Problem 6 - Entrance Test

Let the centers of the two circles be O1 and O2, and their radii be r1 = 5 and r2 = 3, respectively. The distance between their centers is d = O1O2 = 7. The common chord AB is perpendicular to the line segment O1O2 and is bisected by it. Let M be the midpoint of AB. So, AB = 2 * AM. Consider the triangle O1AO2. Its sides are r1=5, r2=3, and d=7. The area of this triangle can be calculated using Heron's formula. First, find the semi-perimeter s = (r1 + r2 + d) / 2 = (5 + 3 + 7) / 2 = 15/2. Area of O1AO2 = sqrt(s * (s-r1) * (s-r2) * (s-d)) Area = sqrt((15/2) * (15/2 - 5) * (15/2 - 3) * (15/2 - 7)) Area = sqrt((15/2) * (5/2) * (9/2) * (1/2)) Area = sqrt((15 * 5 * 9 * 1) / (2 * 2 * 2 * 2)) Area = sqrt(675 / 16) Area = (sqrt(225 * 3)) / 4 = (15 * sqrt(3)) / 4. We can also express the area of triangle O1AO2 in terms of its base O1O2 and height AM. The height corresponding to the base O1O2 is AM, since AM is the altitude from A to O1O2. Area of O1AO2 = 1/2 * base * height = 1/2 * O1O2 * AM = 1/2 * 7 * AM. Equating the two expressions for the area: (15 * sqrt(3)) / 4 = 1/2 * 7 * AM Multiply both sides by 2: (15 * sqrt(3)) / 2 = 7 * AM AM = (15 * sqrt(3)) / (2 * 7) = (15 * sqrt(3)) / 14. The length of the common chord AB is 2 * AM: AB = 2 * (15 * sqrt(3)) / 14 = (15 * sqrt(3)) / 7.