Difference between revisions of "1971 AHSME Problems/Problem 5"
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− | + | == 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, and that the measure of </math>Q<math> equals </math>\widehat{AC}<math>. | ||
+ | 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 22:54, 23 July 2020
Problem 5
Points , and
lie on the circle shown and the measures of arcs
and
are
and
respectively. The sum of the measures of angles
and
is
Solution 1
We see that the measure of equals
Q
\widehat{AC}
\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}, the sum of the measures of
and
is $\widehat{BD} \div 2 = 80^{\circ} \div 2 = 40^{\circ} \Longrightarrow \textbf{(C) }.