There is another version of the Farey diagram in which the vertex labeled p/q is placed at the point (q,p) in the plane, so p/q is the slope of the line through the origin and (q,p). The edges of this new Farey diagram are straight line segments connecting the pairs of vertices that are connected in the original Farey diagram. For example there is a triangle with vertices (1,0), (0,1), and (1,1) corresponding to the big triangle in the upper half of the circular Farey diagram. With this model of the Farey diagram the operation of forming the mediant of two fractions just corresponds to standard vector addition (a,b)+(c,d)=(a+c,b+d).
What you are asked to do in this problem is just to draw the portion of the new Farey diagram consisting of all the triangles whose vertices (q,p) satisfy 0≤q≤5 and 0≤p≤5. Note that since fractions p/q labeling vertices are always in lowest terms, the points (q,p) such that q and p have a common divisor greater than 1 are not vertices of the diagram.