2021 AMC 10A Problems/Problem 24

Problem 24

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

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 $\boxed{D}$