2008 iTest Problems/Problem 49

Revision as of 23:03, 5 August 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 49 -- doesn't matter if there's 14 or 4000 students at the table)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy’s friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched out a blue piece of yarn between them. Then Wendy and Erin stretched out a red piece of yarn between them at about the same height so that the yarns would intersect if possible. If all possible positions of the students around the table are equally likely, let $m/n$ be the probability that the yarns intersect, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

Solution

[asy] draw(circle((0,0),65)); dot((39,52)); dot((-52,39)); dot((25,-60)); dot((-60,-25));  label("W",(-52,39),NW); [/asy]

First, let Wendy pick a seat, and let that point be point $W$. Also, pick three arbitrary points clockwise from Wendy. The four points form a cyclic quadrilateral, and the only way for two lines to intersect strictly inside the circle is if the two lines are diagonals. Given one point, only one of the three points make a diagonal of the cyclic quadrilateral, so $\tfrac{m}{n} = \tfrac{1}{3}.$ Thus, $m+n = \boxed{4}.$

See Also

2008 iTest (Problems)
Preceded by:
Problem 48
Followed by:
Problem 50
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100