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 5 of 5: Reach the maximum
300+10+5=315300+10+5=\boxed{315}
Detailed analysis

Thus the number is at most 300+10+5=315300+10+5=315. A generic placement of the five points avoids every additional accidental concurrence or parallelism, so all counted ordinary intersections are distinct and the bound is attained. The maximum is 315\boxed{315}.