Maths Olympiad Prep

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, 2010

Geometry Difficulty 5.1 AIME, harder Prove it Estonia

Juku drew a regular hexagon and chose three triangles with different areas whose vertices were among the vertices of the hexagon. Prove that the sum of the areas of the triangles is equal to the area of the hexagon.

Solution

Any triangle whose vertices are among the vertices of a regular hexagon is one of the following:
* a triangle Δ1\Delta_1 whose vertices are three consecutive vertices of the hexagon;
* a triangle Δ2\Delta_2 whose two vertices are adjacent vertices of the hexagon and the third one is adjacent to none of the first two;
* a triangle Δ3\Delta_3 where any two vertices are not adjacent vertices of the hexagon.

Since the areas of the chosen triangles are different, the triangles must be equal to the triangles Δ1\Delta_1, Δ2\Delta_2, Δ3\Delta_3. The hexagon can be divided into four parts (Fig. 9): the triangle Δ3\Delta_3 surrounded by three triangles Δ1\Delta_1. The area of the triangle Δ2\Delta_2 (marked by a dotted line in Fig. 9) is twice the area of the triangle Δ1\Delta_1 because they have the same base but the height of Δ2\Delta_2 is twice the height of Δ1\Delta_1.

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