Hello, Natasha!

A fascinating problem . . . but I don't understand your answers.

Are you sure you understand "irregular"?

A

*regular* tetrahedron has all four faces congruent.

. . It is composed of four equilateral triangles.

And these are not counted.

Using 4 triangles of the same shape, you may

*or may not* get a tetrahedron.

Here are the four shapes

Code:

` *`

* * * _

* * * * √2

* * 1 * *

* E1 * * R *

* * * * * * * * * * *

1 1

* *

* *

* * _ * * _

√2 * * √2

* I * * E2 *

* *

* * * * * * * *_* * *

1 √2

I've named the first and last triangles "E" for Equilateral.

The second is "R" for Right triangle.

The third is "I" for Isosceles.

The best approach is to make cardboard triangles and play with them.

. . I have done this . . . I suggest you do the same.

Here's what I've come up with:

two E1's and two R's

one E1 and three I's

one E1, two R's and one I

one R, two I's and one E2

three R's and one E2

two R's and two I's

two I's and 2 E2's

four I's

I hope I didn't miss any . . .

And

*please* don't ask me to draw the tetrahedrons!