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

(→‎Solution 2)
(→‎Solution 2)
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==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} = \boxed{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 (\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} = \boxed{026}</math>.
 +
  
 
== Solution 2 ==
 
== Solution 2 ==
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(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)\\
 
a^3b^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>
 
Adding gets
 
Adding gets
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a^2b+ab^2=ab(a+b)&=2400+5760\\
 
a^2b+ab^2=ab(a+b)&=2400+5760\\
 
240(a+b)&=240\cdot(10+24)\\
 
240(a+b)&=240\cdot(10+24)\\
−
a+b&=34\\
+
a+b&=34
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
Let <math>h</math> be the hypotenuse then
 
Let <math>h</math> be the hypotenuse then
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&=\boxed{26}
 
&=\boxed{26}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
~ Nafer
  
 
== See also ==
 
== See also ==

Revision as of 12:48, 21 August 2019

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 $\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi$ 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\\ a^3b^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

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions
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