Difference between revisions of "2014 AIME II Problems/Problem 8"
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+ | ==Problem== | ||
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+ | Circle <math>C</math> with radius 2 has diameter <math>\overline{AB}</math>. Circle D is internally tangent to circle <math>C</math> at <math>A</math>. Circle <math>E</math> is internally tangent to circle <math>C</math>, externally tangent to circle <math>D</math>, and tangent to <math>\overline{AB}</math>. The radius of circle <math>D</math> is three times the radius of circle <math>E</math>, and can be written in the form <math>\sqrt{m}-n</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m+n</math>. | ||
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==Solution 1== | ==Solution 1== | ||
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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 circle E will be concurrent because C and E intersect the tangent line at a right angle, implying they must be on the same line. | ||
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+ | ==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>. | ||
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+ | 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> | ||
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+ | 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>. | ||
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+ | == See also == | ||
+ | {{AIME box|year=2014|n=II|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
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 circle 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.