Difference between revisions of "1985 AIME Problems/Problem 2"
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+ | == Solution 3(Ratios) == | ||
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+ | Let <math>a</math> and <math>b</math> be the two legs of the equation. We can find <math>\frac{a}{b}</math> by doing <math>\frac{1920\pi}{800\pi}</math>. This simplified is <math>\frac{12}{5}</math>. We can represent the two legs as <math>12x</math> and <math>5x</math> for <math>a</math> and <math>b</math> respectively. | ||
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+ | Since the volume of the first cone is <math>800\pi</math>, we use the formula for the volume of a cone and get <math>100\pi x^3=800</math>. Solving for <math>x</math>, we get <math>x=2</math>. | ||
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+ | Plugging in the side lengths to the Pythagorean Theorem, we get an answer of <math>\boxed{026}</math>. | ||
== See also == | == See also == |
Revision as of 23:59, 24 August 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 .
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 |