Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
Line 11: Line 11:
 
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==
 
−
 
−
With a bit of cognitive effort, it is easy to visualize that readjusting the side lengths of any odd-number sided polygon will change the angle measures no matter what (unless you are upscaling all the side lengths equally, but we are working with a fixed perimiter). Since the star is equiangular, this obviously means there is only one possible perimeter for the star, so the difference between the maximum and minimum perimeters is <math>\boxed{\textbf{(A)} \ 0}</math>.
 
−
 
−
-Joeya
 
  
 
== See also ==
 
== See also ==

Revision as of 23:14, 17 January 2021

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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_16&oldid=142584"