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

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Let one of the sides of the pentagon have length <math>x_1</math> (and the others <math>x_2, x_3, x_4, x_5</math>). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length <math>\frac{x_1}{2} \sec 72^\circ</math>, and so the two sides together have length <math>x_1 \sec 72^\circ</math>. 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 <math>(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ</math> (because the perimeter of the pentagon is <math>1</math>). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is <math>\boxed{\textbf{(A)} \ 0}</math>.
 
Let one of the sides of the pentagon have length <math>x_1</math> (and the others <math>x_2, x_3, x_4, x_5</math>). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length <math>\frac{x_1}{2} \sec 72^\circ</math>, and so the two sides together have length <math>x_1 \sec 72^\circ</math>. 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 <math>(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ</math> (because the perimeter of the pentagon is <math>1</math>). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is <math>\boxed{\textbf{(A)} \ 0}</math>.
  
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==Solution 2==
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The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with <math>\boxed{\textbf{(A)} \ 0}</math>.
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}

Latest revision as of 22:10, 1 April 2022

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 $\boxed{\textbf{(A)} \ 0}$.

Solution 2

The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with $\boxed{\textbf{(A)} \ 0}$.

See also

2013 AMC 12B (Problems • Answer Key • Resources)
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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