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Difference between revisions of "1992 AHSME Problems/Problem 20"

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== Solution ==
 
== Solution ==
  
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If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 <math>\fbox{D}</math>  pairs of A,
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If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 <math>\fbox{D}</math>  pairs of A,B.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1992|num-b=19|num-a=21}}   
 
{{AHSME box|year=1992|num-b=19|num-a=21}}   
  
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[[Category: Introductory Geometry Proble]]
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[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:28, 1 May 2021

Problem

[asy] draw((1,0)--(2*cos(pi/8),2*sin(pi/8))--(cos(pi/4),sin(pi/4))--(2*cos(3*pi/8),2*sin(3*pi/8))--(cos(pi/2),sin(pi/2))--(2*cos(5*pi/8),2*sin(5*pi/8))--(cos(3*pi/4),sin(3*pi/4))--(2*cos(7*pi/8),2*sin(7*pi/8))--(-1,0),black+linewidth(.75)); MP("A_1",(2*cos(5*pi/8),2*sin(5*pi/8)),N);MP("A_2",(2*cos(3*pi/8),2*sin(3*pi/8)),N);MP("A_3",(2*cos(1*pi/8),2*sin(1*pi/8)),N); MP("A_n",(2*cos(7*pi/8),2*sin(7*pi/8)),N); MP("B_1",(cos(4*pi/8),sin(4*pi/8)),S);MP("B_2",(cos(2*pi/8),sin(2*pi/8)),S);MP("B_n",(cos(6*pi/8),sin(6*pi/8)),S); [/asy] Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,A_2,\cdots,A_n$ are congruent, and angles $B_1,B_2,\cdots,B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1$, then $n=$

$\text{(A) } 12\quad \text{(B) } 18\quad \text{(C) } 24\quad \text{(D) } 36\quad \text{(E) } 60$

Solution

If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 $\fbox{D}$ pairs of A,B.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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