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 4 of 4: Cover the remaining integers
Previous stepNext step
2m+n2−3and2m+n2−4(m,n≥3),6,7,8,9,10 by the explicit base diagrams.2m+n^2-3\quad\text{and}\quad2m+n^2-4\quad(m,n\ge3),\qquad6,7,8,9,10\text{ by the explicit base diagrams}.
Detailed analysis

The second standard arrangement in the official solution gives two convex-hexagon families with counts 2m+n^2-3 and 2m+n^2-4 for m,n>=3; these generate every integer at least 11. The official base diagrams give convex hexagons with 6,7,8,9,10 congruent triangles. Together with the arithmetic families, this proves the assertion for every integer n>=6.