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

(→‎Solution 2)
(→‎Solution 2)
Line 10: Line 10:
 
Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets
 
Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets
 
<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) \\
+
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
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
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
−
<math></math>\begin{align*}
+
<cmath>\begin{align*}
 
h&=\sqrt{a^2+b^2}\\
 
h&=\sqrt{a^2+b^2}\\
−
&=s
+
&=sqrt{(a+b)^2-2ab}\\
 +
&=sqrt{34^2-2\cdot240}\\
 +
&=\sqrt{676}\\
 +
&=\boxed{26}
 +
\end{align*}</cmath>
  
 
== 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*}

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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