# Red, blue and yellow points

In the plane, there are $r+b+y=15$ points, of which $r\ge2$ points are colored red, $b\ge2$ points are colored blue, and $y\ge2$ points are colored yellow.

• If we consider all pairs of red and blue points and add up their distances then the sum is $51$.
• If we consider all pairs of red and yellow points and add up their distances then the sum is $39$.
• If we consider all pairs of yellow and blue points and add up their distances then the sum is $1$.

Determine all possible values for $r$ and $b$ and $y$.

• "all pairs of red and blue points" - does this include red-red, blue-blue and red-blue, or only red-blue? Apr 9, 2016 at 16:48
• @astralfenix: "all pairs of red and blue points" only includes red-blue (but not red-red and blue-blue). Apr 9, 2016 at 17:00
• Are the points allowed to coincide? Apr 9, 2016 at 17:07

$r = 8$
$b = 4$
$y = 3$

I'm making this answer a community wiki, since I made a mistake at the end of my original answer and it was Van.Graaf who delivered the final punch.

Reasoning:

Let's consider all the tri-colored triangles (e.g. one vertex red, one vertex blue, one vertex yellow) and apply triangle inequalities. The sum of lengths of all the red-blue segments must be no greater than the sum of all lengths of all the red-yellow segments and all the blue-yellow ones.

What is the sum of lengths of all the red-blue segments? It is $51y$ (each red-blue pair is counted $y$ times and the sum of pair lengths is $51$). Of red-yellow segments — $39b$, of blue-yellow — $r$. Therefore, the inequalities are:

$51y \le 39b + r$
$39b \le 51y + r$
$r \le 51y + 39b$
$r + b + y = 15$

The only triple $(r, b, y)$ which satisfies the above set of constraints is $(8, 4, 3)$.

Here's the space of possible solutions, with blue on the $x$ axis and yellow on the $y$ axis: The graph in Desmos: https://www.desmos.com/calculator/per0tkuthz

A set of points with this property can be realized

in something of a trivial way: construct a triangle $RBY$ with sides $RB = 51/32$, $RY = 39/24$, and $BY = 1/12$. Note that these lengths satisfy the triangle inequalities. Then place all eight red points at $R$, all four blue points at $B$, and all three yellow points at $Y$. A scale drawing of this configuration is below; the points used are $(0,0)$, $(0,1/12)$, and $(\sqrt{5247935}/1536, 991/1536)$. • "The sum of lengths of all the red-blue segments must be no greater than the sum of all lengths of all the red-yellow segments and all the blue-yellow ones." This is not true. Consider 100 red points and one yellow, all in the same place and one blue segment 1 unit away. Then the lengths of red-blue segments add up to 100, while the lengths of yellow-blue and red-yellow ones add up to 1. Apr 9, 2016 at 18:47
• @BaSzAt If you consider all possible triangles (there are 100 of those, choose one of the reds, the blue and the yellow), then the sum of red-blue and yellow-blue over all triangles is the same: 100. PS: Bolo is too fast, I was about to post the same thing.
– ffao
Apr 9, 2016 at 18:51
• @ffao I think I understand now. Thanks for clarifying. Apr 9, 2016 at 18:57
• @BaSzAt Consider that if a >= b and c >= d then a+c >= b+d. This means that if you add the red-blue segments of the first 2 triangles, the result must be no greater than the result of adding the other segments. This can be extended for 3, 4, 5 etc up to all the triangles Apr 9, 2016 at 18:57

$b = 4, ~~ r = 8, ~~ y = 3$.

Reasoning:

Consider all tri-colored triangles and apply triangle inequalities. The sum of lengths of all the red-blue segments must be no greater than the sum of all lengths of all the red-yellow segments and all the blue-yellow ones.

The sum of lengths of all red-blue segments is $51y$. The sum of lengths of all red-yellow segments is $39b$. The sum of lengths of all blue-yellow segments is $r$.

Therefore, the inequalities are:
$51y \le 39b + r$
$39b \le 51y + r$
$r \le 51y + 39b$
$r + b + y = 15$

The integer solutions are $b = 4, ~~ r = 8, ~~ y = 3$ and $b = 1,~~ r = 13, ~~ y = 1$. The first solution has $b\ge2$.

• I can vouch for the correctness of the calculations, that is, those are indeed the only integer solutions to the system of inequalities and equality. That said, I still think it should be shown whether or not the diophantine solution has a geometric realisation on the plane. Apr 11, 2016 at 0:46
• @Fimpellizieri: I added a (semi-trivial) geometric realization of these points in the community wiki answer. Apr 13, 2016 at 17:10
• @MichaelSeifert Are they allowed to coincide? That's a bit of a bummer, haha. Apr 13, 2016 at 17:49
• @Fimpellizieri: Yeah, it's not terribly interesting. I suspect, however, that this solution could be continuously deformed to another one where the points aren't all degenerate. After all, we have 30 free variables (x and y coords for 15 points) and only 3 constraints, so one would expect the solution space to be 27-dimensional. Apr 13, 2016 at 17:52