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Difference between revisions of "1983 AIME Problems/Problem 14"
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== Problem ==
 
== Problem ==
 
In the adjoining figure, two circles with radii <math>6</math> and <math>8</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in sich a way that the chords <math>QP</math> and <math>PR</math> have equal length. (<math>P</math> is the midpoint of <math>QR</math>) Find the square of the length of <math>QP</math>.  
 
In the adjoining figure, two circles with radii <math>6</math> and <math>8</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in sich a way that the chords <math>QP</math> and <math>PR</math> have equal length. (<math>P</math> is the midpoint of <math>QR</math>) Find the square of the length of <math>QP</math>.  
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== Solution ==
 
== Solution ==

Revision as of 21:40, 23 December 2006

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 $QP$.

1983problem14.JPG

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 $PQ^2 = 4x^2 = 130.$

See also

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