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Difference between revisions of "2004 AMC 12B Problems/Problem 19"

(→‎Solution)
(→‎Solution #1)
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<center><asy>
 
<center><asy>
 
import olympiad;
 
import olympiad;
−
size(450);
+
size(220);
 
defaultpen(0.7);
 
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 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);

Revision as of 11:30, 12 November 2018

Problem

A truncated cone has horizontal bases with radii $18$ and $2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$

Solution #1

Consider a trapezoidal (label it $ABCD$ 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]

Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$, so $BC = 20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$:

[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, 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]

By the Pythagorean Theorem, \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{6} \Rightarrow \mathrm{(A)}.\]

Solution #2

Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $\overline{AD}$ and $\overline{BC}$ from the solution above and label the point at where they meet $H$. Because $\frac{\overline{GC}}{\overline{BE}}$ = $\frac{1}{9}$, $\frac{\overline{HG}}{\overline{HE}}$ = $\frac{1}{9}$. Let $\overline{HG} = x$ and $\overline{GE} = 8x$.

Because these are radii, $\overline{GO} = \overline{OE} = \overline{OF} = 4x$. $\overline{OF} \perp \overline{BH}$ so $\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2$. Plugging in, we get $4x^2 + \overline{FH}^2 = 5x^2$ so $\overline{FH} = 3x$.Triangles $OFH$ and $BEH$ are similar so $\frac{\overline{OF}}{\overline{BE}} = \frac{\overline{FH}}{\overline{EH}}$ which gives us $\frac{4x}{18} = \frac{3x}{9x}$. Solving for x, we get \[x = 1.5\] and \[4x = \boxed{6} \Rightarrow \mathrm{(A)}.\]

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

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