1985 AIME Problems/Problem 2

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 $a^2b=2400, ab^2=5760$ . Multiplying gets $\begin{align*} (a^2b)(ab^2)&=2400\cdot5760 \\ a^3b^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5) \\ ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \\ \end{align*}$ Adding gets $\begin{align*} a^2b+ab^2&=ab(a+b)=2400+5760\\ 240(a+b)&=240\cdot(10+24)\\ a+b&=34\\ \end{align*}$ Let $h$ be the hypotenuse then $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*} h&=\sqrt{a^2+b^2}\\ &=s

1985 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1985 AIME Problems/Problem 2

Contents

Problem

Solution

Solution 2

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