1983 AIME Problems/Problem 14

Problem

In the adjoining figure, two circles with radii $6$ and $8$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in sich a way that the chords $QP$ and $PR$ have equal length. ( $P$ is the midpoint of $QR$ ) Find the square of the length of $QP$ .

Solution

First, notice that if we reflect $R$ over $P$ we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths:

Since $P$ is the midpoint of segment $BC$ , $AP$ is a median of triangle $ABC$ . Because we know that $AB=12$ , $BP=PC=6$ , and $AP=8$ , we can find the third side of the triangle using stewarts or whatever else you like. We get $AC = \sqrt{56}$ . So now we have a kite $AQCP$ with $AQ=AP=8$ , $CQ=CP=6$ , and $AC=\sqrt{56}$ , and all we need is the length of the other diagonal $PQ$ . The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$ . Then

$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}$ .

Doing routine algebra on the above equation, we find that $x^2=\frac{65}{2}$ , so $PQ^2 = 4x^2 = 130.$

Solution II

This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoints of QP be $M_1$ , and the midpoint of PR be $M_2$ . Let x be the length of AM_1, and y that of BM_2

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

1983 AIME Problems/Problem 14

Contents

Problem

Solution

Solution II

See also