# 2004 AMC 12B Problems/Problem 19

## 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

### Solution 1

Consider a trapezoid (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 $$4x = \boxed{6} \Rightarrow \mathrm{(A)}.$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 