xy平面上の半円周 $C: x^2 + y^2 = 1, y \ge 0$上に2点$A(1,0), B(-1,0)$と2点$S(\cos\theta, \sin\theta), T(\cos t, \sin t) (0 < \theta < t < \pi)$がある。 (1) 点Tを固定し、点Sが弧AT上を動くとき、弦ASの長さと弦STの長さの和の最大値を$t$を用いて表せ。 (2) 2点S, Tが動くとき、3つの弦AS, ST, TBの長さの和Lの最大値と、それを与える$\theta$と$t$の値をそれぞれ求めよ。

1. 問題の内容

xy平面上の半円周 $C: x^2 + y^2 = 1, y \ge 0$上に2点$A(1,0), B(-1,0)$と2点$S(\cos\theta, \sin\theta), T(\cos t, \sin t) (0 < \theta < t < \pi)$がある。

(1) 点Tを固定し、点Sが弧AT上を動くとき、弦ASの長さと弦STの長さの和の最大値を$t$を用いて表せ。

(2) 2点S, Tが動くとき、3つの弦AS, ST, TBの長さの和Lの最大値と、それを与える$\theta$と$t$の値をそれぞれ求めよ。

2. 解き方の手順

(1)

点Tを固定し、点Sが弧AT上を動くとき、AS + STの最大値を求める。

$AS = \sqrt{(\cos\theta - 1)^2 + (\sin\theta - 0)^2} = \sqrt{\cos^2\theta - 2\cos\theta + 1 + \sin^2\theta} = \sqrt{2 - 2\cos\theta} = \sqrt{2}\sqrt{1 - \cos\theta} = \sqrt{2}\sqrt{2\sin^2\frac{\theta}{2}} = 2|\sin\frac{\theta}{2}| = 2\sin\frac{\theta}{2}$ ($0 < \theta < t < \pi$より$\sin(\theta/2) > 0$)

$ST = \sqrt{(\cos t - \cos\theta)^2 + (\sin t - \sin\theta)^2} = \sqrt{\cos^2t - 2\cos t\cos\theta + \cos^2\theta + \sin^2t - 2\sin t\sin\theta + \sin^2\theta} = \sqrt{2 - 2(\cos t\cos\theta + \sin t\sin\theta)} = \sqrt{2 - 2\cos(t-\theta)} = \sqrt{2}\sqrt{1 - \cos(t-\theta)} = \sqrt{2}\sqrt{2\sin^2\frac{t-\theta}{2}} = 2|\sin\frac{t-\theta}{2}| = 2\sin\frac{t-\theta}{2}$ ($0 < \theta < t < \pi$より$\sin((t-\theta)/2) > 0$)

$AS + ST = 2\sin\frac{\theta}{2} + 2\sin\frac{t-\theta}{2}$

$AS + ST = 2\sin\frac{\theta}{2} + 2\sin\frac{t}{2}\cos\frac{\theta}{2} - 2\cos\frac{t}{2}\sin\frac{\theta}{2} = 2(1-\cos\frac{t}{2})\sin\frac{\theta}{2} + 2\sin\frac{t}{2}\cos\frac{\theta}{2}$

$\frac{d}{d\theta}(AS + ST) = 2(\frac{1}{2})\cos\frac{\theta}{2} - 2(\frac{1}{2})\cos\frac{t-\theta}{2} = \cos\frac{\theta}{2} - \cos\frac{t-\theta}{2} = 0$

$\cos\frac{\theta}{2} = \cos\frac{t-\theta}{2}$

$\frac{\theta}{2} = \frac{t-\theta}{2}$

$\theta = t-\theta$

$2\theta = t$

$\theta = \frac{t}{2}$

$AS + ST = 2\sin\frac{t}{4} + 2\sin\frac{t-\frac{t}{2}}{2} = 2\sin\frac{t}{4} + 2\sin\frac{t}{4} = 4\sin\frac{t}{4}$

(2)

$AS = 2\sin\frac{\theta}{2}$

$ST = 2\sin\frac{t-\theta}{2}$

$TB = \sqrt{(\cos t - (-1))^2 + (\sin t - 0)^2} = \sqrt{(\cos t + 1)^2 + \sin^2t} = \sqrt{\cos^2t + 2\cos t + 1 + \sin^2t} = \sqrt{2 + 2\cos t} = \sqrt{2}\sqrt{1 + \cos t} = \sqrt{2}\sqrt{2\cos^2\frac{t}{2}} = 2|\cos\frac{t}{2}| = 2\cos\frac{t}{2}$ ($0 < t < \pi$より$\cos(t/2) > 0$)

$L = AS + ST + TB = 2\sin\frac{\theta}{2} + 2\sin\frac{t-\theta}{2} + 2\cos\frac{t}{2}$

$L = 2\sin\frac{\theta}{2} + 2\sin\frac{t}{2}\cos\frac{\theta}{2} - 2\cos\frac{t}{2}\sin\frac{\theta}{2} + 2\cos\frac{t}{2} = 2(1-\cos\frac{t}{2})\sin\frac{\theta}{2} + 2\sin\frac{t}{2}\cos\frac{\theta}{2} + 2\cos\frac{t}{2}$

$\frac{\partial L}{\partial \theta} = 2(\frac{1}{2})(1-\cos\frac{t}{2})\cos\frac{\theta}{2} - 2(\frac{1}{2})\sin\frac{t}{2}\sin\frac{\theta}{2} = (1-\cos\frac{t}{2})\cos\frac{\theta}{2} - \sin\frac{t}{2}\sin\frac{\theta}{2} = 0$

$(1-\cos\frac{t}{2})\cos\frac{\theta}{2} = \sin\frac{t}{2}\sin\frac{\theta}{2}$

$\tan\frac{\theta}{2} = \frac{1-\cos\frac{t}{2}}{\sin\frac{t}{2}} = \frac{2\sin^2\frac{t}{4}}{2\sin\frac{t}{4}\cos\frac{t}{4}} = \tan\frac{t}{4}$

$\frac{\theta}{2} = \frac{t}{4}$

$\theta = \frac{t}{2}$

$L = 2\sin\frac{t}{4} + 2\sin\frac{t-\frac{t}{2}}{2} + 2\cos\frac{t}{2} = 2\sin\frac{t}{4} + 2\sin\frac{t}{4} + 2\cos\frac{t}{2} = 4\sin\frac{t}{4} + 2\cos\frac{t}{2} = 4\sin\frac{t}{4} + 2(1-2\sin^2\frac{t}{4}) = -4\sin^2\frac{t}{4} + 4\sin\frac{t}{4} + 2$

$x = \sin\frac{t}{4}$とおくと、$L = -4x^2 + 4x + 2 = -4(x^2 - x) + 2 = -4(x - \frac{1}{2})^2 + 1 + 2 = -4(x - \frac{1}{2})^2 + 3$

$x = \frac{1}{2}$のとき最大値3をとる。

$\sin\frac{t}{4} = \frac{1}{2}$

$\frac{t}{4} = \frac{\pi}{6}$

$t = \frac{2\pi}{3}$

$\theta = \frac{t}{2} = \frac{\pi}{3}$

3. 最終的な答え

(1) $4\sin\frac{t}{4}$

(2) Lの最大値は3。そのときの$\theta = \frac{\pi}{3}, t = \frac{2\pi}{3}$