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We know that, there exist integer triangles with three rational medians. The smallest has sides ($68, 85, 87$). Others include ($127, 131, 158$), ($113, 243, 290$), ($145, 207, 328$) and ($327, 386, 409$). For example, How to find coordinates of vertices of an integer triangle has sides ($68, 85, 87$)? Please help me to get them. I don't know how to start?

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I don't see what rational medians has to do with it. If there is a triangle with sides $a,b,c$ (integer or not), you can put one vertex at $(0,0)$, a second one at $(a,0)$, and then the third at $(x,y)$, and solve the equations $x^2+y^2=b^2$, $(x-a)^2+y^2=c^2$.

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Myyerson Thank you very much. – minthao_2011 Nov 5 '12 at 2:23
I see at about The smallest has sides (68,85,87). Others include (127,131,158), (113,243,290), (145,207,328) and (327,386,409). – minthao_2011 Nov 5 '12 at 8:26
Yes, that's already in the statement of the question. – Gerry Myerson Nov 5 '12 at 11:33

You can just place two vertices, say one at $(0,0)$ and one at $(0,68)$. Then if the other one is at $(x,y)$ we know $x^2+y^2=85^2, (x-68)^2+y^2=87^2$ You can expand the square and subtract the equations to get $-2\cdot 68 x+68^2=87^2-85^2$, which is a linear equation.

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Thank you very much. – minthao_2011 Nov 5 '12 at 2:24
When I solve $x^2+y^2=85^2, (x-68)^2+y^2=87^2$, I have $x = 535/17$ and $y = 30 \sqrt{2002}/17$. But the triangle with vertices $(0,0)$, $(0,68)$ and $(535/17,30\sqrt{2002}/17)$ have not sides 68, 85, 87. You – minthao_2011 Feb 1 '13 at 14:50
@minthao_2011: I agree with your solution, but I get the proper sides – Ross Millikan Feb 1 '13 at 15:43

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