2005 AIME II Problems/Problem 8
Problem
Circles and
are externally tangent, and they are both internally tangent to circle
The radii of
and
are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of
is also a common external tangent of
and
Given that the length of the chord is
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
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 be the centers and
the radii of the circles
. Let
be the points of tangency from the common external tangent of
, respectively, and let the extension of
intersect the extension of
at a point
. Let the endpoints of the chord/tangent be
, and the foot of the perpendicular from
to
be
. From the similar right triangles
,
It follows that , and that
. By the Pythagorean Theorem on
, we find that
and the answer is .
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 |