Let PQR be an equilateral triangle, centered at O. A point X is chosen at random inside the triangle. Find the probability that X is closer to O than to any of the sides. (In other words, find the probability that XO is shorter than XA, XB, and XC.)
The probability that X is closer to O than to any of the sides is 2/3. This is because the area of the triangle is divided into three equal parts by the lines connecting O to the vertices of the triangle. Therefore, the probability that X is in one of the two parts closer to O is 2/3.
