Question: How can I get coordinates of triangle with integer sides and three integral medians?

In the book "Challenges in Geometry" of the author Christopher J. Bradley at p. 32, the triangle with three sides a := 136, b := 170, c := 174 has three medians ma := 158, mb := 131, mc := 127. I checked

restart:

a:=2*68;

b:=2*85;

c:=2*87;

ma:=sqrt((b^2+c^2)/2-a^2/4);

mb:=sqrt((a^2+c^2)/2-b^2/4);

mc:=sqrt((b^2+a^2)/2-c^2/4);

Now I want to find coordinates of vertices of a triangle like that (in plane). I tried

restart;
DirectSearch:-SolveEquations([(x2-x1)^2+(y2-y1)^2 = 136^2,
(x3-x2)^2+(y3-y2)^2 = 170^2, (x3-x1)^2+(y3-y1)^2 = 174^2], {abs(x1) <= 30, abs(x2) <= 30, abs(y1) <= 30, abs(y2) <= 30, abs(x3) <= 30, abs(y3) <= 30}, assume = integer, AllSolutions, solutions = 5);

but my computer runs too long. I think, there is not a triangle with integer coordiantes. 

How can I get  a triangle  with coordinates of vertices are rational numbers?