We will try and get a Diophantine equation you can factor. Let \(x\) and \(y\) be the 2 shorter sides of the right angle triangle.

\[x+y+\sqrt{x^2+y^2}=60\]

\[60-x-y=\sqrt{x^2+y^2}\]

\[3600+x^2+y^2-120x-120y+2xy=x^2+y^2\]

\[xy-60x-60y+1800=0\]

\[(x-60)(y-60)=1800\]

We also have \(x, y < 60\) since the perimeter must be 60. So, both brackets must be negative. Also, \(x, y > 0\) since our triangle is non-degenerate. This implies that \(-60<x-60<0, -60<y-60<0\). We can use this to remove certain factors of 1800.

Now we list all factors of 1800 which its complement that also satisfy the above ranges.

\[1800 = (-36, -50), (-40, -45)\]

From this, we get \((x, y)\) values of \((24, 10)\) and \((20, 15)\). Thus, the only triangles which satisfy are the 10-24-26 triangle and the 15-20-25 triangle.

Note: I didn't bother with solutions \((10, 24)\) and \((15, 20)\) since they gave the same triangles.

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TopNewestWe will try and get a Diophantine equation you can factor. Let \(x\) and \(y\) be the 2 shorter sides of the right angle triangle.

\[x+y+\sqrt{x^2+y^2}=60\]

\[60-x-y=\sqrt{x^2+y^2}\]

\[3600+x^2+y^2-120x-120y+2xy=x^2+y^2\]

\[xy-60x-60y+1800=0\]

\[(x-60)(y-60)=1800\]

We also have \(x, y < 60\) since the perimeter must be 60. So, both brackets must be negative. Also, \(x, y > 0\) since our triangle is non-degenerate. This implies that \(-60<x-60<0, -60<y-60<0\). We can use this to remove certain factors of 1800.

Now we list all factors of 1800 which its complement that also satisfy the above ranges.

\[1800 = (-36, -50), (-40, -45)\]

From this, we get \((x, y)\) values of \((24, 10)\) and \((20, 15)\). Thus, the only triangles which satisfy are the 10-24-26 triangle and the 15-20-25 triangle.

Note: I didn't bother with solutions \((10, 24)\) and \((15, 20)\) since they gave the same triangles.

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