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IMC / 2006 / Problems / Day 2, P7

IMC 2006 · Day 2 · P7

easy

Let VV be a convex polygon with nn vertices.

(a) Prove that if nn is divisible by 3 then VV can be triangulated (i.e. dissected into non-overlapping triangles whose vertices are vertices of VV) so that each vertex of VV is the vertex of an odd number of triangles.

(b) Prove that if nn is not divisible by 3 then VV 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 nn. For the initial cases n=3,4,5n = 3, 4, 5, chose

the triangulations shown in the Figure to prove the statement.

Now assume that the statement is true for some n=kn = k and consider the case n=k+3n = k + 3. Denote the vertices of VV by P1,,Pk+3P_1, \dots, P_{k+3}. Apply the induction hypothesis on the polygon P1P2PkP_1 P_2 \dots P_k; in this triangulation each of vertices P1,,PkP_1, \dots, P_k belong to an odd number of triangles, except two vertices if nn is not divisible by 3. Now add triangles P1PkPk+2P_1 P_k P_{k+2}, PkPk+1Pk+2P_k P_{k+1} P_{k+2} and P1Pk+2Pk+3P_1 P_{k+2} P_{k+3}. This way we introduce two new triangles at vertices P1P_1 and PkP_k so parity is preserved. The vertices Pk+1P_{k+1}, Pk+2P_{k+2} and Pk+3P_{k+3} share an odd number of triangles. Therefore, the number of vertices shared by even number of triangles remains the same as in polygon P1P2PkP_1 P_2 \dots P_k.

How the field did

contestants scored
237
average (of 20)
18.40
solved (≥ 80%)
87.8%
near-0 (≤ 10%)
3.4%
discrimination
0.35

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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