From p40 of Gardner's New Mathematical Diversions:

The question arises: Can any obtuse triangle be dissected into sevein acute isosceles triangles? The answer is no. Verner E. Hoggatt, Jr., and Russ Denman (American Mathematical Monthly, November 1961, pages 912-913) proved that eight such triangles are sufficient for all obtuse triangles, and Free Jamison (ibid., June-July 1962, pages 550-552) proved that eight are also necessary. These articles can be consulted for details as to conditions under which less than eight-piece patterns are possible. A right triangle and an acute nonisosceles triangle can each be cut into nine acute isosceles triangles, and an acute isosceles triangle can be cut into four congruent acute isosceles triangles similar to the original. So, can someone scan in (the picture(s) from) that '61 article? Or just sketch the solution? Amazingly, Verner Hoggatt co-founded the Fibonacci Quarterly (with Brother Alfred Brousseau), and published on reciprocal Fibonacci sums, i.e., the Lambert series of another current math-fun thread. --rwg On Tue, Feb 21, 2012 at 6:18 PM, Bill Gosper <billgosper@gmail.com> wrote:

Ah, thank you both. But alas, the problem has all along been to do it with acute *isosceles* triangles. I believe Neil has a different Gardner book claiming the answer is still nine, but no picture. I'm beginning to wonder if it's an error. --rwg

Stuart>

I am advised (but cant verify);

It's a little further along (page 258) in the answers section. ;)

Hans>

http://chesswanks.com/txt/Acute&IsoscelesTriangles.pdf