1974 AMC 12 Problem 23

Question

In the adjoining figure

TP

and

T'Q

are parallel tangents to a circle of radius

r,

with

T

and

T'

the points of tangency.

PT''Q

is a third tangent with

T''

as point of tangency. If

TP=4

and

T'Q=9

then

r

is

\frac{25}{6}

6

\frac{25}{4}

a number other than

\frac{25}{6},

6,

or

\frac{25}{4}

not determinable from the given information

The center lies on each angle bisector formed by two tangents from the same external point

Show that

\angle POQ=90^\circ

and use the altitude-to-hypotenuse theorem at

T''

The radii to the tangency points make congruent right triangles from each external point, so

OP

and

OQ

bisect the two tangent angles. Because the other two tangents are parallel, these half-angles add to

90^\circ;

hence

\angle POQ=90^\circ.

Thus

OT''=r

is the altitude to the hypotenuse

PQ.

Equal tangent segments give

PT''=PT=4

and

QT''=QT'=9.

The altitude theorem yields

r^2=(PT'')(QT'')=4\cdot9=36,

so

r=6.

Therefore, the correct answer is B.

Problem 23 in other years