1972 AMC 12 Problem 32

Question

Chords

AB

and

CD

in the circle shown intersect at

E

and are perpendicular to each other. If segments

AE,

EB,

and

ED

have measures

2,

6,

and

3

respectively, then the length of the diameter of the circle is:

4\sqrt5

\sqrt{65}

2\sqrt{17}

3\sqrt7

6\sqrt2

First use

AE\cdot EB=CE\cdot ED

Place

E

at the origin; the center is at the intersection of the two chord perpendicular bisectors

The intersecting-chords theorem gives

2\cdot6=CE\cdot3,

so

CE=4.

Put

E=(0,0),

A=(-2,0),

B=(6,0),

C=(0,4),

and

D=(0,-3).

The perpendicular bisectors of

AB

and

CD

meet at

O=\left(2,\frac12\right).

Thus

r^2=OA^2=4^2+\left(\frac12\right)^2=\frac{65}{4},

and the diameter is

2r=\sqrt{65}.

Therefore, the correct answer is B.

Problem 32 in other years