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Difference between revisions of "1985 AIME Problems/Problem 2"

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Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>.  When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>.  Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>.  If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>.  Then <math>\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>.  Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = 026</math>.
 
Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>.  When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>.  Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>.  If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>.  Then <math>\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>.  Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = 026</math>.
  
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==See also==
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== See also ==
 
{{AIME box|year=1985|num-b=1|num-a=3}}
 
{{AIME box|year=1985|num-b=1|num-a=3}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 14:31, 6 May 2007

Problem

When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$. What is the length (in cm) of the hypotenuse of the triangle?

Solution

Let one leg of the triangle have length $a$ and let the other leg have length $b$. When we rotate around the leg of length $a$, the result is a cone of height $a$ and radius $b$, and so of volume $\frac 13 \pi ab^2 = 800\pi$. Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\frac13 \pi b a^2 = 1920 \pi$. If we divide this equation by the previous one, we get $\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}$, so $a = \frac{12}{5}b$. Then $\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$. Then by the Pythagorean Theorem, the hypotenuse has length $\sqrt{a^2 + b^2} = 026$.

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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