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

Revision as of 14:14, 8 August 2018

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

@@ Line 1: / Line 1: @@
 Since <math>\angle CAP = \angle CBP = 10^\circ</math>, quadrilateral <math>ABPC</math> is cyclic.
-[asy]
+<asy>
 import geometry;
 import graph;
@@ Line 23: / Line 23: @@
 draw(circumcircle(A,B,C),dashed);
-label("<math>A</math>", A, W);
+label("$A$", A, W);
-label("<math>B</math>", B, E);
+label("$B$", B, E);
-label("<math>C</math>", C, S);
+label("$C$", C, S);
-label("<math>M</math>", M, SW);
+label("$M$", M, SW);
-label("<math>N</math>", N, SE);
+label("$N$", N, SE);
-label("<math>P</math>", P, S);
+label("$P$", P, S);
-[/asy]
+</asy>
 Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so <math>\angle ABP = 40^\circ</math>. Then <math>\angle ABC = \angle ABP - \angle CBP = 40^

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

Revision as of 14:14, 8 August 2018