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Difference between revisions of "1985 AIME Problems/Problem 2"

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<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)\\
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(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>
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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.  
 
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</math>. Solving for <math>x</math>, we get <math>x=2</math>.  
+
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>.
 
Plugging in the side lengths to the Pythagorean Theorem, we get an answer of <math>\boxed{026}</math>.
 +
 +
~bobthegod78
  
 
== See also ==
 
== See also ==

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 $800\pi \;\textrm{ cm}^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$. What is the length (in cm) of the hypotenuse of the triangle?

Solution

Let one leg of the triangle have length $a$ and let the other leg have length $b$. When we rotate around the leg of length $a$, the result is a cone of height $a$ and radius $b$, and so of volume $\frac 13 \pi ab^2 = 800\pi$. Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\frac13 \pi b a^2 = 1920 \pi$. If we divide this equation by the previous one, we get $\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}$, so $a = \frac{12}{5}b$. Then $\frac{1}{3} \pi \left(\frac{12}{5}b\right)b^2 = 800\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$. Then by the Pythagorean Theorem, the hypotenuse has length $\sqrt{a^2 + b^2} = \boxed{026}$.


Solution 2

Let $a$, $b$ be the $2$ legs, we have the $2$ equations \[\frac{a^2b\pi}{3}=800\pi,\frac{ab^2\pi}{3}=1920\pi\] Thus $a^2b=2400, ab^2=5760$. Multiplying gets \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*} Adding gets \begin{align*} a^2b+ab^2=ab(a+b)&=2400+5760\\ 240(a+b)&=240\cdot(10+24)\\ a+b&=34 \end{align*} Let $h$ be the hypotenuse then \begin{align*} h&=\sqrt{a^2+b^2}\\ &=\sqrt{(a+b)^2-2ab}\\ &=\sqrt{34^2-2\cdot240}\\ &=\sqrt{676}\\ &=\boxed{26} \end{align*}

~ Nafer

Solution 3(Ratios)

Let $a$ and $b$ be the two legs of the equation. We can find $\frac{a}{b}$ by doing $\frac{1920\pi}{800\pi}$. This simplified is $\frac{12}{5}$. We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively.

Since the volume of the first cone is $800\pi$, we use the formula for the volume of a cone and get $100\pi x^3=800 \pi$. Solving for $x$, we get $x=2$.

Plugging in the side lengths to the Pythagorean Theorem, we get an answer of $\boxed{026}$.

~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
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