IMC 2009 · Day 1 · P5

We will prove this by induction on $n$, starting with $n = 1$. In that case we are given an interval $[a, b]$ with a point $p \in (a, b)$, and we have to verify $$(b - p) \frac{b + p}{2} + (p - a) \frac{p + a}{2} = (b - a) \frac{b + a}{2},$$ which is true.

Now let assume the result is true for $n - 1$ and prove it for $n$. We have to show that the point $$X = \sum_{j=0}^{n} \frac{v(S_j)}{v(S)} O(S_j)$$ has the same distance to all the points $P_0, P_1, \dots, P_n$. Let $i \in \{0, 1, 2, \dots, n\}$ and define the sets $M_i = \{P_0, P_1, \dots, P_{i-1}, P_{i+1}, \dots, P_n\}$. The set of all points having the same distance to all points in $M_i$ is a line $h_i$ orthogonal to the hyperplane $E_i$ determined by the points in $M_i$. We are going to show that $X$ lies on every $h_i$. To do so, fix some index $i$ and notice that $$X = \frac{v(S_i)}{v(S)} O(S_i) + \frac{v(S) - v(S_i)}{v(S)} \cdot \underbrace{\sum_{j \ne i} \frac{v(S_j)}{v(S) - v(S_i)} O(S_j)}_{Y}$$ and $O(S_i)$ lies on $h_i$, so that it is enough to show that $Y$ lies on $h_i$.

A map $f : \mathbb{R}_{>0} \to \mathbb{R}^n$ will be called affine if there are points $A, B \in \mathbb{R}^n$ such that $f(\lambda) = \lambda A + (1 - \lambda) B$.

Consider the ray $g$ starting in $P_i$ and passing through $P$. For $\lambda > 0$ let $P_\lambda = (1 - \lambda) P_i + \lambda P$, so that $P_\lambda$ is an affine function describing the points of $g$. For every such $\lambda$ let $S_j^\lambda$ be the $n$-simplex obtained from $S$ by replacing the $j$-th vertex by $P_\lambda$. The point $O(S_j^\lambda)$ is the intersection of the fixed line $h_j$ with the hyperplane orthogonal to $g$ and passing through the midpoint of the segment $P_i P_\lambda$ which is given by an affine function. This implies that also $O(S_j^\lambda)$ is an affine function. We write $\varphi_j = \frac{v(S_j)}{v(S) - s(S_i)}$, and then $$Y_\lambda = \sum_{j \ne i} \varphi_j O(S_j^\lambda)$$ is an affine function. We want to show that $Y_\lambda \in h_i$ for all $\lambda$ (then specializing to $\lambda = 1$ gives the desired result). It is enough to do this for two different values of $\lambda$.

Let $g$ intersect the sphere containing the vertices of $S$ in a point $Z$; then $Z = P_\lambda$ for a suitable $\lambda > 0$, and we have $O(S_j^\lambda) = O(S)$ for all $j$, so that $Y_\lambda = O(S) \in h_i$. Now let $g$ intersect the hyperplane $E_i$ in a point $Q$; then $Q = P_\lambda$ for some $\lambda > 0$, and $Q$ is different from $Z$. Define $T$ to be the $(n-1)$-simplex with vertex set $M_i$, and let $T_j$ be the $(n-1)$-simplex obtained from $T$ by replacing the vertex $P_j$ by $Q$. If we write $v'$ for the volume of $(n-1)$-simplices in the hyperplane $E_i$, then $$\frac{v'(T_j)}{v'(T)} = \frac{v(S_j^\lambda)}{v(S)} = \frac{v(S_j^\lambda)}{\sum_{k \ne i} v(S_k^\lambda)} = \frac{\lambda v(S_j)}{\lambda \sum_{k \ne i} v(S_k)} = \frac{v(S_j)}{v(S) - v(S_i)} = \varphi_j.$$ If $p$ denotes the orthogonal projection onto $E_i$ then $p(O(S_j^\lambda)) = O(T_j)$, so that $p(Y_\lambda) = \sum_{j \ne i} \varphi_j O(T_j)$ equals $O(T)$ by induction hypothesis, which implies $Y_\lambda \in p^{-1}(O(T)) = h_i$, and we are done.

For $n = 1$, the statement is checked easily.

Assume $n \ge 2$. Denote $O(S_j) - O(S)$ by $q_j$ and $P_j - P$ by $p_j$. For all distinct $j$ and $k$ in the range $0, \dots, n$ the point $O(S_j)$ lies on a hyperplane orthogonal to $p_k$ and $P_j$ lies on a hyperplane orthogonal to $q_k$. So we have $$\begin{cases} \langle p_i, q_j - q_k \rangle = 0 \\ \langle q_i, p_j - p_k \rangle = 0 \end{cases}$$ for all $j \ne i \ne k$. This means that the value $\langle p_i, q_j \rangle$ is independent of $j$ as long as $j \ne i$, denote this value by $\lambda_i$. Similarly, $\langle q_i, p_j \rangle = \mu_i$ for some $\mu_i$. Since $n \ge 2$, these equalities imply that all the $\lambda_i$ and $\mu_i$ values are equal, in particular, $\langle p_i, q_j \rangle = \langle p_j, q_i \rangle$ for any $i$ and $j$.

We claim that for such $p_i$ and $q_i$, the volumes $$V_j = |\det(p_0, \dots, p_{j-1}, p_{j+1}, \dots, p_n)|$$ and $$W_j = |\det(q_0, \dots, q_{j-1}, q_{j+1}, \dots, q_n)|$$ are proportional. Indeed, first assume that $p_0, \dots, p_{n-1}$ and $q_0, \dots, q_{n-1}$ are bases of $\mathbb{R}^n$, then we have $$\begin{align*} V_j &= \frac{1}{|\det(q_0, \dots, q_{n-1})|} \left| \det \Bigl( \bigl( \langle p_k, q_l \rangle \bigr)_{\substack{k \ne j \\ l < n}} \Bigr) \right| = \\ &= \frac{1}{|\det(q_0, \dots, q_{n-1})|} \left| \det \Bigl( \bigl( \langle p_k, q_l \rangle \bigr)_{\substack{l \ne j \\ k < n}} \Bigr) \right| = \frac{|\det(p_0, \dots, p_{n-1})|} {|\det(q_0, \dots, q_{n-1})|}\, W_j. \end{align*}$$ If our assumption did not hold after any reindexing of the vectors $p_i$ and $q_i$, then both $p_i$ and $q_i$ span a subspace of dimension at most $n - 1$ and all the volumes are 0.

Finally, it is clear that $\sum_j q_j W_j / \det(q_0, \dots, q_n) = 0$: the weight of $p_j$ is the height of 0 over the hyperplane spanned by the rest of the vectors $q_k$ relative to the height of $p_j$ over the same hyperplane, so the sum is parallel to all the faces of the simplex spanned by $q_0, \dots, q_n$. By the argument above, we can change the weights to the proportional set of weights $V_j / \det(p_0, \dots, p_n)$ and the sum will still be 0. That is, $$0 = \sum \frac{V_j}{\det(p_0, \dots, p_n)}\, q_j = \sum \frac{v(S_j)}{v(S)} (O(S_j) - O(S)) =$$ $$= \frac{1}{v(S)} \left( \sum O(S_j) v(S_j) - O(S) \sum v(S_j) \right) = \frac{1}{v(S)} \left( \sum O(S_j) v(S_j) - O(S) v(S) \right),$$ q.e.d.