Difference between revisions of "1985 AIME Problems/Problem 2"
Bobthegod78 (talk | contribs) (→Solution 3(Ratios)) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(a^2b)(ab^2)&=2400\cdot5760\ | (a^2b)(ab^2)&=2400\cdot5760\ | ||
− | + | (ab)^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> |
Latest revision as of 14:10, 4 September 2020
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
~ Nafer
Solution 3(Ratios)
Let and
be the two legs of the equation. We can find
by doing
. This simplified is
. We can represent the two legs as
and
for
and
respectively.
Since the volume of the first cone is , we use the formula for the volume of a cone and get
. Solving for
, we get
.
Plugging in the side lengths to the Pythagorean Theorem, we get an answer of .
~bobthegod78
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 |