1961 AHSME Problems/Problem 11

Problem

Two tangents are drawn to a circle from an exterior point $A$; they touch the circle at points $B$ and $C$ respectively. A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$, and touches the circle at $Q$. If $AB=20$, then the perimeter of $\triangle APR$ is

$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 39\frac{7}{8} \qquad \textbf{(E)}\ \text{not determined by the given information}$

Solution

1961 AHSME Problem 11.png

Draw the diagram as shown. Note that the two tangent lines from a single outside point of a circle have the exact same length, so $AB = AC = 20$, $BP = PQ$, and $QR = CR$.

The perimeter of the triangle is $AP + PR + AR$. Note that $PQ + QR = PR$, so from substitution, the perimeter is \[AP + PQ + QR + AR\] \[AP + BP + CR + AR\] \[AB + AC\] Thus, the perimeter of the triangle is $40$, so the answer is $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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