Library / /1156 of 1394

, 2016

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

For positive integers $n$ , let $S_{n}$ be the set of integers $x$ such that $n$ distinct lines, no three concurrent, can divide a plane into $x$ regions (for example, $S_{2} = \{3, 4\}$ , because the plane is divided into 3 regions if the two lines are parallel, and 4 regions otherwise). What is the minimum $i$ such that $S_{i}$ contains at least 4 elements?

Solution

Solution:

For $S_{3}$ , either all three lines are parallel (4 regions), exactly two are parallel (6 regions), or none are parallel (6 or 7 regions, depending on whether they all meet at one point), so $|S_{3}| = 3$ .

Then, for $S_{4}$ , either all lines are parallel (5 regions), exactly three are parallel (8 regions), there are two sets of parallel pairs (9 regions), exactly two are parallel (9 or 10 regions), or none are parallel ( $8, 9, 10$ , or 11 regions), so $|S_{4}| = 4$ .

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.