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Difference between revisions of "1983 AIME Problems/Problem 14"

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== Solution II==
 
== Solution II==
 
This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoints of QP be <math>M_1</math>, and the midpoint of PR be <math>M_2</math>. Let x be the length of AM_1, and y that of BM_2
 
This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoints of QP be <math>M_1</math>, and the midpoint of PR be <math>M_2</math>. Let x be the length of AM_1, and y that of BM_2
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* [[1983 AIME Problems/Problem 13|Previous Problem]]
 
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* [[1983 AIME Problems/Problem 15|Next Problem]]
 
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* [[1983 AIME Problems|Back to Exam]]
 
  
 
== See also ==
 
== See also ==
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{{AIME box|year=1983|num-b=13|num-a=15}}
 
* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]

Revision as of 13:14, 6 May 2007

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

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

See also

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
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