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

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

@@ Line 2: / Line 2: @@
 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 \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==
-*[[1985 AIME Problems/Problem 1|Previous Problem]]
-*[[1985 AIME Problems/Problem 3|Next Problem]]
+== Solution 2 ==
-*[[1985 AIME]]
+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\\
+(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 Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
+[[Category:Intermediate Geometry Problems]]

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