2023 AMC 10A Problems/Problem 11

Problem

A square of area $2$ is inscribed in a square of area $3$ , creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle? $[asy] size(200); defaultpen(linewidth(0.6pt)+fontsize(10pt)); real y = sqrt(3); pair A,B,C,D,E,F,G,H; A = (0,0); B = (0,y); C = (y,y); D = (y,0); E = ((y + 1)/2,y); F = (y, (y - 1)/2); G = ((y - 1)/2, 0); H = (0,(y + 1)/2); fill(H--B--E--cycle, gray); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); [/asy]$

$\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1$

Solution

Note that each side length is $\sqrt{2}$ and $\sqrt{3}.$ Let the shorter side of our triangle be $x$ , thus the longer leg is $\sqrt{3}-x$ . Hence, by the Pythagorean Theorem, we have $(x-\sqrt{3})^2+x^2=2$ $2x^2-2x\sqrt{3}+1=0$ .

By the quadratic formula, we find $x=\frac{\sqrt{3}\pm1}{2}$ . Hence, our answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

~SirAppel ~ItsMeNoobieboy

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 10 Problems and Solutions

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2023 AMC 10A Problems/Problem 11

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