*Spoiler for New Scientist Enigma 1758: “Path-o-logical” *(Follow the link to see the puzzle.)

If* a* is a north-south side of the park and *b* is the length of a path then *a* and *b* are a side and hypotenuse of a right triangle. Call its other side *x*. Likewise *b* and *c*, where *c* is an east-west side of the park, must be a side and hypotenuse of another right triangle. Now, if you draw a perpendicular to the south side of the park through the path intersection, then the second of these triangles is split into two. All these triangles are similar and in fact the one in the southeast corner is congruent to the one in the northwest, so that perpendicular line segment must also be *x *and *c*, not *a*, is the long side of the park: *c* = *a* + 25.* *Then the southwest triangle has sides *x* and 25. Call its hypotenuse (the distance from the southwest corner to the path intersection) *y*.

Summarizing, the sides of these three similar triangles are:

*a : x : b**x :*25*: y**b : y : a*+25

*a* and *b* must be integers (but nothing says *x* or *y* must be).

From similarity of the first two triangles we get *x*^{2} = 25*a*. From the Pythagorean Theorem applied to the first triangle, *b*^{2} – *a*^{2} = *x*^{2} = 25*a*. If we write *b* = *a* + δ then 2δ+δ^{2}/*a* = 25. δ then must be an integer in the range [1, 12] and of these only 12 and 10 give integer values for *a* in that equation, namely 144 and 20, with corresponding values for *b* of 156 and 30.

Then *x* = 60 or 10√5 respectively. Using the Pythagorean Theorem for the second triangle, *y* = 65 or 15√5 ~ 33.54 respectively. But for the path intersection to be inside the park we need *y* < *b*. So the only solution that works is: *a* = 144, *b* = 156: