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2023 AMC 10A Problems/Problem 11

Revision as of 20:32, 9 November 2023 by Ghfhgvghj10 (talk | contribs) (→‎Solution)

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

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

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