Q27
Q27 - In △ A B C \triangle ABC , A B = 5 AB = 5 , B C = 7 BC = 7 , and C A = 8 CA = 8 . Let E E and F F be the feet of the altitudes from B B and C C , respectively, and let M M be the midpoint of B C BC . The area of triangle M E F MEF can be expressed as a b c \frac{a\sqrt{b}}{c} for positive integers a a , b b , and c c such that the greatest common divisor of a a and c c is 1 and b b is not divisible by the square of any prime. Compute a + b + c a + b + c . Solution: We first observe that quadrilateral EFBC is cyclic with circumcenter M(Thales theorem,BC diameter subtending 90 on circumference) since ∠ B E C = ∠ C F B = 90 ∘ \angle BEC = \angle CFB = 90^\circ . Thus, M B = M F = M E = M C = B C / 2 = 7 / 2 MB = MF = ME = MC = BC/2 = 7/2 as these segments are radii of the circumscribed circle of EFBC, so triangles △ M B F \triangle MBF , △ M E C \triangle MEC and △ M E F \triangle MEF are isosceles. From these observations, we reduce that ∠ B F M = ∠ B \angle BF...