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

Revision as of 17:27, 1 August 2023

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, $r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12$ Therefore, the answer is \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 $4x = \boxed{6} \Rightarrow \mathrm{(A)}.$

@@ Line 58: / Line 58: @@
 By the [[Pythagorean Theorem]],
-<cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12
+<cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12</cmath>
 Therefore, the answer is \boxed{6}
-\Rightarrow \mathrm{(A)}.</cmath>
+<cmath>\Rightarrow \mathrm{(A)}.</cmath>
 === Solution 2 ===

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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