Difference between revisions of "2014 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
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| + | By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers <math>2</math> disjoint areas of the <math>12</math>-gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between <math>1</math> and <math>11</math>, there are <math>12</math> possible subsets like this (this is true because we can pick any of the <math>12</math> vertices as a "starting" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all <math>12</math> vertices, as well as the empty set. Thus, the total number is <math>12\cdot11 + 2 = \boxed{134}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=4|num-a=6}} | {{AIME box|year=2014|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:18, 15 January 2022
Problem 5
Let the set 
consist of the twelve vertices of a regular 
-gon. A subset 
of 
is called "communal" if there is a circle such that all points of are inside the circle, and all points of not in are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)
Solution
By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers 
disjoint areas of the -gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between 
and 
, there are possible subsets like this (this is true because we can pick any of the vertices as a "starting" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all vertices, as well as the empty set. Thus, the total number is 
.
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
