Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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Note that <math>OX</math> and <math>OY</math> are the same length, which is also the radius <math>R</math> we want. Using the law of cosines on <math>\triangle OYZ</math>, we have <math>11^2=R^2+7^2-2\cdot 7 \cdot R \cdot \cos\theta</math>, where <math>\theta</math> is the angle formed by <math>\angle{OYZ}</math>. Since <math>\angle{OYZ}</math> and <math>\angle{OXZ}</math> are supplementary, <math>\angle{OXZ}=\pi-\theta</math>. Using the law of cosines on <math>\triangle OXZ</math>, <math>11^2=13^2+R^2-2 \cdot 13 \cdot R \cdot \cos(\pi-\theta)</math>. As <math>\cos(\pi-\theta)=-\cos\theta</math>, <math>11^2=13^2+R^2+\cos\theta</math>. Solving for theta on the first equation and substituting gives <math>\frac{72-R^2}{14R}=\frac{48+R^2}{26R}</math>. Solving for R gives <math>R=\textbf{(E)}\ \boxed{\sqrt{30}} </math>. | Note that <math>OX</math> and <math>OY</math> are the same length, which is also the radius <math>R</math> we want. Using the law of cosines on <math>\triangle OYZ</math>, we have <math>11^2=R^2+7^2-2\cdot 7 \cdot R \cdot \cos\theta</math>, where <math>\theta</math> is the angle formed by <math>\angle{OYZ}</math>. Since <math>\angle{OYZ}</math> and <math>\angle{OXZ}</math> are supplementary, <math>\angle{OXZ}=\pi-\theta</math>. Using the law of cosines on <math>\triangle OXZ</math>, <math>11^2=13^2+R^2-2 \cdot 13 \cdot R \cdot \cos(\pi-\theta)</math>. As <math>\cos(\pi-\theta)=-\cos\theta</math>, <math>11^2=13^2+R^2+\cos\theta</math>. Solving for theta on the first equation and substituting gives <math>\frac{72-R^2}{14R}=\frac{48+R^2}{26R}</math>. Solving for R gives <math>R=\textbf{(E)}\ \boxed{\sqrt{30}} </math>. | ||
+ | ==Solution 8== | ||
+ | |||
+ | We first note that <math>C_2</math> is the circumcircle of both <math>\triangle XOZ</math> and <math>\triangle OYZ</math>. Thus the circumradius of both the triangles are equal. We set the radius of <math>C_1</math> as <math>r</math>, and noting that the circumradius of a triangle is <math>\frac{abc}{4A}</math> and that the area of a triangle by Heron's formula is <math>\sqrt{(S)(S-a)(S-b)(S-c)}</math> with <math>S</math> as the semi-perimeter we have the following, <cmath>\begin{align*}\dfrac{r \cdot 13 \cdot 11}{4\sqrt{(12 + \frac{r}{2})(12 - \frac{r}{2})(1 + \frac{r}{2})(\frac{r}{2} - 1)}} &= \dfrac{r \cdot 7 \cdot 11}{4\sqrt{(9 + \frac{r}{2})(9 - \frac{r}{2})(2 + \frac{r}{2})(\frac{r}{2} - 2)}} \\ \dfrac{13}{\sqrt{(144- \frac{r^2}{4})(\frac{r^2}{4} - 1)}} &= \dfrac{7}{\sqrt{(81- \frac{r^2}{4})(\frac{r^2}{4} - 4)}} \\ 169 \cdot (81 - \frac{r^2}{4})(\frac{r^2}{4} - 4) &= 49 \cdot (144 - \frac{r^2}{4})(\frac{r^2}{4} - 1) .\end{align*}</cmath> | ||
+ | Now substituting <math>a = \frac{r^2}{4}</math>, <cmath>\begin{align*}169a^2 - (85) \cdot 169 a + 4 \cdot 81 \cdot 169 &= 49a^2 - (145) \cdot 49 a + 144 \cdot 49 \\ 120a^2 - 7260a + 47700 &= 0 \\ 2a^2 - 121a + 795 &= 0 \\ (2a-15)(a-53) &= 0 \\ a = \frac{15}{2}, 53.\end{align*}</cmath> | ||
+ | This gives us 2 values for <math>r</math> namely <math>r = \sqrt{4 \cdot \frac{15}{2}} = \sqrt{30}</math> and <math>r = \sqrt{4 \cdot 53} = 2\sqrt{53}</math>. | ||
+ | |||
+ | Now notice that we can apply Ptolemy's theorem on <math>XOYZ</math> to find <math>XY</math> in terms of <math>r</math>. We get <cmath>\begin{align*}11 \cdot XY &= 13r + 7r \\ XY &= \frac{20r}{11}.\end{align*}</cmath> | ||
+ | Here we substitute our <math>2</math> values of <math>r</math> receiving <math>XY = \frac{20\sqrt{30}}{11}, \frac{40\sqrt{53}}{11}</math>. Notice that the latter of the <math>2</math> cases does not satisfy the triangle inequality for <math>\triangle XYZ</math> as <math>\frac{40\sqrt{53}}{11} \approx 26.5 > 7 + 13 = 20</math>. But the former does thus our answer is <math>\textbf{(E)}\ \boxed{\sqrt{30}}</math>. | ||
+ | |||
+ | ~Aaryabhatta1 | ||
== See Also == | == See Also == | ||
Revision as of 14:44, 25 January 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 ?
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 |
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