Problem 10 - Entrance Test

Let the angles be A, B, C. Since they are in arithmetic progression, we can write them as B-d, B, B+d for some common difference d. The sum of angles in a triangle is 180 degrees: (B-d) + B + (B+d) = 180. 3B = 180, so B = 60 degrees. Let the sides opposite angles A, B, C be a, b, c respectively. Given that the area of the triangle is 18. Area = (1/2)ac sin(B) = 18. Since B = 60 degrees, sin(B) = sin(60) = sqrt(3)/2. (1/2)ac (sqrt(3)/2) = 18 ac (sqrt(3)/4) = 18 ac = 72 / sqrt(3) = 72 * sqrt(3) / 3 = 24sqrt(3). Given that the perimeter is 18. a + b + c = 18. Now, use the Law of Cosines for side b: b^2 = a^2 + c^2 - 2ac cos(B) b^2 = a^2 + c^2 - 2ac cos(60) b^2 = a^2 + c^2 - 2ac (1/2) b^2 = a^2 + c^2 - ac. From the perimeter, we know a + c = 18 - b. Square both sides: (a+c)^2 = (18-b)^2 a^2 + c^2 + 2ac = (18-b)^2 a^2 + c^2 = (18-b)^2 - 2ac. Substitute this expression for (a^2 + c^2) into the Law of Cosines equation: b^2 = (18-b)^2 - 2ac - ac b^2 = (18-b)^2 - 3ac. We know ac = 24sqrt(3). b^2 = (18-b)^2 - 3(24sqrt(3)) b^2 = 324 - 36b + b^2 - 72sqrt(3). The b^2 terms cancel out: 0 = 324 - 36b - 72sqrt(3) 36b = 324 - 72sqrt(3) Divide by 36: b = (324 - 72sqrt(3)) / 36 b = 324/36 - 72sqrt(3)/36 b = 9 - 2sqrt(3). The length of the side opposite angle B is 9 - 2sqrt(3).