Maximum distance between points in a triangle

An equilateral triangle has sides of unit length.
a)Show that if five points lie in/on the triangle, then at least two of the points lie no farther than 0.5 units apart.
b)Show that 0.5 cannot be replaced by a smaller number even if there are six points.
c)If there are eight points, can 0.5 be replaced by a smaller number?

a) was simple, just divided the triangle into four equilateral triangles and used PHP. b) followed with a counterexample by placing the points on the vertices and midpoints, but c) has me stumped. I tried an approach similar to the one and think that the answer should be Yes, but can't prove it.

PS. How can we rigorously prove that two points in one of the sub-triangles differ by at most 0.5 unit. The distance "should" be maximal when the points are on the vertices, but can't prove it.

Solution 1:

Re your PS: Pick two points $P,Q$ in an equilateral triangle $A,B,C$. project $A,B,C$ to the line $PQ$, giving $A',B',C'$. Say $A'$ is the "leftmost" and $B'$ the "rightmost" of the projections. Then $|PQ|\le |A'B'|\le |AB|$. (We see that by the same argument, two points in a polygon are at most as far apart as two vertices, i.e., as the maximal edge-or.diagonal.

Now to c: Let $ABC$ be our triangle. Pick suitable $r=\frac{\sqrt3-1}{2}\approx 0.366<\frac12$. Draw the circles with radius $r$ around $A,B,C$. In the "middle" there is room for a small equilateral triangle with vertices on these circles and side length $1-r\sqrt 3=r$. Verify that all seven parts in the figure have diameter $r$. Therefore, the pigeons strike back if we have eight points.

To visualize, in the following illustration all blue circles have radius $r$ and centre one of the nodes:

Solution 2:

To answer your 'PS': Consider two distinct points in an equilateral triangle. Draw the line joining them and let this line intersect the triangle in $P$ and $Q$. It will do so in two distinct points because the original points are themselves distinct. Obviously, the distance between $P$ and $Q$ is not less than that between the two original points. Now, move $P$ along the circumference of the triangle until it is as far as possible from $Q$.

$P$ will end up at a vertex of the triangle. To prove this, suppose it ended up at a point $P^{\prime}$ on the edge. Now, drop a perpendicular from $Q$ on to the edge containing $P^{\prime}$, at $D$. ${QP^{\prime}}^2 = QD^2 + {DP^{\prime}}^2$, by Pythagoras' Theorem, but the distance to the further vertex $V$ on the edge, $DV$, is not less than $DP^{\prime}$; thus ${QP^{\prime}}^2 = QD^2 + {DP^{\prime}}^2 \leq QD^2 + DV^2 = QV^2$.

Repeating the above argument for $Q$ as well, we see that both $P$ and $Q$ end up at vertices of the triangle and you cannot move them further apart. Note that the symmetry of an equilateral triangle is implicitly used in the proof, when we talk about dropping the perpendicular and pushing the points away.

To answer 'c': if we draw the four sub-triangles again, we can place six points at the vertices and thus achieve the maximum space between them, but now, a seventh point will definitely be at a distance less than $0.5$ from one of the six points. In fact, the maximum distance will be achieved when the seventh point is at the centroid of a sub-triangle, at a distance of $\frac{1}{\sqrt{3}}$ from a vertex.