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

m (Added a little bit more explanation to the solution)
m (→‎Problem)
 
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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
 
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
  
−
[[File:pdfresizer.com-pdf-convert-q30.png]]
+
<asy>
 +
import geometry;
 +
import graph;
 +
 
 +
unitsize(2 cm);
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pair A, B, C, M, N, P;
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M = (-1,0);
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N = (1,0);
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C = (0,0);
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A = dir(140);
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B = dir(20);
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P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B));
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draw(M--N);
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draw(arc(C,1,0,180));
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draw(A--C--B);
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draw(A--P--B);
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label("$A$", A, NW);
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label("$B$", B, E);
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label("$C$", C, S);
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label("$M$", M, SW);
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label("$N$", N, SE);
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label("$P$", P, S);
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</asy>
  
 
<math>\textbf{(A)}\ 10^{\circ}\qquad
 
<math>\textbf{(A)}\ 10^{\circ}\qquad
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draw(circumcircle(A,B,C),dashed);
 
draw(circumcircle(A,B,C),dashed);
  
−
label("$A$", A, W);
+
label("$A$", A, NW);
 
label("$B$", B, E);
 
label("$B$", B, E);
 
label("$C$", C, S);
 
label("$C$", C, S);
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</asy>
 
</asy>
  
−
Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so, using the fact that angles in a cyclic quadrilateral sum to <math>180^{\circ}</math>, we have <math>\angle ABP = 40^\circ</math>. Hence <math>\angle ABC = \angle ABP - \angle CBP = 40^
+
Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so, using the fact that opposite angles in a cyclic quadrilateral sum to <math>180^{\circ}</math>, we have <math>\angle ABP = 40^\circ</math>. Hence <math>\angle ABC = \angle ABP - \angle CBP = 40^
 
\circ - 10^\circ = 30^\circ</math>.
 
\circ - 10^\circ = 30^\circ</math>.
  
 
Since <math>CA = CB</math>, triangle <math>ABC</math> is isosceles, with <math>\angle BAC = \angle ABC = 30^\circ</math>. Now, <math>\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ</math>. Finally, again using the fact that angles inscribed in the same arc are equal, we have <math>\angle BCP = \angle BAP = \boxed{\textbf{(C)}\ 20^{\circ}}</math>.
 
Since <math>CA = CB</math>, triangle <math>ABC</math> is isosceles, with <math>\angle BAC = \angle ABC = 30^\circ</math>. Now, <math>\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ</math>. Finally, again using the fact that angles inscribed in the same arc are equal, we have <math>\angle BCP = \angle BAP = \boxed{\textbf{(C)}\ 20^{\circ}}</math>.
 +
 +
==See Also==
 +
{{AHSME box|year=1983|num-b=29|after=Last Problem}}
 +
 +
{{MAA Notice}}

Latest revision as of 22:42, 2 December 2021

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

[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); label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]

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

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

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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