Warm up problem 8: largest polygon
A regular polygon of n sides is drawn on each side of a square. What is the maximum number of sides that these polygons can have without overlapping with one of the other polygons?
Assume:
We are in a 2D plane
The polygon and the square have the same side length
The answer is 8.
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1 solution
I got the answer is two ways:
METHOD 1:
This method just took me 5 seconds to approach the correct answer! Try to hallucinate the diagram, a square with a polygon on each side.
when the polygon has 4 sides each, they are way far from each other, when 5 somewhat closer, 6 ... some more but 7 is still more convincing but wait what about 8. Yeah, 8 is the best of all.
Thus, the polygons surrounding the square have 8 sides.
METHOD 2:
This is a more mathematical approach, some may ask why the answer is not 9, 10 or any number, why only 8!!
Here is the solution:
Each corner of a square have an angle of 90 . Thus the outer part or 270 is left for the polygons.
Since two polygons meet with each other, 270* will be equally divided into 2 [135* 's].
Since it is a regular polygon, it will have all angles equal to 135.
According to the formula,
( n − 2 ) × 1 8 0 = Sum of interior angles, where n is the number of sides.
= ( n − 2 ) × 1 8 0 = 135(n)
=180(n) - 135(n) = 360
=45(n) = 360
=n = 8