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Difference between revisions of "1976 AHSME Problems/Problem 26"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(C) }\text{always equal to the length of an external common tangent}}</math>.
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By the [[Two Tangent Theorem]], the segments drawn from point <math>P</math> tangent to circle <math>O</math> are congruent. Let them have length <math>p</math>. Similarly, the two segments drawn from point <math>Q</math> tangent to circle <math>O^{\prime}</math> are congruent. Let them have length <math>q</math>.
  
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Let the length of the two external common tangents be <math>x</math>. We know that the two segments drawn from point <math>P</math> tangent to circle <math>O^{\prime}</math> are congruent. One of these tangents (the one on the external common tangent) has length <math>x-p</math>, so the other (the one on the internal common tangent) also has length <math>x-p</math> Doing the same thing for point <math>Q</math> and circle <math>O</math> shows that the segments drawn from <math>Q</math> tangent to circle <math>O</math> has length <math>x-q</math>. Thus, <math>PQ=\tfrac{x-p+x-q+p+q}2=\tfrac{2x}2=x</math>, because the numerator of this fraction counts the length of <math>\overline{PQ}</math> twice. Because <math>x</math> is the length of the external common tangent, our answer is <math>\boxed{\textbf{(C) }\text{always equal to the length of an external common tangent}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:00, 9 August 2024

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

By the Two Tangent Theorem, the segments drawn from point $P$ tangent to circle $O$ are congruent. Let them have length $p$. Similarly, the two segments drawn from point $Q$ tangent to circle $O^{\prime}$ are congruent. Let them have length $q$.

Let the length of the two external common tangents be $x$. We know that the two segments drawn from point $P$ tangent to circle $O^{\prime}$ are congruent. One of these tangents (the one on the external common tangent) has length $x-p$, so the other (the one on the internal common tangent) also has length $x-p$ Doing the same thing for point $Q$ and circle $O$ shows that the segments drawn from $Q$ tangent to circle $O$ has length $x-q$. Thus, $PQ=\tfrac{x-p+x-q+p+q}2=\tfrac{2x}2=x$, because the numerator of this fraction counts the length of $\overline{PQ}$ twice. Because $x$ is the length of the external common tangent, our answer is $\boxed{\textbf{(C) }\text{always equal to the length of an external common tangent}}$.

See Also

1976 AHSME (Problems • Answer Key • Resources)
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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