Solution to Problem #10

Philippe Fondanaiche of Paris (France), Dan Dima of Constanta (Romania), Larry Chickola of Yardley PA, Kirk Bresniker of Roseville CA, Ignacio Larrosa of Caņestro A Coruņa (Spain), David Snook of Port Coquitlam (British Columbia), Al Zimmermann of New York NY, and Jean Jacquelin of France solved the problem.

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 + 3mn4n5

r = 54m4n + 20m2n3 − 2n5


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