- Find the smallest square which can be divided into unique (non-identical)
right triangles, such that all sides of the triangles are integers.

- Find the smallest square which can be divided into non-similar right
triangles, such that all sides of the triangles are integers.

- Find the smallest square which can be divided into 8 non-similar right triangles, such that all sides of the triangles are integers.

Solutions were received from Denis Borris and Philippe Fondanaiche. Philippe's solutions follow.

**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

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

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

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...

We use again the same approach as in Q1 and Q2 and we get a square of side equal to

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....

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