1970 AMC 12 Problem 12

Question

A circle with radius

r

is tangent to sides

AB,

AD,

and

CD

of rectangle

ABCD

and passes through the midpoint of diagonal

AC.

The area of the rectangle, in terms of

r,

is:

4r^2

6r^2

8r^2

12r^2

20r^2

Tangency to the two opposite sides makes one side of the rectangle equal to

2r

The line from the third tangency point through the center bisects the diagonal, making the corresponding chord a diameter

Let the circle be tangent to the parallel sides

AB

and

CD.

Their separation is its diameter, so

AD=BC=2r.

Let

R

be the tangency point on

AD,

let

Q

be the center, and let

M

be the midpoint of

AC.

The line

RQ

is parallel to

AB

and halfway between

AB

and

CD,

so

R

is the midpoint of

AD

and the line contains

M.

Since

M

lies on the circle and

R,Q,M

are collinear,

RM=2r.

Thus

RM

is a midsegment of triangle

ADC,

so

DC=2RM=4r.

Therefore the area is

(2r)(4r)=8r^2.

Therefore, the correct answer is C.

Problem 12 in other years