Challenge Problem – Polygon Diagonal Formula

Here’s a challenge problem for anyone who’d like to try it.

On Monday I will post the answer and the names of the first five people who got this right. So good luck, everyone.

A regular polygon is a polygon all of whose sides are congruent and all of whose  angles are congruent. For any polygon, a “diagonal” is defined as a line segment that runs from one vertex of the polygon to another, and which runs through the interior of the polygon.

Find a formula that tells how to determine the number of diagonals there are in any regular convex polygon with n sides.

Once you have the formula, use it to figure out the number of diagonals in a regular convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

Good luck!

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One Response to “Challenge Problem – Polygon Diagonal Formula”

  1. Asil says:

    Федя: спасибо за решение. Вы правы, конечно, это чистая комбинаторика, просто рисование треугольников (а у меня еще и “дополнительное построение” имеется) напомнило уроки школьной геометрии. Мое решение немножко отличается в деталях и, мне кажется, несколько попроще (но это мелочи, конечно). 1 –> 2 For every diagonal d of P, take as A and B the mid-points of two sides of P that are inndeict to one end-point of d, and as C and D the mid-points of two sides inndeict to another end-point of d. Then a diagonal is crossed by d if and only if it is corralled by A,B,C,D. 2 –> 1 Let E (resp. F) be the endpoint of some diagonal corralled by A,B,C,D, such that E is between B and C (resp. between A and D) and is closest to B (resp. to A). Then the diagonal EF belongs to T, and there is a (unique) triangle EFG of T such that G is between A and B. Similarly, let E’F’G’ be a triangle of T such that E’F’ is a diagonal of T corralled by A,B,C,D and closest to the segment CD, and G’ is between C and D. Clearly, a diagonal is corralled by A,B,C,D if and only if it is crossed by the diagonal GG’.