## [問題]IMO HK PRELIM 2003

### [問題]IMO HK PRELIM 2003

NO.20

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5/7

scsnake

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scsnake

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i also think the answer is 5/7.
however i think i have used some 茅招...

yptsoi

scsnake

Using AutoCAD to measure, DF:FE=5:7

AB+CA=12 ---(1)

AC斜率=(√35sinθ-0)/(6cosθ+1)=√35sinθ/(6cosθ+1)
AB斜率=(√35sinθ-0)/(6cosθ-1)=√35sinθ/(6cosθ-1)

tan2α=√35sinθ/(6cosθ+1)
2tanα/(1-tan2α)=√35sinθ/(6cosθ+1)

tan2β=√35sinθ/(6cosθ-1)
2tanβ/(1-tan2β)=√35sinθ/(6cosθ-1)

MC方程式為:y=(√35/7)tan(θ/2)(x+1) ---(4)
MB方程式為:y=-(√35/7)cot(θ/2)(x-1) ---(5)

BC的中點為O(0,0)，因此OA方程式為:y=(√35/6)xtanθ ---(6)
DE方程式為: x=cosθ ---(7)

DF=(5/42)√35sinθ，FE=(1/6)√35sinθ
DF/FE=(5/42)/(1/6)=5/7

BC 若為 2, 那周長就是 14 .

(14-2-AC)2 = 22 + AC2

∠C 的分角線方程 y = x    .... (∠C)

sin(∠B) = (35/6) / (37/6) = 35/37
cos(∠B) = 2/(37/6) = 12/37

tan(∠B/2) = (1 - cos(∠B)) / sin(∠B)
= (1 - (12/37)) / (35/37)
= 5/7

10x + 14y = 20
10x + 14x = 20   ← 拿 (∠C) 式來代入, 找兩分角線的交點
x = 20/24 = 5/6  → 代入 (∠C) 式
y = 5/6

35x + 6y = 35
35(5/6) + 6y = 35  ← 拿 (DE) 式來代入, 找交點.
y = 35/36, 也是 FE 的長度, F (x,y) 即 (5/6, 35/36)

DF 長 = (直徑-FE) = (10/6 - 35/36) =  25/36
DF:FE = 25/36 : 35/36 = 5:7

Errfree