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Difference between revisions of "2023 AMC 10A Problems/Problem 11"
(Added See Also and fixed diagram)
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==Problem==
 
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?
 
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]
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<asy>
 
size(200);
 
size(200);
 
defaultpen(linewidth(0.6pt)+fontsize(10pt));
 
defaultpen(linewidth(0.6pt)+fontsize(10pt));
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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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[/asy]
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</asy>
  
 
<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>
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<cmath>2x^2-2x\sqrt{3}+1=0</cmath>.
 
<cmath>2x^2-2x\sqrt{3}+1=0</cmath>.
  
−
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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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}=\boxed{\textbf{(C) }2-\sqrt3}.</math>
  
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~SirAppel
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~SirAppel ~ItsMeNoobieboy
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 21:11, 9 November 2023

Problem

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]

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

Solution

Note that each side length is $\sqrt{2}$ and $\sqrt{3}.$ Let the shorter side of our triangle be $x$, thus the longer leg is $\sqrt{3}-x$. Hence, by the Pythagorean Theorem, we have \[(x-\sqrt{3})^2+x^2=2\] \[2x^2-2x\sqrt{3}+1=0\].

By the quadratic formula, we find $x=\frac{\sqrt{3}\pm1}{2}$. Hence, our answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

~SirAppel ~ItsMeNoobieboy

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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