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

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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.
 
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>
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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>.[https://latex.artofproblemsolving.com/d/5/8/d5896d95c00fde8b69428d09a084959a86e83dfa.png Diagram A]
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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>[https://latex.artofproblemsolving.com/d/5/8/d5896d95c00fde8b69428d09a084959a86e83dfa.png Diagram A]
 
<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>[https://latex.artofproblemsolving.com/d/9/d/d9d46af414fdc1eca2b350c8eb991be067717b49.png Diagram B]
 
<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>
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If circle <math>Q</math> is outside 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>.
+
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>
 
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>.
 
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>.

Revision as of 15:04, 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 \[\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\] 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 $\boxed{E}$.

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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