Question 1
Let a square ABCD.
The first natural method to divide this square in right triangles is to try to
find:
1) a point E on AB such that ADE and BCE are right triangles and the triangle
CED is split in two right
triangles with the height drawn from C (or from D). Let AB=n and AE=a with a
and n integers a<>n/2.
We have to find a and n such that a^2+n^2 is a square and (n-a)^2+n^2 is also a
square.
There is no solution for n<5000 and probably whatever n.
or
2) a point M within the square such that MA,MB,MC and MD are integers. If this
point exists, it is easy
to divide the square in 4 pairs of identical right triangles and to subdivide
one triangle of each pair into two
unique right triangles.
Such a point M does not exist....
So we have considered many other approaches. Only one of them provides a
significant number of possible
solutions and consists of finding 4 points E,F,G,H respectively on AB,BC,CD and
DE such that the triangles
AEH, BEF, CFG and DGH are Pythagorean and the diagonal EG or HF is an integer
(see the diagram hereafter)
With a square of side equal to 468, we obtain 8 eight right triangles
with the following measures:
The sides of the right triangles are a1<a2<a3.
If the ratios a1/a2 are the same, the corresponding triangles are similar.
triangles
a1
a2
a3
a1/a2
AEH
78
104
130
0,75
BEF
208
390
442
0,5333..
CFG
195
260
325
0,6
DGH
273
364
455
0,75
EFI
280
342
442
0,81871...
FGI
165
280
325
0,589285...
EHJ
66
112
130
0,589285...
GHJ
112
441
455
0,253968..
Question 2
With
the same approach as above but with the constraint that all the ratios a1/a2
are different, we get
a square of length n=3060 which can be divided into 7 right triangles as
EGF is already a right triangle.
triangles
a1
a2
a3
a1/a2
AEH
153
680
697
0,225
BEF
2380
2907 3757
0,81871..
CFG
680
1275
1445
0,5333..
DGH
1785
2380
2975
0,75
EHI
455
528
697
0,861742...
EFG
1445
3468
3757
0,41666..
GHI
455
2940
2975
0,154761..
We can take the opportunity to look for a square divided into a minimum of
right triangles.
In this case, the pattern is defined hereafter with a point E on AB and a point
F on AD. We succeed
to divide the square into only 5 right triangles. The side of the square is n=9000.
triangles
a1
a2
a3
a1/a2
AEF
3400
6375 7225
0,53333...
BEC
5600
9000 10600
0,62222...
CDF
2625
9000
9375
0,291666..
EFG
1479
7072 7225
0,2091346...
CFG
7072
7896 10600
0,895643...
Question 3
We use again the same approach as in Q1 and Q2 and we get a square of side
equal to 3276
divided into 8 non-similar right triangles with the following diagram and
measures:
triangles
a1
a2
a3
a1/a2
AEH
364
1248 1300
0,291666..
BEF
1716
2912 3380
0,589285..
CFG
1547
1560 2197
0,991666...
DGH
1729
2028 2665
0,852564....
EHJ
816
1012 1300
0,806324...
EFI
832
3276 3380
0,253968...
EGI
1365
3276 3549
0,416666...
GHJ
816
2537
2665 0,321639....