Difference between revisions of "2004 AMC 12B Problems/Problem 19"
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== Problem == | == Problem == | ||
− | A truncated | + | 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 | <math>\mathrm{(A)}\ 6 | ||
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\qquad\mathrm{(E)}\ 6\sqrt{3}</math> | \qquad\mathrm{(E)}\ 6\sqrt{3}</math> | ||
− | == Solution | + | == Solution == |
+ | === Solution 1 === | ||
Consider a [[trapezoid]] (label it <math>ABCD</math> as follows) cross-section of the truncate cone along a diameter of the bases: | Consider a [[trapezoid]] (label it <math>ABCD</math> as follows) cross-section of the truncate cone along a diameter of the bases: | ||
<center><asy> | <center><asy> | ||
Line 57: | Line 58: | ||
By the [[Pythagorean Theorem]], | By the [[Pythagorean Theorem]], | ||
− | <cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{ | + | <cmath>r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12</cmath> |
+ | Therefore, the answer is <math>\boxed{{A (6)}}.</math> | ||
− | == Solution | + | === 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>. | 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 18:12, 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:
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 :
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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