1976 AMC 12 Problem 26

Attempt Problem 26 of the 1976 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1976 AMC 12 solutions, or check the answer key.

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26.

In the adjoining figure, every point of circle OO' is exterior to circle O.O. Let PP and QQ be the points of intersection of an internal common tangent with the two external common tangents. Then the length of PQPQ 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 OO and OO' 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 circlestangent linehomothety
Difficulty rating: 2200
Small Hint:

Mark the tangency point on each of the three tangent lines

Big Hint:

From either PP or Q,Q, tangent segments to the same circle have equal lengths

Solution:

Let the internal tangent touch OO and OO' at RR and S.S. Let the external tangent through PP touch them at XX and Y,Y, and let the one through QQ touch them at VV and W.W. Equal tangent segments from a point give PR=PX,PS=PY,QR=QV,QS=QW. \begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW. \end{gathered} Along the internal tangent, PR+QR=PQPR+QR=PQ and PS+QS=PQ.PS+QS=PQ. Adding, PR+QR+PS+QS=2PQ. \begin{aligned} PR+QR+PS+QS=2PQ. \end{aligned} Along the external tangents, PX+PY=XYPX+PY=XY and QV+QW=VW.QV+QW=VW. Therefore 2PQ=XY+VW.2PQ=XY+VW. The two external common tangent segments have equal length, so XY=VWXY=VW and hence PQ=XY=VW.PQ=XY=VW.

Therefore, the correct answer is C.

← Problem 25#25
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