View source for 1971 AHSME Problems/Problem 5

== Problem 5 ==

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> 
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

<math>\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} </math>

<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 <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 <math>\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \boxed{\textbf{(C) }40^{\circ}}</math>.

== Solution 2 ==
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>.

== See Also ==
{{AHSME 35p box|year=1971|num-b=4|num-a=6}}
{{MAA Notice}}
[[Category:Introductory Geometry Problems]]