Blue x Red = Unimaginable

Probability Level 4

5 points X 1 , X 2 , . . . , X 5 X_1, X_2, ..., X_5 are chosen on a plane. A blue line is drawn through each pair of points.

Now suppose the 5 points are chosen in such a way that any 2 blue lines are neither parallel nor perpendicular to each other.

Now for every point X i , 1 ≤ i ≤ 5 , X_i,\,1\le i \le 5, and every blue line L L not passing through X i X_i , a red line is drawn through X i X_i and perpendicular to L L .

(Let's take X 1 X_1 and the line X 2 X 3 X_2X_3 as an example. X 2 X 3 X_2X_3 doesn't pass through X 1 X_1 since no blue lines are parallel. Then we draw a red line that pass through the point X 1 X_1 and perpendicular the line X 2 X 3 X_2X_3 )

Find the maximum number of intersections formed by the red lines.


The answer is 315.

Raymond Chan by Raymond Chan , 20, USA

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1 solution

Jeremy Galvagni
Jul 7, 2018

For 5 starting points there will be 5C2=10 blue lines, but 4 of them will pass through it so only 6 will not. This means there are 5*6=30 red lines.

in general 30 lines can have up to 30C2=435 intersections, but this number will be smaller if there are any parallel lines or sets of 3 or more lines that coincide. There are 3 such families:

Because the red lines are constructed as perpendicular to blue lines, some will be parallel. Each blue line contains two of the points which means there are 3 points to construct perpendicular. So the red lines come in 10 sets of 3 that are parallel. Had these lines not been parallel there could have been 3 crossings. So we must deduct 10*3=30 potential crossings.

Red lines coincide at the original points in sets of 6, one for each blue line that does not pass through it. Where there could have been 6C2=15 crossings there is just 1. We lose 14 for each original point. Deduct another 14*5=70.

The process of creating red lines for each triangle is the same as for constructing the orthocenter of a triangle. These make sets of 3 lines that coincide. Where there could have been 3C2=3 crossing there is just one. We lose 2 for each triangle and there are 5C3=10 such triangles. Deduct another 2*10=20

435 − 30 − 70 − 20 = 315 \large 435-30-70-20=\boxed{315}

Incidentally, with 5 starting points there are 12 red lines and the above reasoning gives the calculation 66-6-8-8=44 which I was able to draw and count.

Also, for the record, a formula for n ≥ 4 n\ge4 starting points is ( 3 n 6 − 21 n 5 + 51 n 4 − 47 n 3 + 6 n 2 + 32 n ) / 24 (3n^{6}-21n^{5}+51n^{4}-47n^{3}+6n^{2}+32n)/24

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