Using straight lines, dissect an isosceles triangle with sides 42, 42, and 21, into six smaller triangles, all similar to the original but with no two of equal sizes.

Good problem!

Equivalently, demonstrate that a large triangle of side (2x),(2x),(x) can be built from 6 smaller similar ones:

Erect the first triangle with the long sides upwards, construct a copy twice the size on the right-hand long side, then the third, again twice the size, on the upper side of the second, the fourth, again twice the size, on the right-hand long side of that, and the fifth, again twice the size, on the upper side of the fourth. It can easily be shown that the upper sides of the second and fourth triangles are parallel; hence the upper long side of the last triangle fits snugly along the collinear lower long sides of the second and fourth triangles.

The particular significance of the 42,42, 21 dimensions is that this results in all triangles having integral sides, with the smallest being 2,2,1; however,the same construction seems to work for any even number, say n, greater than 6, as well, since the pattern of the first 5 triangles can be repeated indefinitely.

Addendum: it does work. The base length of the largest upright triangle is 2^(n-2)x, while the base length of the triangle along the sides of the second , fourth etc triangles is 1/6(2^n-4)x, so the two cannot be the same.

Using these lengths:

2*2^(n-2)x+1/6(2^n-4)x=2/3(2^n-1); and

1/3(2^n-1)x=1/3(2^n-4)x+1;

which kind of explains why it 'works'.

Edited on February 15, 2012, 2:57 pm

Posted by broll on 2012-02-15 00:17:20