Here is problem A1 of 25th Putnam 1964.
Let $A_1, A_2, A_3, A_4, A_5, A_6$ be distinct points in the plane. Let $D$ be the longest distance between any pair, and $d$ the shortest distance. Show that $D/d ≥ \sqrt3$.
The following solution is taken from this page out of John Scholes's collection of math problems and solutions.
Let ABC be a triangle with sides $a = BC ≥ b = CA ≥ c = AB$. Suppose that angle $A ≥ 2π/3$. It follows immediately from the cosine formula we have $a^2 = b^2 + c^2 - 2bc \cos A ≥ b^2 + c^2 + bc ≥ 3c^2$, and hence that $a/c ≥ \sqrt 3$. So it is sufficient to find a triangle with an angle of at least $2π/3$.
If the 6 points form a convex hexagon, then the six angles of the hexagon sum to 4π so at least one is at least 2π/3. If not then one point is inside the convex hull of the others. Draw diagonals to triangulate the hull. Then the inside point must lie inside (or on) one of the triangles. But if P lies inside (or on) the triangle ABC, then at least one of the angles APB, BPC, CPA is at least 2π/3.
I don't understand how this proof shows that $D/d \ge \sqrt 3$. This proof seems to just show that a triangle with angle greater than $2\pi/3$ exists in the triangulation of the hull of the set. How does this show that $D/d \ge \sqrt 3$? The two points $\{d_1, d_2\}$ with distance $d$ might not have any point in common with points $\{D_1, D_2\}$ that have distance $D$.
What am I missing here?
When the solution "draws diagonals to triangulate the hull", it means to triangulate the convex hull using only the given $6$ points as vertices of the triangles. No other points should be introduced. Hence every triangle obtained can be used to show $D/d\ge\sqrt3$.
To convince yourself that triangulation is possible, play with some random $6$ points of your choice. The vertices of the convex hull will be among the $6$ points. Whether the convex hull is a triangle or quadrilateral or pentagon or hexagon, you will be able to triangulate it easily without introducing new points as vertices.