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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==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> | ||
Latest revision as of 22:04, 11 May 2021
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 . .
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
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.