# For any eight points on an equilateral triangle of side $1$, there's at least one pair of those points at most $1/3$ apart

Let's say we have an equilateral triangle of side length $$1$$. Show that for any configuration of eight points on this triangle (on the sides or in the interior), there is at least one pair of from these eight points such that the distance between the two points in the pair is less than or equal to $$1/3$$ cm.

My idea was to draw out the triangle and put three points on each edge, such that each point lies on the trisection of a side, but I'm not quite sure if this is rigorous enough.

Also, as an extension question, can we say anything for the more generalized version of this problem with an equilateral triangle of side length $$n$$ and $$m$$ points being chosen?

• – Blue Mar 6 '19 at 1:41
• Is it an homework ? If so, why are you asked to prove something false (and known to be false : see the reference I give and the excellent answer by @FredH) ? – Jean Marie Mar 6 '19 at 21:13

It turns out that it is possible to place eight points in an equilateral triangle of side $$1$$ so that the minimum distance is greater than $$\frac13$$. The diagram below shows how the points can be arranged. Points $$A$$, $$B$$ and $$C$$ are the vertices of the equilateral triangle. Point $$D$$ is the midpoint of $$AB$$. Points $$E$$ and $$F$$ are on sides $$AC$$ and $$BC$$, respectively, each at a distance $$d$$ (to be determined) from $$C$$.

Points $$G$$ and $$H$$ are the circumcenters of triangles $$ADE$$ and $$BDF$$, respectively. The value of $$d$$ will be chosen so that the circumradius of each circle is also $$d$$, which will then be the minimum distance between any pair of the eight points.

It remains to compute $$d$$. Since $$\angle DAE = \frac{\pi}{3}$$, the central angle $$\angle DGE$$ subtending the same arc is $$\frac{2\pi}{3}$$. $$GD = GE = d$$, so an application of the law of cosines to $$\angle DGE$$ finds that $$DE = \sqrt{3}\,d$$. $$AE = 1 - d$$ and $$AD = \frac12$$, so using the law of cosines at $$\angle DAE$$ gives $$(1-d)^2 + (\frac12)^2 - 2(1-d)(\frac12)\cos\frac{\pi}{3} = (\sqrt{3}\,d)^2,$$ which simplifies to $$8d^2 + 6d - 3 = 0$$. The positive root is $$d = \dfrac{\sqrt{33}-3}{8} \approx 0.34307$$. Since this is greater than $$\frac13$$, this arrangement satisfies the claimed condition.

(I first found a different arrangement that worked, although with a smaller minimum distance. It then occurred to me to checking the best-known packings of circles into an equilateral triangle at Erich Friedman's Packing Center. The packing for eight circles, when adapted to points, was both better and easier to explain than the one I had found earlier.)

• One can find also in quora.com/… the fact that this distance is slightly bigger than 1/3 – Jean Marie Mar 6 '19 at 21:09