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
.