Triangles in a Complete Graph

A straight-line drawing of a graph G G is a drawing of G G on the plane such that all its edges are straight line segments.

The maximum number of triangles that can be found in a straight-line drawing of the complete graph K 4 K_4 is 8, as seen in the image. For K 5 K_5 , there can be up to 35 triangles.

What is the maximum number of triangles that can be found in a straight-line drawing of the complete graph K 10 K_{10} ?


The answer is 2430.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Daniel Liu
Jul 15, 2015

We have four cases to consider based on the triangle's vertices.

Case 1: All vertices are vertices of G G . In this case, there are simply ( 10 3 ) = 120 \dbinom{10}{3}=120

Case 2: Two vertices are vertices of G G . In this case, the last vertex can be determined by the intersection of two diagonals, which means it can be determined with 4 points. Thus, there are ( 10 4 ) \dbinom{10}{4} ways. However, we can choose any two consecutive points in the 4 4 points as the other two vertices, bringing the total up to 4 ( 10 4 ) = 840 4\cdot \dbinom{10}{4} = 840

Case 3: One vertex is a vertex of G G . In this case, the two vertices are determined with the intersection of a separate diagonal with two diagonals coming out of the one vertex that is a part of G G , so this case can be determined with 5 vertices, or ( 10 5 ) \dbinom{10}{5} ways. However, we can orient the one vertex that has two diagonals coming from it as any of the 5 vertices, giving the total to be 5 ( 10 5 ) = 1260 5\cdot \dbinom{10}{5}=1260

Case 4: No vertex is a vertex of G G . In this case, the triangle can be determined by 6 vertices by connecting opposite vertices together with a diagonal. In this case there are ( 10 6 ) = 210 \dbinom{10}{6}=210 ways.

In total, there are 120 + 840 + 1260 + 210 = 2430 120+840+1260+210=\boxed{2430} triangles.

Could you explain what Kn denotes? Thanks.

Richard Zhou - 5 years, 10 months ago

In case 4, what if the three diagonals concur? Don't you need to subtract this case then? Or do you draw the complete graph so that this does not occur?

Andrew Lin - 5 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...