Among the right-angled triangles with equal perimeter, determine the one into which the largest incircle can be drawn. What is the radius of the circle if the perimeter of the triangle is $k$?
If the radius of the incircle is $r$, then, as is known,

$$
r=\frac{2 t}{k}
$$

Since $k$ is a constant, $r$ will be the largest when $t$, the area of the triangle, is the largest. We know that among right-angled triangles with equal perimeters, the isosceles one has the largest area (Math. Practice Book, I. 449.) and thus

$$
2 t=a^{2}
$$

but then

$$
k=2 a+a \sqrt{2}=a(2+\sqrt{2})
$$

from which

$$
a=\frac{k}{2+\sqrt{2}}=\frac{k}{2}(2-\sqrt{2})
$$

Substituting (3) into (2) and this into (1), we get:

$$
r=\frac{a^{2}}{k}=\frac{k}{4}(6-4 \sqrt{2})=\frac{k}{2}(3-2 \sqrt{2})
$$