2023 AMC 12A Problems/Problem 9

The following problem is from both the 2023 AMC 10A #11 and 2023 AMC 12A #9, so both problems redirect to this page.

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, we know that the square inscribed in the square with area $3$ has area $2$ . Thus, the area outside of the small square is $3-2=1.$ This area is composed of $4$ congruent triangles, so we know that each triangle has an area of $\dfrac14$ .

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

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

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)

(Clarity & formatting edits by Technodoggo)

Solution 3

Let $x$ be the ratio of the shorter leg to the longer leg, and $y$ be the length of longer leg. The length of the shorter leg will be $xy$ .

Because the sum of two legs is the length of a side of the outside square, we have $xy + y = \sqrt{3}$ , which means $(xy)^2 + y^2 + 2xy^2 = 3$ . Using the Pythagorean Theorem for the shaded right triangle, we also have $(xy)^2 + y^2 = 2$ . Solving both equations, we get $2xy^2 = 1$ . Using $y^2=\frac{1}{2x}$ to substitute $y$ in the second equation, we get $x^2\cdot \frac{1}{2x} + \frac{1}{2x} = 2$ . Hence, $x^2 - 4x + 1 = 0$ . By using the quadratic formula, we get $x=2\pm \sqrt3$ . Because $x$ be the ratio of the shorter leg to the longer leg, it is always less than $1$ . Therefore, the answer is $\boxed{\textbf{(C) }2-\sqrt3}$ .

~sqroot

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.

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2023 AMC 12A Problems/Problem 9

Contents

Problem

Solution

Solution 2 (Area Variation of Solution 1)

Solution 3

See Also