1953 AHSME Problems/Problem 14

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Problem 14

Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q$. Draw $PQ$. Which of the following statements is false?

$\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\  \textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\  \textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\\  \textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}$

Solution

We will test each option to see if it can be true or not. Links to diagrams are provided. \[\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\] Let circle $Q$ be inside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR = p$, and $QR = q$, so $PR - QR = PQ = p-q.$Diagram A \[\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\] Let circle $Q$ be outside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR = p$, and $QR = q$, so $PR + QR = PQ = p+q.$Diagram B \[\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\] Let circle $Q$ be outside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ respectively. $PR = p$ and $QS = q$, and $PR + QS < PQ$, so $p+q < PQ.$ Diagram C \[\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\] Let circle $Q$ be inside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ respectively. $PR = p$ and $QS = q$, and $QS < QR$, so $PR - QS < PR - QR$, and $PR - QR = PQ$, so $p-q < PQ.$ Diagram D Since options A, B, C, and D can be true, the answer must be $\boxed{E}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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