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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Consider a reflection of circle <math>E</math> over diameter <math>\overline{AB}</math>. 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 <math>r</math>, <math>r</math>, and <math>3r</math>, and the big circle has radius <math>2</math>. | ||
+ | |||
+ | Descartes' Circle Theorem gives <math>(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12)^2 = 2((\frac{1}{r})^2+(\frac{1}{r})^2+(\frac{1}{3r})^2+(-\frac12)^2)</math> | ||
+ | |||
+ | Note that the big circle has curvature <math>-\frac12</math> because it is internally tangent. | ||
+ | Solving gives <math>3r=\sqrt{240}-14</math> for a final answer of <math>\boxed{254}</math>. | ||
== See also == | == See also == |
Revision as of 18:06, 16 February 2017
Contents
[hide]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
.
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.