Track / Stage 4 / 145 of 340 #405 of 1964
Problem 405 AMC 12 late, AIME early Difficulty 4.7 Multiple choice
As shown in Figure 2, in the right trapezoid A B C D A B C D A B C D ∠ B = \angle B= ∠ B = ∠ C = 90 ∘ , A B = B C \angle C=90^{\circ}, A B=B C ∠ C = 9 0 ∘ , A B = B C E E E B C B C B C △ A D E \triangle A D E △ A D E △ A D E \triangle A D E △ A D E A B C D A B C D A B C D
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A 3 − 1 \sqrt{3}-1 3 − 1 B 3 2 \frac{\sqrt{3}}{2} 2 3 C 2 3 \frac{2}{3} 3 2 D ( 3 − 1 ) 2 (\sqrt{3}-1)^{2} ( 3 − 1 ) 2
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Official solution 6. D.
As shown in Figure 6, extend trapezoid A B C D ABCD A B C D A B C F ABCF A B C F C D = C E = x CD = CE = x C D = C E = x B E = F D = 1 − x .
BE = FD = 1 - x.
B E = F D = 1 − x .
By the Pythagorean theorem, we have 2 x 2 = 1 + ( 1 − x ) 2 2x^2 = 1 + (1 - x)^2 2 x 2 = 1 + ( 1 − x ) 2 x x x x = 3 − 1 x = \sqrt{3} - 1 x = 3 − 1 Therefore, S △ A D E S trapezoid A B C D = 3 4 × 2 x 2 1 2 ( 1 + x ) × 1 = ( 3 − 1 ) 2 .
\text{Therefore, } \frac{S_{\triangle ADE}}{S_{\text{trapezoid } ABCD}} = \frac{\frac{\sqrt{3}}{4} \times 2x^2}{\frac{1}{2}(1 + x) \times 1} = (\sqrt{3} - 1)^2.
Therefore, S trapezoid A B C D S △ A D E = 2 1 ( 1 + x ) × 1 4 3 × 2 x 2 = ( 3 − 1 ) 2 .
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Source: NuminaMath-1.5 ,
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