Difference between revisions of "1985 AIME Problems/Problem 2"
(→Solution 2) |
(→Solution 2) |
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\end{align*}</cmath> | \end{align*}</cmath> | ||
Adding gets | Adding gets | ||
− | <cmath>a^2b+ab^2=ab(a+b)=2400+5760 | + | <cmath>\begin{align*} |
− | + | a^2b+ab^2&=ab(a+b)=2400+5760\ | |
− | + | 240(a+b)&=240\cdot(10+24)\ | |
+ | a+b&=34\ | ||
+ | \end{align*}</cmath> | ||
Let <math>h</math> be the hypotenuse then | Let <math>h</math> be the hypotenuse then | ||
− | <math></math>h= | + | <math></math>\begin{align*} |
+ | h&=\sqrt{a^2+b^2}\ | ||
+ | &=s | ||
== See also == | == See also == |
Revision as of 12:46, 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)\begin{align*}
h&=\sqrt{a^2+b^2}\
&=s
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 |