1999 AHSME Problems/Problem 26
Contents
Problem
Three non-overlaping regular plane polygons, at least two of which are congruent, all have sides of length . The polygons meet at a point in such a way that the sum of the three interior angles at is . Thus the three polygons for a new polygon with as an interior point. What is the largest possible perimeter that this polygon can have?
Solution
We are looking for three regular polygons such that the sum of their internal angle sizes is exactly .
Let the number of sides in our polygons be . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value .
The integral angle of a regular -gon is . Therefore we are looking for integer solutions to:
Which can be simplified to:
Furthermore, we know that two of the polygons are congruent, thus WLOG . Our equation now becomes
Multiply both sides by and simplify to get .
Using the standard technique for Diophantine equations, we can add to both sides and rewrite the equation as .
Remembering that the only valid options for are: , , , and .
These correspond to the following pairs : , , , and .
The perimeters of the resulting polygon for these four cases are , , , and .
Solution 2
We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore degrees, so they are dodecagons. Of all the sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has sides.
See Also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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