# A triangle with its Fermat point and six integer lengths.

Do there exist three positive integers $a_1$,$a_2$ and $a_3$ such that for every $i\neq j$, $a_i^2 +a_ia_j+a_j^2$ are perfect squares.

In other words do there exist an acute triangle $ABC$ in the Euclidean plane with Fermat point $P$ such that segments $AB$, $AC$, $BC$, $AP$, $BP$ and $CP$ all have integer lengths.

• I think need to write more clearly. For example. What system of equations must be solved. You write and will continue to think about the solution. – individ Oct 11 '14 at 18:00
• Project Euler 143 gives an example of a Torricelli triangle, but I had answered this particular question before I saw it. – Henry Oct 12 '14 at 18:57

Yes.

Looking for primitive examples with all three initial values below $10^5$, I came across these

      a       b       c     f(a,b)  f(c,a)  f(b,c)  f triangle type
----    ----    ----    -----   -----   -----   ---------------
1    195     264     325      399     455     511   acute
2    264     325     440      511     616     665   acute
3    384     805    1520     1051    1744    2045   slightly obtuse
4    455    1824    2145     2089    2405    3441   slightly obtuse
5    360    1015    3864     1235    4056    4459   slightly obtuse
6    435    1656    4669     1911    4901    5681   slightly obtuse
7   1272    2065    4928     2917    5672    6223   acute
8    765    1064    5016     1591    5439    5624   acute
9   6307    6765    7208    11323   11713   12103   acute
10   520    3105    8184     3395    8456   10101   slightly obtuse
11  6120    8512    9405    12728   13545   15523   acute


where the $f(x,y)=\sqrt{x^2+xy+y^2}$ are sides of the triangle, and $a,b,c$ are distances from the vertices to a Fermat point.

The five "slightly obtuse" cases are where the largest angle of the triangle is below $\frac23 \pi$ or $120^\circ$, so the Fermat point is not a vertex.

There is a coincidence between cases $1$ and $2$ involving $264$ and $325$.