**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 $?

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