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

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 $\boxed{\textbf{(C)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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1961 AHSME Problems/Problem 11

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