Tonight I joined in on the fun at the Games Night at Math Prize for Girls. Picture a large banquet hall filled with large tables each showcasing a different kind of math-related hands-on activity. Girls came from all over the country as well as Canada to take part. There were lots of volunteers helping from MIT. And, there was a lot of cake!

At one of the tables, girls were building polyhedra out of paper and string. They would roll up the paper into tight cylinders and tape them so they wouldn’t unravel. They’d then run string through the cylinders to hook them together and form edges.

I was talking to one girl named Michelle about these polyhedra and somehow, we ended up deciding to compute the volume of a regular tetrahedron as a function of its side length . One way to proceed would be to use the formula for the volume of a pyramid with base area and height . However, we computed it by exploiting the fact that four vertices of a cube can be chosen as vertices of a regular tetrahedron, as shown below.

Instead of computing the volume directly, we would compute the volume of the cube and subtract off the volumes of the 4 right triangular pyramids that rest on the faces of the regular tetrahedron.

Because the side of the tetrahedron is the diagonal of a face of the cube, the cube has side length . Therefore the cube has volume . The 4 right triangular pyramids that must be carved off the cube to produce the regular tetrahedron each have volume .

Thus, the volume of the regular tetrahedron is .

Maybe because there was a beautiful icosahedron perched in the middle of the table, we discussed how to get the volume of that too. Perhaps you’d like to try to find that volume for yourself?

Tomorrow’s the competition. To all the competitors: Good luck, have fun, and I hope you make many new friends!

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