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

Revision as of 01:49, 20 December 2010

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

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-arccos(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); 
draw((-14,0)--MP("H",H,W)--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(C_3,T,H));
 (Error making remote request. Unknown error_msg)

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 $O_3T = \frac{58}{7}$ . 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 $m+n+p=\boxed{405}$ .

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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Difference between revisions of "2005 AIME II Problems/Problem 8"

Revision as of 01:49, 20 December 2010

Problem

Solution

See also