Difference between revisions of "1983 USAMO Problems/Problem 1"
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| − | + | == Problem == | |
| + | On a given circle, six points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, and <math>F</math> are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles <math>ABC</math> and <math>DEF</math> are disjoint, i.e., have no common points. | ||
| + | == Solution == | ||
| + | First we give the circle an orientation (e.g., letting the circle be the unit circle in polar coordinates). Then, for any set of six points chosen on the circle, there are exactly <math>6!</math> ways to label them one through six. Also, this does not affect the probability we wish to calculate. This will, however, make calculations easier. | ||
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| + | Note that, for any unordered set of six points chosen from the circle boundary, the number of ways to number them such that they satisfy this disjoint-triangle property is constant: there are six ways to choose which triangle will be numbered with the numbers one through three, and there are <math>(3!)^2</math> ways to arrange the numbers one through three and four through six on these two triangles. Therefore, for any given configuration of points, there are <math>6^3=216</math> ways to label them to have this disjoint-triangle property. There are, however, <math>6!=720</math> ways to label the points in all, so given any six unordered points, the probability that when we inflict an ordering we produce the disjoint-triangle property is <math>216/720=3/10</math>. | ||
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| + | Since this probability is constant for any configuration of six unordered points we choose, we must have that <math>3/10</math> is the probability that we produce the disjoint-triangle property if we choose the points as detailed in the problem statement. | ||
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| + | == See Also == | ||
| + | {{USAMO box|year=1983|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Olympiad Combinatorics Problems]] | ||
Latest revision as of 21:29, 17 July 2016
Problem
On a given circle, six points 
, 
, 
, 
, 
, and 
are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles 
and 
are disjoint, i.e., have no common points.
Solution
First we give the circle an orientation (e.g., letting the circle be the unit circle in polar coordinates). Then, for any set of six points chosen on the circle, there are exactly 
ways to label them one through six. Also, this does not affect the probability we wish to calculate. This will, however, make calculations easier.
Note that, for any unordered set of six points chosen from the circle boundary, the number of ways to number them such that they satisfy this disjoint-triangle property is constant: there are six ways to choose which triangle will be numbered with the numbers one through three, and there are 
ways to arrange the numbers one through three and four through six on these two triangles. Therefore, for any given configuration of points, there are 
ways to label them to have this disjoint-triangle property. There are, however, 
ways to label the points in all, so given any six unordered points, the probability that when we inflict an ordering we produce the disjoint-triangle property is 
.
Since this probability is constant for any configuration of six unordered points we choose, we must have that 
is the probability that we produce the disjoint-triangle property if we choose the points as detailed in the problem statement.
See Also
| 1983 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
