Putnam 2013, Problem A-5

For \( m \geq 3 \), a list of \( \binom{m}{3} \) real numbers \( a_{ijk} \) (\( 1 \leq i < j < k \leq m \)) is said to be *area definite* for \( \mathbb{R}^n \) if the inequality
\[
\sum_{1 \leq i < j < k \leq m} a_{ijk} \cdot \text{Area}(\Delta A_i A_j A_k) \geq 0
\]
holds for every choice of \( m \) points \( A_1, \ldots, A_m \) in \( \mathbb{R}^n \). For example, the list of four numbers \( a_{123} = a_{124} = a_{134} = 1 \), \( a_{234} = -1 \) is area definite for \( \mathbb{R}^2 \). Prove that if a list of \( \binom{m}{3} \) numbers is area definite for \( \mathbb{R}^2 \), then it is area definite for \( \mathbb{R}^3 \).

Solution:
Let $A_1, \ldots, A_m$ be points in $\mathbb{R}^3$, and let $\hat{n}_{ijk}$ denote a unit vector normal to $\Delta A_i A_j A_k$ (unless $A_i, A_j, A_k$ are collinear, there are two possible choices for $\hat{n}_{ijk}$). If $\hat{n}$ is a unit vector in $\mathbb{R}^3$, and $\Pi_{\hat{n}}$ is a plane perpendicular to $\hat{n}$, then the area of the orthogonal projection of $\Delta A_i A_j A_k$ onto $\Pi_{\hat{n}}$ is $\text{Area}(\Delta A_i A_j A_k) |\hat{n}_{ijk} \cdot \hat{n}|$. Thus if $\{a_{ijk}\}$ is area definite for $\mathbb{R}^2$, then for any $\hat{n}$,

$$\sum a_{ijk} \text{Area}(\Delta A_i A_j A_k) |\hat{n}_{ijk} \cdot \hat{n}| \geq 0.$$

Note that integrating $|\hat{n}_{ijk} \cdot \hat{n}|$ over $\hat{n} \in S^2$, the unit sphere in $\mathbb{R}^3$, with respect to the natural measure on $S^2$ gives a positive number $c$, which is independent of $\hat{n}_{ijk}$ since the measure on $S^2$ is rotation-independent. Thus integrating the above inequality over $\hat{n}$ gives $c \sum a_{ijk} \text{Area}(\Delta A_i A_j A_k) \geq 0$. It follows that $\{a_{ijk}\}$ is area definite for $\mathbb{R}^3$, as desired.

**Remark:** It is not hard to check (e.g., by integration in spherical coordinates) that the constant $c$ occurring above is equal to $2\pi$. It follows that for any convex body $C$ in $\mathbb{R}^3$, the average over $\hat{n}$ of the area of the projection of $C$ onto $\Pi_{\hat{n}}$ equals $1/4$ of the surface area of $C$.

More generally, let $C$ be a convex body in $\mathbb{R}^n$. For $\hat{n}$ a unit vector, let $\Pi_{\hat{n}}$ denote the hyperplane through the origin perpendicular to $\hat{n}$. Then the average over $\hat{n}$ of the