1976 AHSME Problems/Problem 26

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Problem

[asy] size(150); dotfactor=4; draw(circle((0,0),4)); draw(circle((10,-6),3)); pair O,A,P,Q; O = (0,0); A = (10,-6); P = (-.55, -4.12); Q = (10.7, -2.86); dot("$O$", O, NE); dot("$O'$", A, SW); dot("$P$", P, SW); dot("$Q$", Q, NE); draw((2*sqrt(2),2*sqrt(2))--(10 + 3*sqrt(2)/2, -6 + 3*sqrt(2)/2)--cycle); draw((-1.68*sqrt(2),-2.302*sqrt(2))--(10 - 2.6*sqrt(2)/2, -6 - 3.4*sqrt(2)/2)--cycle); draw(P--Q--cycle); //Credit to happiface for the diagram [/asy]

In the adjoining figure, every point of circle $\mathit{O'}$ is exterior to circle $\mathit{O}$. Let $\mathit{P}$ and $\mathit{Q}$ be the points of intersection of an internal common tangent with the two external common tangents. Then the length of $PQ$ is

$\textbf{(A) }\text{the average of the lengths of the internal and external common tangents}\qquad\\ \textbf{(B) }\text{equal to the length of an external common tangent if and only if circles }\mathit{O}\text{ and }\mathit{O'} \text{have equal radii}\\ \textbf{(C) }\text{always equal to the length of an external common tangent}\qquad\\ \textbf{(D) }\text{greater than the length of an external common tangent}\qquad\\ \textbf{(E) }\text{the geometric mean of the lengths of the internal and external common tangents}$


Solution

$\boxed{\textbf{(C) }\text{always equal to the length of an external common tangent}}$.


See Also

1976 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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