# 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 $a > 0$?

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

## 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)

For the equation $(x+ay)^2 = 4a^2,$ the cases are

(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 $-\frac 1a.$

For the equation $(ax-y)^2 = a^2,$ the cases are

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

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

Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle.

Plugging $a=2$ into the 4 above equations and solving systems of equations for intersecting lines [(1) and (1)*, (1) and (2)*, (2) and (1)*, (2) and (2)*], we get the respective solutions $$(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).$$ Finally, by the Distance Formula, the length and width of the rectangle are $\frac{8\sqrt5}{5}$ and $\frac{4\sqrt5}{5}.$ The area we seek is $$\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.$$ Plugging $a=2$ into the choices gives

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

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

~MRENTHUSIASM

## Solution 3 (Geometry)

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 $2a\cdot \sin A\cdot4\sin A$

= $8a(\sin A)^2$

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

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

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

-fnothing4994

~ pi_is_3.14