PUMaC 2011 · 几何（A 组） · 第 5 题

PUMaC 2011 — Geometry (Division A) — Problem 5

[ 5 ] Let ` and ` be two parallel lines, a distance of 15 apart. Points A and B lie on while 1 2 1 ◦ ◦ points C and D lie on such that ∠ BAC = 30 and ∠ ABD = 60 . The minimum value of 2 √ AD + BC is a b , where a and b are integers and b is squarefree. Find a + b .

解析

Without loss of generality, suppose A lies to the left of B . Let D be the point such that ′ DAD B is a parallelogram. No matter what the positions of A and B are, we have that √ ◦ ◦ ′ ′ BD = 15 / sin(60 ) = 10 3, AC = 15 / sin(30 ) = 30, and ∠ CAD = ∠ CAB + ∠ BAD = √ ◦ ′ ∠ CAB + ∠ DBA = 90 . Thus, CD is always 20 3 as A and B vary. Note that AD + BC = √ ′ ′ BD + BC . By the triangle inequality, this length is no less that CD = 20 3, and equality √ ′ can be achieved by fixing A and moving B to the intersection of CD with ℓ . Thus, 20 3 is 1 the minimum length, so the answer is 20 + 3 = 23 . 3 Figure 6: Problem 5 diagram.