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1985 AIME Problems/Problem 2

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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}.

See also

1985 AIME (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
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