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.


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 1 (Manipulation)

Let $a$ be the length of the shorter leg and $b$ be the longer leg. By the Pythagorean theorem, we can derive that $a^2+b^2=2$. Using area we can also derive that $2ab=1$. $(a+b)^2=3$ as given in the diagram, we can find that $(a-b)^2=1$ because $4ab=2$. This means that $b+a=\sqrt{3}$ and $b-a=1$. Adding the equations gives $b=\frac{\sqrt{3}+1}{2}$ and when $b$ is plugged in $a = \frac{\sqrt{3}-1}{2}$. Rationalizing the denominators gives us $\boxed{\textbf{(C) } 2-\sqrt{3}}$.

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Solution 2 (Area)

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 $\frac{x\left(\sqrt{3} - x\right)}{2} = \frac{1}{4}$.

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 side length 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 $\boxed{\textbf{(C) }2-\sqrt3}$.


Solution 4

The side length of the bigger square is equal to $\sqrt{3}$, while the side length of the smaller square is $\sqrt{2}$. Let $x$ be the shorter leg and $y$ be the longer one. Clearly, $x+y=\sqrt{3}$, and $xy=\frac{1}{2}$. Using Vieta's to build a quadratic, we get \[x^2-\sqrt{3}x+\frac{1}{2}=0.\] Solving, we get $x=\frac{\sqrt{3}-1}{2}$ and $y=\frac{\sqrt{3}+1}{2}$. Thus, we find $\frac{x}{y}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)\cdot(\sqrt{3}-1)}=\frac{4-2\sqrt{3}}{2}=\boxed{\textbf{(C) }2-\sqrt3}$.


Solution 5

Let $\theta$ be the angle opposite the smaller leg. We want to find $\tan\theta$.

The area of the triangle is $\frac{1}{2}\left(\sqrt{2}\sin\theta\right)\left(\sqrt{2}\cos\theta\right)=\frac{1}{2}\sin 2\theta=\frac{1}{4},$ which implies $\sin 2\theta = \frac{1}{2},$ or $\theta=15^\circ$. Therefore $\tan \theta = \boxed{\textbf{(C) }2-\sqrt3}$

Solution 6

Allow a, b to be the sides of a triangle. WLOG, suppose $a > b$We want to find $\frac{b}{a}$. Notice that the area of a triangle is $\frac{3-2}{4}$, which results in $\frac{1}{4}$. Thus, $ab = \frac{1}{2}$. However, the square of the hypotenuse of this triangle is $a^2+b^2$, but also $2$. We can write $b$ as $\frac{1}{2a}$, and then plug it in. We get $a^2+\frac{1}{4a^2} = 2$, so $4a^4-8a^2+1 = 0$. Applying the quadratic formula, $a^2 = \frac{8 \pm 4\sqrt{3}}{2}$, or $4 \pm 2\sqrt3$. However, since $a$ and $b$ must both be solutions of the quadratic, since both equations were cyclic. Since $a>b$, then $a^2 = 4+2\sqrt3$, and $b^2 = 4-2\sqrt3$. To find $\frac{b}{a}$, we can simply find the square root of $\frac{b^2}{a^2}$. This is $\sqrt{\frac{4-2\sqrt3}{4+2\sqrt3}} = \sqrt{\frac{2-\sqrt3}{2+\sqrt3}} = \sqrt{(2-\sqrt3)(2-\sqrt3)} = \boxed {2-\sqrt3}$, so the answer is $\boxed {C}$. - Sepehr2010

Solution 6

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 \[(\sqrt{3}-x)^2+x^2=2\] \[2x^2-2x\sqrt{3}+1=0\].

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

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See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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