Difference between revisions of "2023 AMC 10A Problems/Problem 11"

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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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[quote=fruitmonster97]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?
 
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[asy]
-Jonathan Yu
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size(200);
Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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defaultpen(linewidth(0.6pt)+fontsize(10pt));
 
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real y = sqrt(3);
-Jonathan Yu
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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>[/quote]

Revision as of 16:01, 9 November 2023

[quote=fruitmonster97]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] $\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1$[/quote]