# Difference between revisions of "2021 AMC 10A Problems/Problem 24"

## Problem

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$, where $a$ a positive real number. What is the area of this region in terms of $a$, valid for all $a > 0$?

$\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}$

## Diagram

Graph in Desmos: https://www.desmos.com/calculator/satawguqsc

~MRENTHUSIASM

## Solution 1

The conditions $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$ give $|x+ay| = |2a|$ and $|ax-y| = |a|$ or $x+ay = \pm 2a$ and $ax-y = \pm a$. The slopes here are perpendicular, so the quadrilateral is a rectangle. Plug in $a=1$ and graph it. We quickly see that the area is $2\sqrt{2} \cdot \sqrt{2} = 4$, so the answer can't be $A$ or $B$ by testing the values they give (test it!). Now plug in $a=2$. We see using a ruler that the sides of the rectangle are about $\frac74$ and $\frac72$. So the area is about $\frac{49}8 = 6.125$. Testing $C$ we get $\frac{16}3$ which is clearly less than $6$, so it is out. Testing $D$ we get $\frac{32}5$ which is near our answer, so we leave it. Testing $E$ we get $\frac{16}5$, way less than $6$, so it is out. So, the only plausible answer is $\boxed{D}$ ~firebolt360

## Solution 2 (Casework: Rectangle)

The cases for $(x+ay)^2 = 4a^2$ are $x+ay = \pm2a,$ or two parallel lines. We rearrange each case and construct the table below: $$\begin{array}{c||c|c|c|c} & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-intercept} & \boldsymbol{y}\textbf{-intercept} & \textbf{Slope} \\ [0.5ex] \hline & & & & \\ [-1.5ex] 1 & x+ay-2a=0 & 2a & 2 & -\frac1a \\ [2ex] 2 & x+ay+2a=0 & -2a & -2 & -\frac1a \\ [0.75ex] \end{array}$$ The cases for $(ax-y)^2 = a^2$ are $ax-y=\pm a,$ or two parallel lines. We rearrange each case and construct the table below: $$\begin{array}{c||c|c|c|c} & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-intercept} & \boldsymbol{y}\textbf{-intercept} & \textbf{Slope} \\ [0.5ex] \hline & & & & \\ [-1.5ex] 1* & ax-y-a=0 & 1 & -a & a \\ [2ex] 2* & ax-y+a=0 & -1 & a & a \\ [0.75ex] \end{array}$$ Since the slopes of intersecting lines $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),$ and $(2)\cap(2*)$ are negative reciprocals, we get four right angles, from which this quadrilateral is a rectangle.

### Solution 2.1 (Distance Between Parallel Lines)

Recall that for constants $A,B,C_1$ and $C_2,$ the distance $d$ between parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is $$d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.$$

From this formula:

• The distance between lines $(1)$ and $(2)$ is $\frac{4a}{\sqrt{1+a^2}},$ the length of this rectangle.
• The distance between lines $(1*)$ and $(2*)$ is $\frac{2a}{\sqrt{a^2+1}},$ the width of this rectangle.

The area we seek is $$\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$$

~MRENTHUSIASM

Plugging $a=2$ into the answer choices gives

$\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}$

At $a=2,$ the respective solutions to systems $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*), (2)\cap(2*)$ are $$(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).$$

By the Distance Formula, the length and width of this rectangle are $\frac{8\sqrt5}{5}$ and $\frac{4\sqrt5}{5},$ respectively.

Finally, the area we seek is $$\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5},$$ from which the answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

~MRENTHUSIASM

## Solution 3 (Trigonometry)

Similar to Solution 2, we will use the equations of the four cases:

(1) $x+ay=2a.$ This is a line with $x$-intercept $2a$, $y$-intercept $2$, and slope $-\frac 1a.$

(2) $x+ay=-2a.$ This is a line with $x$-intercept $-2a$, $y$-intercept $-2$, and slope $-\frac 1a.$

(3)* $ax-y=a.$ This is a line with $x$-intercept $1$, $y$-intercept $-a$, and slope $a.$

(4)* $ax-y=-a.$ This is a line with $x$-intercept $-1$, $y$-intercept $a$, and slope $a.$

The area of the rectangle created by the four equations can be written as $2a\cdot \cos A\cdot4\sin A$

= $8a\cos A \cdot \sin A$

= $8a\cdot~\frac{1}{\sqrt{a^2+1}}\cdot~\frac{a}{\sqrt{a^2+1}}$

= $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

(Note: $\tan A=$ slope $a$)

-fnothing4994

## Solution 4 (bruh moment solution)

Trying $a = 1$ narrows down the choices to options $\textbf{(C)}$, $\textbf{(D)}$ and $\textbf{(E)}$. Trying $a = 2$ and $a = 3$ eliminates $\textbf{(C)}$ and $\textbf{(E)}$, to obtain $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$ as our answer. -¢

~ pi_is_3.14