1979 AMC 12 Problem 28

Question

Circles with centers

A,

B,

and

C

each have radius

r,

where

1\lt r\lt2.

The distance between each pair of centers is

2.

If

B'

is the point of intersection of circle

A

and circle

C

which is outside circle

B,

and if

C'

is the point of intersection of circle

A

and circle

B

which is outside circle

C,

then length

B'C'

equals

3r-2

r^2

r+\sqrt{3(r-1)}

1+\sqrt{3(r^2-1)}

none of these

Introduce the analogous point

A'

outside circle

A

; symmetry makes

A'B'C'

equilateral

Use the common centroid and the altitude

\sqrt{r^2-1}

of an isosceles triangle with sides

r,

r,

and

2

Let

A'

be the analogous outer intersection of the circles centered at

B

and

C.

Then

A'B'C'

and

ABC

are concentric equilateral triangles. If

K

is their common centroid and

M

is the midpoint of

BC,

then

\begin{aligned} A'M&=\sqrt{r^2-1},\\ MK&=\frac{\sqrt3}{3},\\ AK&=\frac{2\sqrt3}{3}. \end{aligned}

Similarity gives

\begin{aligned} \frac{B'C'}2&=\frac{A'K}{AK}\\ &=\frac{\sqrt{r^2-1}+\frac{\sqrt3}{3}} {\frac{2\sqrt3}{3}}. \end{aligned}

Hence

B'C'=1+\sqrt{3(r^2-1)}.

Therefore, the correct answer is D.

Problem 28 in other years