Mock Geometry AIME 2011 Problems/Problem 6
Three points are chosen at random on a circle. The probability that there exists a point inside an equilateral triangle such that can be expressed in the form where are relatively prime positive integers. Find
Let , , and be the three distances from point to each of the vertices, with being the longest distance. Let's consider the case in which point is actually on the line:
In this case, we can use Stewart's Theorem to find the relationship between the three variables. This can be manipulated into the Law of Cosines, with an angle of .
In order for point to be inside the equilateral triangle: Thus, Triangle (which has sides of and ) cannot have any angles greater than or equal to . Next, we can fix point to the mark of a unit circle, as shown below:
Thus, the places points and should be on the circle, such that Triangle has all angles less than is OUTSIDE of the red outline, but inside of the green outline shown in the graph below:
The red hexagon has vertices at coordinates and , while the green square has vertices at coordinates and . Each other these coordinates represent positions for points and . For example, means point is at on the unit circle, while point is at on the unit circle.
Therefore, the probability of points and forming a triangle with angles less than is: . And the answer is - Solution by adyj