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Given any set $S$ of $9$ points (no three collinear) within a unit square, there always exist $3$ distinct points in $S$ such that the area of the triangle formed by these $3$ points is less than or equal to $\frac{1}{8}$. Proving existence of a triangle of area $\leq \frac{1}{8}$

That means having 9 points is a sufficient condition for an existing triangle that has area $\leq \frac{1}{8}$. But is it necessary? In other words, if we have less than 9 points, we can always find a placement that all triangle area $> \frac{1}{8}$. Is this true or false?

For example, if I place $8$ points in a unit square (no three collinear), is there any placement that the area of the triangle formed by any $3$ points is larger than $\frac{1}{8}$? If such placement exists, how to find it?

I tried to use Python brute-force method (repeated 20000000 times) but failed to find such a placement.

Any idea?

Hui
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  • Please do not delete then repost questions. – Shaun May 12 '25 at 11:21
  • The problem can be stated in the language of Real Closed Fields, which permits quantifier elimination, and so the problem is decidable. Following Dan's idea, it probably suffices to solve the version of the problem with 6 points, as this reduces the dimensionality to 12. It's possible to eliminate a couple degrees of freedom by assuming some of the points are located on edges. – Jade Vanadium May 19 '25 at 06:14
  • I really don't think this is a combinatorics problem honestly. I'm certain Dan is correct, and it should be provable also, but the proof is just not going to be as simple as the 9 point problem. You need to get into the weeds of the geometry and, with so many degrees of freedom, it's not going to be easy. – Jade Vanadium May 19 '25 at 23:32

1 Answers1

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I don't think I can get to six without running out of room: all six of the triangles in this image have area $1/8$; I can't move any of the points without either leaving the unit square or making one of the triangles smaller. The interior points are at $(1/2, 1/4)$ and $(1/2, 3/4)$.

Six triangles in the unit square: four extending from two interior points to the corners, and two connecting one interior point to a nearby edge of the square

Dan Uznanski
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  • Mmmnnnng. You might be able to beat this with a convex thing with 2-fold (but not 4-fold!) rotational symmetry, but I can't do it in my head. – Dan Uznanski May 11 '25 at 20:26
  • Putting an almost regular hexagon into the square also works. However, I think if you try to put an almost regular heptagon some of the triangles will be a little smaller than $1/8$ but I think such a configuration would get closest. – quarague May 12 '25 at 06:36
  • @quarague Wait, why "almost regular"? Why wouldn't the largest regular hexagon work? – Debalanced May 12 '25 at 14:50
  • @Debalanced You can make an almost regular hexagon with the points (0.5, 0), (0.5, 1), (0, 0.25), (0, 0.75), (1, 0.25), (1, 0.75) that works. I don't think one can fit a bigger regular hexagon into the square. – quarague May 12 '25 at 16:27
  • @quarague I see, I am very confident 7 points doesn’t work, mostly because the tightest circle packing contains two triangles with area approx 0.1245 – Debalanced May 12 '25 at 17:24
  • @quarague that arrangement can be converted into uncountably many equivalent arrangements by applying an area preserving skew transform, namely one which leaves the points at $(0.5,0)$ and $(0.5,1)$ fixed. The largest such skew can be used to produce the configuration $(0,0),(0,0.5),(0.5,0), (1,1),(0.5,1),(1,0.5)$ where the smallest triangles have area $\frac{1}{8}$. – Jade Vanadium May 20 '25 at 14:56