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

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We will test each option to see if it can be true or not. Links to diagrams are provided.
 
We will test each option to see if it can be true or not. Links to diagrams are provided.
 
<cmath>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}</cmath>
 
<cmath>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}</cmath>
−
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>[https://latex.artofproblemsolving.com/d/5/8/d5896d95c00fde8b69428d09a084959a86e83dfa.png Diagram A]
+
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>
 +
<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>
 
<cmath>\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}</cmath>
 
<cmath>\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}</cmath>
−
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>[https://latex.artofproblemsolving.com/d/9/d/d9d46af414fdc1eca2b350c8eb991be067717b49.png Diagram B]
+
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>
 +
<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>
 
<cmath>\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}</cmath>
 
<cmath>\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}</cmath>
−
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>
 +
<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>
 
<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>
−
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>  
+
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>
−
[https://latex.artofproblemsolving.com/2/2/9/229380edd13828236175da31534b31fe96f3b47b.png Diagram D]
+
<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 <math>\boxed{E}</math>.
 
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}}
+
{{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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