1983 AHSME Problems/Problem 30

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

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

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 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 $\angle BCP = \angle BAP = \boxed{\textbf{(C)}\ 20^{\circ}}$ .

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1983 AHSME Problems/Problem 30

Problem

Solution