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Difference between revisions of "2005 AIME II Problems/Problem 8"

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<cmath> \frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}. </cmath>
 
<cmath> \frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}. </cmath>
  
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It follows that <math>HO_1 = \frac{28}{3}</math>, and that <math>O_3T = \frac{58}{7}</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that  
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It follows that <math>HO_1 = \frac{28}{3}</math>, and that <math>O_3T = \frac{58}{7}\dagger</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that  
  
 
<cmath>AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}</cmath>
 
<cmath>AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}</cmath>
  
 
and the answer is <math>m+n+p=\boxed{405}</math>.
 
and the answer is <math>m+n+p=\boxed{405}</math>.
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<math>\dagger</math> Alternatively, drop an altitude from <math>O_1</math> to <math>O_3T</math> at <math>O_3'</math>, and to <math>O_2T_2</math> at <math>O_2'</math>. Then, <math>O_2O_2'=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</math> so <math>O_3T=4+\frac{30}{7}=\frac{58}{7}</math>.
  
 
==Solution 2==
 
==Solution 2==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:03, 28 December 2021

Problem

Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

Solution

[asy] pointpen = black; pathpen = black + linewidth(0.7); size(200); pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); draw(cir1); draw(cir2); draw(cir3); draw((14,0)--(-14,0)); draw(A--B); MP("H",H,W); draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); draw(MP("O_1",C1)); draw(MP("O_2",C2)); draw(MP("O_3",C3)); draw(MP("T",T,N)); draw(MP("A",A,NW)); draw(MP("B",B,NE)); draw(C1--MP("T_1",T1,N)); draw(C2--MP("T_2",T2,N)); draw(C3--T); draw(rightanglemark(C3,T,H)); [/asy]

Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$. Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$, respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$. Let the endpoints of the chord/tangent be $A,B$, and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$. From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$,

\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]

It follows that $HO_1 = \frac{28}{3}$, and that $O_3T = \frac{58}{7}\dagger$. By the Pythagorean Theorem on $\triangle ATO_3$, we find that

\[AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}\]

and the answer is $m+n+p=\boxed{405}$.

$\dagger$ Alternatively, drop an altitude from $O_1$ to $O_3T$ at $O_3'$, and to $O_2T_2$ at $O_2'$. Then, $O_2O_2'=10-4=6$, and $O_1O_2=14$. But $O_1O_3O_3'$ is similar to $O_1O_2O_2'$ so $O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}$. From rectangles, $O_3'T=O_1T_1=4$ so $O_3T=4+\frac{30}{7}=\frac{58}{7}$.

Solution 2

Call our desired length $x$. Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$, we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$. Then, extend $\overline{O_3T}$ until it reaches the circle on both sides; call them $P,Q$. By Power of a Point, $\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}$. Since $\overline{AT}=\overline{TB}=\frac{1}{2}x$, \[(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2\] \[\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2\] After solving for $x$, we get $x=\frac{8\sqrt{390}}{7}$, so our answer is $8+390+7=\boxed{405}$

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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