Anyway, you know the drill by now. First correct response in the comments wins a copy of my Math Guide. Full contest rules here.
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AF, and E is the midpoint of DF. If f(x) = px2 + 8 and the area of concave quadrilateral BDCE is equal to 8, what is p?
UPDATE: Peter got it first. Solution below the cut.
The first thing to focus on here is the equation, f(x) = px2 + 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:
The only other thing we have to work with is that 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.
Let's focus on △DEC.
If △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:
Anyway, now we know FC = 4. Since F is the origin and C is on the x-axis, that means point C is (4, 0).
And we can use that to solve for 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) = px2 + 8
f(4) = p(4)2 + 8 = 0
16p + 8 = 0
16p = –8
p = –0.5
And there you have it! Pretty awesome question, if I do say so myself.