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| − | A square of area <math>2</math> is inscribed in a square of area <math>3</math>, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
| + | #redirect[[2023 AMC 12A Problems/Problem 9]] |
| − | [asy]
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| − | size(200);
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| − | defaultpen(linewidth(0.6pt)+fontsize(10pt));
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| − | real y = sqrt(3);
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| − | pair A,B,C,D,E,F,G,H;
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| − | A = (0,0);
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| − | B = (0,y);
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| − | C = (y,y);
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| − | D = (y,0);
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| − | E = ((y + 1)/2,y);
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| − | F = (y, (y - 1)/2);
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| − | G = ((y - 1)/2, 0);
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| − | H = (0,(y + 1)/2);
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| − | fill(H--B--E--cycle, gray);
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| − | draw(A--B--C--D--cycle);
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| − | draw(E--F--G--H--cycle);
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| − | [\asy]
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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>
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| − | ==Solution==
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| − | Note that each side length is <math>\sqrt{2}</math> and <math>\sqrt{3}.</math> Let the shorter side of our triangle be <math>x</math>, thus the longer leg is <math>\sqrt{3}-x</math>. Hence, by the Pythagorean Theorem, we have <cmath>(x-\sqrt{3})^2+x^2=2</cmath>
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| − | <cmath>2x^2-2x\sqrt{3}+1=0</cmath>.
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| − | By the quadratic formula, we find <math>x=\frac{\sqrt{3}\pm1}{2}</math>. Hence, our answer is <math>\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}.</math>
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| − | ~SirAppel
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