Difference between revisions of "1985 AIME Problems/Problem 2"
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When a [[right triangle]] is rotated about one leg, the [[volume]] of the [[cone]] produced is <math>800\pi \;\textrm{ cm}^3</math>. When the [[triangle]] is rotated about the other leg, the volume of the cone produced is <math>1920\pi \;\textrm{ cm}^3</math>. What is the length (in cm) of the [[hypotenuse]] of the triangle? | When a [[right triangle]] is rotated about one leg, the [[volume]] of the [[cone]] produced is <math>800\pi \;\textrm{ cm}^3</math>. When the [[triangle]] is rotated about the other leg, the volume of the cone produced is <math>1920\pi \;\textrm{ cm}^3</math>. What is the length (in cm) of the [[hypotenuse]] of the triangle? | ||
==Solution== | ==Solution== | ||
− | Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>. When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>. Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>. If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>. Then <math>\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>. Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = 026</math>. | + | Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>. When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>. Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>. If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>. Then <math>\frac{1}{3} \pi \left(\frac{12}{5}b\right)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>. Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = \boxed{026}</math>. |
− | ==See also== | + | |
+ | == Solution 2 == | ||
+ | |||
+ | Let <math>a</math>, <math>b</math> be the <math>2</math> legs, we have the <math>2</math> equations | ||
+ | <cmath>\frac{a^2b\pi}{3}=800\pi,\frac{ab^2\pi}{3}=1920\pi</cmath> | ||
+ | Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets | ||
+ | <cmath>\begin{align*} | ||
+ | (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 | ||
+ | \end{align*}</cmath> | ||
+ | Adding gets | ||
+ | <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 | ||
+ | <cmath>\begin{align*} | ||
+ | h&=\sqrt{a^2+b^2}\\ | ||
+ | &=\sqrt{(a+b)^2-2ab}\\ | ||
+ | &=\sqrt{34^2-2\cdot240}\\ | ||
+ | &=\sqrt{676}\\ | ||
+ | &=\boxed{26} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 3(Ratios) == | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 \pi</math>. Solving for <math>x</math>, we get <math>x=2</math>. | ||
+ | |||
+ | Plugging in the side lengths to the Pythagorean Theorem, we get an answer of <math>\boxed{026}</math>. | ||
+ | |||
+ | ~bobthegod78 | ||
+ | |||
+ | == See also == | ||
{{AIME box|year=1985|num-b=1|num-a=3}} | {{AIME box|year=1985|num-b=1|num-a=3}} | ||
+ | * [[AIME Problems and Solutions]] | ||
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 15:10, 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 |