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Difference between revisions of "1953 AHSME Problems/Problem 14"

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Let circle <math>Q</math> be outside circle <math>P</math> and not tangent to circle <math>P</math>, and the intersection of <math>\overline{PQ}</math> with the circles be <math>R</math> and <math>S</math> respectively. <math>PR = p</math> and <math>QS = q</math>, and <math>PR + QS < PQ</math>, so <math>p+q < PQ.</math> [https://latex.artofproblemsolving.com/2/0/d/20d001912b80a7a6ef4c267abdd424da9ca14784.png Diagram C]
 
Let circle <math>Q</math> be outside circle <math>P</math> and not tangent to circle <math>P</math>, and the intersection of <math>\overline{PQ}</math> with the circles be <math>R</math> and <math>S</math> respectively. <math>PR = p</math> and <math>QS = q</math>, and <math>PR + QS < PQ</math>, so <math>p+q < PQ.</math> [https://latex.artofproblemsolving.com/2/0/d/20d001912b80a7a6ef4c267abdd424da9ca14784.png Diagram C]
 
<cmath>\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}</cmath>
 
<cmath>\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}</cmath>
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If circle <math>Q</math> is inside circle <math>P</math> and it is not tangent to circle <math>P</math>, then <math>PQ</math> is greater than <math>p-q</math>.
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Let circle <math>Q</math> be inside circle <math>P</math> and not tangent to circle <math>P</math>, and the intersection of <math>\overline{PQ}</math> with the circles be <math>R</math> and <math>S</math> respectively. <math>PR = p</math> and <math>QS = q</math>, and <math>QS < QR</math>, so <math>PR - QS < PR - QR</math>, and <math>PR - QR = PQ</math>, so <math>p-q < PQ.</math>  
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[https://latex.artofproblemsolving.com/2/2/9/229380edd13828236175da31534b31fe96f3b47b.png Diagram D]
 
Since options A, B, C, and D can be true, the answer must be <math>\boxed{E}</math>.
 
Since options A, B, C, and D can be true, the answer must be <math>\boxed{E}</math>.
  

Revision as of 15:10, 15 July 2018

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 (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

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