Difference between revisions of "2013 AMC 12B Problems/Problem 16"

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==Problem==
 
==Problem==
  
Let <math>ABCDE</math> be an equiangular convex pentagon of perimeter <math>1</math>. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let <math>s</math> be the perimeter of this star. What is the difference between the maximum and the minimum possible values of <math>s</math>.
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Let <math>ABCDE</math> be an equiangular convex pentagon of perimeter <math>1</math>. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let <math>s</math> be the perimeter of this star. What is the difference between the maximum and the minimum possible values of <math>s</math>?
  
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\  \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\  \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}</math>
  
==Solution==
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==Solution 1==
  
 
The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure <math>\frac{180^\circ (5-2)}{5}=108^\circ</math>, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure <math>180^\circ - 108^\circ = 72^\circ</math>. The base angles are equal, so the triangles must be isosceles.
 
The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure <math>\frac{180^\circ (5-2)}{5}=108^\circ</math>, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure <math>180^\circ - 108^\circ = 72^\circ</math>. The base angles are equal, so the triangles must be isosceles.
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 23:46, 23 January 2020

Problem

Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\  \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$

Solution 1

The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure $180^\circ - 108^\circ = 72^\circ$. The base angles are equal, so the triangles must be isosceles.

Let one of the sides of the pentagon have length $x_1$ (and the others $x_2, x_3, x_4, x_5$). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length $\frac{x_1}{2} \sec 72^\circ$, and so the two sides together have length $x_1 \sec 72^\circ$. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get $(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ$ (because the perimeter of the pentagon is $1$). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is $\boxed{\textbf{(A)} \ 0}$.

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AMC 12 Problems and Solutions

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