Philippe Fondanaiche writes:
Let ABC the triangle with BC = a, CA = b and AB = c Let P,Q,R the midpoints of BC, CA and AB with AP = p, BQ = q and CR = r. It is known that the medians can be expressed from the sides of the triangles with the following identities: 4p^2 = 2*b^2 + 2c^2 - a^2 4q^2 = 2c^2 + 2a^2 - b^2 4r^2 = 2a^2 + 2b^2 - c^2 We can consider WLOG a<=b<=c. We have three diophantine equations with c < a+b. A very simple program written in BASIC gives the first solution a=136, b=170, c=174 ==> p=158, q=131 and r=127
Additional solutions are:
a | b | c | p | q | r |
226 | 486 | 580 | 523 | 367 | 244 |
254 | 262 | 316 | 261 | 255 | 204 |
290 | 414 | 656 | 529 | 463 | 142 |
318 | 628 | 650 | 619 | 404 | 377 |
466 | 510 | 884 | 683 | 659 | 208 |
554 | 892 | 954 | 881 | 640 | 569 |
654 | 772 | 818 | 725 | 632 | 587 |
Euler gave the following (non-exhaustive) parametric solution:
a = 18m5 − 18m4n + 52m3n2 + 12m2n3 + 2mn4 − 2n5
b = 18m5 + 18m4n + 52m3n2 − 12m2n3 + 2mn4 + 2n5
c = 36m5 − 40m3n2 − 12mn4
p = 27m5 + 9m4n − 18m3n2 + 26m2n3 + 3mn4 + n5
q = 27m5 − 9m4n − 18m3n2 − 26m2n3 + 3mn4 − n5
r = 54m4n
+ 20m2n3
− 2n5