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 4 of 5: Add forced intersections
(53)=10and5\binom53=10\quad\text{and}\quad5
Detailed analysis

Each choice of three original points gives an orthocenter, hence (53)=10\binom53=10 orthocenters. The five original vertices are also intersections of perpendiculars based at the same vertex. These were excluded from the ordinary count.