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

(Solution)
m (Solution)
Line 16: Line 16:
 
== Solution ==
 
== 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 <math>\fbox{D}</math>  pairs of A,B.
+
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,
  
 
== See also ==
 
== See also ==

Revision as of 12:49, 31 December 2020

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,

See also

1992 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS