Red, blue and yellow points

Question

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 $y$.

Answer 1

The updated answer is:

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

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 $(\sqrt{5247935}/1536, 991/1536)$.

Answer 2

The answer is:

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

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 $b\ge2$.