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

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\textbf{(C)}\ p+q\text{ can be less than }\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}  </math>
 
\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}  </math>
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==Solution==
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We will test each option to see if it can be true or not. Links to diagrams are provided.
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<cmath>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}</cmath>
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Let circle <math>Q</math> be inside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR - QR = PQ = p-q.</math>
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<asy> pair P, Q, R; P = (0,0); Q = (3,0); R = (4,0); draw(Circle(P,4)); draw(Circle(Q,1)); draw(P--R); dot(P); dot(Q); dot(R); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,E); </asy>
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<cmath>\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}</cmath>
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Let circle <math>Q</math> be outside circle <math>P</math> and tangent to circle <math>P</math>, and the point of tangency be <math>R</math>. <math>PR = p</math>, and <math>QR = q</math>, so <math>PR + QR = PQ = p+q.</math>
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<asy> pair P, Q, R; P = (0,0); Q = (5,0); R = (4,0); draw(Circle(P,4)); draw(Circle(Q,1)); draw(P--Q); dot(P); dot(Q); dot(R); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); </asy>
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<cmath>\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}</cmath>
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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>
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<asy> pair P, Q, R, SS; P = (0,0); Q = (5,0); R = (3,0); SS = (4,0); draw(Circle(P,3)); draw(Circle(Q,1)); draw(P--Q); dot(P); dot(Q); dot(R); dot(SS); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); label("$S$",SS,SW); </asy>
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<cmath>\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}</cmath>
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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> as shown in the diagram. <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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<asy> pair P, Q, R, SS; P = (0,0); Q = (3,0); R = (6,0); SS = (4.5,0); draw(Circle(P,6)); draw(Circle(Q,1.5)); draw(P--R); dot(P); dot(Q); dot(R); dot(SS); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); label("$S$",SS,SW); </asy>
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Since options A, B, C, and D can be true, the answer must be <math>\boxed{E}</math>.
  
 
==See Also==
 
==See Also==
  
−
{{AHSME 50p box|year=1953|num-b=12|num-a=14}}
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{{AHSME 50p box|year=1953|num-b=13|num-a=15}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:23, 19 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 $PR - QR = PQ = p-q.$ [asy] pair P, Q, R; P = (0,0); Q = (3,0); R = (4,0); draw(Circle(P,4)); draw(Circle(Q,1)); draw(P--R); dot(P); dot(Q); dot(R); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,E); [/asy] \[\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\] 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 $PR + QR = PQ = p+q.$ [asy] pair P, Q, R; P = (0,0); Q = (5,0); R = (4,0); draw(Circle(P,4)); draw(Circle(Q,1)); draw(P--Q); dot(P); dot(Q); dot(R); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); [/asy] \[\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\] 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 $p+q < PQ.$ [asy] pair P, Q, R, SS; P = (0,0); Q = (5,0); R = (3,0); SS = (4,0); draw(Circle(P,3)); draw(Circle(Q,1)); draw(P--Q); dot(P); dot(Q); dot(R); dot(SS); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); label("$S$",SS,SW); [/asy] \[\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\] 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 $S$ as shown in the diagram. $PR = p$ and $QS = q$, and $QS < QR$, so $PR - QS < PR - QR$, and $PR - QR = PQ$, so $p-q < PQ.$ [asy] pair P, Q, R, SS; P = (0,0); Q = (3,0); R = (6,0); SS = (4.5,0); draw(Circle(P,6)); draw(Circle(Q,1.5)); draw(P--R); dot(P); dot(Q); dot(R); dot(SS); label("$P$",P,S); label("$Q$",Q,S); label("$R$",R,SW); label("$S$",SS,SW); [/asy] 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 13 		Followed by
Problem 15
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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