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

Revision as of 21:54, 23 July 2020

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}$. Since$\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}, the sum of the measures of $P$ and $Q$ is $\widehat{BD} \div 2 = 80^{\circ} \div 2 = 40^{\circ} \Longrightarrow \textbf{(C) }.

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