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, $\angle BCP = \angle BAP = \boxed{20^\circ}$ .

Page

Toolbox

Search

1983 AHSME Problems/Problem 30