: All the eels are electric except 12. The non-electric eels are M + F =7 +5=12. Correct.Second statement: All are moray except 14. Non-moray are E + F=9+5=14. Correct.Third statement: All are freshwater except 16. Non-freshwater are E + M=9+7=16. Correct.So, the total is 21, which is option D. Hmm, but the options include 21 as D. Wait, but let me check if the initial equations were correct.Wait, another way to think: Maybe the problem says "except 12" meaning that the number of electric eels is T -12. Wait, but that would be the same as what I did. Because "all are electric except 12" means that the non-electric eels are 12. So, yes, M + F =12. So equations are correct.Alternatively, someone might misinterpret the statements as the electric eels being T -12, but that's the same as we have here.So, I think the answer is D)21.Wait, but let me check the options again. The choices are A)42, B)33, C)24, D)21, E)20.Yes, 21 is option D. So that's the answer. But let me check once more to avoid any mistake.Total eels =9+7+5=21.Check each condition:- All except 12 are electric: non-electric is 12 (7+5=12). Correct.- All except 14 are moray: non-moray is14 (9+5=14). Correct.- All except 16 are freshwater: non-fresh is16 (9+7=16). Correct.Everything checks out. So the answer is D)21.

✅:To determine the total number of eels in the tank, we translate the given statements into equations:1. Non-electric eels (moray + freshwater) = 12: ( M + F = 12 ) 2. Non-moray eels (electric + freshwater) = 14: ( E + F = 14 ) 3. Non-freshwater eels (electric + moray) = 16: ( E + M = 16 )Let ( T = E + M + F ) represent the total number of eels. Solving the system of equations:1. From ( M + F = 12 ), express ( M = 12 - F ). 2. From ( E + F = 14 ), express ( E = 14 - F ). 3. Substitute ( M ) and ( E ) into ( E + M = 16 ): [ (14 - F) + (12 - F) = 16 implies 26 - 2F = 16 implies F = 5. ]Substitute ( F = 5 ) back into earlier equations: - ( E = 14 - 5 = 9 ), - ( M = 12 - 5 = 7 ).Total eels: [T = E + M + F = 9 + 7 + 5 = 21.]Verification: - Non-electric eels: ( 7 + 5 = 12 ) ✔️ - Non-moray eels: ( 9 + 5 = 14 ) ✔️ - Non-freshwater eels: ( 9 + 7 = 16 ) ✔️ Answer: (boxed{D})

🔑:Given the problem, we need to determine the total number of eels in the tank, which contains three types of eels: electric, moray, and freshwater eels. Let's use the following variables:- Let x be the number of electric eels.- Let y be the number of moray eels.- Let z be the number of freshwater eels.The notice provides the following information:1. All the eels are electric eels except 12.2. All the eels are moray eels except 14.3. All the eels are freshwater eels except 16.These statements translate into the following three equations depicting the relationships between the different types of eels:1. y + z = 122. x + z = 143. x + y = 16To find the total number of eels, x + y + z, let's add the three equations together:[(y + z) + (x + z) + (x + y) = 12 + 14 + 16]Simplifying the left side, we get:[2x + 2y + 2z = 42]Dividing both sides of the equation by 2:[x + y + z = 21]Therefore, the total number of eels in the tank is:[boxed{21}]

🧠:Okay, so I need to find the value of 'a' such that when the polynomial ( x^{6} - 8x^{3} + 6 ) is divided by ( (x - 1)(x - 2) ), the remainder is ( 7x - a ). Hmm, let's think about how to approach this.First, I remember that when a polynomial P(x) is divided by a divisor D(x), the remainder R(x) will have a degree less than the degree of D(x). In this case, the divisor is ( (x - 1)(x - 2) ), which is a quadratic polynomial. So, the remainder should be a linear polynomial, which matches the given remainder ( 7x - a ).Another thing I recall is the Remainder Theorem, which states that the remainder of a polynomial P(x) divided by (x - c) is P(c). But here, the divisor is a product of two linear factors, so maybe I can use the Chinese Remainder Theorem or set up equations based on evaluating the polynomial at the roots of the divisor.Since the divisor is (x - 1)(x - 2), the roots are x = 1 and x = 2. Therefore, when the polynomial ( x^{6} - 8x^{3} + 6 ) is divided by this divisor, the remainder ( 7x - a ) must satisfy the same conditions as the original polynomial at x = 1 and x = 2. That is, the remainder when evaluated at x = 1 should equal P(1), and similarly at x = 2, it should equal P(2).So, maybe I can compute P(1) and P(2) and set them equal to the remainder evaluated at those points, then solve for 'a'. Let me try that.First, compute P(1):( P(1) = 1^{6} - 8 times 1^{3} + 6 = 1 - 8 + 6 = (1 + 6) - 8 = 7 - 8 = -1 )Then, the remainder when evaluated at x = 1 is:( R(1) = 7 times 1 - a = 7 - a )Since P(1) = R(1), we have:( -1 = 7 - a )Solving for 'a' gives:( -1 - 7 = -a )( -8 = -a )( a = 8 )Wait, hold on, let me check that again. If I plug x = 1 into the remainder, it's 7(1) - a = 7 - a, and this equals P(1) which is -1. So:7 - a = -1Subtract 7 from both sides:-a = -8Multiply both sides by -1:a = 8Okay, that seems straightforward. Now let's check x = 2 to make sure.Compute P(2):( P(2) = 2^{6} - 8 times 2^{3} + 6 = 64 - 8 times 8 + 6 )Calculate each term:2^6 = 648 * 2^3 = 8 * 8 = 64So, P(2) = 64 - 64 + 6 = 0 + 6 = 6Then, the remainder evaluated at x = 2 is:R(2) = 7 * 2 - a = 14 - aSince P(2) = R(2), we have:14 - a = 6Solving for 'a':14 - 6 = aa = 8So both x = 1 and x = 2 give a = 8. That seems consistent. Therefore, the value of 'a' is 8.Wait, but just to be thorough, maybe there's another way to check this? For example, using polynomial division or expressing the original polynomial in terms of the divisor.Alternatively, since the divisor is (x - 1)(x - 2), any polynomial can be written as:( P(x) = (x - 1)(x - 2)Q(x) + R(x) )Where Q(x) is the quotient and R(x) is the remainder, which is linear: R(x) = 7x - a.Therefore, substituting x = 1 and x = 2 into P(x) gives:At x = 1: P(1) = 0*Q(1) + R(1) = R(1) = 7(1) - aSimilarly, at x = 2: P(2) = 0*Q(2) + R(2) = R(2) = 7(2) - aWhich is exactly what we did earlier, so this confirms the method.But just to make sure there's no mistake in calculations, let's recheck P(1) and P(2):For P(1):1^6 = 18*(1)^3 = 8*1 = 8So, 1 - 8 + 6 = (1 + 6) - 8 = 7 - 8 = -1. Correct.For P(2):2^6 = 648*(2)^3 = 8*8 = 6464 - 64 + 6 = 0 + 6 = 6. Correct.Then R(1) = 7 - a = -1 => a = 8R(2) = 14 - a = 6 => a = 8Consistent in both cases. Therefore, a = 8.Alternatively, if I were to approach this problem by expressing the remainder as R(x) = ax + b (but here it's given as 7x - a, so maybe their notation is different). Wait, in the problem statement, the remainder is given as 7x - a, so the coefficient of x is 7, and the constant term is -a. So perhaps in standard terms, if we let R(x) = mx + c, then m = 7 and c = -a, so we need to find c, which is -a. But since they directly give R(x) = 7x - a, then solving for a is straightforward as above.Another way is to perform polynomial division, but that might be more time-consuming for a degree 6 polynomial divided by a quadratic. But maybe we can use the method of equating coefficients.Let me try that approach as a cross-check.We have:( x^{6} - 8x^{3} + 6 = (x - 1)(x - 2)Q(x) + 7x - a )Let me denote Q(x) as the quotient polynomial, which will be of degree 4 (since divisor is degree 2 and dividend is degree 6). But expanding the right-hand side would be complicated. However, since we know that when x = 1 or x = 2, the term (x - 1)(x - 2)Q(x) becomes zero, so we can use the two equations from x = 1 and x = 2 as before. So maybe this is redundant, but let's see.Alternatively, we can write the polynomial identity:For all x, ( x^{6} - 8x^{3} + 6 = (x - 1)(x - 2)Q(x) + 7x - a )If we take derivatives of both sides, maybe we can get more equations? Wait, but that might complicate things. Alternatively, if we know that the remainder when divided by (x - 1)(x - 2) is 7x - a, then the remainder must satisfy both:R(1) = P(1) and R(2) = P(2). Since we already did that, and both give a = 8, I think that's sufficient.But just to ensure that there's no error, maybe test another value? Wait, but since the remainder is determined uniquely by its values at two distinct points (since it's a linear polynomial), two equations are sufficient to determine the two coefficients. In this case, the remainder is given as 7x - a, so the coefficient of x is fixed at 7, and only the constant term is variable. Therefore, only one equation is needed. Wait, but in reality, the remainder is of the form R(x) = ax + b, so two coefficients. But in the problem, they specify the remainder as 7x - a, so the coefficient of x is 7, and the constant term is -a, meaning only one unknown: a. Hence, we need two equations to solve for a? Wait, no, because if we know the coefficient of x is 7, then the remainder is 7x + c, and then we need to find c. Since we have two points x=1 and x=2, we can set up two equations:At x=1: 7*1 + c = P(1) => 7 + c = -1 => c = -8At x=2: 7*2 + c = P(2) => 14 + c = 6 => c = -8Therefore, c = -8, which means the remainder is 7x - 8, so -a = -8 => a = 8.Therefore, same result. So regardless of approach, the answer is 8.Therefore, I can be confident that a = 8.Final AnswerThe value of ( a ) is boxed{8}.

✅:To find the value of ( a ) such that the remainder when ( x^6 - 8x^3 + 6 ) is divided by ( (x-1)(x-2) ) is ( 7x - a ), we use the Remainder Theorem. The polynomial ( P(x) = x^6 - 8x^3 + 6 ) must satisfy the conditions that the remainder ( R(x) = 7x - a ) equals ( P(x) ) at ( x = 1 ) and ( x = 2 ).First, we evaluate ( P(1) ):[P(1) = 1^6 - 8 cdot 1^3 + 6 = 1 - 8 + 6 = -1]The remainder ( R(1) ) is:[R(1) = 7 cdot 1 - a = 7 - a]Setting ( P(1) = R(1) ):[-1 = 7 - a implies -a = -8 implies a = 8]Next, we evaluate ( P(2) ):[P(2) = 2^6 - 8 cdot 2^3 + 6 = 64 - 64 + 6 = 6]The remainder ( R(2) ) is:[R(2) = 7 cdot 2 - a = 14 - a]Setting ( P(2) = R(2) ):[6 = 14 - a implies -a = -8 implies a = 8]Both evaluations confirm that ( a = 8 ).Thus, the value of ( a ) is (boxed{8}).

🔑:1. Define the given polynomial function: Let ( f(x) = x^6 - 8x^3 + 6 ).2. Recognize the divisor: We need to find the remainder when ( f(x) ) is divided by ((x - 1)(x - 2)). 3. Polynomial division and remainders: According to the polynomial division theorem, the remainder of ( f(x) ) when divided by ((x - 1)(x - 2)) will be a linear polynomial of the form ( R(x) = bx + c ). 4. Given remainder polynomial: The problem states that the remainder is ( 7x - a ), therefore ( R(x) = 7x - a ).5. Determine coefficients using values ( x = 1 ) and ( x = 2 ): - Substitute ( x = 1 ) into the equation: [ f(1) = 1^6 - 8 cdot 1^3 + 6 = 1 - 8 + 6 = -1 ] Since ( f(1) = R(1) ): [ R(1) = 7 cdot 1 - a = 7 - a ] Thus, we have: [ 7 - a = -1 implies a = 8 ] - Substitute ( x = 2 ) extra to cross-check: [ f(2) = 2^6 - 8 cdot 2^3 + 6 = 64 - 64 + 6 = 6 ] The remainder at ( x = 2 ): [ R(2) = 7 cdot 2 - a = 14 - a ] Since ( f(2) = R(2) ): [ 6 = 14 - a implies a = 8 ] From both substitutions, we find that ( a ) must be ( 8 ). Conclusion:[boxed{8}]

🧠:Okay, so I have this geometry problem here: Given a triangle ABC where the ratio of BC to AC is less than 1/2, I need to prove that angle A is less than 30 degrees. Hmm, let's see. Let me start by understanding the problem properly.First, let's label the triangle. Let me consider triangle ABC with vertices A, B, and C. The sides opposite these angles are BC, AC, and AB, respectively. Wait, in standard notation, side opposite angle A is BC, side opposite angle B is AC, and side opposite angle C is AB. So the problem states that BC/AC < 1/2. So BC is shorter than half of AC. And we need to prove that angle A is less than 30 degrees.Alright, angle A is at vertex A, between sides AB and AC. So angle A is opposite side BC. Wait, no. Wait, angle A is at vertex A, so the sides adjacent to angle A are AB and AC, and the side opposite angle A is BC. So, in triangle ABC, side BC is opposite angle A, side AC is opposite angle B, and side AB is opposite angle C.Given that BC/AC < 1/2, so BC < (1/2)AC. So BC is less than half of AC. So, the side opposite angle A is less than half the length of side AC, which is opposite angle B. Hmm.Now, I need to connect the ratio of the sides to the measure of angle A. In triangles, the larger side is opposite the larger angle. So if BC is shorter than half of AC, then angle A, which is opposite BC, should be smaller than angle B, which is opposite AC. But how does that relate to angle A being less than 30 degrees?Maybe using the Law of Sines? The Law of Sines states that BC/sin A = AC/sin B = AB/sin C. So, if BC/AC < 1/2, then (BC/AC) = (sin A)/sin B < 1/2. So, sin A / sin B < 1/2. That gives sin A < (1/2) sin B.But how does that help me? Well, angle A and angle B are two angles in a triangle, so they add up to less than 180 degrees. Since the sum of all three angles is 180 degrees. So, angle A + angle B + angle C = 180. If angle A is less than 30 degrees, then angle B + angle C > 150 degrees. But I'm not sure how this helps directly.Alternatively, maybe using the Law of Cosines. Let's see. Let me denote the sides as follows: Let’s call BC = a, AC = b, AB = c. Then, according to the problem, a/b < 1/2. We need to prove angle A < 30 degrees.By the Law of Cosines, angle A can be expressed as:cos A = (b² + c² - a²) / (2bc)But we need to find a relationship that can give us an upper bound on angle A. Alternatively, using the Law of Sines:sin A / a = sin B / b = sin C / cGiven that a/b < 1/2, so a = BC, b = AC.So, a/b < 1/2 => sin A / sin B < 1/2. So, sin A < (1/2) sin B.Since in a triangle, angles are between 0 and 180 degrees, so sin B is positive. Also, since angle B is part of triangle ABC, angle B can be expressed as 180 - angle A - angle C. Therefore, angle B = 180 - A - C.But maybe this is getting too abstract. Let's try to consider specific cases or perhaps use some inequalities.Suppose angle A is 30 degrees. Then, if we can show that under this assumption, the ratio BC/AC must be at least 1/2, which would contradict the given condition. Hence, angle A must be less than 30 degrees.Let’s try that approach. Assume angle A is 30 degrees. Then, using the Law of Sines:BC / AC = sin A / sin B = sin 30° / sin B = (1/2) / sin BGiven that angle B is part of the triangle, angle B = 180° - angle A - angle C = 150° - angle C. Since angle C must be positive, angle B < 150°. Therefore, sin B has a maximum value of 1 (when angle B is 90°), but in reality, angle B can be at most less than 150°, so sin B is at most sin 90° = 1, and at minimum greater than sin(0° + ε) which approaches 0.Wait, but if angle A is 30°, then BC / AC = (1/2) / sin B. To find the minimum value of BC / AC when angle A is 30°, we need to find the maximum value of sin B. Since sin B is maximum when angle B is 90°, then BC / AC would be (1/2)/1 = 1/2. So, if angle A is 30°, the ratio BC / AC can be as small as 1/2 (when angle B is 90°), but if angle B is smaller, sin B would be smaller, making BC / AC larger. Wait, that seems contradictory. Let me check.Wait, if angle A is 30°, then angle B can vary. If angle B is 90°, then BC / AC = 1/2. If angle B is smaller, say angle B is 45°, then sin B is √2/2 ≈ 0.707, so BC / AC = (1/2)/0.707 ≈ 0.707/2 ≈ 0.3535, which is less than 1/2. Wait, but that contradicts the previous thought. Wait, maybe I made a mistake here.Wait, let's clarify. If angle A is 30°, angle B can range from just above 0° to 150°, but in reality, angle B must satisfy angle C = 150° - angle B > 0°, so angle B < 150°. Therefore, angle B can be between 0° and 150°, but angle C must be positive.But according to the Law of Sines, BC / AC = sin A / sin B. If angle A is 30°, then BC / AC = sin 30° / sin B = (1/2)/sin B. So, as angle B increases from near 0° to 90°, sin B increases from near 0 to 1, so BC / AC decreases from infinity to 1/2. Then, as angle B increases beyond 90° up to 150°, sin B decreases from 1 to sin 150° = 1/2, so BC / AC increases from 1/2 to (1/2)/(1/2) = 1.Wait, that makes sense. So, if angle A is 30°, then BC / AC can be as small as 1/2 (when angle B is 90°) or as large as 1 (when angle B is 150°, making angle C 0°, which is degenerate). Therefore, the minimal ratio BC / AC when angle A is 30° is 1/2. Therefore, if in our problem, BC / AC < 1/2, then angle A must be less than 30°, because if angle A were 30°, the ratio BC / AC cannot be less than 1/2. Hence, angle A has to be smaller than 30° to make BC even shorter relative to AC.That seems like a solid argument. Let me verify this again. Suppose angle A is 30°, then the minimal BC / AC ratio is 1/2. If angle A were larger than 30°, say 45°, then BC would be longer (since BC is opposite angle A), so the ratio BC / AC would be even larger. Wait, but in our problem, the ratio is given to be smaller than 1/2. Therefore, angle A cannot be 30° or more. Thus, angle A must be less than 30°.Alternatively, let's approach this with the Law of Cosines. Suppose we have triangle ABC with sides BC = a, AC = b, AB = c. Given that a/b < 1/2, need to prove angle A < 30°.By Law of Cosines:a² = b² + c² - 2bc cos AWe need to relate a and b. Since a < (1/2)b, let's substitute a with (1/2)b - ε where ε is a positive number.But maybe instead, we can consider using the Law of Sines again. From the Law of Sines:a / sin A = b / sin BTherefore, (a/b) = sin A / sin BGiven that a/b < 1/2, so sin A / sin B < 1/2So sin A < (1/2) sin BBut in the triangle, angles A + B + C = 180°, so B = 180° - A - CBut perhaps we can relate angles A and B. Let's assume angle A is 30°, then sin A = 1/2. Then, according to the inequality, 1/2 < (1/2) sin B => sin B > 1, which is impossible. Therefore, angle A cannot be 30°, so angle A must be less than 30°.Wait, wait. Let me rephrase that. If angle A were equal to 30°, then sin A = 1/2. Then, the given ratio a/b = sin A / sin B = (1/2)/sin B. But according to the problem, a/b < 1/2. Therefore, (1/2)/sin B < 1/2 => 1/sin B < 1 => sin B > 1. But sin B cannot exceed 1. Therefore, angle A cannot be 30°, so angle A must be less than 30°.That's a nice contradiction. So if angle A is 30°, then to satisfy a/b < 1/2, sin B must be greater than 1, which is impossible. Therefore, angle A must be less than 30°.That seems to be a straightforward proof. Let me check again.Given a/b < 1/2. From Law of Sines, a/b = sin A / sin B. Therefore, sin A / sin B < 1/2. Suppose angle A is 30°, then sin A = 1/2. Then 1/2 / sin B < 1/2 => 1/sin B < 1 => sin B > 1. Impossible. Therefore, angle A cannot be 30°, and since angles in triangle are related such that if angle A increases, keeping other angles in check, the ratio a/b would increase. Thus, angle A must be less than 30° to maintain a/b < 1/2.Alternatively, angle A being less than 30° would mean sin A is less than sin 30°, which is 1/2. So, if angle A is less than 30°, sin A < 1/2. Then, sin A / sin B < 1/(2 sin B). But we need sin A / sin B < 1/2. So, combining these, we need 1/(2 sin B) < 1/2, which would imply sin B > 1. Wait, that seems conflicting. Maybe this approach is not correct.Wait, perhaps better to think in terms of if angle A is increasing, then side BC (opposite angle A) increases, so the ratio BC/AC increases. Therefore, to have BC/AC as small as less than 1/2, angle A must be smaller than the angle that would give BC/AC = 1/2. From the earlier argument, when angle A is 30°, the minimal BC/AC is 1/2. Therefore, if BC/AC is less than 1/2, angle A must be less than 30°.Alternatively, imagine constructing triangle ABC where BC/AC = 1/2. In such a triangle, angle A is 30°. Then, if we make BC even shorter, angle A would have to decrease to keep the ratio BC/AC below 1/2. That seems intuitive.Another way is to use the area formula. The area of the triangle can be expressed as (1/2)ab sin C. But not sure if that helps here.Alternatively, using the tangent function. Suppose we drop a perpendicular from B to AC, forming a right triangle. Let me try to visualize.Let’s consider triangle ABC. If we drop a perpendicular from B to AC, let’s call the foot D. Then BD is the height of the triangle with respect to base AC. Then, BD = AB sin A = BC sin C.But maybe this complicates things. Let me think.Alternatively, consider the triangle where angle A is 30°, BC = 1, AC = 2. Then, by Law of Sines, 1/sin 30° = 2/sin B => 1/(1/2) = 2/sin B => 2 = 2/sin B => sin B = 1. Therefore, angle B = 90°, angle C = 60°. So, this is a 30-60-90 triangle. Here, BC/AC = 1/2. So, this is the case where angle A is 30°, and BC/AC = 1/2. If BC/AC is less than 1/2, then angle A must be less than 30°, because if you decrease BC while keeping AC the same, angle A has to decrease to accommodate the shorter side.Alternatively, think in terms of constructing triangles. If BC is shorter, with AC fixed, then point B is closer to point C along the base AC. Thus, angle at A would be smaller.Alternatively, use the Law of Cosines to set up an inequality. Let’s denote AC = b, BC = a, AB = c. Given that a < b/2. Need to show that angle A < 30°.From Law of Cosines:a² = b² + c² - 2bc cos ABut we need to relate angle A to a and b. Let’s rearrange:cos A = (b² + c² - a²) / (2bc)We need to find an upper bound for angle A, so a lower bound for cos A (since cosine is decreasing in [0°, 180°]).But since a < b/2, perhaps substitute a with b/2 - ε where ε > 0.But this might get messy. Let me think differently. Suppose we fix AC = b, then BC = a < b/2. Let’s try to maximize angle A given a < b/2.What's the maximum possible angle A when a < b/2? If angle A is maximum, then side BC is as long as possible, i.e., a approaches b/2. So, the maximum angle A occurs when a = b/2. As shown earlier, in that case, angle A is 30°. Therefore, for a < b/2, angle A must be less than 30°.Therefore, the conclusion holds.Wait, that seems like a good approach. If a <= b/2, then angle A <= 30°, with equality when a = b/2. Therefore, if a < b/2, angle A must be strictly less than 30°. Hence, proved.Another approach: Using the Law of Sines, as before.Let’s denote angle A = θ. Then, from Law of Sines:a / sin θ = b / sin BSo, a / b = sin θ / sin BGiven that a/b < 1/2, so sin θ / sin B < 1/2 => sin θ < (1/2) sin B.But in a triangle, angles θ + B + C = 180°, so B = 180° - θ - C. Since C > 0°, B < 180° - θ.But not sure. Alternatively, consider the maximum possible value of sin B. The maximum value of sin B is 1 (when B = 90°). Therefore, sin θ < (1/2) * 1 => sin θ < 1/2. Therefore, θ < 30° or θ > 150°. But since in a triangle, angles are less than 180°, and angle A is one angle, so θ must be less than 30°.Wait, this is a key point. If sin θ < 1/2, then θ is either in (0°, 30°) or (150°, 180°). But angle A is part of a triangle, so it must be less than 180°, and the other angles must also be positive. If angle A were greater than 150°, then the sum of angles B and C would be less than 30°, which would make both B and C less than 30°, but then side BC (opposite angle A) would be the largest side, but in our problem, BC is smaller than half of AC. So, if angle A were greater than 150°, BC would be the largest side, which contradicts BC < AC/2. Hence, angle A cannot be in (150°, 180°), so θ must be in (0°, 30°).Therefore, angle A < 30°. That seems like another valid argument.So, summarizing: From the Law of Sines, a/b = sin θ / sin B < 1/2. Thus, sin θ < (1/2) sin B. The maximum possible value of sin B is 1, so sin θ < 1/2, which implies θ < 30° or θ > 150°, but θ > 150° is impossible in a triangle with BC < AC/2. Hence, θ < 30°.Therefore, angle A is less than 30 degrees.I think that covers multiple approaches: using contradiction, Law of Sines, considering maximum angles, and so on. All lead to the conclusion that angle A must be less than 30 degrees when BC/AC is less than 1/2.Final Answerboxed{angle A < 30^circ}

✅:Given a triangle (ABC) with (frac{BC}{AC} < frac{1}{2}), we need to prove that (angle A < 30^circ).1. Using the Law of Sines: - The Law of Sines states (frac{BC}{sin A} = frac{AC}{sin B}). - Given (frac{BC}{AC} < frac{1}{2}), we can rewrite this as (frac{sin A}{sin B} < frac{1}{2}). - This implies (sin A < frac{1}{2} sin B).2. Contradiction Approach: - Assume (angle A = 30^circ). Then (sin A = frac{1}{2}). - Substituting into the inequality, we get (frac{1}{2} < frac{1}{2} sin B), which simplifies to (sin B > 1). This is impossible since (sin B) cannot exceed 1. - Therefore, (angle A) cannot be (30^circ).3. Bounding (sin A): - From (sin A < frac{1}{2} sin B) and knowing the maximum value of (sin B) is 1, we get (sin A < frac{1}{2}). - The angles where (sin A < frac{1}{2}) are (0^circ < A < 30^circ) or (150^circ < A < 180^circ). - Since (A) is an angle in a triangle, it must be less than (180^circ). If (A) were greater than (150^circ), the other two angles would sum to less than (30^circ), making (BC) the longest side, which contradicts (BC < frac{1}{2} AC).4. Conclusion: - Therefore, (angle A) must be less than (30^circ).[boxed{angle A < 30^circ}]

🔑:Given triangle ( ABC ), we know that ( frac{BC}{AC} < frac{1}{2} ). We want to prove that ( angle A < 30^circ ).1. Consider the Law of Sines, which states: [ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ] where ( a = BC ), ( b = AC ), and ( c = AB ).2. Using the given inequality ( frac{BC}{AC} < frac{1}{2} ), we can rewrite it in terms of ( a ) and ( b ): [ frac{a}{b} < frac{1}{2} ] Thus, ( a < frac{b}{2} ).3. Applying the Law of Sines to sides ( a ) and ( b ): [ frac{a}{sin A} = frac{b}{sin B} ] Rearrange to relate ( sin A ) and ( sin B ): [ sin A = frac{a sin B}{b} ]4. Since ( a < frac{b}{2} ): [ sin A < frac{frac{b}{2} sin B}{b} = frac{1}{2} sin B ] 5. Recall that ( sin B le 1 ), hence: [ sin A < frac{1}{2} ]6. The sine function reaches ( frac{1}{2} ) at an angle of ( 30^circ ). Therefore: [ sin 30^circ = frac{1}{2} ]7. Knowing ( sin A < frac{1}{2} ) implies: [ A < 30^circ ]Thus, we have shown that:[boxed{angle A < 30^circ}]

🧠:Okay, so I need to prove that if you take two parallelograms, ABCD and A'B'C'D', in the plane, then the midpoints of the segments AA', BB', CC', and DD' form another parallelogram. Hmm, let me think about how to approach this.First, I remember that a quadrilateral is a parallelogram if both pairs of opposite sides are parallel and equal in length, or if the midpoints of the sides form some particular properties. Maybe using vectors would be a good approach here since we're dealing with midpoints and parallelograms in a plane. Coordinates might also work, but vectors might be more straightforward.Let me recall some properties of parallelograms. In a parallelogram, the opposite sides are equal and parallel, which in vector terms means that vectors AB and DC are equal, and vectors AD and BC are equal. Also, the diagonals bisect each other. So the midpoint of diagonal AC is the same as the midpoint of diagonal BD.Now, considering the midpoints of AA', BB', CC', DD'. Let me denote these midpoints as M1, M2, M3, M4 respectively. So M1 is the midpoint of AA', M2 of BB', and so on. The goal is to show that M1M2M3M4 is a parallelogram.To prove that M1M2M3M4 is a parallelogram, I need to show that either the opposite sides are equal and parallel, or that the midpoints satisfy some vector relationships. Let me try using vectors.Let me assign position vectors to the points. Let’s denote the position vectors of A, B, C, D as vectors a, b, c, d respectively, and similarly for A', B', C', D' as vectors a', b', c', d'.Since ABCD is a parallelogram, the position vectors satisfy certain conditions. In a parallelogram, the vector from A to B is equal to the vector from D to C. So, vector AB = vector DC. Which translates to b - a = c - d. Similarly, vector AD = vector BC, so d - a = c - b.Similarly, for the parallelogram A'B'C'D', the same relationships hold: b' - a' = c' - d' and d' - a' = c' - b'.Now, let's express the midpoints M1, M2, M3, M4 in terms of these position vectors.Midpoint M1 of AA' is (a + a')/2.Similarly, M2 is (b + b')/2, M3 is (c + c')/2, M4 is (d + d')/2.To show that M1M2M3M4 is a parallelogram, we can show that the vectors M1M2 and M1M4 are equal to vectors M3M4 and M3M2 respectively, or that the vectors M1M2 and M3M4 are equal, and M2M3 and M1M4 are equal. Wait, actually, in a parallelogram, opposite sides are equal and parallel, so vector M1M2 should be equal to vector M3M4, and vector M2M3 should be equal to vector M1M4.Let me compute vector M1M2: this is M2 - M1 = [(b + b')/2 - (a + a')/2] = [(b - a) + (b' - a')]/2.Similarly, vector M3M4 is M4 - M3 = [(d + d')/2 - (c + c')/2] = [(d - c) + (d' - c')]/2.But since ABCD is a parallelogram, we know that b - a = c - d. Wait, let me check that again. In ABCD, AB = DC, so vector AB = vector DC. Vector AB is b - a, vector DC is c - d. Therefore, b - a = c - d. So d - c = a - b. Similarly, in A’B’C’D’, b’ - a’ = c’ - d’ => d’ - c’ = a’ - b’.So substituting back into vector M3M4: [(d - c) + (d’ - c’)]/2 = [(a - b) + (a’ - b’)]/2.But vector M1M2 is [(b - a) + (b’ - a’)]/2. Comparing this with vector M3M4, which is [(a - b) + (a’ - b’)]/2. These look like negatives of each other. So M1M2 = - M3M4. Hmm, but in a parallelogram, the opposite sides should be equal and parallel, which would mean that M1M2 is equal to M3M4 in vector terms. But here they are negatives. Wait, but if vectors are negatives, that just means they are equal in magnitude and opposite in direction. But in terms of the sides of the quadrilateral, if M1M2 is a vector and M3M4 is its negative, then the sides M1M2 and M3M4 are actually parallel and equal in length but pointing in opposite directions. However, in the quadrilateral M1M2M3M4, the sides M1M2 and M3M4 are opposite sides. So if their vectors are negatives, that actually still satisfies the condition for being a parallelogram because the opposite sides are parallel and equal. Because in the quadrilateral, the vector from M3 to M4 is the same as the vector from M1 to M2. Wait, maybe I made a miscalculation here.Wait, let's clarify. The vector M1M2 is M2 - M1, and the vector M3M4 is M4 - M3. So if we can show that M2 - M1 = M4 - M3, then sides M1M2 and M3M4 are equal and parallel. But according to our previous calculation:M2 - M1 = [(b - a) + (b’ - a’)]/2M4 - M3 = [(d - c) + (d’ - c’)]/2But in parallelogram ABCD, we have b - a = c - d (since AB and DC are equal and opposite). So c - d = -(b - a). Similarly, in A’B’C’D’, c’ - d’ = -(b’ - a’). Therefore:(d - c) = -(c - d) = (b - a)Similarly, (d’ - c’) = -(c’ - d’) = (b’ - a’)Therefore, substituting back into M4 - M3:[(d - c) + (d’ - c’)]/2 = [(b - a) + (b’ - a’)]/2 = M2 - M1Therefore, M4 - M3 = M2 - M1. Hence, vectors M1M2 and M3M4 are equal. So the sides M1M2 and M3M4 are equal and parallel.Similarly, we need to check the other pair of opposite sides: M2M3 and M1M4.Compute vector M2M3: M3 - M2 = [(c + c')/2 - (b + b')/2] = [(c - b) + (c’ - b’)]/2Vector M1M4: M4 - M1 = [(d + d')/2 - (a + a')/2] = [(d - a) + (d’ - a’)]/2Again, using properties of the parallelograms. In ABCD, vector AD = vector BC. Vector AD is d - a, vector BC is c - b. So d - a = c - b. Similarly, in A’B’C’D’, vector A’D’ = vector B’C’, so d’ - a’ = c’ - b’.Therefore:[(d - a) + (d’ - a’)]/2 = [(c - b) + (c’ - b’)]/2 = M3 - M2Therefore, vector M1M4 = vector M2M3. Hence, sides M2M3 and M1M4 are equal and parallel.Therefore, since both pairs of opposite sides of quadrilateral M1M2M3M4 are equal and parallel, M1M2M3M4 is a parallelogram.Wait, let me just verify that again. So M1M2 is equal to M3M4, and M2M3 is equal to M1M4. Therefore, opposite sides are equal and parallel, which is the definition of a parallelogram. That seems to check out.Alternatively, another approach could be using coordinate geometry. Let me try that as well to confirm.Suppose we place the first parallelogram ABCD in the plane with coordinates. Let me assign coordinates to simplify calculations. Let’s let point A be at the origin (0,0). Since ABCD is a parallelogram, we can let B be at (b, 0), D be at (d1, d2), then C would be at (b + d1, d2) because in a parallelogram, the vector from A to B plus the vector from A to D gives the vector from A to C.Similarly, for parallelogram A’B’C’D’, let’s assign coordinates: A’ at (a’, e’), B’ at (a’ + b’, e’), D’ at (a’ + d1’, e’ + d2’), then C’ would be at (a’ + b’ + d1’, e’ + d2’). Wait, this might get complicated. Maybe it's better to use vectors.Alternatively, since the problem is affine (doesn't depend on specific coordinates), maybe using vectors is better. Wait, but let's see.Alternatively, take complex numbers. Let me think. Points in the plane can be represented as complex numbers. The midpoint of AA' is (A + A')/2, etc. Then, the quadrilateral formed by these midpoints would have vertices at (A + A')/2, (B + B')/2, (C + C')/2, (D + D')/2.To prove that these form a parallelogram, we can show that the midpoints satisfy the parallelogram condition. For example, the vector from M1 to M2 should equal the vector from M4 to M3.Wait, in complex numbers, the vector from M1 to M2 is M2 - M1 = [(B + B')/2 - (A + A')/2] = (B - A + B' - A')/2.Similarly, the vector from M4 to M3 is M3 - M4 = [(C + C')/2 - (D + D')/2] = (C - D + C' - D')/2.Since ABCD is a parallelogram, B - A = C - D. Similarly, in A’B’C’D’, B’ - A’ = C’ - D’.Therefore, (B - A + B' - A') = (C - D + C' - D'). Therefore, (B - A + B' - A')/2 = (C - D + C' - D')/2. Hence, M2 - M1 = M3 - M4, which implies that the vector M1M2 is equal to the vector M4M3, which in terms of the quadrilateral, means that side M1M2 is equal and parallel to side M4M3. Similarly, for the other pair of sides, we can do the same.Alternatively, another approach: the midpoints form a parallelogram if the midpoints of the diagonals of the quadrilateral M1M2M3M4 coincide. Wait, but maybe that's more involved. Wait, in any quadrilateral, the midpoints of the diagonals coincide if and only if the quadrilateral is a parallelogram. So if we compute the midpoints of the diagonals M1M3 and M2M4, they should be the same if it's a parallelogram.Let me compute the midpoint of M1M3: which is [(M1 + M3)/2] = [( (A + A')/2 + (C + C')/2 ) / 2 ] = (A + A' + C + C')/4.Similarly, the midpoint of M2M4: [(M2 + M4)/2] = [( (B + B')/2 + (D + D')/2 ) / 2 ] = (B + B' + D + D')/4.For these midpoints to coincide, we need A + A' + C + C' = B + B' + D + D'.But in parallelogram ABCD, we have A + C = B + D. Because in a parallelogram, the diagonals bisect each other, so the midpoint of AC is the same as the midpoint of BD. So (A + C)/2 = (B + D)/2 => A + C = B + D.Similarly, in A’B’C’D’, we have A’ + C’ = B’ + D’.Therefore, A + C + A’ + C’ = B + D + B’ + D’ => (A + C) + (A’ + C’) = (B + D) + (B’ + D’). Since A + C = B + D and A’ + C’ = B’ + D’, this equality holds. Therefore, the midpoints of the diagonals M1M3 and M2M4 coincide. Hence, quadrilateral M1M2M3M4 is a parallelogram.That's another way to see it. So this method shows that the midpoints of the diagonals of M1M2M3M4 coincide, which is a property of parallelograms. Therefore, the quadrilateral must be a parallelogram.Alternatively, another approach could be using affine transformations. Since any parallelogram can be mapped to any other parallelogram via an affine transformation, perhaps we can reduce the problem to a simpler case, like when one parallelogram is a translation of the other or something. But maybe that's complicating.Wait, but perhaps considering specific cases could help. For example, take ABCD as a square at the origin, and A’B’C’D’ as another parallelogram somewhere else. Compute the midpoints and see if they form a parallelogram. But this might not be a proof, but just verification for a specific case.But let's try. Let me take ABCD as a square with coordinates A(0,0), B(1,0), C(1,1), D(0,1). Let’s take another parallelogram A’B’C’D’ with coordinates A’(2,2), B’(3,3), C’(4,4), D’(3,3). Wait, but is that a parallelogram? Wait, let me check. In a parallelogram, the vectors A’B’ and A’D’ should define the sides. A’B’ is (3-2, 3-2) = (1,1). A’D’ should be equal to B’C’. Wait, if D’ is (3,3), then A’D’ is (1,1). Then B’C’ would be (4-3,4-3) = (1,1). So yes, A’B’C’D’ with A’(2,2), B’(3,3), C’(4,4), D’(3,3) is a parallelogram? Wait, but D’ is (3,3), so connecting A’(2,2) to D’(3,3) is a vector (1,1), and B’(3,3) to C’(4,4) is also (1,1). Then, connecting D’(3,3) to C’(4,4) is (1,1), but in a parallelogram, opposite sides should be equal and parallel. Wait, maybe my coordinates are not correct. Let me correct that.Let me choose A’(2,2), B’(4,2), C’(5,4), D’(3,4). Then check if it's a parallelogram. Vector A’B’ is (2,0), vector A’D’ is (1,2). Then vector B’C’ should be (5-4,4-2)=(1,2), and vector D’C’ is (5-3,4-4)=(2,0). So yes, opposite sides are equal and parallel. So ABCD is the unit square, and A’B’C’D’ is this other parallelogram.Compute midpoints:M1: midpoint of AA’ is ((0+2)/2, (0+2)/2) = (1,1)M2: midpoint of BB’ is ((1+4)/2, (0+2)/2) = (2.5,1)M3: midpoint of CC’ is ((1+5)/2, (1+4)/2) = (3,2.5)M4: midpoint of DD’ is ((0+3)/2, (1+4)/2) = (1.5,2.5)Now, check if these four points form a parallelogram.Compute the vectors between consecutive points:M1(1,1), M2(2.5,1), M3(3,2.5), M4(1.5,2.5)Compute M1M2: (2.5 - 1, 1 - 1) = (1.5, 0)Compute M2M3: (3 - 2.5, 2.5 - 1) = (0.5, 1.5)Compute M3M4: (1.5 - 3, 2.5 - 2.5) = (-1.5, 0)Compute M4M1: (1 - 1.5, 1 - 2.5) = (-0.5, -1.5)Wait, in a parallelogram, opposite sides should be equal and parallel. Let's check M1M2 and M3M4: M1M2 is (1.5, 0), M3M4 is (-1.5, 0). These are negatives, so they are parallel and same length but opposite direction. Similarly, M2M3 is (0.5, 1.5), M4M1 is (-0.5, -1.5). Again, negatives. So in terms of the quadrilateral, the sides are M1M2, M2M3, M3M4, M4M1. Wait, but this seems to form a parallelogram? Wait, no, if opposite sides are equal and parallel. Wait, but M1M2 and M3M4 are equal in magnitude and opposite in direction, which would mean they are parallel and equal. Similarly, M2M3 and M4M1 are also equal in magnitude and opposite in direction. So the quadrilateral is actually a parallelogram.Wait, but in the coordinate system, plotting these points:M1(1,1), M2(2.5,1), M3(3,2.5), M4(1.5,2.5). Connecting M1 to M2 to M3 to M4 to M1. This should form a parallelogram. Let me check the vectors again.Alternatively, compute vectors of opposite sides:Vector M1M2: (2.5 - 1, 1 - 1) = (1.5, 0)Vector M3M4: (1.5 - 3, 2.5 - 2.5) = (-1.5, 0) = - (1.5, 0). So they are scalar multiples (scalar -1), hence parallel and same length.Vector M2M3: (3 - 2.5, 2.5 - 1) = (0.5, 1.5)Vector M4M1: (1 - 1.5, 1 - 2.5) = (-0.5, -1.5) = - (0.5, 1.5). Again, scalar multiple, so parallel and same length.Therefore, even though the vectors are negatives, in terms of the sides of the quadrilateral, opposite sides are equal and parallel. Therefore, the quadrilateral is a parallelogram. So this example checks out.Therefore, combining the vector approach and the coordinate example, it seems that the midpoints do form a parallelogram. Thus, the proof seems solid.Another way to think about this is using the concept of midpoint polygons. The midpoints of corresponding vertices of two polygons form a new polygon, and in the case of parallelograms, this midpoint polygon inherits some properties. Since parallelograms are closed under addition and scalar multiplication (they form a vector space), the average of two parallelograms (which is what the midpoints represent) should also be a parallelogram. But this is more abstract.Alternatively, think of the translation between the two parallelograms. If you consider the vector from A to A', B to B', etc., but since the original figures are parallelograms, the differences between corresponding vertices might have some structure. However, the previous vector approach seems sufficient.Therefore, summarizing the key steps:1. Express the midpoints M1, M2, M3, M4 in terms of position vectors of the original parallelograms.2. Show that the vectors M1M2 and M3M4 are equal by utilizing the properties of the original parallelograms (i.e., the relationships between their sides).3. Similarly, show that the vectors M2M3 and M1M4 are equal.4. Conclude that since both pairs of opposite sides are equal and parallel, the quadrilateral formed by the midpoints is a parallelogram.Alternatively, using the midpoint of diagonals approach:1. Show that the midpoints of the diagonals of the midpoint quadrilateral coincide, which is a property of parallelograms.2. Use the fact that the sum of the position vectors of opposite vertices in a parallelogram are equal (A + C = B + D and A’ + C’ = B’ + D’).3. Therefore, the midpoints of the diagonals of M1M2M3M4 are the same, confirming it's a parallelogram.Both methods lead to the same conclusion. Therefore, the proof is established.

✅:To prove that the midpoints of segments ( AA' ), ( BB' ), ( CC' ), and ( DD' ) of two parallelograms ( ABCD ) and ( A'B'C'D' ) form a parallelogram, we use vector analysis:1. Define Position Vectors: Let the position vectors of ( A, B, C, D ) be ( mathbf{a}, mathbf{b}, mathbf{c}, mathbf{d} ), and those of ( A', B', C', D' ) be ( mathbf{a'}, mathbf{b'}, mathbf{c'}, mathbf{d'} ).2. Midpoints: The midpoints ( M_1, M_2, M_3, M_4 ) of ( AA', BB', CC', DD' ) are: [ M_1 = frac{mathbf{a} + mathbf{a'}}{2}, quad M_2 = frac{mathbf{b} + mathbf{b'}}{2}, quad M_3 = frac{mathbf{c} + mathbf{c'}}{2}, quad M_4 = frac{mathbf{d} + mathbf{d'}}{2}. ]3. Vectors Between Midpoints: - Vector ( M_1M_2 = frac{mathbf{b} - mathbf{a} + mathbf{b'} - mathbf{a'}}{2} ). - Vector ( M_3M_4 = frac{mathbf{d} - mathbf{c} + mathbf{d'} - mathbf{c'}}{2} ). Since ( ABCD ) and ( A'B'C'D' ) are parallelograms: [ mathbf{b} - mathbf{a} = mathbf{c} - mathbf{d} quad text{and} quad mathbf{b'} - mathbf{a'} = mathbf{c'} - mathbf{d'}. ] Substituting these into ( M_3M_4 ): [ M_3M_4 = frac{(mathbf{c} - mathbf{d}) + (mathbf{c'} - mathbf{d'})}{2} = frac{mathbf{b} - mathbf{a} + mathbf{b'} - mathbf{a'}}{2} = M_1M_2. ]4. Second Pair of Sides: - Vector ( M_2M_3 = frac{mathbf{c} - mathbf{b} + mathbf{c'} - mathbf{b'}}{2} ). - Vector ( M_1M_4 = frac{mathbf{d} - mathbf{a} + mathbf{d'} - mathbf{a'}}{2} ). Using parallelogram properties: [ mathbf{d} - mathbf{a} = mathbf{c} - mathbf{b} quad text{and} quad mathbf{d'} - mathbf{a'} = mathbf{c'} - mathbf{b'}. ] Thus: [ M_1M_4 = frac{mathbf{c} - mathbf{b} + mathbf{c'} - mathbf{b'}}{2} = M_2M_3. ]5. Conclusion: Since ( M_1M_2 parallel M_3M_4 ) and ( M_2M_3 parallel M_1M_4 ), with equal lengths, quadrilateral ( M_1M_2M_3M_4 ) is a parallelogram.Final Answer: The midpoints of segments ( AA' ), ( BB' ), ( CC' ), and ( DD' ) indeed form a parallelogram, as their opposite sides are equal and parallel. Thus, the proof is complete.boxed{text{The midpoints form a parallelogram.}}

🔑:1. Let us denote the midpoints of the segments AA', BB', CC', DD' by A'', B'', C'', D'', respectively. Additionally, let O, O' and O'' be the intersection points of the diagonals AC and BD, A'C' and B'D', and the diagonals A''C'' and B''D'', respectively.2. According to the definition of the midpoints and the properties of parallelograms, we know that the midpoints of the diagonals of a parallelogram coincide with each other. This means that O and O' are the midpoints of their respective diagonals.3. We need to show that O'' also bisects the diagonals A''C'' and B''D'', implying that A''B''C''D'' is a parallelogram.4. Recall that for any quadrilateral MPRT, the midpoints N, Q, S, U of the sides MP, PR, RT, TM respectively, form a parallelogram. This holds true even if the quadrilateral is concave or self-intersecting. Furthermore, if a quadrilateral's diagonals bisect each other, the figure formed by the midpoints of the sides is also a parallelogram.5. Applying the above property to quadrilaterals AC C' A' and BD D' B', which may be concave or self-intersecting as well, we obtain that the quadrilaterals O C'' O' A'' and O D'' O' B'' are parallelograms.6. Consequently, the point O'' which is the common midpoint of O O', must also lie on the diagonals C''A'' and D''B'' at their midpoints. Hence, O'' bisects both C''A'' and D''B''. Conclusion:Since O'' bisects the diagonals A''C'' and B''D'', this implies that A''B''C''D'' is a parallelogram by definition. Thus, the midpoints of the segments joining corresponding vertices of two parallelograms also form a parallelogram.boxed{text{Proved}}