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

Revision as of 12:29, 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 $\frac{a^2b\pi}{3}=800\pi$ $\frac{ab^2\pi}{3}=1920\pi$ Thus $a^2b=2400, ab^2=5760$

@@ Line 9: / Line 9: @@
 <cmath>\frac{a^2b\pi}{3}=800\pi</cmath>
 <cmath>\frac{ab^2\pi}{3}=1920\pi</cmath>
-Thus
+Thus <math>a^2b=2400, ab^2=5760</math>
-<cmath>a^2b=2400, ab^2=5760</cmath>
 == See also ==

1985 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1985 AIME Problems/Problem 2"

Revision as of 12:29, 21 August 2019

Contents

Problem

Solution

Solution 2

See also