29. [15] Let ABC be a triangle such that AB = 3, AC = 4, and ∠ BAC = 75 . Square BCDE is 2 2 constructed outside triangle ABC . Compute AD + AE . Proposed by: Pitchayut Saengrungkongka √ √ Answer: 75 + 24 2 = 75 + 1152 Solution: X A B C E D

∼ Construct point X such that △ CBD ∼ △ CXA . Then, △ CBX △ CAD . Thus, AD = BX . We

√ ◦ have AB = 3, AX = 4 2, and ∠ BAX = 120 , so law of cosine gives √ 2 √ √ 2 2 2 ◦ AD = BX = 3 + 4 2 − 2 · 3 · (4 2) · cos 120 = 41 + 12 2 . Similarly, we may compute 2 √ √ √ 2 2 ◦ AE = 4 + 3 2 − 2 · 4 · (3 2) · cos 120 = 34 + 12 2 , √ so the answer is 75 + 24 2.