2023 AMC 12A Problems/Problem 9

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)

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

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

Page

Toolbox

Search

2023 AMC 12A Problems/Problem 9

Contents

Problem

Solution

Solution 2 (Area Variation of Solution 1)

See Also