把起點記為 O O O dist ( O , 邊界 ) = 1 \operatorname{dist}(O,\text{邊界})=1 dist ( O , 邊界 ) = 1
1) 等價改寫：邊界＝單位圓的某條切線
以 O O O
x 2 + y 2 = 1. x^2+y^2=1. x 2 + y 2 = 1. 任何距離 O O O 找一條從 O O O 。
把切線用角度參數化：對每個 α ∈ [ 0 , 2 π ) \alpha\in[0,2\pi) α ∈ [ 0 , 2 π ) ( cos α , sin α ) (\cos\alpha,\sin\alpha) ( cos α , sin α )
L α : x cos α + y sin α = 1. L_\alpha:\quad x\cos\alpha+y\sin\alpha=1. L α : x cos α + y sin α = 1. 2) 走法（構造一條保證相交的路徑）
取
θ = π 6 ( = 30 ∘ ) . \theta=\frac{\pi}{6}\;(=30^\circ). θ = 6 π ( = 3 0 ∘ ) . (A) 先直走到 P P P x x x
O P = sec θ = 1 cos θ = 2 3 OP=\sec\theta=\frac{1}{\cos\theta}=\frac{2}{\sqrt3} OP = sec θ = cos θ 1 = 3 2 到 P = ( 2 3 , 0 ) P=(\frac{2}{\sqrt3},0) P = ( 3 2 , 0 )
(B) 右轉 120 ∘ 120^\circ 12 0 ∘ A A A P P P A A A O A P OAP O A P
P A = tan θ = sin θ cos θ = 1 3 , A = ( cos θ , sin θ ) = ( 3 2 , 1 2 ) . PA=\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1}{\sqrt3},
\quad
A=(\cos\theta,\sin\theta)=\left(\frac{\sqrt3}{2},\frac12\right). P A = tan θ = cos θ sin θ = 3 1 , A = ( cos θ , sin θ ) = ( 2 3 , 2 1 ) . （這一步等價於你說的「右轉 120 ∘ 120^\circ 12 0 ∘ 1 / 3 1/\sqrt3 1/ 3 Reddit
(C) 沿單位圓走一段圓弧到 C C C
φ = 3 π 2 − θ = 4 π 3 ( = 240 ∘ ) \varphi=\frac{3\pi}{2}-\theta=\frac{4\pi}{3}\;(=240^\circ) φ = 2 3 π − θ = 3 4 π ( = 24 0 ∘ ) 的點
C = ( cos φ , sin φ ) = ( − 1 2 , − 3 2 ) . C=(\cos\varphi,\sin\varphi)=\left(-\frac12,-\frac{\sqrt3}{2}\right). C = ( cos φ , sin φ ) = ( − 2 1 , − 2 3 ) . 這段圓弧的圓心角為
φ − θ = ( 3 π 2 − θ ) − θ = 3 π 2 − 2 θ = 3 π 2 − π 3 = 7 π 6 , \varphi-\theta=\left(\frac{3\pi}{2}-\theta\right)-\theta=\frac{3\pi}{2}-2\theta
=\frac{3\pi}{2}-\frac{\pi}{3}=\frac{7\pi}{6}, φ − θ = ( 2 3 π − θ ) − θ = 2 3 π − 2 θ = 2 3 π − 3 π = 6 7 π , 所以弧長就是 7 π 6 \frac{7\pi}{6} 6 7 π
(D) 從 C C C C C C L − θ L_{-\theta} L − θ D D D
3) 為什麼一定會碰到任意邊界切線 L α L_\alpha L α
把 α ∈ [ 0 , 2 π ) \alpha\in[0,2\pi) α ∈ [ 0 , 2 π )
(i) α ∈ [ θ , φ ] = [ π 6 , 4 π 3 ] \alpha\in[\theta,\varphi]=\left[\frac{\pi}{6},\frac{4\pi}{3}\right] α ∈ [ θ , φ ] = [ 6 π , 3 4 π ]
你在 (C) 走的圓弧包含了圓上所有角度從 θ \theta θ φ \varphi φ L α L_\alpha L α ( cos α , sin α ) (\cos\alpha,\sin\alpha) ( cos α , sin α ) L α L_\alpha L α
(ii) α ∈ [ 0 , θ ] ∪ [ 2 π − θ , 2 π ) \alpha\in[0,\theta]\cup[2\pi-\theta,2\pi) α ∈ [ 0 , θ ] ∪ [ 2 π − θ , 2 π ) x x x 30 ∘ 30^\circ 3 0 ∘
看你走的切線段 P A PA P A n α = ( cos α , sin α ) n_\alpha=(\cos\alpha,\sin\alpha) n α = ( cos α , sin α ) X ( t ) = ( 1 − t ) P + t A X(t)=(1-t)P+tA X ( t ) = ( 1 − t ) P + t A t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ]
f ( t ) = n α ⋅ X ( t ) f(t)=n_\alpha\cdot X(t) f ( t ) = n α ⋅ X ( t ) 是 t t t
f ( 0 ) = n α ⋅ P = sec θ cos α ≥ sec θ cos θ = 1 f(0)=n_\alpha\cdot P=\sec\theta\cos\alpha\ge \sec\theta\cos\theta=1 f ( 0 ) = n α ⋅ P = sec θ cos α ≥ sec θ cos θ = 1 （因為此段 α \alpha α cos α ≥ cos θ \cos\alpha\ge\cos\theta cos α ≥ cos θ
f ( 1 ) = n α ⋅ A = cos ( α − θ ) ≤ 1. f(1)=n_\alpha\cdot A=\cos(\alpha-\theta)\le 1. f ( 1 ) = n α ⋅ A = cos ( α − θ ) ≤ 1. 所以 f ( 0 ) ≥ 1 ≥ f ( 1 ) f(0)\ge 1\ge f(1) f ( 0 ) ≥ 1 ≥ f ( 1 ) f f f t t t f ( t ) = 1 f(t)=1 f ( t ) = 1 X ( t ) ∈ L α X(t)\in L_\alpha X ( t ) ∈ L α L α L_\alpha L α P A PA P A
(iii) α ∈ [ φ , 2 π − θ ] = [ 4 π 3 , 11 π 6 ] \alpha\in[\varphi,2\pi-\theta]=\left[\frac{4\pi}{3},\frac{11\pi}{6}\right] α ∈ [ φ , 2 π − θ ] = [ 3 4 π , 6 11 π ]
同理看最後那段切線段 C D CD C D α \alpha α L α L_\alpha L α C C C D D D = 1 =1 = 1 C D CD C D L α L_\alpha L α Reddit
綜合 (i)(ii)(iii)，任意切線 L α L_\alpha L α 都會跟你的路徑相交；而真正的邊界直線就是某個 L α L_\alpha L α
4) 總路程
O P + P A + 弧長 + C D = 2 3 + 1 3 + 7 π 6 + 1 = 3 + 7 π 6 + 1 = 1 + 3 + π ⋅ 7 6 . OP+PA+\text{弧長}+CD
=\frac{2}{\sqrt3}+\frac{1}{\sqrt3}+\frac{7\pi}{6}+1
=\sqrt3+\frac{7\pi}{6}+1
=1+\sqrt3+\pi\cdot\frac{7}{6}. OP + P A + 弧長 + C D = 3 2 + 3 1 + 6 7 π + 1 = 3 + 6 7 π + 1 = 1 + 3 + π ⋅ 6 7 . 這就證明：走 1 + 3 + 7 π 6 \;1+\sqrt3+\frac{7\pi}{6}\; 1 + 3 + 6 7 π Reddit +1
