Cross Points in a Polygon

Let P be a point in the plain of polygon A₀A₁...A_n-1, n an odd integer. The side A_kA_k+1 is said to be opposite vertex A_m, if k ≡ m + ⌊n/2⌋ mod n. Cross points are the intersections of lines PA_m with the side opposite to A_m. Prove that the number of cross points is always odd.

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Proof

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Proof

Observe that the notion of being "opposite" is - in a sense - reciprocal. If A_kA_k+1 is opposite A_m, then side A_mA_m+1 is opposite A_k+1. Indeed, if k = m + ⌊n/2⌋ mod n, then

(k + 1) + ⌊n/2⌋ = (m + ⌊n/2⌋) + 1 + ⌊n/2⌋ = m + 1 + n - 1 ≡ m (mod n).

Assume PA_m crosses the side A_kA_k+1 opposite A_m. There are two ways to PA_m may cross the extension of A_kA_k+1, instead. It's either on the side of A_k or on the side of A_k+1. Imagine P move so that the cross point moves in the interior of A_kA_k+1 towards A_k. To move the cross point over beyond A_k, P needs to cross the diagonal A_kA_m. When this happens, both PA_k and PA_m coincide with the diagonal so their cross points both change from the interior of the opposite side to the exterior, or vice versa. In any case, the total number of cross points changes by 2, implying that the parity of the number of cross points is invariant, i.e., independent of the position of P. Now, with P outside the polygon, the number of cross points is 1; it remains odd wherever P changes its location.

Note: there could be complications with the above argument where several vertices are collinear, or when a vertex lies on an extension of the opposite side. In those cases, the mere definition of a cross point becomes uncertain. With the eclusion of these case, the argument goes through even when the polygon is not convex.

There is an extension of Ceva's theorem to n-gons, with n odd. Let's denote the cross point on side A_kA_k+1 B_k. Think of the polygon as being oriented. The orientation of the polygon induces an orientation on every side and side line. For two points X, Y on a side line of the polygon, XY is thought to be positive if the direction from X to Y follows the induced orientation on the side line. It is thought to be negative otherwise.

Extension of Ceva's Theorem to n-gons, n odd

Let, for a given point P, B_m are the cross points on the side line of side A_kA_k+1 opposite vertex A_m, k = 0, ..., n-1. Then the product of all fractions A_kB_m/B_mA_k+1 equals 1.

Observe that the ratio A_kB_m/B_mA_k+1 is positive if and only if B_m is an internal point of segment A_kA_k+1. For the whole product to be positive, let alone 1, the number of positive ratios needs to be odd, so that the number of negative ratios would be even.

For a proof, I'll use the sine formula for the area of a triangle:

A_kB_m/B_mA_k+1 = Area(A_kB_mP)/Area(B_mA_k+1) = [PA_k·sin(∠A_kPB_m)] / [PA_k+1·sin(∠B_mPA_k+1)],

where we assume that the angles inherit the signs of their bases.

Note that, say, ∠A_kPB_m = ∠B_kPA_m and ∠B_mPA_k+1 = ∠A_mPB_k+1 - because of the reciprocity. For this reason, all the terms on the right in the product of all the ratios A_kB_m/B_mA_k+1 cancel out, thus proving the statement.

Menelaus and Ceva

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