# Cross Points in a Polygon

Let P be a point in the plain of polygon A_{0}A_{1}...A_{n-1}, n an odd integer. The side A_{k}A_{k+1} is said to be opposite vertex A_{m}, if _{m} with the side opposite to A_{m}. Prove that the number of cross points is always odd.

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Copyright © 1996-2018 Alexander Bogomolny
Let P be a point in the plain of polygon A_{0}A_{1}...A_{n-1}, n an odd integer. The side A_{k}A_{k+1} is said to be opposite vertex A_{m}, if _{m} with the side opposite to A_{m}. Prove that the number of cross points is always odd.

What if applet does not run? |

### Proof

Observe that the notion of being "opposite" is - in a sense - reciprocal. If A_{k}A_{k+1} is opposite A_{m}, then side A_{m}A_{m+1} is opposite A_{k+1}. Indeed, if

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

Assume PA_{m} crosses the side A_{k}A_{k+1} opposite A_{m}. There are two ways to PA_{m} may cross the extension of A_{k}A_{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_{k}A_{k+1} towards A_{k}. To move the cross point over beyond A_{k}, P needs to cross the diagonal A_{k}A_{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_{k}A_{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

_{m}are the cross points on the side line of side A

_{k}A

_{k+1}opposite vertex A

_{m},

_{k}B

_{m}/B

_{m}A

_{k+1}equals 1.

Observe that the ratio A_{k}B_{m}/B_{m}A_{k+1} is positive if and only if B_{m} is an internal point of segment A_{k}A_{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_{k}B_{m}/B_{m}A_{k+1} = Area(A_{k}B_{m}P)/Area(B_{m}A_{k+1}) = [PA_{k}·sin(∠A_{k}PB_{m})] / [PA_{k+1}·sin(∠B_{m}PA_{k+1})],

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

Note that, say, ∠A_{k}PB_{m} = ∠B_{k}PA_{m} and ∠B_{m}PA_{k+1} = ∠A_{m}PB_{k+1} - because of the *reciprocity*. For this reason, all the terms on the right in the product of all the ratios A_{k}B_{m}/B_{m}A_{k+1} cancel out, thus proving the statement.

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Copyright © 1996-2018 Alexander Bogomolny