设BC与x轴正向的最小夹角（沿逆时针方向）为α,则点C的坐标为C(acosα,asin(α+π/3)),即C(acosα,asinα/2+a√3cosα/2),所以
OC²=a²[cos²α+(sinα/2+√3cosα/2)²]
=a²[cos²α+sin²α/4+3cos²α/4+2√3sinαcosα/4]
=a²[4cos²α+sin²α+3cos²α+2√3sinαcosα]/4
=a²[4cos²α+(1-cos²α)+3cos²α+2√3sinαcosα]/4
=a²[6cos²α+1+2√3sinαcosα]/4,
4OC²=a²[6cos²α+1+2√3sinαcosα],
4OC²-a²(6cos²α+1)=a²2√3sinαcosα,两边分别平方得
16(OC²)²+(a²)²[36(cos²α)²+12cos²α+1]-8OC²a²(6cos²α+1)=12(a²)²sin²αcos²α
=12(a²)²(1-cos²α)cos²α
=12(a²)²cos²α-12(a²)²(cos²α)²,即
16(OC²)²+(a²)²[36(cos²α)²+12cos²α+1]-8OC²a²(6cos²α+1)=12(a²)²cos²α-12(a²)²(cos²α)²,
整理得
48(a²)²(cos²α)²-48OC²a²cos²α+16(OC²)²-8OC²a²+(a²)² =0.
因为cos²α为实数,故
(48OC²a²)²-4×48(a²)²[16(OC²)²-8OC²a²+(a²)²]≥0,
即(24OC²)²-48[16(OC²)²-8OC²a²+(a²)²]≥0,
(6OC²)²-3[16(OC²)²-8OC²a²+(a²)²]≥0,
36(OC²)²-3[16(OC²)²-8OC²a²+(a²)²]≥0,
12(OC²)²-[16(OC²)²-8OC²a²+(a²)²]≥0,
12(OC²)²-16(OC²)²+8OC²a²-(a²)²≥0,
4(OC²)²-8OC²a²+(a²)²≤0,
(OC²-a²-√3a²/2)(OC²-a²+√3a²/2)≤0,
(1-√3/2)a²≤OC²≤(1+√3/2)a²,又OC＞0,故OC≤(√3+1)a/2,所以OC最大值为OC=(1+√3)a/2.
稍后.