3
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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

2
$\begingroup$

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$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.