Difference between revisions of "2023 AMC 10A Problems/Problem 11"
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draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(E--F--G--H--cycle); | draw(E--F--G--H--cycle); | ||
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<math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> | <math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> | ||
Revision as of 21:00, 9 November 2023
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
.
By the quadratic formula, we find . Hence, our answer is
~SirAppel