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Let's say there is a regular hexagon with center at point O.

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Question 1. How many triangles can you obtain using the 6 vertices and its center?

Question 2. What is the largest number of different triangles you can make?

Update

Two triangles will be called different if at least one element differs in them: the angle value or the side length.

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    $\begingroup$ Hi Nick, I've corrected some spelling/grammar here, but I wasn't sure at all how to reword Q2 - as it stands it is very tricky to understand what it wants... Do you mean "How many different such arrangements of triangles are there?" or is it more "What's the largest number of different triangles you can make while doing this?" or something else altogether? Thanks. $\endgroup$
    – Stiv
    Commented Nov 6, 2020 at 9:39
  • $\begingroup$ @Stiv, thank you for corrections spelling/grammar. I have rewrite the Q2, but hexomino gives the answer on first version. $\endgroup$
    – Nick
    Commented Nov 6, 2020 at 10:36
  • $\begingroup$ @JaapScherphuis, I have corrected the Q2 again, but heximino's answer is very close. $\endgroup$
    – Nick
    Commented Nov 6, 2020 at 14:57

2 Answers 2

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Question 1

I create a triangle by choosing three vertices from the seven given.
The only time I get a degenerate triangle is when the three vertices are in a straight line which happens in three cases - $\{A,O,D\}, \{B,O,E\}, \{C,O,F\}$.
Subtracting these cases I get $$\binom{7}{3} - 3 = 32$$

Question 2

I think there are five essentially different triangles all obtained by rotation or inversion of the following triangles $$ AOB, AFB, AEB, AOC, AEC $$ and the number of times each of these triangles appears is as follows $$6,6,12,6,2 $$ which adds up to $32$, as desired.

Jean Hominal makes an important point in the comments

Triangles $AFB$ and $AOC$ are congruent so could be obtained one from the other by an extra translation although we'd lose the overall symmetry of the hexagon. My answer above assumes that we can rotate and invert while maintaining the hexagonal symmetry but if you would like to consider congruent triangles as being the same then the answer would be four.

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    $\begingroup$ Are AOC and AFB not triangles that are essentially the same? In a regular hexagon, the distance between the center and any summit is the same as the distance between two summits. $\endgroup$
    – Jean Hominal
    Commented Nov 6, 2020 at 13:26
  • $\begingroup$ @JeanHominal You make an important point, I've edited my answer to address this concern. $\endgroup$
    – hexomino
    Commented Nov 6, 2020 at 15:56
  • $\begingroup$ There's one fewer if you match on scaling, too. $\endgroup$
    – Ian MacDonald
    Commented Nov 6, 2020 at 18:14
  • $\begingroup$ Technically, the degenerate triangle AED might add one to the "different triangles" list, if it counted. The question does nto state definitively that it does not. $\endgroup$
    – Ben Barden
    Commented Nov 6, 2020 at 19:37
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We can obtain four triangles, specifically two equilaterals ABG and ECG, one isosceles triangle EFD and one right angle triangle ABC. If we draw the other four missing chords and the one missing radius, we obtain too many triangles to count (I stopped at thirty).

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