Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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===Solution 4=== | ===Solution 4=== | ||
− | Let <math>P = XY \cap OZ</math>. Consider an inversion about <math>C_1 \implies C_2 \to XY \implies Z \to P \implies OP \cdot OZ = r^2 \implies OP = r^2/11 \implies OZ = \dfrac{121 - r^2}{11}</math>. Using <math>\triangle YPZ \sim OXZ \implies | + | Let <math>P = XY \cap OZ</math>. Consider an inversion about <math>C_1 \implies C_2 \to XY \implies Z \to P \implies OP \cdot OZ = r^2 \implies OP = r^2/11 \implies OZ = \dfrac{121 - r^2}{11}</math>. Using <math>\triangle YPZ \sim OXZ \implies r = \sqrt{30} \implies \boxed{E}</math>. |
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+ | -Solution by '''IDMsaterz''' | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2012|ab=A|num-b=15|num-a=17}} |
Revision as of 20:44, 6 February 2013
Contents
Problem
Circle has its center
lying on circle
. The two circles meet at
and
. Point
in the exterior of
lies on circle
and
,
, and
. What is the radius of circle
?
Solution
Solution 1
Let denote the radius of circle
. Note that quadrilateral
is cyclic. By Ptolemy's Theorem, we have
and
. Let t be the measure of angle
. Since
, the law of cosines on triangle
gives us
. Again since
is cyclic, the measure of angle
. We apply the law of cosines to triangle
so that
. Since
we obtain
. But
so that
.
.
Solution 2
Let us call the the radius of circle
, and
the radius of
. Consider
and
. Both of these triangles have the same circumcircle (
). From the Extended Law of Sines, we see that
. Therefore,
. We will now apply the Law of Cosines to
and
and get the equations
,
,
respectively. Because , this is a system of two equations and two variables. Solving for
gives
.
.
Solution 3
Let denote the radius of circle
. Note that quadrilateral
is cyclic. By Ptolemy's Theorem, we have
and
. Consider isosceles triangle
. Pulling an altitude to
from
, we obtain
. Since quadrilateral
is cyclic, we have
, so
. Applying the Law of Cosines to triangle
, we obtain
. Solving gives
.
.
-Solution by thecmd999
Solution 4
Let . Consider an inversion about
. Using
.
-Solution by IDMsaterz
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |