20.64 A circle with radius $r$ is inscribed in a right isosceles triangle, and a circle with radius $R$ is circumscribed around this triangle. Then $\frac{R}{r}$ equals
(A) $1+\sqrt{2}$.
(B) $\frac{2+\sqrt{2}}{2}$.
(C) $\frac{\sqrt{2}-1}{2}$.
(D) $\frac{1+\sqrt{2}}{2}$.
(E) $2(2-\sqrt{2})$.
(25th American High School Mathematics Examination, 1974)
[Solution] As shown in the figure, $\triangle ABC$ is an isosceles right triangle, where $\angle BAC = 90^{\circ}$, $AB = AC$. It is inscribed in a circle with center $O$ and radius $R$.

The segment $AO$ is of length $R$, and it bisects segment $BC$ and $\angle BAC$. Another circle with center $O'$ (located on $AO$) and radius $r$ is inscribed in $\triangle ABC$, and the points of tangency with $AB$ and $AC$ are $T$ and $T'$, respectively.
$\because \triangle ATO$ has angles of $45^{\circ}-45^{\circ}-90^{\circ}$,
and $O'T = r$, $AT = r$, so $O'A = r \sqrt{2}$.
$\therefore R = r + r \sqrt{2}$, which means $\frac{R}{r} = 1 + \sqrt{2}$.
Therefore, the answer is $(A)$.