Difference between revisions of "2018 AMC 12B Problems/Problem 25"
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Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersection of lines <math>O_1P_1</math> and <math>O_2P_2</math>. Because <math>\angle P_1P_2P_3 = 60^\circ</math>, it follows that <math>\triangle P_2KP_1</math> is a <math>30-60-90^\circ</math> triangle. Let <math>d=P_1K</math>; then <math>P_2K = 2d</math> and <math>P_1P_2 = \sqrt 3d</math>. The Law of Cosines in <math>\triangle O_1KO_2</math> gives <cmath>8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\cos 60^\circ,</cmath>which simplifies to <math>3d^2 - 12d - 16 = 0</math>. The positive solution is <math>d = 2 + \tfrac23\sqrt{21}</math>. Then <math>P_1P_2 = \sqrt 3 d = 2\sqrt 3 + 2\sqrt 7</math>, and the required area is <cmath>\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.</cmath>The requested sum is <math>300 + 252 = \boxed{552}</math>. | Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersection of lines <math>O_1P_1</math> and <math>O_2P_2</math>. Because <math>\angle P_1P_2P_3 = 60^\circ</math>, it follows that <math>\triangle P_2KP_1</math> is a <math>30-60-90^\circ</math> triangle. Let <math>d=P_1K</math>; then <math>P_2K = 2d</math> and <math>P_1P_2 = \sqrt 3d</math>. The Law of Cosines in <math>\triangle O_1KO_2</math> gives <cmath>8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\cos 60^\circ,</cmath>which simplifies to <math>3d^2 - 12d - 16 = 0</math>. The positive solution is <math>d = 2 + \tfrac23\sqrt{21}</math>. Then <math>P_1P_2 = \sqrt 3 d = 2\sqrt 3 + 2\sqrt 7</math>, and the required area is <cmath>\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.</cmath>The requested sum is <math>300 + 252 = \boxed{552}</math>. | ||
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== Solution 2 == | == Solution 2 == |
Revision as of 19:39, 16 April 2019
Problem
Circles ,
, and
each have radius
and are placed in the plane so that each circle is externally tangent to the other two. Points
,
, and
lie on
,
, and
respectively such that
and line
is tangent to
for each
, where
. See the figure below. The area of
can be written in the form
for positive integers
and
. What is
?
Solution 1
Let be the center of circle
for
, and let
be the intersection of lines
and
. Because
, it follows that
is a
triangle. Let
; then
and
. The Law of Cosines in
gives
which simplifies to
. The positive solution is
. Then
, and the required area is
The requested sum is
.
And we are done.
Solution 2
Let and
be the centers of
and
respectively and draw
,
, and
. Note than
and
are both right. Furthermore, since
is equilateral,
and
. Mark
as the base of the altitude from
to
. By special right triangles,
and
. since
and
, we can find find
. Thus,
. This makes
. This makes the answer
.
Solution 3
Let be the center of circle
for
. Let
be the centroid of
, which also happens to be the centroid of
. Because
and
,
.
is
the height of
, thus
is
.
Applying cosine law on , one finds that
. Multiplying by
to solve for the height of
, one gets
. Simply multiplying by
and then calculating the equilateral triangle's area, one would get the final result of
.
This makes the answer .
~AlbeePach~
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.