Track / Stage 4 / 290 of 340 #550 of 1964
Problem 550 AMC 12 late, AIME early Difficulty 4.9 Multiple choice
18.73 A regular octagon is formed by cutting off equal isosceles right triangles from the corners of a square. If the side length of the square is 1, then the length of the legs of these triangles is
Pick one
A 2 + 2 3 \frac{2+\sqrt{2}}{3} 3 2 + 2 B 2 − 2 2 \frac{2-\sqrt{2}}{2} 2 2 − 2 C 1 + 2 2 \frac{1+\sqrt{2}}{2} 2 1 + 2 D 1 + 2 3 \frac{1+\sqrt{2}}{3} 3 1 + 2 E 2 − 2 3 \frac{2-\sqrt{2}}{3} 3 2 − 2
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Official solution [Solution] Let the length of one leg of the cut-off isosceles right triangle be x x x 2 x \sqrt{2} x 2 x x + 2 x + x = 1.
x+\sqrt{2} x+x=1 .
x + 2 x + x = 1.
We getx = 2 − 2 2 .
x=\frac{2-\sqrt{2}}{2} \text {. }
x = 2 2 − 2 .
Therefore, the answer is ( B ) (B) ( B )
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Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
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