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?