# Pheno Menon and his coloured flags

Professor Pheno Menon has sent for me for helping with some decorations. When I reached his home, he explained:

"Look here, I've got $6$ flags: $2$ red, $2$ blue and $2$ green. I want to pin them in my garden such that no three are collinear."

"I am assuming there's some catch?", I said.

"Indeed, there is." he replied, "I want them to be so arranged that all the possible triangles formed with $3$ differently-coloured vertices are similar to each other. I've trying to properly place them for some hours, but can't quite make it work. I am beginning to think it's not even possible; in that case, I'd like a proof of that. I thought you might be able to help me."

But I am stumped as well. Maybe you can help?

Adapted from a problem by Morteza Saghafian in Iranian Geometry Olympiad 2016.

• I imagine a straight line counts as a triangle here? – Jonathan Allan Oct 3 '16 at 6:13
• Hah, @Jonathan Allan, you're wondering about a 1-D garden. How about a 3-D garden, perhaps the entire planet? – humn Oct 3 '16 at 6:16
• @humn I was actually thinking that in 2D a trivial solution would be all flags in a straight line making no triangles; or a slightly less trivial equilateral triangle RGB with each side's midpoint as the third colour making 5 similar triangles and three straight lines. – Jonathan Allan Oct 3 '16 at 6:21
• @JonathanAllan yep, degenerate triangles are triangles. – Ankoganit Oct 3 '16 at 6:52
• @justhalf Not all degenerate triangles are similar. For example a 2,1,1 triangle is not similar to a 7,4,3 triangle. – Ankoganit Oct 3 '16 at 10:49

There is an error in user3445853's impossibility proof:

"either abc are all in the interior of ABC, all in the exterior (which is the same case as previous but reversed)" No, they could both be exterior to each other.

I offer: (DELETED: NOPE, close but not quite) Am I missing something? (EDIT: YES. Yes I was.)

I offer:

Well the angles don't work out nicely like they did in my wrong answer, so this time I'll give co-ordinates and distances. Put two red flags at $(0,\pm 2)$, two green flags at $(\pm \sqrt{7}, 1)$ and two blue flags at $(\pm \sqrt{7}, -1)$. The distance between a red flag and non-red flag is $\sqrt{8}$ or $4$, the distance between a green and a blue flag is $2$ or $\sqrt{32}$. The red-green-blue triangles are then $2:\sqrt{8}:4$ or $\sqrt{8}:4:\sqrt{32}$, so they are similar. NOW am I missing something?

• I'm not sure I understood your construction, do you mean something like this? – Ankoganit Oct 6 '16 at 4:40
• that's what I meant, yes. – deep thought Oct 6 '16 at 4:43
• In that case, I don't think $R_1B_1G_2$ is similar to $R_1B_2G_2$. :( – Ankoganit Oct 6 '16 at 4:48
• Well.... it isn't. But it was half an hour ago, I swear it! – deep thought Oct 6 '16 at 4:52
• I think this works, and is very nice! I've added a figure for clarity. +1 for now, I'll accept it after some more time...hope you don't mind. :) – Ankoganit Oct 6 '16 at 10:11

You can arrange them by forming 2 pyramids with the same square base.
The base will consist of the blue and green flags arranged in such a way that the diagonals have the same colors on the ends.
And the red flags would be the top of the pyramids

Something like this:

The triangles are:

$R_1B_1G_1$
$R_1B_1G_2$
$R_1B_2G_1$
$R_1B_2G_2$
$R_2B_1G_1$
$R_2B_1G_2$
$R_2B_2G_1$
$R_2B_2G_2$

They are all similar because they contains one of the edges $B_1G_1$ or $B_1G_2$ or $B_2G_1$ or $B_2G_2$ which are the same length (they are edges of a square). And the edge going from $R_1$ or $R_2$ to the any $B_x$ or $G_x$ point have the same length.
They can even all be constructed to be equilateral (but I didn't find a proper picture on the web).

• Uh....Are you sure that works? – Ankoganit Oct 3 '16 at 6:54
• Indeed, we've got a 2d plane to work with, so 3d is not allowed. Sorry if this wasn't clear from the OP. – Ankoganit Oct 3 '16 at 7:06
• Well you got yourself an upvote from me anyway :) I do think the answer will be a proof of impossibility to be honest. – Jonathan Allan Oct 3 '16 at 7:09
• I think the idea is to have distinct points. Otherwise you can just pin all the flags on top of each other. – Marius Oct 3 '16 at 8:48
• @ΈρικΚωνσταντόπουλος I am the OP (Original Poster), and I wrote the OP (Original Post). – Ankoganit Oct 3 '16 at 14:20

I don't think I understand the question right, because it seems trivial.

Lets mark one set of flags A,B,C the second set a,b,c with A and a one color, B and b the next color, C and c the remaining color (colors respectively red, green, blue, or whatever permutation). Arrange the first set A,B,C into an equilateral triangle. Mark the middle of each side, and put the remaining flag of the opposing corner color there: So the main big triangle A-B-C now has sides with middle points A-c-B, B-a-C, C-b-A.

That solves the situation because

All triangles that can occur are equilateral triangles; all non-equilateral triangles -- say aBb --- have two the same colors.

This is provable by contradiction because

ONE or more original/first-set vertices are involved (otherwise it's the inner equilateral triangle abc you deal with), assume A (re-labeling makes it hold for the others, like I relabeled colors anyway to start); and ONE or more second-set vertices must be involved (otherwise it's the original equilateral triangle ABC you deal with), assume this is b (again, b by relabeling if necessary). Now the third vertex must be c or C; it cannot be C as that's collinear/ not a triangle, so we're left with Abc which is equilateral.

Or did you want to allow for "collinear triangles" (which are not triangles)? EDIT: Bof, if you allow for those (which by now, reading http://mathworld.wolfram.com/Triangle.html it satisfies that definition for a limit case where the angles tend to 0,0,180), then any collinear points (A,B,C) with B between A and C form a (degenerate) triangle with the angles at A and C being zero, at C it's 180 (or pi). But according to the definition I'm following here --- http://mathworld.wolfram.com/SimilarTriangles.html --- any three collinear points gives a similar triangle because it's the angles of the vertices NOT the length ratios of the sides. So ABC collinear whether a limit of ABC isosceles (so |AB|=|BC| in the limit) or not, that's all similar because the same angles. So arrange all six in any order on a single line, and Pheno Menon's question is answered by example. [Note that then they're also inversely similar as you can reach the limit from the mirrored triangle.]

If now somebody invokes some further overlooked rule why they would not be similar, and only collinear triplets at the same ratio |AB|:|BC| are similar, then clearly the arrangement with all six collinear is ruled out. Which means

there cannot be any collinear triplets, and no equilateral triangles either; then you start with a (non-degenerate) triangle ABC and see that either abc are all in the interior of ABC, all in the exterior (which is the same case as previous but reversed), or one in the interior two exterior, or the reverse. Here it's quick to see they cannot give you the same triplets of angles.

• Jonathan Allan pointed out this in the comments.. As I said there, degenerate triangles need to be regarded as triangles, so we need, for example, ABC~ABc, which is not the case here. – Ankoganit Oct 3 '16 at 11:14
• Since the inclusion of degenerate triangles is causing some confusion, I have removed it altogether. I earnestly apologize for any inconvenience caused. – Ankoganit Oct 3 '16 at 13:24

I don't think it is possible, and here is a partial proof:

Suppose I get rid of one of the red flags for now. The remaining five flags form four triangles, all involving the one red flag. These four triangles are supposed to be similar, and lets call the three angles of these similar triangles a, b, and c.
At the red flag, four angles come together and must add to 360 degrees. We know that a+b+c=180, so if at least one of each type of angle occurs at the red flag, the fourth angle would be 180 degrees, leading to a degenerate triangle.

If the same angle meets at the red flag 4 times, then it is a right angle. The four triangles then form either a rhombus or a particular type of kite shape. In either case the red flag is at the intersection of the diagonals, and there is no other point where the red flag could be. Therefore the second red flag can only be planted in the same spot as the first red flag.

If only two types of angle meet at the red flag, then we can assume it is either 2a+2b or 3a+b. In the first case we get a+b=180, again leading to a degenerate triangle.

The only remaining case that needs a proof is when three of the same type of angle meet at the red flag. This is the most complex of the cases so far and I don't have time to work it out now, but I will come back to it later.

• I read your attempt only in part, but I believe the angles can overlap at the red flag. – Matsmath Oct 3 '16 at 14:45
• @Matsmath: You're right. I implicitly assumed the other flags alternated (e.g. GBGB clockwise around the red flag) whereas they need not do so. Also, I did not properly deal with the possibility that the four flags all lie to one side of the red one. This is going to need some work. – Jaap Scherphuis Oct 3 '16 at 14:51