| Toothpick Polygon |
A regular polygon is created by placing N toothpicks of equal size end to end.
The polygon is then altered by moving exactly two of the toothpicks. The resulting figure is still a polygon, but it has 31/3% less interior diagonals.
What is the value of N? |
Problem Moderated by: Douglas |
| Problem Solution |
First, remember that the number of diagonals in a convex polygon is given by N(N-3)/2, where N is the number of sides.
However, if we move two toothpicks in such a way that the new figure is still a polygon, but with fewer interior diagonals, then the new polygon must not be convex.
The diagram below shows an example, where N=6. The black lines represent the original polygon; the red lines represent the new positions of two of the toothpicks.
Although the new polygon has the same number of diagonals as the original polygon, it has three less interior diagonals. The diagram above shows the three non-interior diagonals in green.
Now, since 31/3% = 1/30, we have the following equation:
(29/30)(N(N-3)/2) = N(N-3)/2 - 3
29(N(N-3)/2) = 30(N(N-3)/2 - 3)
29N(N-3) = 30(N(N-3) - 3)
90 = N(N-3)
N2 - 3N - 90 = 0
(N - 15)(N + 12) = 0
Since N must be positive, N = 15
|
|
|
