返回题库

HMMT 二月 2016 · 代数 · 第 10 题

HMMT February 2016 — Algebra — Problem 10

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let a, b and c be positive real numbers such that 2 2 a + ab + b = 9 2 2 b + bc + c = 52 2 2 c + ca + a = 49 . 2 2 49 b − 33 bc + 9 c Compute the value of . 2 a
解析
  1. Let a, b and c be positive real numbers such that 2 2 a + ab + b = 9 2 2 b + bc + c = 52 2 2 c + ca + a = 49 . 2 2 49 b − 33 bc + 9 c Compute the value of . 2 a Proposed by: Alexander Katz Answer: 52 Consider a triangle ABC with Fermat point P such that AP = a, BP = b, CP = c . Then 2 2 2 ◦ AB = AP + BP − 2 AP · BP cos(120 ) by the Law of Cosines, which becomes 2 2 2 AB = a + ab + b √ and hence AB = 3. Similarly, BC = 52 and AC = 7. Furthermore, we have 2 2 2 BC = 52 = AB + BC − 2 AB · BC cos ∠ BAC 2 2 = 3 + 7 − 2 · 3 · 7 cos ∠ BAC = 58 − 42 cos ∠ BAC 1 And so cos ∠ BAC = . 7 ′ ′ ′ Invert about A with arbitrary radius r . Let B , P , C be the images of B, P, C respectively. Since ′ ′ ◦ ′ ′ ◦ ′ ′ ′ ◦ ∠ AP B = ∠ AB P = 120 and ∠ AP C = ∠ AC P = 120 , we note that ∠ B P C = 120 − ∠ BAC , and so ′ ′ ′ ◦ cos ∠ B P C = cos(120 − ∠ BAC ) ◦ ◦ = cos 120 cos ∠ BAC − sin 120 sin ∠ BAC ( ) √ √ ( ) 1 1 3 4 3 = − + 2 7 2 7 11 = 14 Furthermore, using the well-known result 2 r BC ′ ′ B C = AB · AC for an inversion about A , we have 2 BP r ′ ′ B P = AB · AP 2 br = a · 3 2 br = 3 a √ 2 2 cr r 52 ′ ′ ′ ′ ′ ′ ′ and similarly P C = , B C = . Applying the Law of Cosines to B P C gives us 7 a 21 ′ ′ 2 ′ ′ 2 ′ ′ 2 ′ ′ ′ ′ ◦ B C = B P + P C − 2 B P · P C cos(120 − ∠ BAC ) 4 2 4 2 4 4 52 r b r c r 11 bcr = ⇒ = + − 2 2 2 2 21 9 a 49 a 147 a 2 2 52 b c 11 bc = ⇒ = + − 2 2 2 2 21 9 a 49 a 147 a 2 2 52 49 b − 33 bc + 9 c = ⇒ = 2 2 2 21 21 a 2 2 49 b − 33 bc +9 c and so = 52 . 2 a Motivation: the desired sum looks suspiciously like the result of some Law of Cosines, so we should try 7 b 3 c 33 bc 11 building a triangle with sides and . Getting the − term is then a matter of setting cos θ = . a a a 14 Now there are two possible leaps: noticing that cos θ = cos(120 − ∠ BAC ) , or realizing that it’s pretty 7 b b difficult to contrive a side of – but it’s much easier to contrive a side of . Either way leads to the a 3 a natural inversion idea, and the rest is a matter of computation.