Weighted Points and Lines in Projective Plane of Order 17

In the current research ,the points and lines of the projective plane of the seventeenth (17th)order were constructed ,this was followed by examining the arcs designated (k,n;f)- in the plane type (m,n).As a result, it is concluded that these arcs do exist having all their points of the order one or two ,but the order of their lines are m or n only .A further result was arriving at proving the theorems being concluded. The research also included a study of Monoidal arcs with some given examples.

In this paper, we studied the existance of(k,n;f)-arcs in projective plane of order seventeen .
To explain this work we need first the following introduction about the projective plane  .
If f is a function from the set of points of the projective plane  in to the set of natural number N, the value f(p) is called the weight point p and F is a function from the set of lines of  in to N, the value F(L) is called the weight line L, i.e F(L)=   L p f(p).
A (k,n;f)-arc Kof  is a set of k points such that K does not contain any points of weight zero.
The line L of  is called i-secant if the total weight of L is i, Xj denotes the number of the points having weight j for j=0, 1, 2,……, v (where v = k p max  f(p)) and we used Wi j for the number of lines of weight i through a point of weight j; we also denote that the number of lines of weight i is zi , the integers zi are called the characters of K . If the points in the plane are only of weight zero and one, then K is a (k, n)-arc.
Let V denote the total weight of K, so by [2] we have: Arcs for which equality holds on the left are called minimal and arcs for which equality holds on the right are called maximal also [2] has proved to be a necessary condition for the existence of a (k,n;f)-arc K of type (m, n), 0<m<n is that q ≡ 0 mod (n-m) ------(ii) and v = n-m ------(iii) 2-(k,n;f)-arc of type (n-17,n)with Li>0, i=0, 1, 2, Lj = 0, j=3,.....,17.
Simillarly we can find all others 305 lines of this plane by adding in each step one to get the other line.

4-(150,29;f)-arc of type (12,29) in PG(2,17).
To find this arc we need to find the value of X0 which makes equation (7) be a square and then find from it n and δ i.e we try to check X0 from 1 to 307 and after a large number of efforts we found that, when X0=157, we get δ=34 and n=29, and from (6) we have q=17, which means that (k, n; f)-arc of type (m, n) exist in PG (2,17). From (4) and (5), we get X1= 84 and X2= 66 Which implies that this arc is (150, 29; f)-arc of type (12, 29) and the point of weight 0 form (157, 12)-arc in PG (2, 17).

Proof:
Let L be 13-secant having on it 13 points of weight 0, then the remaining points of L is five, if all these points of weight 2 we get the weight of the line L is ten which is a contraduction . Since the weight of L is 12.#

6-Monoidal arc:
In this section we consider (k,n;f)-arc which have only one point of weight greater than one , such arcs are called monoidal arcs, i.e a (k, n; f)-arc is called monoidal if Imf={0,1,v}and Xv=1.

Proof:
Let L be 17-secants of (272, 17)-arc and having two points of weight 2, then the weight of L because 4 and this is a contraduction because the weight of all lines of the plane is either 2 or 19.# Corollary(6-2-1): There is no point of weight 1 on each 17-secants of (272,17)-arc.

Proof:
Let L is 16-secant line the remaining points of L is two .If one of these two points of weight 2 means that the weight of L is 3, which is a contraduction as in the previous lemma.# Proof:From lemma (2-1), and type (2,19) of these lines, we know that the weight of each 0-secant line is 19 , this means that each 0-secant line must be contain one point of weight 2 and 17 points of weight 1 .# see table -2.
In this table we have: 1-The star points are points of weight 2. 2-The underlined blodface points are points of weight 1. 3-The other points are of weight zero. Table (1)