1960 AHSME Problems/Problem 35

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Problem 35

From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn. Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$. It is found that $t_1$, the length of the tangent, is the mean proportional between $m$ and $n$. If $m$ and $t$ are integers, then $t$ may have the following number of values:

$\textbf{(A)}\ \text{zero}\qquad \textbf{(B)}\ \text{one}\qquad \textbf{(C)}\ \text{two}\qquad \textbf{(D)}\ \text{three}\qquad \textbf{(E)}\ \text{infinitely many}$

Solution

By definition of mean proportional, $t = \sqrt{mn}$. Since $m+n=10$, $t = \sqrt{m(10-m)}$.

With trial and error, note that when $m=9$, $t=3$ and when $m=2$, $t=4$. These values work since another tangent line can be drawn from $P$, and the angle between the tangent and secant can decrease to match the values of $m$ and $n$. Thus, the answer is $\boxed{\textbf{(C)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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