## Tuesday, October 3, 2006

### Conehead. Topic: Geometry. Level: AMC.

Problem: (1998 Putnam - A1) A right circular cone has base of radius $1$ and height $3$. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side length of the cube?

Solution: Suppose the cube has side length $s$. Then the top face of the cube has to fit into a circle of radius

$1-\frac{s}{3}$

by proportions ($3-s$ of the total height of $3$ will still be above the cube). So its side length can be at most

$\left(1-\frac{s}{3}\right)\sqrt{2}$

since a square with $\sqrt{2}$ times the radius of a circle is perfectly inscribed. Then we have

$\left(1-\frac{s}{3}\right)\sqrt{2} \ge s \Rightarrow s \le \frac{9\sqrt{2}-6}{7}$,

so the side length must be the maximum $\frac{9\sqrt{2}-6}{7}$ in order to be inscribed. QED.

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Comment: This problem was not difficult at all. Just drawing a picture and using some ratios it's easy to find the inequality and then solve for $s$ knowing that equality must hold for inscribed things.

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Practice Problem: Can you do this for a general cone with base radius $a$ and height $b$?