1961 AHSME Problems/Problem 11


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

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


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

Solution 2 (Non-rigorous)

Since $Q$ can be anywhere on the circle between $A$ and $B$, it can basically be "on top" of $A$. Then $R$ will be at the same point as $A$, so $APR$ form a degenerate triable with side length $AB=20$. So its perimeter will be $40$. Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in length, $PR=PQ+QR$ will "grow" to compensate, so the perimeter will stay constant with a value of $\boxed{\textbf{(C)}\ 40}$.

We can also skip verifying that the perimeter will stay constant, since it seems unlikely that MAA would create a question with $\text{not determined by the given information}$ as the answer.


See Also

1961 AHSC (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png