Difference between revisions of "2012 AMC 12A Problems/Problem 16"
(→Solution) |
(Tag: Undo) |
||
(25 intermediate revisions by 15 users not shown) | |||
Line 5: | Line 5: | ||
<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> | <asy> | ||
size(8cm,8cm); | size(8cm,8cm); | ||
Line 52: | Line 22: | ||
draw(O--Z); | draw(O--Z); | ||
draw(x[1]--Z); | draw(x[1]--Z); | ||
+ | draw(O--x[0]); | ||
draw(circ1); | draw(circ1); | ||
draw(circ2); | draw(circ2); | ||
Line 57: | Line 28: | ||
draw(x[1]--O--y[0]); | draw(x[1]--O--y[0]); | ||
label("$O$",O,NE); | label("$O$",O,NE); | ||
− | label("$Y$",x[0], | + | label("$Y$",x[0],SE); |
label("$X$",x[1],NW); | label("$X$",x[1],NW); | ||
label("$Z$",Z,S); | label("$Z$",Z,S); | ||
label("$A$",y[0],SW); | label("$A$",y[0],SW); | ||
label("$B$",B,SW);</asy> | label("$B$",B,SW);</asy> | ||
− | |||
− | ===Solution 6=== | + | ==Solution 1== |
+ | Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Let <math>t</math> be the measure of angle <math>YOX</math>. Since <math>YO=OX=r</math>, the law of cosines on triangle <math>YOX</math> gives us <math>\cos t =-79/121</math>. Again since <math>ZYOX</math> is cyclic, the measure of angle <math>YZX=180-t</math>. We apply the law of cosines to triangle <math>ZYX</math> so that <math>XY^2=7^2+13^2-2(7)(13)\cos(180-t)</math>. Since <math>\cos(180-t)=-\cos t=79/121</math> we obtain <math>XY^2=12000/121</math>. But<math> XY^2=400r^2/121</math> so that <math>r=\boxed{(E)\sqrt{30}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let us call the <math>r</math> the radius of circle <math>C_1</math>, and <math>R</math> the radius of <math>C_2</math>. Consider <math>\triangle OZX</math> and <math>\triangle OZY</math>. Both of these triangles have the same circumcircle (<math>C_2</math>). From the Extended Law of Sines, we see that <math>\frac{r}{\sin{\angle{OZY}}} = \frac{r}{\sin{\angle{OZX}}}= 2R</math>. Therefore, <math>\angle{OZY} \cong \angle{OZX}</math>. We will now apply the Law of Cosines to <math>\triangle OZX</math> and <math>\triangle OZY</math> and get the equations | ||
+ | |||
+ | <math>r^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cdot \cos{\angle{OZX}}</math>, | ||
+ | |||
+ | <math>r^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos{\angle{OZY}}</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== | ||
+ | |||
+ | Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Consider isosceles triangle <math>XOY</math>. Pulling an altitude to <math>XY</math> from <math>O</math>, we obtain <math>\cos(\angle{OXY}) = \frac{10}{11}</math>. Since quadrilateral <math>ZYOX</math> is cyclic, we have <math>\angle{OXY}=\angle{OZY}</math>, so <math>\cos(\angle{OXY}) = \cos(\angle{OZY})</math>. Applying the Law of Cosines to triangle <math>OZY</math>, we obtain <math>\frac{10}{11} = \frac{7^2+11^2-r^2}{2(7)(11)}</math>. Solving gives <math>r=\sqrt{30}</math>. <math>\boxed{E}</math>. | ||
+ | |||
+ | -Solution by '''thecmd999''' | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>P = XY \cap OZ</math>. Consider an inversion about <math>C_1 \implies C_2 \to XY, Z \to P</math>. So, <math>OP \cdot OZ = r^2 \implies OP = r^2/11 \implies PZ = \dfrac{121 - r^2}{11}</math>. Using <math>\triangle YPZ \sim OXZ \implies r = \sqrt{30} \implies \boxed{E}</math>. | ||
+ | |||
+ | |||
+ | -Solution by '''IDMasterz''' | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Use the diagram above. 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>. Consider the power of point <math>Z</math> with respect to Circle <math>O,</math> we have <math>13\cdot 7 = (11 + r)(11 - r) = 11^2 - r^2,</math> which gives <math>r=\boxed{\sqrt{30}}.</math> | ||
+ | |||
+ | ==Solution 7 (Only Law of Cosines)== | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==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 19:13, 21 September 2024
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.