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Difference between revisions of "1971 AHSME Problems/Problem 5"
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== Solution ==  
 
== Solution ==  
  
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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} \div 2</math>.
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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>.
−
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 $\widehat{BD} \div 2 = 80^{\circ} \div 2 = 40^{\circ} \Longrightarrow \textbf{(C) }.
+
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 \textbf{(C) }</math>.

Revision as of 21:56, 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

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 $\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \textbf{(C) }$.

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