IMC / 2006 / Problems / Day 2, P7
IMC 2006 · Day 2 · P7
easyLet be a convex polygon with vertices.
(a) Prove that if is divisible by 3 then can be triangulated (i.e. dissected into non-overlapping triangles whose vertices are vertices of ) so that each vertex of is the vertex of an odd number of triangles.
(b) Prove that if is not divisible by 3 then can be triangulated so that there are exactly two vertices that are the vertices of an even number of the triangles.
Solution (official)
Apply induction on . For the initial cases , chose
the triangulations shown in the Figure to prove the statement.
Now assume that the statement is true for some and consider the case . Denote the vertices of by . Apply the induction hypothesis on the polygon ; in this triangulation each of vertices belong to an odd number of triangles, except two vertices if is not divisible by 3. Now add triangles , and . This way we introduce two new triangles at vertices and so parity is preserved. The vertices , and share an odd number of triangles. Therefore, the number of vertices shared by even number of triangles remains the same as in polygon .
How the field did
Score distribution (field cohort)
Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.