Above you can see the problem where an equilateral triangle of side length $4$ coins is made up of a total of $T_4$ coins. And it can be seen in the solution that the minimal number of coin moves needed to be done to flip the triangle upside-down is $3$ coins.

Prove (or disprove):

The minimal (optimal) number of coin moves $m$ needed to be done to flip an equilateral triangle of side length $n$ coins made up of $T_n$ number of coins is given by:

$\textrm{optimal }m=\left\lfloor\frac{T_n}{3}\right\rfloor$

**Details and Assumptions:**

- $T_n$ is used to denote the $n^{\textrm{th}}$ triangular number.

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## Comments

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TopNewestWell I have different approach for this problem .. and it's easily done on equilateral triangles with a regular hexagon core (It's doable also with other irregular hexagons) here is a link

28 coins equilateral triangle

you can see that i made a different color inside for a 10 coins triangle as it will follow the same movement

it's safe to say it will always be an odd number and it'll follow you rule as well

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That's a particular case you've shown. This note is about a general proof for the case of $T_n$ coins. You've only shown that it works for the case of $T_8$.

Hence, not a proof.

Actually, I have seen a proof of this statement but I don't think it's rigorous enough. So, I posted this note here to see if someone from the community comes up with a full and rigorous proof of this.

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It actually presents two cases of regular hexagon core $T_{10}$ and $T_{28}$

And yes it's not a rigid proof .. but I thought I would share the idea in my mind as it involves angles in a way so it might help someone to come up with a solid proof as I really sadly don't have the time (due to my work) to put my mind in it; so let me explain what I have in mind in words

I believe the maximum regular Hexagon that can be drawn inside the equilateral triangle is the key to prove this statement, why? because it will give you the minimum area left over of the triangle which you can use to rearrange this triangle in any direction

So we can use the triangles with regular coin hexagon to do so that's a simplification of course cause we can have a coin triangle of any size (number of coins) so in some cases you will find yourself having a hexagon with part coins out, that's why i used $T_{28}$ and $T_{10}$ so I can explain the idea in more simple form

I hope that would help

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I appreciate that you took your time to share your ideas with us. And yes, your hexagon core idea is actually the key point in the proof and hence your informal explanation is correct. Note that your argument is equivalent to what Daniel Liu stated in his comment since the "overlaying" he mentioned gives us the hexagon core and the remaining coins outside the core are the ones that we need to move.

The proof that I have seen also uses the same idea but the conclusion in that proof is not derived completely using mathematical means but instead using computational means (computer-assisted, I think). Here's the original discussion post that I saw and the answer posted there is the proof I spoke of. Feel free to check it out. :)

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here after I saw the challenge master's comment and link to your note

Sorry I didn't notice Daniel's comment he even put in fewer clear words :) .. as I actually got here fromThanks for your appreciation and for the link as well :) I'll check it,

I'll also try to figure out a formula for this or prove this one but after Ramadan so I can drink a cup of tea while thinking :D

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And yes, tea/coffee is very essential when working on math problems. As Alfréd Rényi and Paul Erdős used to say,

"

A mathematician is a device for turning coffee into theorems."Log in to reply

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The key step is to overlay the beginning and end positions to find which coins need to be moved and which coins can stay where they are. With this idea in mind, just consider the three cases $n\equiv 0, 1, 2\pmod{3}$ respectively and you can prove it.

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I tried doing that for a few values of n but I couldn't find any pattern. Can you explain a bit more?

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This is a little hand-wavy, but we want as much overlap as possible so we want the centers of the two triangles to be as close as possible. So try reflecting the original triangle across the line passing through or near its center.

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