2023 AMC 10A Problems/Problem 11
A square of area 
is inscribed in a square of area 
, 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]
Solution
Note that each side length is 
and 
Let the shorter side of our triangle be 
, thus the longer leg is 
. Hence, by the Pythagorean Theorem, we have ![\[(x-\sqrt{3})^2+x^2=2\]](http://latex.artofproblemsolving.com/4/3/b/43bfb185f856b6399445c3f2a3ffe675ce459426.png)
![\[2x^2-2x\sqrt{3}+1=0\]](http://latex.artofproblemsolving.com/a/3/d/a3d3d78940ddfb7a6a316dac8be206a3c8edb50d.png)
.
By the quadratic formula, we find 
. Hence, our answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}}=2-\sqrt{3}.$ (Error compiling LaTeX. Unknown error_msg)
~SirAppel
