Would anyone be able to help me with the following problem or give me a push in the right direction? I am not entirely sure where to start and I have been looking at this problem for hours... Any help is appreciated.

Let n ≥ 3. Suppose an n- sided convex polygon is formed by taking n points on a circle, and joining adjacent points by straight lines. A diagonal of such a polygon is a straight line between two corners, that is not one of the sides of the original polygon.

Prove that the maximum number of diagonals that can be drawn in such a convex n- sided polygon, so that no two diagonals meet except at a corner, is n - 3.

  • $\begingroup$ Start small. What if $n=4$? $\endgroup$ – Lee Mosher Sep 16 '14 at 12:52
  • 2
    $\begingroup$ Last phrase: "is $~n\ge 3$" or "is $~n-3$" must be ? $\endgroup$ – Oleg567 Sep 16 '14 at 12:58

Denote maximal number of such diagonals as $D(n)$.

For triangle it is obvious: it has $0$ diagonals: $D(3)=0$.

For quadrilateral it is obvious: it can has only $1$ diagonal: $D(4)=1$.

Now use math. induction.

Suppose that for any $n<n_0$ this statement is true: $D(n)=n-3$.
Prove that this statement is true for $n=n_0$ (prove that $D(n_0)=n_0-3$).

Any diagonal cuts $n_0$-gon into $2$ parts:

  • convex $j$-gon,
  • convex $(n_0-j+2)$-gon.

($3\le j < n_0$).

Apply statement for $j$-gon and for $(n_0-j+2)$-gon separately.

You'll obtain:

$$ D(n_0) = 1+(j-3)+(n_0-j-1)=n_0-3. $$

$\qquad \qquad$ Polygon


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