Round 5: Problem 5
Each of nine lines cut a unit square into two quadrilaterals whose areas are in a ratio of 2:1. Prove that at least three lines must be concurrent (i.e. they intersect at a common point).
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***Spoilers***
Take a square ABCD.
A line must go through two opposite sides to make two quadrilaterals. Assume that such a line passes through sides AB and CD. Connect the midpoints of BC and AD by a line segment, called EF.
The line must intersect EF at some point. Because the ratio is fixed, it can be shown that the line splits EF into a ratio of 2:1. There are only two points on EF that split it into 2:1.
If the line passed through BC and AD instead, then it must intersect the midpoint connector of AB and CD instead, giving us two more points.
So every line has to pass through exactly one of four points. Since there are nine lines passing through four points, at least one point must be on at least three lines.
Franklin Lee
May 13, 2009 at 17:30