Right-Angled Isosceles Triangle covering puzzle Consider a RAIT (right-angled isosceles triangle), from which we remove a RAIT smaller than half its area by a cut perpendicular to the hypotenuse, like this:

How many RAITs are required to cover the white part at the right?
The covering RAITs may overlap each other, but they may NOT overlap the black RAIT at the left, and they MUST be contained within the white part.
I managed to find a solution with 3 RAITs:

This solution works when the black RAIT at the left is sufficiently small. If it is larger, then there is a different solution with 3 RAITs:

Can you find a solution for these two cases which requires only 2 RAITs?
 A: The figure in question has adjacent angles of $90-135-90$ degrees.  Two adjacent angles of a right-angled isoceles triangle have $135$ degrees total.  If you put an acute angle of one triangle adjacent to the right angle of another, you're still $45$ degrees short on one angle.  Therefore, there can't be any solutions with $2$ triangles.
A: Suppose that the biggest RAIT has three edges whose lengths are $1,1,\sqrt 2$ and that the black RAIT has three edges whose lengths are $a/\sqrt 2,a/\sqrt 2,a$ with $0\lt a\lt 1$.
Then, I've got the following two :


*

*If $a\ge 1/2$, then your second solution works because 
$$\frac{a}{\sqrt 2}+\frac{a}{\sqrt 2}+\frac{1}{\sqrt 2}\ge \sqrt 2\iff a\ge \frac 12.$$

*If $a\le 2/3$, then your first solution works because
$$(1-a)+\left(1-\frac a2\right)\ge 1\iff a\le\frac 23.$$
These imply that you need at most three RAITs for any size of the black RAIT.
In the following, let us prove that there is no solution with $2$ RAITs.
Since the 'white part at the right' you mention is a quadrilateral, the sum of the inner angles is $360^\circ=180^\circ\times 2$. Hence, if there is a solution with $2$ RAITs, then the two RAITs cannot overlap each other. 
By the way, one of the RAITs has to exist at the right-most corner as the following pictures (red ones or blue ones), but in either case, the left part (white part) cannot be in the form of a RAIT (because the black RAIT is smaller than the half of the biggest RAIT). Hence, there is no solution with $2$ RAITs.


As a result, we can see that the minimum number of the RAITs one need is $3$ for any size of the black RAIT.
A: In addition to the previous answers, I found proof that a covering with 2 RAITs does not exist. Consider the three segments in the picture below:

Any covering of the white part must cover all these three segments. Now:


*

*The red and the green cannot be covered by a single RAIT because the angle between them is 135;

*The green and the blue cannot be covered by a single RAIT because the angle between them is 90 and the only triangle having a 90-degree angle at the top necessarily breaks into the black area.

*The blue and the red cannot be covered by a single RAIT from similar considerations.


Hence we need at least 3 RAITs - one per segment.
