You know, It’s funny. When I sit down to write one of these, I usually have a rough sketch in my mind of how the question’s going to go. But half the time, the problem completely evolves into something else by the time I’m done with it. I’m not sure why that is. It could just be that I’m capricious, but it’s probably because math comes into much sharper focus once it’s written down. Which I only mention to gently encourage you to write everything down when you take tests.

Anyway, you know the drill by now. First correct response in the comments wins a copy of my Math Guide. Full contest rules here.

**Note:** If you’re new to the site, your comment might not appear right away. Don’t panic—I receive comments in my email in the order they’re submitted. If you’re the first to get it right, you’ll win even if your comment doesn’t immediately appear.

In the figure above,

Ais the vertex off(x),BandCare thex-intercepts off(x),Fis the origin,Dis the midpoint ofAF, andEis the midpoint ofDF. Iff(x) =px^{2}+ 8 and the area of concave quadrilateralBDCEis equal to 8, what isp?

Good luck!

UPDATE: Peter got it first. Solution below the cut.

The first thing to focus on here is the equation, *f*(*x*) = *px*^{2} + 8, which tells you a very important thing: the *y*-intercept of *f*(*x*) is 8. This is one of the few things you need to keep in mind about parabolas—that the constant in a parabola equation tells you its *y*-intercept. This makes sense, of course, because the *y*-intercept is when *x* = 0, and f(0) = *p*(0)^{2} + 8 = 8. So yeah. Point *A* is (0, 8).

We know that *D* and *E* are midpoints, so we now know a few distances. Let’s throw them on the figure:

*BDCE*has an area of 8. There are a few ways one might use that information—and please, as always, leave your own solutions in the comments—but I recommend recognizing that, because parabolas are symmetrical, everything on this figure is symmetrical about the

*y*-axis. So if

*BDCE*has an area of 8, then △

*DEB*and △

*DEC*each have areas of 4. We can work more easily with triangles.

*DEC*.

*DEC*has an area of 8, and a base of

*DE*= 2, we can solve its height,

*FC*.* We do this with the handy-dandy triangle area formula:

*FC*= 4. Since

*F*is the origin and

*C*is on the

*x*-axis, that means point

*C*is (4, 0).

*p*. All we need to do is plug the point we know is on the parabola,

*C*, into the parabola’s equation. We know (4, 0) is on the parabola. (Another way of saying this:

*f*(4) = 0.)

*f*(

*x*) =

*px*

^{2}+ 8

*f*(4) =

*p*(4)

^{2}+ 8 = 0

*p*+ 8 = 0

*p*=

*–8*

*p*= –0.5

4. Total guess

-1/2

p=-1/2

Nice work!

I get -1/8 but I am out of practice. Kudos for keeping this blog going strong. It’s as good today as the day I found it almost two years ago. One of if not the best free resource on the net for SAT test prep. Hope things are going well with book sales and all else…

Regards,

JD

Thanks JD–great to hear from you. Things are good here…hope the same is true for you!

-0.5

p is -2

im gonna correct myself…it is -0.5

Well, as long as we are correcting ourselves I will change to -1/2 as well..

the base of the triangles is 8 so the x-interecepts are -4 and +4…

so px^2 must equal -8 so that f(x)=0…

if x=4 or -4, p must = -1/2…

-0.5