Difference between revisions of "2004 AMC 12B Problems/Problem 19"
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By the [[Pythagorean Theorem]], | By the [[Pythagorean Theorem]], | ||
| − | <cmath>r = \frac{ | + | <cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{6} \Rightarrow \mathrm{(A)}.</cmath> |
== See also == | == See also == | ||
Revision as of 19:28, 22 December 2009
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(220); 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/7/9/b/79b290d343dc7c6049f7d45f5f1abc749c2af3d2.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(450); 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, O=(E+G)/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);dot(O); 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("\(O\)",O,NE); label("\(2\)",P,N); label("\(12\)",Q,W); label("\(18\)",R,S); label("\(16\)",T,S); label("\(20\)",(A+D)/2,NW); label("\(r\)",(O+E)/2,SE); [/asy]](http://latex.artofproblemsolving.com/a/7/c/a7c2a0d6df966f5c1a8d81064b9a12c58faa12ac.png)
By the Pythagorean Theorem,
![\[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{6} \Rightarrow \mathrm{(A)}.\]](http://latex.artofproblemsolving.com/d/f/3/df3b7879ecc2ac8d3fff4c2b6c95853efdf10988.png)
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 | |
