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{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12</cmath> | <cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12</cmath> | ||
− | Therefore, the answer is <math>\boxed{A (6)}.</math> | + | Therefore, the answer is <math>\boxed{{A (6)}}.</math> |
=== Solution 2 === | === Solution 2 === | ||
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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>. | 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 = | + | 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 == | == See also == |
Latest revision as of 19:12, 20 November 2023
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
Solution 1
Consider a trapezoid (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(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]](http://latex.artofproblemsolving.com/d/f/5/df55cbef93e6c3485f8c0d77176293f0fe2bcb3a.png)
By the Pythagorean Theorem,
Therefore, the answer is
Solution 2
Create a trapezoid with inscribed circle exactly like in Solution #1, and extend lines
and
from the solution above and label the point at where they meet
. Because
=
,
=
. Let
and
.
Because these are radii, .
so
. Plugging in, we get
so
.Triangles
and
are similar so
which gives us
. Solving for x, we get
and
. Thus, the answer is
.
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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