Track / Stage 4 / 205 of 340 #465 of 1964
Problem 465 AMC 12 late, AIME early Geometry Difficulty 4.8 Find the answer
81 Given a rhombus A B C D A B C D A B C D with side length a a a and ∠ A = π 3 \angle A=\frac{\pi}{3} ∠ A = 3 π , the rhombus A B C D A B C D A B C D is folded along the diagonal to form a dihedral angle θ \theta θ , where θ ∈ [ π 3 , 2 π 3 ] \theta \in\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] θ ∈ [ 3 π , 3 2 π ] . Then the maximum distance between the two diagonals is A. 3 2 a \frac{3}{2} a 2 3 a B. 3 4 a \frac{\sqrt{3}}{4} a 4 3 a C. 3 2 a \frac{\sqrt{3}}{2} a 2 3 a D. 3 4 a \frac{3}{4} a 4 3 a
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Official solution 81 D. If folded along B D B D B D , let the midpoint of A C A C A C be F F F , and the midpoint of B D B D B D be E E E , then A E = 3 2 a A E=\frac{\sqrt{3}}{2} a A E = 2 3 a ,E F = A E cos θ 2 ⩽ 3 2 a cos π 6 = 3 4 a .
E F=A E \cos \frac{\theta}{2} \leqslant \frac{\sqrt{3}}{2} a \cos \frac{\pi}{6}=\frac{3}{4} a .
E F = A E cos 2 θ ⩽ 2 3 a cos 6 π = 4 3 a .
If folded along A C A C A C , let the midpoint of A C A C A C be F F F , and the midpoint of B D B D B D be E E E , then B D = 1 2 a , E F = B D cos θ 2 ⩽ B D=\frac{1}{2} a, E F=B D \cos \frac{\theta}{2} \leqslant B D = 2 1 a , E F = B D cos 2 θ ⩽ 3 a 4 .
\frac{\sqrt{3} a}{4} .
4 3 a .
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Source: NuminaMath-1.5 ,
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