Difference between revisions of "2014 AIME II Problems/Problem 8"
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Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives <math>3r=\sqrt{240}-14</math> for a final answer of <math>\boxed{254}</math>. | Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives <math>3r=\sqrt{240}-14</math> for a final answer of <math>\boxed{254}</math>. | ||
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+ | *Notice that C, E and the point of tangency to circle C for triangle E will be concurrent because C and E intersect the tangent line at a right angle, implying they must be on the same line. | ||
==Solution 2== | ==Solution 2== |
Revision as of 22:53, 7 October 2017
Contents
Problem
Circle with radius 2 has diameter
. Circle D is internally tangent to circle
at
. Circle
is internally tangent to circle
, externally tangent to circle
, and tangent to
. The radius of circle
is three times the radius of circle
, and can be written in the form
, where
and
are positive integers. Find
.
Solution 1
Using the diagram above, let the radius of be
, and the radius of
be
. Then,
, and
, so the Pythagorean theorem in
gives
. Also,
, so
Noting that
, we can now use the Pythagorean theorem in
to get
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives for a final answer of
.
- Notice that C, E and the point of tangency to circle C for triangle E will be concurrent because C and E intersect the tangent line at a right angle, implying they must be on the same line.
Solution 2
Consider a reflection of circle over diameter
. By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii
,
, and
, and the big circle has radius
.
Descartes' Circle Theorem gives
Note that the big circle has curvature because it is internally tangent.
Solving gives
for a final answer of
.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.