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I want to maximize the number of similar triangles with vertices from three fixed sets, one vertex from each set. For example, if you fix two points $X$, $Y$ (i.e. two sets with only one member), then there are at most 12 points $Z_1,Z_2,\ldots,Z_{12}$ such that all the triangles $XYZ_i$ are similar.

This is the main question:

Let $A$, $B$, $C$ be three sets of points in the plane with the property that all of the triangles $XYZ$ with $X \in A$, $Y \in B$, $z \in C$ are similar. Is it true that $|A| \cdot |B| \cdot |C| \leq 12$?

For the $2\times2\times2$ case, can you find six points $X_1, X_2, Y_1, Y_2, Z_1, Z_2$ such that all eight triangles $X_iY_jZ_k$ are similar?

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  • $\begingroup$ Regarding the 2x2x2 case: It might be a mess to do by hand, but there is a way of coding this. Specifically, have the computer go through all possible angle permutations of 8 triangles (there are $6^7$ possibilities, since you can fix the angles of the first one). You can use symmetry of the problem to reduce this a bit further, if needed. This gives you equations among the squares of the sides $|X_iY_j|^2,|Y_iZ_j|^2,|Z_jX_i|^2$. Add to these the equations of planarity (mathworld.wolfram.com/Cayley-MengerDeterminant.html) and see if the only solution is all zeros by Groebner bases. $\endgroup$
    – Lev Borisov
    Commented Aug 12, 2014 at 19:06
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    $\begingroup$ I suppose for the $2\times2\times2$ you insist on planarity, else the vertices of a regular octahedron will do. $\endgroup$
    – Gerry Myerson
    Commented Aug 12, 2014 at 23:18
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    $\begingroup$ @Gerry Myerson Good example for 3D, but Of course planarity is important, if not you can find infinitely many three colored triangles in the 3 dimensional space. ($1\times1\times\infty$) $\endgroup$
    – Morteza
    Commented Aug 14, 2014 at 6:02

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Here is a construction for the $2\times2\times2$ case. Perhaps, it is not unique.

Choose $t<1$ such that in a triangle $XYZ$ with $XY=1$, $XZ=t^3$, $YZ=t^2$ we have $2\angle X+3\angle Y=\pi$. Such $t$ exists, since for the minimal value of $t$ (i.e. $t^2+t^3=1$) we have $2\angle X+3\angle Y=0<\pi$, while for $t=1$ we have $2\angle X+3\angle Y=5\pi/3>\pi$.

Then the configuration below is what you need.

On this picture, the angles equal to $\angle X$ are marked with an arc, those equal to $\angle Y$ are marked with a double arc. If a segment is marked with a roman numeral $n$, its length is $t^n$; an unmarked segment has a unit length.

2x2x2 configuration

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Not an answer, just an illustration of Morteza's $12$ points $Z_1,\ldots,Z_{12}$ and $12$ similar triangles:

          Similar12

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