2009 AMC 12A Problems/Problem 19
Problem
Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were and , respectively. Each polygon had a side length of . Which of the following is true?
Solution
In any regular polygon with side length , consider the isosceles triangle formed by the center of the polygon and two consecutive vertices and . We are given that . Obviously , where is the radius of the circumcircle. Let be the midpoint of . Then , and , where is the radius of the incircle.
Applying the Pythagorean theorem on the triangle , we get that .
Then the area between the circumcircle and the incircle can be computed as .
Hence , , and therefore .
Alternate Solution (Applying Basic Trig)
Similar to the first solution, consider the isosceles triangle formed by each polygon. If you drop an altitude to the side of each polygon, you get in both polygons a right triangle with base of . For both the pentagon and heptagon, the hypotenuse of these right triangles is the radii of the larger circles and the apothems (the altitude we dropped to the side of each polygon) are the radii of the smaller circles.
Label the apothem of the pentagon and the apothem of the heptagon . Label the hypotenuse in the pentagon and the hypotenuse in the heptagon .
Now, finding the angles in the triangles and applying trig functions to find these radii, we get the following:
=
=
=
=
Now, the areas in between the circles are:
=
=
Note the trig identity , from which we can easily get that
Thus, the area between the circles in both the heptagon and the pentagon are equivalent to , and therefore the answer is .
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.