Math Gold Medalist

Lor

2023 AMC 12A 

Problem 9

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]

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

Pythagorean Theorem

Quadratic Equation

Congruent Triangles

x^2 + (sqrt(3)-x)^2 = 2

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   Solution