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2005 AIME II Problems/Problem 8

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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 ...

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}.

See also

2005 AIME II (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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