- 09 Jan '05 22:45

I think I know what the answer is, I'll pm you my solution to give the others a chance*Originally posted by ranjan sinha***You have a square figure. The puzzle is to think of cutting it into Acute Angled Triangles( AAT's) by using straight line cuts. The problem is , what is the MINIMUM number of AAT's , into which you can cut the square ; And indicate how will you do it .**

- 10 Jan '05 08:09

Is cutting and sticking or moving or removing the dissected parts permitted ?*Originally posted by ranjan sinha***You have a square figure. The puzzle is to think of cutting it into Acute Angled Triangles( AAT's) by using straight line cuts. The problem is , what is the MINIMUM number of AAT's , into which you can cut the square ; And indicate how will you do it .** - 11 Jan '05 13:33

By 'straight line cuts', do you mean cuts that go right through the square? In this case, I don't think it's possible. However, I suspect it is possible to assemble a square out of AATs, but I can't think of a formal argument in that case.*Originally posted by ranjan sinha***You have a square figure. The puzzle is to think of cutting it into Acute Angled Triangles( AAT's) by using straight line cuts. The problem is , what is the MINIMUM number of AAT's , into which you can cut the square ; And indicate how will you do it .** - 12 Jan '05 08:07 / 3 edits

I have found a way . I can do it with 10 AAT's.*Originally posted by Jay Peatea***Hopefully I've got the right answer this time PM on its way**

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. Draw isosceles triangles APQ & CPR such that Q lies on AD with AQ=AP and R lies on CD with CR=CP.

Let QS and RS be the internal bisectors of angles /_PQD & /_PRD respectively. Obviously , due to symmetry the point S lies on the diagonal BD.

Now with S as vertex draw an equilateral triangle STU such that T lies on PQ, U lies on PR, and TU is perpendicular to the diagonal BD.

Join SQ, SR.

Thus, the 10 triangles namely PBC, PBA, PQA, PRC, PTU, STU, STQ, SQD, SDR and SRU are all acute angled triangles.

I don't know whether 10 is the required minimum number or it can be improved ! - 12 Jan '05 08:25

No, The straight line cuts may not go right across the square.*Originally posted by Acolyte***By 'straight line cuts', do you mean cuts that go right through the square? In this case, I don't think it's possible. However, I suspect it is possible to assemble a square out of AATs, but I can't think of a formal argument in that case.** - 12 Jan '05 10:15 / 1 edit

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 ;*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 !

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. - 15 Jan '05 12:09 / 1 edit

But 10 is NOT the minimum number.*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 !

It is possible to do it with less than 10 AAT's.

It can be improved. - 28 Jan '05 10:41 / 1 edit

I was thinking thus far that the minimum number of AAT's is 9*Originally posted by ranjan sinha***But 10 is NOT the minimum number.**

It is possible to do it with less than 10 AAT's.

It can be improved.

But Jay Peatea has shown that it can be done with 8 AAT's.

Excellent Jay Peatea, your method is really the work of a genius.

Hey, Jay Peatea, why don't you post your detailed solution?

I think 8 is 'the' minimum number. Any takers , who can further improve it ? - 28 Jan '05 21:58 / 1 edit

Wow praise indeed!*Originally posted by ranjan sinha***I was thinking thus far that the minimum number of AAT's is 9**

But Jay Peatea has shown that it can be done with 8 AAT's.

Excellent Jay Peatea, your method is really the work of a genius.

...[text shortened]... minimum number. Any takers , who can further improve it ?

But it must be said that I wouldn't have done it without your 9 aats solution, with the pentagon in the corner. I went down that route myself after seeing the 10 aat solution posted earlier, but I couldn't figure how to do it. It was only after I accurately drew your solution (and checked it with a protractor) that I was inspired to try again but this time much more accurately. What amazes' me is that the 8 aat solution is quite simple, and none of us thought of it before.

Anyway here it is:-

Take the square from your answer including the points M,N. Then draw an imaginary line parallel to the base which crosses line MN at a distance of 1.5/8 of MN. (The imaginary line should be closer to the side AD). Draw two isosceles triangles from corners AMY & DMZ. The third corner of each triangle (i.e Y & Z)should fall on the imaginary line. Now draw the following lines YZ, YB, YM, ZM & ZC

that should give you 8 aats.

Maybe it can be beaten ?