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Difference between revisions of "1971 AHSME Problems/Problem 5"

(Created page with "Solution 1: Let the measure of P equal (BD-AC)/2. BD = 80. Let the measure of Q equal 1/2 the measure of AC. Add the measures of P and Q to get (80-AC)/2 + AC/2 Therefore,...")
 
(added second solution)
 
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Solution 1:
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== Problem 5 ==
  
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Let the measure of P equal (BD-AC)/2. BD = 80.
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Points <math>A,B,Q,D</math>, and <math>C</math> lie on the circle shown and the measures of arcs <math>\widehat{BQ}</math> and <math>\widehat{QD}</math>
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Let the measure of Q equal 1/2 the measure of AC.
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are <math>42^\circ</math> and <math>38^\circ</math> respectively. The sum of the measures of angles <math>P</math> and <math>Q</math> is
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Add the measures of P and Q to get (80-AC)/2 + AC/2  
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Therefore, the sum of P and Q is 40 as AC cancels out.
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<math>\textbf{(A) }80^\circ\qquad
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\textbf{(B) }62^\circ\qquad
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\textbf{(C) }40^\circ\qquad
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\textbf{(D) }46^\circ\qquad
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\textbf{(E) }\text{None of these} </math>
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<asy>
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size(3inch);
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draw(Circle((1,0),1));
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pair A, B, C, D, P, Q;
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P = (-2,0);
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B=(sqrt(2)/2+1,sqrt(2)/2);
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D=(sqrt(2)/2+1,-sqrt(2)/2);
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Q = (2,0);
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A = intersectionpoints(Circle((1,0),1),B--P)[1];
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C = intersectionpoints(Circle((1,0),1),D--P)[0];
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draw(B--P--D);
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draw(A--Q--C);
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label("$A$",A,NW);
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label("$B$",B,NE);
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label("$C$",C,SW);
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label("$D$",D,SE);
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label("$P$",P,W);
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label("$Q$",Q,E);
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//Credit to chezbgone2 for the diagram</asy>
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== Solution 1 ==  
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We see that the measure of <math>P</math> equals <math>(\widehat{BD}-\widehat{AC})/2</math>, and that the measure of <math>Q</math> equals <math>\widehat{AC}/2</math>.
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Since <math>\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}</math>, the sum of the measures of <math>P</math> and <math>Q</math> is <math>\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \boxed{\textbf{(C) }40^{\circ}}</math>.
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== Solution 2 ==
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Arcs are measured by the angle measures of their corresponding [[central angle|central angles]]. Thus, the [[inscribed angle]] <math>\measuredangle BAQ = \tfrac{42^{\circ}}2 = 21^{\circ}</math>, and, likewise, <math>\measuredangle QCD = \tfrac{38^{\circ}}2 = 19^{\circ}</math>. Thus, by [[supplementary]] angles, <math>\measuredangle PAQ = 180^{\circ} - \measuredangle BAQ = 159^{\circ}</math>, and <math>\measuredangle PCQ = 180^{\circ} - \measuredangle QCD = 161^{\circ}</math>. Because the sum of the [[interior angle]] measures of a quadrilateral add to <math>360^{\circ}</math>, we see that <math>\measuredangle P + \measuredangle Q = 360^{\circ} - \measuredangle PAQ - \measuredangle PCQ = 360^{\circ} - 159^{\circ} - 161^{\circ} = 40^{\circ}</math>. Thus, our answer is <math>\boxed{\textbf{(C) }40^{\circ}}</math>.
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=4|num-a=6}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 09:03, 1 August 2024

Problem 5

Points $A,B,Q,D$, and $C$ lie on the circle shown and the measures of arcs $\widehat{BQ}$ and $\widehat{QD}$ are $42^\circ$ and $38^\circ$ respectively. The sum of the measures of angles $P$ and $Q$ is

$\textbf{(A) }80^\circ\qquad \textbf{(B) }62^\circ\qquad \textbf{(C) }40^\circ\qquad \textbf{(D) }46^\circ\qquad \textbf{(E) }\text{None of these}$

[asy] size(3inch); draw(Circle((1,0),1)); pair A, B, C, D, P, Q; P = (-2,0); B=(sqrt(2)/2+1,sqrt(2)/2); D=(sqrt(2)/2+1,-sqrt(2)/2); Q = (2,0); A = intersectionpoints(Circle((1,0),1),B--P)[1]; C = intersectionpoints(Circle((1,0),1),D--P)[0]; draw(B--P--D); draw(A--Q--C); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SW); label("$D$",D,SE); label("$P$",P,W); label("$Q$",Q,E); //Credit to chezbgone2 for the diagram[/asy]


Solution 1

We see that the measure of $P$ equals $(\widehat{BD}-\widehat{AC})/2$, and that the measure of $Q$ equals $\widehat{AC}/2$. Since $\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}$, the sum of the measures of $P$ and $Q$ is $\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \boxed{\textbf{(C) }40^{\circ}}$.

Solution 2

Arcs are measured by the angle measures of their corresponding central angles. Thus, the inscribed angle $\measuredangle BAQ = \tfrac{42^{\circ}}2 = 21^{\circ}$, and, likewise, $\measuredangle QCD = \tfrac{38^{\circ}}2 = 19^{\circ}$. Thus, by supplementary angles, $\measuredangle PAQ = 180^{\circ} - \measuredangle BAQ = 159^{\circ}$, and $\measuredangle PCQ = 180^{\circ} - \measuredangle QCD = 161^{\circ}$. Because the sum of the interior angle measures of a quadrilateral add to $360^{\circ}$, we see that $\measuredangle P + \measuredangle Q = 360^{\circ} - \measuredangle PAQ - \measuredangle PCQ = 360^{\circ} - 159^{\circ} - 161^{\circ} = 40^{\circ}$. Thus, our answer is $\boxed{\textbf{(C) }40^{\circ}}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
All AHSME Problems and Solutions

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