Since you guys tore up my last challenge question so quickly, I figured I’d make this one a tiny bit harder. Let’s see if I can’t stump you for more than 8 minutes.

First correct response in the comments gets a free Math Guide. Usual contest rules apply. Ready? Here we go!

In the figure above, triangle ABC is inscribed in a circle with center P. If the area of triangle ABC is equal to $\frac{x^2-y^2}{2}$, and AC = xy, what is the area of the circle in terms of x and y? (Be careful typing your answer—parentheses matter!)

MUAHAHAHA. Good luck!

UPDATE: Tuấn got it first, although a others came VERY close (careful with parentheses!!!). Solution below the cut.

It might not be immediately clear what to do here, but as is true of most geometry questions, it’s a good idea to start by listing everything you know. First, you know that triangle ABC is a right triangle, because BC is a diameter, and point A is on the circle. You might recall from geometry class that if an angle lies on a circle, its measure is half of the measure of the arc it corresponds to. Since A is on the circle and corresponds to a 180º arc, the measure of angle A is 90º. You probably won’t see that rule tested on the SAT, but it’s not outside the realm of possibility. So here’s what we know:

Of course, we also know that the formula for the area of a triangle is ½bh. So let’s figure out AB:

\begin{align*}\frac{1}{2}(AB)(x-y)&=\frac{x^2-y^2}{2}\\(AB)(x-y)&=x^2-y^2\\(AB)(x-y)&=(x-y)(x+y)\\AB&=x+y\end{align*}
Oh look. Difference of two squares. Crazy how often that appears!
Now we can solve for BC, using the Pythagorean Theorem:
\begin{align*}(x+y)^2(x-y)^2&=(BC)^2\\(x^2+2xy+y^2)+(x^2-2xy+y^2)&=(BC)^2\\2x^2+2y^2&=(BC)^2\\\sqrt{2(x^2+y^2)}&=BC\end{align*}
Nice. Here’s what we’ve got now:
Of course, BC is the diameter of the circle, and to find the circle’s area we want the radius, which is half of BC:
$r=\frac{\sqrt{2(x^2+y^2)}}{2}$
\begin{align*}A&=\pi\left(\frac{\sqrt{2(x^2+y^2)}}{2}\right)^2\\A&=\pi\left(\frac{2(x^2+y^2)}{4}\right)\\A&=\frac{\pi(x^2+y^2)}{2}\end{align*}