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

Revision as of 23:47, 11 March 2009

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

Solution 1

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 Stewart's Theorem or similar approaches. 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 $PQ^2 = 4x^2 = \boxed{130}.$

Solution 2

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 $BM_2$ .

Template:Incomplete

Solution 3

Let $QP=PR=x$ . Angles $QPA$ , $APB$ , and $BPR$ must add up to $180^{\circ}$ . By the Law of Cosines, $\angle APB=\cos^{-1}(-11/24)$ . Also, angles $QPA$ and $BPR$ equal $\cos^{-1}(x/16)$ and $\cos^{-1}(x/12)$ . So we have

$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$

Taking the $\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$ .

Solution 4

Let the circles of radius $8$ and $6$ be centered at $A$ and $B,$ respectively. Let the midpoints of $QP$ and $PR$ be $N$ and $O.$ Dropping a perpendicular from $B$ to $AN$ (let the point be $K$ ) gives a rectangle.

Now note that triangle $ABK$ is right. Let the midpoint of $AB$ (segment of length $12$ ) be $M.$ Hence, $KM = 6 = BM = BP.$

By now obvious similar triangles, $3BO = 3KN = AN,$ so it's a quick system of two linear equations to solve for the desired length.

Template:Incomplete

@@ Line 23: / Line 23: @@
 === Solution 4 ===
-Let the circles of radius <math>8</math> and <math>6</math> be centered at <math>A</math> and <math>B,</math> respectively. Let the midpoints of <math>QP</math> and <math>PR</math> be <math>N</math> and <math>O.</math> Dropping a perpendicular from <math>B</math> to <math>AN</math> (let the point be <math>K?</math>) gives a rectangle.
+Let the circles of radius <math>8</math> and <math>6</math> be centered at <math>A</math> and <math>B,</math> respectively. Let the midpoints of <math>QP</math> and <math>PR</math> be <math>N</math> and <math>O.</math> Dropping a perpendicular from <math>B</math> to <math>AN</math> (let the point be <math>K</math>) gives a rectangle.
 Now note that triangle <math>ABK</math> is right. Let the midpoint of <math>AB</math> (segment of length <math>12</math>) be <math>M.</math> Hence, <math>KM = 6 = BM = BP.</math>
 By now obvious [[similar triangles]], <math>3BO = 3KN = AN,</math> so it's a quick system of two linear equations to solve for the desired length.
+{{incomplete|solution}}
 == 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

Page

Toolbox

Search