I found this problem in a section of an old notebook, where I used to write down weird problems I came across and that I didn't know how to solve. Long story short, I rediscovered this notebook a week ago and managed to solve most geometry problems, with the exception of the following.

Let $n\ge 4$, and let $A_1, A_2,\dotsc, A_n$ be concyclic. Let $h$ be the set of the orthocenters of the triangles determined by these points, and let us label its elements (the orthocenters) as $H_1, H_2, \dotsc$

Show that: $$\sum_{H_a,H_b∈h} H_aH_b ≥\frac{(n-2)(n-3)}{2} \sum_{1\leq i,j\leq n} A_iA_j,$$ and determine when equality takes place.

Note: $H_aH_b$ and $A_iA_j$ are the length of the segments.

I am completely stumped by this problem.

  • $\begingroup$ What is your product of points here? Is that meant to be a dot product? $\endgroup$ Nov 21, 2020 at 17:27
  • $\begingroup$ H1H2 and AiAj refer to the length of the segments $\endgroup$
    – cristi0p
    Nov 21, 2020 at 17:35
  • $\begingroup$ Not clear what you mean by $H_1H_2\in h$, since $h$ consists of points, not segments. Also, $H_1$ and $H_2$ denote two fixed elements of $h$ by the sentence before the display, so you might want to use different symbols in the sum. $\endgroup$
    – GH from MO
    Nov 21, 2020 at 17:54
  • $\begingroup$ @GH: there should probably be a comma between $H_1$ and $H_2$ there. Should probably be $\sum_{H_i,H_j,\; i\neq j}H_i H_j$ or something $\endgroup$
    – Qfwfq
    Nov 21, 2020 at 17:55
  • 1
    $\begingroup$ I think it would be cleaner to write $a,b\in\{1,2,\dotsc,\binom{n}{3}\}$ instead of $H_a,H_b\in h$. This is subjective of course. $\endgroup$
    – GH from MO
    Nov 21, 2020 at 18:20

1 Answer 1


If the origin is circumcenter of the triangle $A_iA_jA_k$, then orthocenter of this triangle is $A_i+A_j+A_k$.

Hence the distance of two specified orthocenters is $|A_i+A_j+A_k - (A_m+A_j+A_k)| = |A_i-A_m|$. Note that choice number for $j,\ k$ is $ \ _{n-2}C_2$.


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