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

Latest revision as of 18:12, 20 November 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 $\boxed{{A (6)}}.$

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 $4x =6$ . Thus, the answer is $\boxed{{A (6)}}$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-A truncated [[cone]] has horizontal bases with radii <math>18</math> and <math>2</math>. A [[sphere]] is tangent to the top, bottom, and lateral surface of the truncated cone. What is the [[radius]] of the sphere?
+A truncated cone has horizontal bases with radii <math>18</math> and <math>2</math>. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
 <math>\mathrm{(A)}\ 6
@@ Line 9: / Line 9: @@
 == Solution ==
-Consider a [[trapezoid]]al (label it <math>ABCD</math> as follows) cross-section of the truncate cone along a diameter of the bases:
+=== Solution 1 ===
+Consider a [[trapezoid]] (label it <math>ABCD</math> as follows) cross-section of the truncate cone along a diameter of the bases:
 <center><asy>
 import olympiad;
@@ Line 31: / Line 32: @@
 <center><asy>
 import olympiad;
-size(450);
+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);
@@ Line 57: / Line 58: @@
 By the [[Pythagorean Theorem]],
-<cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{6} \Rightarrow \mathrm{(A)}.</cmath>
+<cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12</cmath>
+Therefore, the answer is <math>\boxed{{A (6)}}.</math>
+=== Solution 2 ===
+Create a trapezoid with inscribed circle <math>O</math> exactly like in Solution #1, and extend lines <math>\overline{AD}</math> and <math>\overline{BC}</math> from the solution above and label the point at where they meet <math>H</math>. Because <math>\frac{\overline{GC}}{\overline{BE}}</math> = <math>\frac{1}{9}</math>, <math>\frac{\overline{HG}}{\overline{HE}}</math> = <math>\frac{1}{9}</math>. Let <math>\overline{HG} = x</math> and <math>\overline{GE} = 8x</math>.
+Because these are radii, <math>\overline{GO} = \overline{OE} = \overline{OF} = 4x</math>. <math>\overline{OF} \perp \overline{BH} </math> so <math>\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2</math>. Plugging in, we get <math>4x^2 + \overline{FH}^2 = 5x^2</math> so <math>\overline{FH} = 3x</math>.Triangles <math>OFH</math> and <math>BEH</math> are similar so <math>\frac{\overline{OF}}{\overline{BE}}  = \frac{\overline{FH}}{\overline{EH}}</math> which gives us <math>\frac{4x}{18} = \frac{3x}{9x}</math>. Solving for x, we get <cmath>x = 1.5</cmath> and <cmath>4x =6</cmath>. Thus, the answer is <math>\boxed{{A (6)}}</math>.
 == 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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