Problem 13 - Entrance Test

We are given the equation (a+b+c)(a+b-c) = 3ab. Let's expand the left side of the equation. We can group (a+b) as a single term: ( (a+b) + c ) ( (a+b) - c ) = 3ab. This is in the form (X+Y)(X-Y) = X^2 - Y^2, where X = (a+b) and Y = c. So, (a+b)^2 - c^2 = 3ab. Expand (a+b)^2: a^2 + 2ab + b^2 - c^2 = 3ab. Rearrange the terms to isolate a^2 + b^2 - c^2: a^2 + b^2 - c^2 = 3ab - 2ab a^2 + b^2 - c^2 = ab. Now, recall the Law of Cosines, which states that for a triangle with sides a, b, c and angle C opposite to side c: c^2 = a^2 + b^2 - 2ab * cos(C). We can rearrange the Law of Cosines to express a^2 + b^2 - c^2: a^2 + b^2 - c^2 = 2ab * cos(C). Now we have two expressions for a^2 + b^2 - c^2. Equate them: ab = 2ab * cos(C). Since a and b are side lengths of a triangle, they must be positive, so ab cannot be zero. We can divide both sides by ab: 1 = 2 * cos(C) cos(C) = 1/2. The angle C in a triangle must be between 0 and 180 degrees. The angle C for which cos(C) = 1/2 is 60 degrees. Therefore, the measure of angle C is 60 degrees.