2013 AMC 12B Problems/Problem 16
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 ?
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 .
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 .
|2013 AMC 12B (Problems • Answer Key • Resources)|
|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|