Given a polygon with 2016 vertices. Alisa and Basilio play the following game. In each turn a player draws a diagonal of the polygon which intersects the other drawn diagonals or the sides only at the vertices. When the polygon is cut into triangles the game is finished. For each triangle having exactly zero sides among the sides of the initial polygon Alisa is paid 1 cent. For each triangle having exactly two sides among the sides of the initial polygon Basilio is paid 1 cent. Who will get more money and what would be the difference if both are clever players?
Fig. 02
Solution
Let a be the number of the triangles which have 0 sides among the sides of the initial polygon, b be the number of the triangles having 1 such side, and c be the number of the triangles with 2 such sides (fig. 02). Then b+2c=2016 since the polygon has 2016 sides. Also, since we will get 2014 triangles, 2014=a+b+c. Hence c=a+2. So Basilio will get 2 cents more than Alisa with no dependence on how they are playing.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.
