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>. | ||
+ | |||
+ | *Notice that C, E and the point of tangency to circle C for circle E will be collinear 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 <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>\left(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12\right)^2 = 2\left(\left(\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2+\left(\frac{1}{3r}\right)^2+\left(-\frac12\right)^2\right)</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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We use the notation of Solution 1 for triangle <math>\triangle DEC</math> | ||
+ | <cmath>\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.</cmath> | ||
+ | We use Cosine Law for <math>\triangle DEC</math> and get: | ||
+ | <cmath>(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2 </cmath>. | ||
+ | <cmath>(24 + 6 \sqrt{15} ) r^2 = (8 + 4 \sqrt {15})r \implies 3r = 4 \sqrt{15} – 14 \implies \boxed{\textbf{254}}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 4== | ||
+ | This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, <math>(k_1 + k_2 + k_3 + k_4)^{2} = 2(k_1^{2} + k_2^{2} + k_3^{2} + k_4^{2})</math>, where <math>k_i</math> is the curvature of circle <math>i</math>, meaning <math>k_i = \dfrac{1}{r}</math>. When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle <math>E</math> over <math>\overline{AB}</math>. Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle <math>C</math> has radius <math>1</math>, for ease of computation. Let the radius of Circle <math>D</math> be <math>r</math>, so Circle <math>E</math> has radius <math>\dfrac{r}{3}</math>. Then, we have that <math>(-1 + \dfrac{1}{r} + \dfrac{3}{r} + \dfrac{3}{r})^{2} = 2(1 + \dfrac{1}{r^{2}} + \dfrac{9}{r^{2}} + \dfrac{9}{r^{2}})</math>. This simplifies to <math>\dfrac{49}{r^{2}} - \dfrac{14}{r} + 1 = \dfrac{2r^{2} + 38}{r^{2}}</math>. Multiplying both sides by <math>r^{2}</math>, we get that <math>49 - 14r + r^{2} = 2r^{2} + 38</math>, or <math>r^2 + 14r - 11 = 0</math>. We get <math>r = -7 \pm 2\sqrt{15}</math>, but we want the positive solution, which is <math>r = 2\sqrt{15} - 7</math>. We have to rescale back up, so we get <math>r = 4\sqrt{15} - 14 = \sqrt{240} - 14</math>, so we get that our answer is <math>240 + 14 = \boxed{254}</math>. | ||
+ | ~Puck_0 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:31, 9 January 2024
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 collinear 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
.
Solution 3
We use the notation of Solution 1 for triangle
We use Cosine Law for
and get:
.
vladimir.shelomovskii@gmail.com, vvsss
Solution 4
This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, , where
is the curvature of circle
, meaning
. When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle
over
. Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle
has radius
, for ease of computation. Let the radius of Circle
be
, so Circle
has radius
. Then, we have that
. This simplifies to
. Multiplying both sides by
, we get that
, or
. We get
, but we want the positive solution, which is
. We have to rescale back up, so we get
, so we get that our answer is
.
~Puck_0
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.