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 $QP$ . [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=793[/img]

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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