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

(→‎Solution 2)
(→‎Solution 2)
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Let <math>a</math>, <math>b</math> be the <math>2</math> legs, we have the <math>2</math> equations
 
Let <math>a</math>, <math>b</math> be the <math>2</math> legs, we have the <math>2</math> equations
 
<cmath>\frac{a^2b\pi}{3}=2400\pi</cmath>
 
<cmath>\frac{a^2b\pi}{3}=2400\pi</cmath>
 +
<cmath>\frac{ab^2\pi}{3}=1920\pi</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 12:27, 21 August 2019

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 $\sqrt{a^2 + b^2} = \boxed{026}$.

Solution 2

Let $a$, $b$ be the $2$ legs, we have the $2$ equations \[\frac{a^2b\pi}{3}=2400\pi\] \[\frac{ab^2\pi}{3}=1920\pi\]

See also

1985 AIME (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions
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