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 == | ||
+ | |||
+ | 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 \textbf{(C) }</math>. |
Latest revision as of 22:56, 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
We see that the measure of equals , and that the measure of equals . Since , the sum of the measures of and is .