Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

Revision as of 12:43, 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 \[a^2b+ab^2=ab(a+b)=2400+5760\] \[240(a+b)=240\cdot(10+24)\] \[a+b=34\] Let $h$ be the hypotenuse then $$ (Error compiling LaTeX. Unknown error_msg)h=

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_2&oldid=109150"