2023 WSMO Speed Round Problems/Problem 4: Difference between revisions
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==Solution== | ==Solution== | ||
Let <math>s</math> and <math>h</math> denote the sidelength and height of the right prism, respectively. The ratio of the two volumes is equal to <cmath>\frac{\frac{1}{3}\cdot\left(\frac{s}{2}\right)^2\cdot h\pi}{s^2\cdot h} = \frac{\frac{\pi}{12}\cdot s^2\cdot h}{s^2\cdot h} = \frac{\pi}{12}\implies1+12 = \boxed{13}.</cmath> | |||
~pinkpig | |||
Latest revision as of 10:09, 12 September 2025
Problem
A right circular cone is inscribed in a right prism as shown. If the ratio of the volume of the cone to the volume of the prism is 
for relatively prime positive integers 
and 
find 
![[asy] import three; import graph3; defaultpen(linewidth(0.8)); size(200); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); draw((0,0,0)--(0,0,1)); draw((1,0,0)--(1,0,1)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw(Circle((0.5,0.5,0),0.5),dashed); draw((0.5,0.5,1)--(0.5,0,0),dashed); draw((0.5,0.5,1)--(0.5,1,0),dashed); draw((0.5,0.5,1)--(1,0.5,0),dashed); draw((0.5,0.5,1)--(0,0.5,0),dashed); [/asy]](http://latex.artofproblemsolving.com/f/8/8/f88f1aa84eb253bc2a345357f1137195d30e09c7.png)
Solution
Let 
and 
denote the sidelength and height of the right prism, respectively. The ratio of the two volumes is equal to ![\[\frac{\frac{1}{3}\cdot\left(\frac{s}{2}\right)^2\cdot h\pi}{s^2\cdot h} = \frac{\frac{\pi}{12}\cdot s^2\cdot h}{s^2\cdot h} = \frac{\pi}{12}\implies1+12 = \boxed{13}.\]](http://latex.artofproblemsolving.com/3/3/a/33ac14c45b7f5457cea083a9fc4f8c13aafddd18.png)
~pinkpig
