Difference between revisions of "1985 AIME Problems/Problem 2"
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Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets | Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | (a^2b)(ab^2)=2400\cdot5760 \ | + | (a^2b)(ab^2)&=2400\cdot5760 \ |
− | a^3b^3=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5) \ | + | a^3b^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5) \ |
− | ab=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \ | + | ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Adding gets | Adding gets |
Revision as of 12:43, 21 August 2019
Contents
[hide]Problem
When a right triangle is rotated about one leg, the volume of the cone produced is . When the triangle is rotated about the other leg, the volume of the cone produced is . What is the length (in cm) of the hypotenuse of the triangle?
Solution
Let one leg of the triangle have length and let the other leg have length . When we rotate around the leg of length , the result is a cone of height and radius , and so of volume . Likewise, when we rotate around the leg of length we get a cone of height and radius and so of volume . If we divide this equation by the previous one, we get , so . Then so and so . Then by the Pythagorean Theorem, the hypotenuse has length .
Solution 2
Let , be the legs, we have the equations Thus . Multiplying gets Adding gets Let be the hypotenuse then $$ (Error compiling LaTeX. Unknown error_msg)h=
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 |