*Spoiler for New Scientist Enigma 1704:*

Joe drew a right-angled triangle on an A6 file card. He asked Penny to cut it out and then cut it in two to make two right-angled triangles. Then he asked her to cut each triangle in two to make four right-angled triangles in total. Now Joe’s triangle was very special. Penny found that the lengths of all the sides of all the triangles were a whole number of millimetres.

What was the area of the smallest triangle?

*[Edited for clarity; added diagrams]*

I can’t believe how long I messed around with this before seeing how ~~obvious~~ easy it is. When you cut a right triangle into two smaller right triangles (by cutting along the perpendicular to the hypotenuse that goes through the right vertex), both new triangles are similar to the first.

Cut each of the two smaller right triangles into still smaller right triangles and you get four more triangles also similar to the first. In fact two of these four triangles are congruent. So you have a triangle with sides *a*:*b*:*c*, two with sides *d*:*a*:*e*, and one with sides *f*:*d*:*g*, all similar:

The two congruent triangles make a rectangle of size *a* by *d*, where *a*<*d*. Adjacent to the short side is the smallest triangle, *a:b:c*. Then the two triangles *d:a:e* are larger by the ratio *a/b,* and the largest triangle *f:d:g* is larger than that by another factor of *a/b*.

From this we get *d* = *a* (*a*/*b*), *e* = *c* (*a*/*b*), *f* = *a*(*a*/*b*)^{2}, and *g* = *c*(*a*/*b*)^{2}. Now if *a:b:c* are a Pythagorean triple then (multiplying by *b*^{2}) so are *ab*^{2}:*b*^{3}:*cb*^{2} and *db*^{2}:*ab*^{2}:*eb*^{2} = *a*^{2}*b*:*ab*^{2}:*cab* and *fb*^{2}:*db*^{2}:*gb*^{2} = *a*^{3}:*a*^{2}*b*:*ca*^{2}. The lengths of the sides of the original triangle are *b*^{3}+*db*^{2} = *b*^{3}+*a*^{2}*b* and *ab*^{2}+*fb*^{2} = *ab*^{2}+*a*^{3}, and the hypotenuse is *cb*^{2} + *gb*^{2} = *cb*^{2} + *ca*^{2} = *c*^{3}.

If *a:b:c* is the smallest Pythagorean triple, 4:3:5 (recall *a>b*), the sides are 75 and 100 and the hypotenuse is 125. An A6 card is 105 mm by 148 mm, so this triangle fits. The next smallest Pythagorean triple (that is, next smallest value of *c*) is 8:6:10 which gives a hypotenuse of 1000 mm, much too large. So there is only one solution, and the smallest triangle has area *ab*^{5}/2 = 486 square mm.

Thanks for posting this, I’d got that diagram myself and the ratios of the sides of the triangles but then I tied myself in knots writing a program to test given a pythag triple (a,b,c) are all the sides of the triangles e.g. a^3/b^2 for your f. I didn’t think about just rescaling (a,b,c) by b^2 to get rid of fractions! Thank you – Chris

Sorry, correction that should have read: “are all the sides of the triangles (e.g. a^3/b^2 for your f) integers”