Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem: editted in the figure)
(still working on)
Line 12: Line 12:
  
 
Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = 130.</math>
 
Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = 130.</math>
 +
== 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
  
 
----
 
----

Revision as of 22:43, 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 $PQ^2 = 4x^2 = 130.$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_14&oldid=12026"