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Difference between revisions of "1966 AHSME Problems/Problem 21"

(Created page with "== Problem == An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; ...")
 
(→‎Problem)
 
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== Problem ==
 
== Problem ==
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 +
<asy>
 +
draw((0,-5)--(-6,10),black+dashed+linewidth(1));
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draw((-6,10)--(10,0),black+dashed+linewidth(1));
 +
draw((10,0)--(-10,0),black+dashed+linewidth(1));
 +
draw((-10,-0)--(10,10),black+dashed+linewidth(1));
 +
draw((10,10)--(0,-5),black+dashed+linewidth(1));
 +
draw((-2,0)--(-10/3,10/3),black+linewidth(2));
 +
draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2));
 +
draw((10/9,50/9)--(90/17,50/17),black+linewidth(2));
 +
draw((90/17,50/17)--(10/3,0),black+linewidth(2));
 +
draw((10/3,0)--(-2,0),black+linewidth(2));
 +
MP("1", (1,0), N);  MP("2", (-2.5,2), E);  MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N);
 +
</asy>
 +
 
An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>).
 
An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>).
  
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== Solution ==
 
== Solution ==
−
 
+
<math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 02:03, 23 September 2014

Problem

[asy] draw((0,-5)--(-6,10),black+dashed+linewidth(1)); draw((-6,10)--(10,0),black+dashed+linewidth(1)); draw((10,0)--(-10,0),black+dashed+linewidth(1)); draw((-10,-0)--(10,10),black+dashed+linewidth(1)); draw((10,10)--(0,-5),black+dashed+linewidth(1)); draw((-2,0)--(-10/3,10/3),black+linewidth(2)); draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2)); draw((10/9,50/9)--(90/17,50/17),black+linewidth(2)); draw((90/17,50/17)--(10/3,0),black+linewidth(2)); draw((10/3,0)--(-2,0),black+linewidth(2)); MP("1", (1,0), N); MP("2", (-2.5,2), E); MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N); [/asy]

An "$n$-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively $1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5$; for all $n$ values of $k$, sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2$; prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5$).

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals:

$\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)$

Solution

$\fbox{E}$

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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