1983 AHSME Problems/Problem 30

Since $\angle CAP = \angle CBP = 10^\circ$ , quadrilateral $ABPC$ is cyclic.

[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 $\angle ABP = 40^\circ$ . Then $\angle ABC = \angle ABP - \angle CBP = 40^ \circ - 10^\circ = 30^\circ$ .

Since $CA = CB$ , triangle $ABC$ is isosceles, and $\angle BAC = \angle ABC = 30^\circ$ . Then $\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ$ . Hence, $\angle BCP = \angle BAP = \boxed{20^\circ}$ .

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1983 AHSME Problems/Problem 30