2004 AMC 12B Problems/Problem 19
Problem
A truncated cone has horizontal bases with radii and
. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
Solution
Consider a trapezoidal (label it as follows) cross-section of the truncate cone along a diameter of the bases:
![[asy] import olympiad; size(250); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2; draw(A--B--C--D--cycle); draw(circumcircle(E,F,G)); dot(E); dot(F); dot(G); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,NE); label("\(G\)",G,N); [/asy]](http://latex.artofproblemsolving.com/b/9/8/b989bdd1f86df9e3e2736a92e1211c61688c556f.png)
Above, and
are points of tangency. By the Two Tangent Theorem,
and
, so
. We draw
such that it is the foot of the altitude
to
:
![[asy] import olympiad; size(400); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2, H=(16,0); pair P=(D+G)/2, Q=(D+H)/2, R=(B+E)/2, T=(A+H)/2; draw(A--B--C--D--cycle); draw(G--E--H--D); draw(circumcircle(E,F,G)); dot(E); dot(F); dot(G); dot(H); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,NE); label("\(G\)",G,N); label("\(H\)",H,S); label("\(O\)",(E+G)/2,E); label("\(2\)",P,N); label("\(12\)",Q,W); label("\(18\)",R,S); label("\(16\)",T,S); label("\(20\)",(A+D)/2,NW); [/asy]](http://latex.artofproblemsolving.com/b/3/5/b3523f82ad1f6e3c1ed0a0c36880fe8f8388cd5f.png)
By the Pythagorean Theorem,
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |