Answer: Four Random Points on a Sphere: Tetrahedron Contains the Center

Key Idea / Intuition

Choose a point on the sphere in two stages. First pick a random diameter. Then flip a fair coin to decide which endpoint to use. Fix the diameters for $P_1, P_2, P_3$ and the point $P_4$. Among the $2^3 = 8$ equally likely ways of choosing endpoints, exactly one gives a tetrahedron that contains the center. The probability is therefore $1/8$, and no integration is needed. The same argument gives $1/4$ on the circle and $1/2^n$ for $n+1$ points on $S^{n-1}$.

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Formal Proof / Solution

Answer: $\dfrac{1}{8}$.

Step 1: Re-randomize. Let $Q_1, Q_2, Q_3$ be independent uniform points on $S^2$. Let $\varepsilon_1, \varepsilon_2, \varepsilon_3$ be independent fair $\pm 1$ signs, and let $P_4$ be uniform. All of these are independent. Set $P_i = \varepsilon_i Q_i, \qquad i = 1, 2, 3.$ The uniform measure on $S^2$ is invariant under $x \mapsto -x$, so each $P_i$ is again uniform. All four points are independent, so $(P_1, \dots, P_4)$ has exactly the distribution in the problem.

We condition on $Q_1, Q_2, Q_3, P_4$. We will show that, almost surely, exactly one of the 8 sign patterns $(\varepsilon_1, \varepsilon_2, \varepsilon_3)$ produces a tetrahedron containing the origin $O$.

Step 2: Reformulate "contains the center". With probability $1$:

• $Q_1, Q_2, Q_3$ are linearly independent, and
• no three of the four points lie on a plane through $O$.

Both failures happen on measure-zero sets. From now on we assume this generic situation.

$O$ lies in the tetrahedron exactly when there are weights $\lambda_1, \dots, \lambda_4 \ge 0$ with $\sum \lambda_i = 1$ and $\sum \lambda_i P_i = 0$.

• All weights are positive. If some $\lambda_j = 0$, then $O$ lies in the convex hull of the other three points, so those three lie on a plane through $O$. This is excluded by genericity, so every $\lambda_i > 0$.
• Divide by $\lambda_4$. Then $O$ lies in the tetrahedron iff $-P_4 = a_1 P_1 + a_2 P_2 + a_3 P_3 \quad \text{for some } a_1, a_2, a_3 > 0.$
• Converse. If such $a_i > 0$ exist, then $P_4 + \sum a_i P_i = 0$. Dividing by $1 + a_1 + a_2 + a_3$ gives valid convex weights.

Step 3: Exactly one sign pattern works. Since $Q_1, Q_2, Q_3$ form a basis of $\mathbb{R}^3$, there is a unique expansion $-P_4 = c_1 Q_1 + c_2 Q_2 + c_3 Q_3.$ Each $c_i \neq 0$. Indeed, if $c_3 = 0$, then $P_4$ would lie in the plane spanned by $Q_1, Q_2$, which passes through $O$. That is excluded by genericity, and the same holds for $c_1, c_2$.

Now substitute $P_i = \varepsilon_i Q_i$: $-P_4 = \sum_i (c_i \varepsilon_i)\, P_i,$ and by uniqueness these are the only possible coefficients. By Step 2, the tetrahedron contains $O$ iff all $c_i \varepsilon_i > 0$, that is, iff $\varepsilon_i = \operatorname{sign}(c_i) \quad \text{for } i = 1, 2, 3.$ Exactly one of the 8 sign patterns satisfies this.

Step 4: Conclude. Conditionally on $Q_1, Q_2, Q_3, P_4$, the signs are uniform on $\{\pm 1\}^3$, so the conditional probability is $1/8$. Averaging over the conditioning variables gives $\Pr(O \in \text{tetrahedron}) = \frac{1}{8}.$

Warm-up (circle). The identical argument with two diameters and one free point gives $1/2^2 = 1/4$.

Generalization. For $n+1$ uniform points on $S^{n-1} \subset \mathbb{R}^n$, the simplex contains the center with probability $1/2^n$.

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Why it's beautiful.

• A direct computation would mean integrating areas of spherical triangles, which is messy. The symmetry $x \mapsto -x$ turns the problem into counting sign patterns.
• Linear algebra (a unique expansion in a basis) does all the geometric work.
• The answer generalizes to every dimension at once.

Difficulty: 3/5.

Source: Putnam 1992, Problem A6 (see materials/putnam/1992.pdf).