1983 AIME Problem 4

Question

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is

\sqrt{50}

cm, the length of

AB

is

6

cm, and that of

BC

is

2

cm. The angle

ABC

is a right angle. Find the square of the distance (in centimeters) from

B

to the center of the circle.

Answer

Put

B=(0,0),

A=(0,6),

and

C=(2,0)

The center lies on the perpendicular bisector of

AC

at distance

\sqrt{40}

from its midpoint

Put

B=(0,0),

A=(0,6),

and

C=(2,0).

The midpoint of

AC

is

M=(1,3),

and

AC=\sqrt{40}.

If

O

is the center, then

\begin{aligned} OM&=\sqrt{OA^2-AM^2}\\ &=\sqrt{50-10}=\sqrt{40}. \end{aligned}

A vector perpendicular to

AC=(2,-6)

is

(6,2),

which already has length

\sqrt{40}.

Thus the two possible centers are

\begin{aligned} M+(6,2)&=(7,5),\\ M-(6,2)&=(-5,1). \end{aligned}

The pictured notched circle has its center on the side opposite the notch, so

O=(-5,1).

Therefore

BO^2=(-5)^2+1^2=26.

Problem 4 in other years