**ANSWER TO FRIDAY’S PROBLEM:**

The problem, once again:

For any polygon, a “diagonal” is defined as a line segment that runs from one vertex to another, running through the polygon’s interior. Find a formula that determines the number of diagonals in any convex polygon with n sides. Once you have the formula, use it to figure out the number of diagonals in a convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

The winning answer was provided by Chris Mark. The formula, for a convex polygon with n sides, is this: Number of Diagonals = [n(n– 3)]/2. For n = 1000, the number of diagonals = 498,500.

The reasoning behind the formula. A polygon has as many vertices as sides. So a polygon with n sides also has n vertices. Now, consider any vertex of the n-gon. From that vertex the number of diagonals that can be drawn is (n – 3). That is because we cannot draw a diagonal to 3 vertices: the vertex chosen, and the two adjacent vertices. So the expression (n – 3) = the number of diagonals that can be drawn from any vertex. We multiply (n – 3) by n to obtain the total number of diagonals that can be drawn from all n vertices. But if we simply multiply n by (n – 3), we’d be counting each diagonal twice. To eliminate that problem, we divide the product, [ n(n – 3)], by 2, and that provides the correct formula:

Diagonals = [n(n – 3)]/2

Applying this formula to a convex polygon with 1,000 sides, we see that the number of diagonals = 498,500.

In addition to providing the answer, Chris pointed out that the problem need not be restricted to regular polygons, as it was when posted. This formula works for all convex polygons, regular or not.

Thanks, Chris. And thanks to everyone who submitted answers.

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