HMMT 十一月 2021 · 冲刺赛 · 第 23 题

HMMT November 2021 — Guts Round — Problem 23

[12] Side AB of 4 ABC is the diameter of a semicircle, as shown below. If AB = 3 + 3, BC = 3 2, √ √ a + b + c d π ( ) and AC = 2 3, then the area of the shaded region can be written as , where a, b, c, d, e are e integers, e is positive, d is square-free, and gcd( a, b, c, e ) = 1. Find 10000 a + 1000 b + 100 c + 10 d + e . C A B

解析

[12] Side AB of 4 ABC is the diameter of a semicircle, as shown below. If AB = 3 + 3, BC = 3 2, √ √ a + b + c d π ( ) and AC = 2 3, then the area of the shaded region can be written as , where a, b, c, d, e are e integers, e is positive, d is square-free, and gcd( a, b, c, e ) = 1. Find 10000 a + 1000 b + 100 c + 10 d + e . C A B Proposed by: David Vulakh Answer: 147938 Solution: Drop an altitude to point D on AB from C and let x = AD . Solving for x , we find ( ) ( ) √ 2 √ √ 2 2 12 − x = 18 − 3 + 3 − x ⇒ 12 = 18 − 9 − 6 3 − 3 + 2 3 + 3 x − x ( ) √ √ ⇒ 6 + 6 3 = 6 + 2 3 x √ ⇒ x = 3 √ ◦ So AC = 2 AD , from which we have ∠ CAD = 60 . Also, CD = AD 3 = 3 and BD = AB − AD = √ √ ◦ 3 + 3 − 3 = 3, so ∠ DBC = 45 . Then, if E is the intersection of the circle with AC , F is the ◦ ◦ intersection of the circle with BC , and O is the midpoint of AB , ∠ AOE = 60 and ∠ BOF = 90 . AB Then, letting r = , we get that the area of the part of 4 ABC that lies inside the semicircle is 2 √ ( ) 1 1 1 1 1 1 3 1 2 2 2 ◦ 2 ◦ 2 2 2 πr − + πr + r sin 60 + r sin 90 = πr + r + r 2 4 6 2 2 12 4 2 ( ) √ 1 2 = π + 3 3 + 6 r 12 So the desired area is √ ( ) ( ) ( ) √ √ √ 1 9 + 3 3 1 2 3 r − π + 3 3 + 6 r = − π + 3 3 + 6 2 + 3 12 2 8 ( ) ( ) ( ) √ √ √ 1 1 1 = 9 + 3 3 − 2 + 3 π − 21 + 12 3 2 8 8 ( √ ) 15 − 2 + 3 π = . 8