Interesting Diagonal Discovery

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Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all!

Consider an n-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for n=4,5,6,7,8.

Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one)

Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one)

Based on above experiment, I observed:

The number of new diagonals contributed by each vertex of a n-sided polygon follows the sequence: (n-3),(n-3),(n-4),\ldots, 1,0,0

Now let’s try to  prove this observation:

Since we can’t draw diagonals to the adjacent vertices, each vertex will have (n-1)-2 = (n-3) diagonals.

Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is (n-3)).

For first vertex, we have no restriction, so it will contribute (n-3) diagonals.

Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute (n-3) diagonals.

But, starting with third vertex we observe that first vertex has already taken one of the diagonals from its maximum contribution (second vertex can’t affect its contribution count since it’s adjacent vertex), thus it contributes (new) (n-3)-1 = (n-4) diagonals.

Continuing same way, for k^{th} vertex, consider the restriction to contribution caused by 1^{st} to (k-2)^{th} vertices. Thus for 1<k<n we get the number of new diagonals contributed by k^{th} vertex equal to (n-3)-(k-2) = (n-k-1).

Since new contribution can’t be negative and for (n-1)^{th} vertex we end up with zero (new) contribution, n^{th} vertex will also contribute zero diagonals.

Combining all of the above arguments I complete the proof of my observation and call it “New Diagonal Contribution Theorem” (NDCT).

4 responses

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