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

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==Problem==
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Distinct points <math>A</math> and <math>B</math> are on a semicircle with diameter <math>MN</math> and center <math>C</math>.
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The point <math>P</math> is on <math>CN</math> and <math>\angle CAP = \angle CBP = 10^{\circ}</math>. If <math>\stackrel{\frown}{MA} = 40^{\circ}</math>, then <math>\stackrel{\frown}{BN}</math> equals
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[[File:pdfresizer.com-pdf-convert-q30.png]]
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<math>\textbf{(A)}\ 10^{\circ}\qquad
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\textbf{(B)}\ 15^{\circ}\qquad
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\textbf{(C)}\ 20^{\circ}\qquad
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\textbf{(D)}\ 25^{\circ}\qquad
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\textbf{(E)}\ 30^{\circ} </math>
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==Solution==
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Since <math>\angle CAP = \angle CBP = 10^\circ</math>, quadrilateral <math>ABPC</math> is cyclic.
 
Since <math>\angle CAP = \angle CBP = 10^\circ</math>, quadrilateral <math>ABPC</math> is cyclic.
  

Revision as of 18:05, 27 January 2019

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.

[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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