*Originally posted by CoolPlayer*

** I have found a way . I can do it with 10 AAT's.
**

Let ABCD be the square with side-length x with BC the base (BC=x) . Join BD(diagonal). On the diagonal BD, take apoint P such that BP=3x/4. Join AP & CP. Dr ...[text shortened]... r 10 is the required minimum number or it can be improved !

As mentioned in my earlier post, I too have found a method of dividing a square into 10 AAT's but my method is different. It is like this ;

Let ABCD be the given square . With BC as the base, construct an isosceles triangle PBC in which , each of the angles on the base is 36 degrees i.e. /_PBC = /_PCB = 36 deg. ( The point P lies inside the square.)

Hence vertex angle /_BPC=108 deg.

Now construct a regular equi-pentagon PQRST ,such that Q lies on PB, T lies on PC, base of pentagon RS lies on BC. ( This is possible because each internal angle of a regular pentagon is equal to 108 deg. )

Let O be the centre of this regular pentagon. Join OP, OQ, OR, OS, OT, PA and PD.

Now the 10 triangles PAB, PAD, PDC, BQR, CTS, OPQ, OQR, ORS, OST and OTP are all acute angled triangles.