AIME Problem: The distance AB is 12...

[Trenters4325]

The distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP^2.

How would you do this problem by radii perpendicular to the two chords?

I know how to do it with other methods, so I just need help with a solution that uses the perpendicular radii.

 

[Denis]

"The circle center A radius 8 and the circle center B radius 6 meet at P (and another point)."
"meet" is a bad choice of words: the circles INTERSECT.

This would be clearer: Line QR is drawn going through point P, such that
QP is a chord of larger circle, and PR is a chord of the other, and QP = PR.

"radii perpendicular to the two chords" is quite unclear:
you mean a perpendicular line from A to chord QP and from B to chord PR.
Make these lines AV (V on QP) and BW (W on PR).
Since V and W are midpoints of the chords, then VP = PW;
let u = VP = PW, v = AV and w = BW.

We have right triangle AVP with legs u,v and hypotenuse = 8, and
we have right triangle BWP with legs u,w and hypotenuse = 6;
so u^2 + v^2 = 64 and u^2 + w^2 = 36: leads to v = sqrt(w^2 + 28).

And of course triangle APB has sides = 6,8,12.

That's all you get from me; I'm sure you can finish it off :?:

Hint: drop a perpendicular from P to AB; then you're almost finished...