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> | + | 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== | + | ==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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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>. | ||
− | + | ==Solution 2== | |
− | ==Solution== | + | 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>. |
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== 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 23:10, 1 April 2022
Contents
Problem
Let be an equiangular convex pentagon of perimeter
. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let
be the perimeter of this star. What is the difference between the maximum and the minimum possible values of
?
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 , and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure
. The base angles are equal, so the triangles must be isosceles.
Let one of the sides of the pentagon have length (and the others
). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length
, and so the two sides together have length
. 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
(because the perimeter of the pentagon is
). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is
.
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 .
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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