Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are five points in a plane situated so that no two of the straight lines formed by joining pairs of points are parallel, perpendicular or coincident. From each point perpendiculars are drawn to all the straight lines joining the other 4 points. Find the greatest number of intersections amongst all the perpendiculars drawn.

share|cite|improve this question
Are the "perpendiculars" being drawn supposed to be lines, rays, or line segments? – mjqxxxx Jul 11 '12 at 22:48
up vote 3 down vote accepted


Let's call the original lines $g$ and the perpendiculars $h$. This notation without indices is a bit vague, but will convey the idea, I believe.


  • There are $\binom52=10$ different lines $g$ in your setup. As no pair of $g$s is parallel, they must have $10$ different slopes.
  • For every line $g$, there are exactly $3$ lines $h$ to be drawn: one for each of the points not incident with $g$. So there is a total of $10\times3=30$ lines $h$ in the setup. The three lines $h$ for a single line $g$ will share a common slope, whereas lines $h$ belonging to different lines $g$ will have different slopes.
  • Lines with different slopes must intersect, whereas lines with the same slope will not have a finite point of intersection. For this reason, every line $h$ will intersect $3(10 - 1)=27$ other lines $h$, $3$ for each of the $9$ other slopes.
  • Multiplying the number of lines by the number of intersections per line, we obtain a number of $30\times 27=810$ ordered intersecting pairs, or $\frac{810}{2}=405$ unordered intersecting pairs.
  • Some points are special. The original five points each have $\binom42=6$ lines $h$ passing through them, corresponding to $\binom62=15$ unordered pairs, accounting for a total of $5\times 15=75$ unordered pairs.
  • Every combination of three original points will result in a triangle where the altitudes meet in a point. So there will be $\binom53=10$ such orthocenters, each corresponding to $\binom32=3$ unordered pairs, accounting for $10\times3=30$ such pairs.
  • The remaining $405-75-30=300$ pairs are “ordinary” intersecting pairs where in general only two lines intersect in a single point.


There are 315 points of intersection:

  • The original 5 points with 6 intersecting lines each
  • The 10 triangle orthocenters with 3 intersecting lines each
  • The 300 ordinary points with 2 intersecting lines each


So far the above only proves an upper bound. But a simple example shall suffice to prove that a count of 315 is in fact possible:

$$A=[9,1],\quad B=[0,0],\quad C=[1,8],\quad D=[5,7],\quad E=[2,2]$$

Proof by computation using $\mathbb{ZP}^2$, i.e. homogenous integer coordinates, so that you won't encounter any numeric problems.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.