## Inscribed triangles and tetrahedra

The following problems appeared in The Riddler. They involve randomly picking points on a circle or sphere and seeing if the resulting shape contains the center or not.

Problem 1: Choose three points on a circle at random and connect them to form a triangle. What is the probability that the center of the circle is contained in that triangle?

Problem 2: Choose four points at random (independently and uniformly distributed) on the surface of a sphere. What is the probability that the tetrahedron defined by those four points contains the center of the sphere?

Here is my solution to both problems:
[Show Solution]

## A tetrahedron puzzle

This post is about a 3D geometry Riddler puzzle involving spheres and tetrahedra! Here is the problem:

We want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year’s fidget sphere craze, and we’d like to sell them in sets of four. We also have a lot of extra tetrahedral packaging from last month’s Pyramid Fest. What’s the smallest tetrahedron into which we can pack four spheres?

Here is my solution:
[Show Solution]

## Space race

This Riddler puzzle is about a game involving filling up the space on a square table using coins.

Two players are seated at a square table. The first player places a coin on the table, the second places a coin on the table, and they carry on placing coins one after another, with the only condition being that the coins are not allowed to touch. The winner is the person who places the final coin on the table, meaning that he or she fills the last remaining space between the other coins.

The table has to be larger than a single coin, and all the coins placed must be identically sized. If the players play optimally, is one of the two players guaranteed to win? If so, what is the winning strategy?

Need a hint?
[Show Solution]

Here is my solution:
[Show Solution]

## Cutting polygons in half

This Riddler puzzle is about cutting polygons in half. Here is the problem:

The archvillain Laser Larry threatens to imminently zap Riddler Headquarters (which, seen from above, is shaped like a regular pentagon with no courtyard or other funny business). He plans to do it with a high-powered, vertical planar ray that will slice the building exactly in half by area, as seen from above. The building is quickly evacuated, but not before in-house mathematicians move the most sensitive riddling equipment out of the places in the building that have an extra high risk of getting zapped.

Where are those places, and how much riskier are they than the safest spots? (It’s fine to describe those places qualitatively.)

Extra credit: Get quantitative! Seen from above, how many high-risk points are there? If there are infinitely many, what is their total area?

Here is my solution:
[Show Solution]

And here is a bonus interactive graphic showing the solution

## What if robots cut your pizza?

This Riddler puzzle is about random chords of a circle and the regions they describe.

At RoboPizza™, pies are cut by robots. When making each cut, a robot will randomly (and independently) pick two points on a pizza’s circumference, and then cut along the chord connecting them. If you order a pizza and specify that you want the robot to make exactly three cuts, what is the expected number of pieces your pie will have?

Here is a simple solution, which was pointed out to me in a comment to my original post.
[Show Solution]

The following solution is a bit more complicated, and computes the entire distribution rather than just its expected value.
[Show Solution]

If you’ve already read the solution above and you’re interested in the distribution of pieces for the general case, read on!
[Show Solution]

## A clever integral

I was recently reminded of this problem from one of my favorite books: Problem-Solving Through Problems. The problem originally appeared in the 1980 Putnam Competition.

Evaluate the following definite integral.

$\int_0^{\pi/2} \frac{\mathrm{d}x}{1 + (\tan x)^{\sqrt{2}}}$

The solution:
[Show Solution]