Question: from Matthew Coleman, USA
Three distinct vertices are chosen at random from the vertices of a given regular hendecagon (11-sided polygon). If each vertex is equally likely to be chosen, what is the probability that the center of the hendecagon lies in the interior of the triangle determined by the three randomly chosen vertices? Note: Your answer must include a leading zero.
Answer: There are (11 choose 3) ways to pick three vertices from the hendecagon. (11 choose 3) = 165. We will count the ways to select the vertices such that the center is NOT in the interior of the triangle. These are exactly the ways in which the three chosen vertices are among 5 consecutive vertices. We will count clockwise around the polygon. So there are 11 ways to choose the first vertex. Then there are (4 choose 2) ways to choose the other two vertices. (4 choose 2) = 6. So the probability that the triangle does not contain the center is p = (11 * 6)/165 = 0.4. So the probability that the triangle contains the center is 1 – 0.4 = 0.6.
Solution to bonus question
Using the Pythagorean theorem, the length of MN is sqrt(13) and MP is 2.5. Angle BCD is sin-1 4/5 = 53.13010235 degrees. Using the law of cosines, NO = 2.5. Angle ADC = 180 - 53.13010235 = 126.86989765 degrees. Using the law of cosines, PO = sqrt(13). Thus, MNOP is a rectangle and its perimeter is 5 + 2sqrt(13) = 12.2111 units after rounding to 4 decimals.