Putnam 1997, Problem B-6

The dissection of the $3$-$4$-$5$ triangle shown below (into four congruent right triangles similar to the original) has diameter $5/2$. Find the least diameter of a dissection of this triangle into four parts. (The diameter of a dissection is the least upper bound of the distances between pairs of points belonging to the same part.)


Solution:
The answer is \( 25/13 \). Place the triangle on the cartesian plane so that its vertices are at \( C = (0,0) \), \( A = (0,3) \), \( B = (4,0) \). Define also the points \( D = (20/13,24/13) \), and \( E = (27/13,0) \). We then compute that  
\[
\frac{25}{13} = AD = BE = DE, \quad \frac{27}{13} = BC - CE = BE < BC, \quad \frac{39}{13} = AC < \sqrt{AC^2 + CE^2} = AE, \quad \frac{40}{13} = AB - AD = BD < AB
\]  
and that \( AD < CD \). In any dissection of the triangle into four parts, some two of \( A,B,C,D,E \) must belong to the same part, forcing the least diameter to be at least \( 25/13 \).  

We now exhibit a dissection with least diameter \( 25/13 \). (Some variations of this dissection are possible.) Put \( F = (15/13,19/13) \), \( G = (15/13,0) \), \( H = (0,19/13) \), \( J = (32/15,15/13) \), and divide \( ABC \) into the convex polygonal regions \( ADFH \), \( BEJ \), \( CGFH \), \( DFGEJ \). To check that this dissection has least diameter \( 25/13 \), it suffices (by the following remark) to check that the distances  
\[
AD, AF, AH, BE, BJ, DE, CF, CG, CH, DF, DG, DH, DJ, EF, EG, EJ, FG, FH, FJ, GJ
\]  
are all at most \( 25/13 \). This can be checked by a long numerical calculation, which we omit in favor of some shortcuts: note that \( ADFH \) and \( BEJ \) are contained in circular sectors centered at \( A \) and \( B \), respectively, of radius \( 25/13 \) and angle less than \( \pi/3 \), while \( CGFH \) is a rectangle with diameter \( CF < 25/13 \).  

**Remark.** The preceding argument uses implicitly the fact that for \( P \) a simple closed polygon in the plane, if we let \( S \) denote the set of points on or within \( P \), then the maximum distance between two points of \( S \) occurs between some pair of vertices of \( P \). This is an immediate consequence of the compactness of \( S \) (which guarantees the existence of a maximum) and the convexity of the function taking \( (x,y) \in S \times S \) to the squared distance between \( x \) and \( y \) (which is obvious in terms of Cartesian coordinates).