Difference between revisions of "1997 AIME Problems/Problem 4"
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If (in the diagram above) we draw the line going through the centers of the circles with radii <math>8</math> and <math>\frac mn = r</math>, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii <math>5</math>. Then we form two [[right triangles]], of lengths <math>5, x, 5+r</math> and <math>5, 8+r+x, 13</math>, wher <math>x</math> is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii <math>5</math>. By the [[Pythagorean Theorem]], we now have two equations with two unknowns: | If (in the diagram above) we draw the line going through the centers of the circles with radii <math>8</math> and <math>\frac mn = r</math>, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii <math>5</math>. Then we form two [[right triangles]], of lengths <math>5, x, 5+r</math> and <math>5, 8+r+x, 13</math>, wher <math>x</math> is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii <math>5</math>. By the [[Pythagorean Theorem]], we now have two equations with two unknowns: | ||
− | < | + | <cmath>\begin{eqnarray*} |
5^2 + x^2 &=& (5+r)^2 \\ | 5^2 + x^2 &=& (5+r)^2 \\ | ||
x &=& \sqrt{10r + r^2} \\ | x &=& \sqrt{10r + r^2} \\ | ||
Line 16: | Line 16: | ||
10r+r^2 &=& 16 - 8r + r^2\\ | 10r+r^2 &=& 16 - 8r + r^2\\ | ||
r &=& \frac{8}{9} | r &=& \frac{8}{9} | ||
− | \end{eqnarray*}</ | + | \end{eqnarray*}</cmath> |
So <math>m+n = \boxed{17}</math>. | So <math>m+n = \boxed{17}</math>. |
Revision as of 18:49, 10 March 2015
Problem
Circles of radii and are mutually externally tangent, where and are relatively prime positive integers. Find
Solution
If (in the diagram above) we draw the line going through the centers of the circles with radii and , that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii . Then we form two right triangles, of lengths and , wher is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii . By the Pythagorean Theorem, we now have two equations with two unknowns:
So .
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.