第2章 2.15

(1)

$P_1Q_1 = a\sin30^\circ = \frac{1}{2}a$

$Q_1P_2 = P_1Q_1\cos30^\circ = \frac{1}{2}a\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{4}a$

$P_2Q_2 = Q_1P_2\cos30^\circ = \frac{\sqrt3}{4}a\cdot\frac{\sqrt3}{2} = \frac{3}{8}a$

よってこの級数は初項$\dfrac{1}{2}a$公比$\dfrac{\sqrt3}{2}$の無限等比級数なので

$P_1Q_1+Q_1P_2+P_2Q_2+Q_2P_3+\cdots = \frac{a}{r-1} = \frac{\dfrac{1}{2}a}{\dfrac{\sqrt3}{2}-1} = \frac{a}{\sqrt3-2} = (2+\sqrt3)a$

(2)

$P_1Q_1 = a\sin30^\circ = \frac{1}{2}a$

$P_2Q_2 = Q_1P_2\cos30^\circ = \frac{\sqrt3}{4}a\cdot\frac{\sqrt3}{2} = \frac{3}{8}a$

$P_3Q_3 = Q_2P_3\cos30^\circ = P_2Q_2\cos^2 30^\circ = \frac{3}{8}a\left(\frac{\sqrt3}{2}\right)^2 = \frac{9}{32}a$

よってこの級数は初項$\dfrac{1}{2}a$公比$\dfrac{3}{4}$の無限等比級数なので

$P_1Q_1+P_2Q_2+P_3Q_3+\cdots = \frac{\dfrac{1}{2}a}{1-\dfrac{3}{4}} = 2a$

(3)

$\Delta OP_1Q_1 = a\cos30^\circ \cdot P_1Q_1 \cdot \frac{1}{2} = a\frac{\sqrt{3}}{2}\cdot\frac{1}{2}a\cdot\frac{1}{2} = \frac{\sqrt{3}}{8}a^2$

$\Delta OP_2Q_2 = (a\cos30^\circ - Q_1P_2\sin30^\circ)\cdot P_2Q_2 \cdot \frac{1}{2} = \left(\frac{\sqrt{3}}{2}a - \frac{\sqrt{3}}{4}a\frac{1}{2}\right)\cdot\frac{3}{8}a\frac{1}{2} = \frac{9}{128}\sqrt{3}a^2$

$\Delta OP_3Q_3 = (a\cos30^\circ - Q_1P_2\sin30^\circ - Q_2P_3\sin30^\circ)\cdot P_2Q_2 \cdot \frac{1}{2}$

$= \left(\frac{\sqrt{3}}{2}a - \frac{\sqrt{3}}{4}a\frac{1}{2} - \frac{3\sqrt{3}}{16}a\frac{1}{2}\right)\cdot\frac{9}{32}a\frac{1}{2} = \frac{9\cdot9}{128\cdot16}\sqrt{3}a^2$

よってこの級数は初項$\dfrac{\sqrt{3}}{8}a^2$、公比$\dfrac{9}{16}$の無限等比級数なので

$\Delta OP_1Q_1 + \Delta OP_2Q_2 + \Delta OP_3Q_3 + \cdots = \frac{\dfrac{\sqrt{3}}{8}a^2}{1-\dfrac{9}{16}} = \frac{2}{7}\sqrt{3}a^2$