1974 AMC 12 Problem 16

Question

A circle of radius

r

is inscribed in a right isosceles triangle, and a circle of radius

R

is circumscribed about the triangle. Then

\frac{R}{r}

equals

1+\sqrt2

\frac{2+\sqrt2}{2}

\frac{\sqrt2-1}{2}

\frac{1+\sqrt2}{2}

2(2-\sqrt2)

Let each leg have length

a

and express both radii in terms of

a

For a right triangle, the circumradius is half the hypotenuse; the inradius is

\frac{a+a-a\sqrt2}{2}

With both legs of length

a

and hypotenuse

a\sqrt2,

\begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2}. \end{aligned}

Therefore

\begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2. \end{aligned}

Therefore, the correct answer is A.

Problem 16 in other years