1976 AMC 12 Problem 26

Question

In the adjoining figure, every point of circle

O'

is exterior to circle

O.

Let

P

and

Q

be the points of intersection of an internal common tangent with the two external common tangents. Then the length of

PQ

is

the average of the lengths of the internal and external common tangents

equal to the length of an external common tangent if and only if circles

O

and

O'

have equal radii

always equal to the length of an external common tangent

greater than the length of an external common tangent

the geometric mean of the lengths of the internal and external common tangents

Answer: C

Concepts:tangent circles tangent line homothety

Difficulty rating: 2200

Small hint:

Mark the tangency point on each of the three tangent lines

Big hint:

From either

P

or

Q,

tangent segments to the same circle have equal lengths

Solution:

Let the internal tangent touch

O

and

O'

at

R

and

S.

Let the external tangent through

P

touch them at

X

and

Y,

and let the one through

Q

touch them at

V

and

W.

Equal tangent segments from a point give

\begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW. \end{gathered}

Along the internal tangent,

PR+QR=PQ

and

PS+QS=PQ.

Adding,

\begin{aligned} PR+QR+PS+QS=2PQ. \end{aligned}

Along the external tangents,

PX+PY=XY

and

QV+QW=VW.

Therefore

2PQ=XY+VW.

The two external common tangent segments have equal length, so

XY=VW

and hence

PQ=XY=VW.

Therefore, the correct answer is C.