How Much Can We Squeeze In?

Remember that the radius of the largest inscribable circle in a triangle is equal to the area of the triangle divided by the semi-perimeter. That is $\frac{40 \times 40}{2}$ divided by $\frac{40+40+40\sqrt{2}}{2}$ = $\frac{800}{40+20\sqrt{2}}$ = approx. 11.716. But this is the diameter so we have to multiply it by 2 which yields 23.431 which we round down to $\boxed{23}$ .

I have $(1+\sqrt{2})r=\frac{40}{\sqrt{2}}$ so $r=20(2-\sqrt{2})$ . 23 seems to me to be the diameter.

In a right triangle of legs a & b, the in-radius is given by: ((a+b) -√(a²+b²))/2. So in this case, the in-radius=40 - 20√2 ~11.7157 or the diameter ~ 23.4314, I mistakenly understood the problem as the largest circle that can be fitted into the space left after the in-circle is drawn which I calculated as radius ~ 5.23064. Not having found a suitable alternative, I realized that the question was about finding the diameter of the in-circle of the right isosceles triangle.

23.43 but since you want integer value its 23.