2013 AMC 12B Problems/Problem 16
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
Because this pentagon is equiangular and has a constant perimeter, there is no scaling (stretch) that can change it. This means the one unchanging configuration of points given by the pentagon gives both the max and min perimeters of the star, which are therefore equal. Subtraction of equal terms yields .
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.