Cross Points in a Polygon

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


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Proof

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Copyright © 1996-2017 Alexander Bogomolny

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


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Proof

Observe that the notion of being "opposite" is - in a sense - reciprocal. If AkAk+1 is opposite Am, then side AmAm+1 is opposite Ak+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 PAm crosses the side AkAk+1 opposite Am. There are two ways to PAm may cross the extension of AkAk+1, instead. It's either on the side of Ak or on the side of Ak+1. Imagine P move so that the cross point moves in the interior of AkAk+1 towards Ak. To move the cross point over beyond Ak, P needs to cross the diagonal AkAm. When this happens, both PAk and PAm 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 AkAk+1 Bk. 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, Bm are the cross points on the side line of side AkAk+1 opposite vertex Am, k = 0, ..., n-1. Then the product of all fractions AkBm/BmAk+1 equals 1.

Observe that the ratio AkBm/BmAk+1 is positive if and only if Bm is an internal point of segment AkAk+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:

AkBm/BmAk+1 = Area(AkBmP)/Area(BmAk+1) = [PAk·sin(∠AkPBm)] / [PAk+1·sin(∠BmPAk+1)],

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

Note that, say, ∠AkPBm = ∠BkPAm and ∠BmPAk+1 = ∠AmPBk+1 - because of the reciprocity. For this reason, all the terms on the right in the product of all the ratios AkBm/BmAk+1 cancel out, thus proving the statement.

Menelaus and Ceva

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