How can a cube be inscribed in a right circular cone?

[Ackbach]

Here is this week's POTW:

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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?

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[Ackbach]

Re: Problem Of The Week # 237 - Oct 14, 2016

This was Problem A-1 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to kiwi for his correct solution, which follows:

Spoiler

The top face of the square is located at an elevation equal to the side length, I call this y.

The top face is inscribed in a circular section of the cone with radius $r = 1-y/3.$

The top face is a square with side length $y = r\sqrt{2}.$

Combining these two equations:

\(r=\frac y { \sqrt 2} = 1 - \frac y3\)

so

\(\frac{3y}{ \sqrt{2}} = 3 - y\)

\(y=\frac{3}{\frac{3}{\sqrt{2}}+1}=\frac{3 \sqrt{2}}{3+\sqrt{2}} \approx 0.96\)