MathLabs

Problem 5

Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have.
Step 3 of 5: Correct the pair count
30⋅202=300\frac{30\cdot20}{2}=300
Detailed analysis

Summing the bound over all 3030 perpendiculars counts each ordinary intersection twice, once for each line. Therefore there are at most 300300 ordinary intersections.