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1997 AIME Problems/Problem 4

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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

Image:1997_AIME-4.png

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...

So m+n = \boxed{17}.

See also

1997 AIME (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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