Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30} </math> | <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30} </math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | size(8cm,8cm); | ||
+ | path circ1, circ2; | ||
+ | circ1=circle((0,0),5); | ||
+ | circ2=circle((3,4),3); | ||
+ | pair O, Z; | ||
+ | O=(3,4); | ||
+ | Z=(3,-4); | ||
+ | pair [] x=intersectionpoints(circ1,circ2); | ||
+ | pair [] y=intersectionpoints(x[1]--Z,circ2); | ||
+ | pair B; | ||
+ | B=midpoint(x[1]--y[0]); | ||
+ | draw(B--O); | ||
+ | draw(x[0]--Z); | ||
+ | draw(O--Z); | ||
+ | draw(x[1]--Z); | ||
+ | draw(O--x[0]); | ||
+ | draw(circ1); | ||
+ | draw(circ2); | ||
+ | draw(rightanglemark(Z,B,O,15)); | ||
+ | draw(x[1]--O--y[0]); | ||
+ | label("$O$",O,NE); | ||
+ | label("$Y$",x[0],SE); | ||
+ | label("$X$",x[1],NW); | ||
+ | label("$Z$",Z,S); | ||
+ | label("$A$",y[0],SW); | ||
+ | label("$B$",B,SW);</asy> | ||
==Solution 1== | ==Solution 1== | ||
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respectively. Because <math>\angle{OZY} \cong \angle{OZX}</math>, this is a system of two equations and two variables. Solving for <math>r</math> gives <math>r = \sqrt{30}</math>. <math>\boxed{E}</math>. | respectively. Because <math>\angle{OZY} \cong \angle{OZX}</math>, this is a system of two equations and two variables. Solving for <math>r</math> gives <math>r = \sqrt{30}</math>. <math>\boxed{E}</math>. | ||
+ | |||
+ | ===Note=== | ||
+ | Instead of using the Extended Law of Sines, you can note that <math>OX = OY \implies \text{arc}\ OX =\text{arc}\ OY \implies \angle{OZY} \cong \angle{OZX}</math>, since the angles inscribe arcs of the same length. | ||
==Solution 3== | ==Solution 3== | ||
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==Solution 5== | ==Solution 5== | ||
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Notice that <math>\angle YZO=\angle XZO</math> as they subtend arcs of the same length. Let <math>A</math> be the point of intersection of <math>C_1</math> and <math>XZ</math>. We now have <math>AZ=YZ=7</math> and <math>XA=6</math>. Furthermore, notice that <math>\triangle XAO</math> is isosceles, thus the altitude from <math>O</math> to <math>XA</math> bisects <math>XZ</math> at point <math>B</math> above. By the Pythagorean Theorem, <cmath>\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^2\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math> | Notice that <math>\angle YZO=\angle XZO</math> as they subtend arcs of the same length. Let <math>A</math> be the point of intersection of <math>C_1</math> and <math>XZ</math>. We now have <math>AZ=YZ=7</math> and <math>XA=6</math>. Furthermore, notice that <math>\triangle XAO</math> is isosceles, thus the altitude from <math>O</math> to <math>XA</math> bisects <math>XZ</math> at point <math>B</math> above. By the Pythagorean Theorem, <cmath>\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^2\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math> | ||
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~Aaryabhatta1 | ~Aaryabhatta1 | ||
+ | |||
+ | ==Solution 9 (Similar Triangles)== | ||
+ | |||
+ | We first apply [[Ptolemy's Theorem]] on [[cyclic quadrilateral]] <math>XZYO</math> to get <math>13r+7r = 11 \cdot XY \Longrightarrow XY=\frac{20r}{11}</math>. Since <math>\angle ZXY = \angle ZOY</math> and <math>\angle XZO = \angle XYO</math>. From this, we can see <math>\triangle ZPY \sim \triangle XPO</math> and <math>\triangle ZPX \sim \triangle YPO</math>. That means <math>ZP:PY = 13:r, \: ZP:PX = 7:r, \: XP:PO = 13:r</math>. So, if you let <math>PY=x</math>, you will get <math>ZP = \frac{13x}{r}</math>. Continuing in this fashion, we can get <math>XP = \frac{13x}{r} \cdot \frac{r}{7} = \frac{13x}{7}</math> and <math>PO = \frac{13x}{7} \cdot \frac{r}{13} = \frac{xr}{7}</math>. Since <math>XY = \frac{20r}{11} = XP+PY</math>, we have <math>x+\frac{13x}{7} = \frac{20r}{11}</math> which gives us <math>x=\frac{7r}{11}</math>. Plugging it into <math>ZO = 11 = ZP+PO</math> gives | ||
+ | |||
+ | <cmath>\frac{13x}{r} + \frac{xr}{7} = \frac{13 \cdot \frac{7r}{11}}{r} + \frac{r \cdot \frac{7r}{11}}{7} = \frac{91}{11} + \frac{r^2}{11} = 11.</cmath> | ||
+ | |||
+ | Solving for <math>r</math> yields <math>r=\boxed{\sqrt{30}}</math>. | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
== See Also == | == See Also == | ||
Latest revision as of 02:40, 2 October 2022
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
?
Diagram
Solution 1
Let denote the radius of circle
. Note that quadrilateral
is cyclic. By Ptolemy's Theorem, we have
and
. Let
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
.
.
Note
Instead of using the Extended Law of Sines, you can note that , since the angles inscribe arcs of the same length.
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
. So,
. Using
.
-Solution by IDMasterz
Solution 5
Notice that as they subtend arcs of the same length. Let
be the point of intersection of
and
. We now have
and
. Furthermore, notice that
is isosceles, thus the altitude from
to
bisects
at point
above. By the Pythagorean Theorem,
Thus,
Solution 6
Use the diagram above. Notice that as they subtend arcs of the same length. Let
be the point of intersection of
and
. We now have
and
. Consider the power of point
with respect to Circle
we have
which gives
Solution 7 (Only Law of Cosines)
Note that and
are the same length, which is also the radius
we want. Using the law of cosines on
, we have
, where
is the angle formed by
. Since
and
are supplementary,
. Using the law of cosines on
,
. As
,
. Solving for theta on the first equation and substituting gives
. Solving for R gives
.
Solution 8
We first note that is the circumcircle of both
and
. Thus the circumradius of both the triangles are equal. We set the radius of
as
, and noting that the circumradius of a triangle is
and that the area of a triangle by Heron's formula is
with
as the semi-perimeter we have the following,
Now substituting
,
This gives us 2 values for
namely
and
.
Now notice that we can apply Ptolemy's theorem on to find
in terms of
. We get
Here we substitute our
values of
receiving
. Notice that the latter of the
cases does not satisfy the triangle inequality for
as
. But the former does thus our answer is
.
~Aaryabhatta1
Solution 9 (Similar Triangles)
We first apply Ptolemy's Theorem on cyclic quadrilateral to get
. Since
and
. From this, we can see
and
. That means
. So, if you let
, you will get
. Continuing in this fashion, we can get
and
. Since
, we have
which gives us
. Plugging it into
gives
Solving for yields
.
~sml1809
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.