Let $A_0A_1...A_n$ be a simplex in $\Bbb E^n.$ Let $B_{ij}$ be a point on the edge $A_iA_j,\ 0\le i\not=j\le n.$

Denote by $\beta_i$ the hyperplane passing through the points $B_{i0},$ $B_{i1},$ $B_{ii-1},$ $B_{ii+1},$ $...,$ $B_{in}.$

Assume that the reflection of $A_i$ in the hyperplane $\beta_i$ lies on the hyperplane $A_0A_1...A_{i-i}A_{i+1}...A_n$ for all $i$.

My question. Can we show that $B_{ij}$ must be midpoints of the edges $A_iA_j?$

  • Here is the obvious proof for $n=2$. Then there is a vertex such that all adjacent $B's$ are not further away than the midpoints and at least one of the two points is closer. Then the distance to the edge connecting those two B's is less than half the height. Thus reflecting along it can never hit the opposing edge of the original triangle. – HenrikRüping Jul 6 at 14:18
  • @HenrikRüping: I believe you can have $B_{01}$ close to $A_0$, $B_{12}$ close to $A_1$ and $B_{02}$ close to $A_2$, and your argument does not apply. (And, by the way, even if it did, it would only work for acute triangles, right?) – Mateusz Kwaśnicki Jul 6 at 16:47
  • By the way, numerical experiments strongly suggest that the answer is affirmative for $n = 2$, but I fail to find an elementary proof. – Mateusz Kwaśnicki Jul 6 at 18:01
  • Oops.you are right. – HenrikRüping Jul 7 at 17:03
  • OK here is a similar idea along those lines. Assign to any such configuration of $B$'s the sum of the distances of $A_i$ to $\beta_i$. In the case where all B's are exactly the midpoints that sum is the sum of all heights/2. If those reflections lie on the opposing hyperplanes, then each of the distances has to be at least half the height. Now it seems as if that sum is the absolute maximum, i.e. whenever you choose any of the B_i's differently, you get a smaller sum. I just checked a couple of examples, but with a calculus, it should either be provable or falsifiable. – HenrikRüping Jul 11 at 15:01

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