Maths Olympiad Prep

Library / /108 of 377

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Consider a 20032003-gon inscribed in a circle and a triangulation of it with diagonals intersecting only at vertices. What is the smallest possible number of obtuse triangles in the triangulation?

Solution

Solution:

By induction, it follows easily that any triangulation of an nn-gon inscribed in a circle has n2n-2 triangles. A triangle is obtuse unless it contains the center of the circle in its interior (in which case it is acute) or on one of its edges (in which case it is right). It is then clear that there are at most 22 non-obtuse triangles, and 22 is achieved when the center of the circle is on one of the diagonals of the triangulation. So the minimum number of obtuse triangles is 20012=19992001-2=1999.

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.