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Difference between revisions of "1983 AHSME Problems/Problem 30"

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Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so, using the fact that angles in a cyclic quadrilateral sum to <math>180^{\circ}</math>, we have <math>\angle ABP = 40^\circ</math>. Hence <math>\angle ABC = \angle ABP - \angle CBP = 40^
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Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so, using the fact that opposite angles in a cyclic quadrilateral sum to <math>180^{\circ}</math>, we have <math>\angle ABP = 40^\circ</math>. Hence <math>\angle ABC = \angle ABP - \angle CBP = 40^
 
\circ - 10^\circ = 30^\circ</math>.
 
\circ - 10^\circ = 30^\circ</math>.
  

Revision as of 20:19, 10 September 2020

Problem

Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$. The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$. If $\stackrel{\frown}{MA} = 40^{\circ}$, then $\stackrel{\frown}{BN}$ equals

Pdfresizer.com-pdf-convert-q30.png

$\textbf{(A)}\ 10^{\circ}\qquad \textbf{(B)}\ 15^{\circ}\qquad \textbf{(C)}\ 20^{\circ}\qquad \textbf{(D)}\ 25^{\circ}\qquad \textbf{(E)}\ 30^{\circ}$

Solution

Since $\angle CAP = \angle CBP = 10^\circ$, quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem "angles inscribed in the same arc are equal".

[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B)); draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B); draw(A--B); draw(circumcircle(A,B,C),dashed); label("$A$", A, W); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]

Since $\angle ACM = 40^\circ$, $\angle ACP = 140^\circ$, so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\circ}$, we have $\angle ABP = 40^\circ$. Hence $\angle ABC = \angle ABP - \angle CBP = 40^ \circ - 10^\circ = 30^\circ$.

Since $CA = CB$, triangle $ABC$ is isosceles, with $\angle BAC = \angle ABC = 30^\circ$. Now, $\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ$. Finally, again using the fact that angles inscribed in the same arc are equal, we have $\angle BCP = \angle BAP = \boxed{\textbf{(C)}\ 20^{\circ}}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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