# Difference between revisions of "1997 AIME Problems/Problem 4"

## Problem

Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

## Solution

If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\frac mn = r$, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$. Then we form two right triangles, of lengths $5, x, 5+r$ and $5, 8+r+x, 13$, wher $x$ is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii $5$. By the Pythagorean Theorem, we now have two equations with two unknowns: $\begin{eqnarray*} 5^2 + x^2 &=& (5+r)^2 \\ x &=& \sqrt{10r + r^2} \\ && \\ (8 + r + \sqrt{10r+r^2})^2 + 5^2 &=& 13^2\\ 8 + r + \sqrt{10r+r^2} &=& 12\\ \sqrt{10r+r^2}&=& 4-r\\ 10r+r^2 &=& 16 - 8r + r^2\\ r &=& \frac{8}{9} \end{eqnarray*}$

So $m+n = \boxed{17}$.

## See also

 1997 AIME (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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