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 3 of 4: Fill the fourth residue class
q=n+1,p=n+2,(n+1)p=(n+2)q⟹4n+3.q=n+1,\quad p=n+2,\quad (n+1)p=(n+2)q\quad\Longrightarrow\quad4n+3.
Detailed analysis

To obtain 4n+3 pieces, use the modified block arrangement from the official solution. Its upper and lower parts contain 2n+1 and 2n+2 small triangles; choose q=n+1 and p=n+2, so the joining lengths match because (n+1)p=(n+2)q. The result is convex and contains 4n+3 congruent triangles.