MathLabs

Problem 5

Show that for every integer n>=6 there exists a convex hexagon that can be dissected into exactly n congruent triangles.
Step 2 of 4: Obtain three arithmetic families
L1+nM+R1→4n+2,L1+nM+R2→4n+5,L2+nM+R2→4n+6.L_1+nM+R_1\to4n+2,\quad L_1+nM+R_2\to4n+5,\quad L_2+nM+R_2\to4n+6.
Detailed analysis

With p=q=1, the block counts in the official diagrams are 4n+2 for L_1+nM+R_1, 4n+5 for L_1+nM+R_2, and 4n+6 for L_2+nM+R_2. The stated lower bounds on n make each resulting boundary a convex hexagon.