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Difference between revisions of "2023 AMC 12A Problems/Problem 9"

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do little worthless cheater
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==Problem==
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go grind amc if you wanna suceed
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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?
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best of luck
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<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}=\boxed{\textbf{(C) }2-\sqrt3}.</math>
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~SirAppel ~ItsMeNoobieboy
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==Solution 2 (Area Variation of Solution 1)==
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Looking at the diagram, knowing the square inscribed the square with area 3 is area 2. We would automatically know the area of the triangles are <math>\frac{1}{4}</math>
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From solution 1, the base is <math>x</math> and the height <math>\sqrt{3} - x</math>. Which means <math>\frac{x(\sqrt{3} - x)}{2} = \frac{1}{4}</math>
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We can turn this into a quadratic equation making it <math>x^2-x\sqrt{3}+\frac{1}{2} = 0</math>
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By using the quadratic formula, we get <math>x=\frac{\sqrt{3}\pm1}{2}</math>.Therefore, the answer is <math>\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.</math>
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~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)
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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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{{AMC12 box|year=2023|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 23:44, 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

Solution 2 (Area Variation of Solution 1)

Looking at the diagram, knowing the square inscribed the square with area 3 is area 2. We would automatically know the area of the triangles are $\frac{1}{4}$

From solution 1, the base is $x$ and the height $\sqrt{3} - x$. Which means $\frac{x(\sqrt{3} - x)}{2} = \frac{1}{4}$

We can turn this into a quadratic equation making it $x^2-x\sqrt{3}+\frac{1}{2} = 0$

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

~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)

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
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 12 Problems and Solutions

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