In and out

The hypotenuse of a right triangle circumscribed in a circle will be a diameter of that circle, and so the hypotenuse will have length $2R.$ The midpoint of the hypotenuse corresponds to the center of the circumscribing circle. Since the given triangle is isosceles, the center of the inscribed circle will then be a perpendicular distance $r$ from the center of the circumscribing circle.

We can thus form a right triangle with the vertices being the centers of the two circles in question along with one of the endpoints of the hypotenuse. The legs of this right triangle will then have lengths $r$ and $R,$ with the angle opposite the side length $r$ being $22.5^{\circ},$ since the hypotenuse of this triangle will be the bisector of one of the non-right interior angles of the original right isosceles triangle.

Thus $\dfrac{R}{r} = \cot(22.5^{\circ}).$ Now by the appropriate double-angle formula we know that

$1 = \tan(45^{\circ}) = \dfrac{2\tan(22.5^{\circ})}{1 - \tan^{2}(22.5^{\circ})} \Longrightarrow \tan^{2}(22.5^{\circ}) + 2\tan(22.5^{\circ}) - 1 = 0$

$\Longrightarrow \tan(22.5^{\circ}) = \dfrac{-2 + \sqrt{4 + 4}}{2} = -1 + \sqrt{2},$

where we took the positive roots since $\tan(22.5^{\circ}) \gt 0.$ We then have that

$\dfrac{R}{r} = \dfrac{1}{\tan(22.5^{\circ})} = \dfrac{1}{\sqrt{2} - 1} * \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} = \sqrt{2} + 1 = \boxed{2.414}$ to 3 decimal places.

The formula for the radius of the incircle: r=A/s, where A is the area of the triangle and s is the semiperimeter ( (a+b+c)/2 , where a, b and c are the side lengths.

The formula for the radius of the circumcircle:

R=abc/4A

Therefore: R/r=abcs/4A^2

If we take the right isosceles triangle with the length of the catheti (shorter sides) of 1 unit (the ratio of R:r won't change if the side length of the catheti is x, due to similarity), then we get a=b=1, c=√2, A=0.5 , abc=√2 , s=(2+√2)/2 .

After substituting these values into the above formula for R/r and simplifying, we get R/r = √2+1 = 2.4142 (to 4 decimal places).