1953 AHSME Problems/Problem 14

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 $\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}$ If circle $Q$ is outside circle $P$ and it is not tangent to circle $P$ , then $PQ$ is greater than $p+q$ . $\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}$ If circle $Q$ is inside circle $P$ and it is not tangent to circle $P$ , then $PQ$ is greater than $p-q$ . Since options A, B, C, and D can be true, the answer must be $\boxed{E}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1953 AHSME Problems/Problem 14

Problem 14

Solution

See Also