IAL 2024 June Q10 | 國際數學站

Figure 1

Figure 1 shows a sketch of the curve $C$ with polar equation

\begin{align*} r = 1 + \cos \theta \qquad 0 \leqslant \theta \leqslant \pi \end{align*}

and the line $l$ with polar equation

\begin{align*} r = k\sec \theta \qquad 0 \leqslant \theta < \frac{\pi}{2} \end{align*}

where $k$ is a positive constant.

Given that

$C$ and $l$ intersect at the point $P$
$OP = 1 + \frac{\sqrt{3}}{2}$

(a) determine the exact value of $k$ .

(2)

The finite region $R$ , shown shaded in Figure 1, is bounded by $C$ , the initial line and $l$ .

(b) Use algebraic integration to show that the area of $R$ is

\begin{align*} p\pi + q\sqrt{3} + r \end{align*}

where $p$ , $q$ and $r$ are simplified rational numbers to be determined.

(7)

解答

(a)

The curve $C$ is given by $r = 1 + \cos\theta$ and the line $l$ is $r = k\sec\theta$ . They intersect at point $P$ where $OP = r = 1 + \frac{\sqrt{3}}{2}$ .

First, find the angle $\theta$ at $P$ using the equation for curve $C$ :

\begin{align*} 1 + \cos\theta = &\,1 + \frac{\sqrt{3}}{2}\\[4mm] \cos\theta = &\,\frac{\sqrt{3}}{2}\\[4mm] \theta = &\,\frac{\pi}{6} \end{align*}

Since $P$ also lies on the line $l$ , substitute $r$ and $\theta$ into the line’s polar equation:

\begin{align*} r = &\,k\sec\theta\\[4mm] r\cos\theta = &\,k\\[4mm] \left( 1 + \frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) = &\,k\\[4mm] k = &\,\frac{\sqrt{3}}{2} + \frac{3}{4} \end{align*}

(b)

The region $R$ is bounded by $C$ from $\theta = \frac{\pi}{6}$ to $\theta = \pi$ , and by a triangle formed by the pole, the line $l$ (which is a vertical line $x=k$ ), and the line segment $OP$ .

The area of region $R$ can be found by adding the area under $C$ between $\frac{\pi}{6}$ and $\pi$ to the area of the right-angled triangle formed between $\theta = 0$ and $\theta = \frac{\pi}{6}$ up to the line $x=k$ .

Area under curve $C$ from $\frac{\pi}{6}$ to $\pi$ :

\begin{align*} \frac{1}{2} \int_{\frac{\pi}{6}}^{\pi} r^2 \,\mathrm{d}\theta = &\,\frac{1}{2} \int_{\frac{\pi}{6}}^{\pi} (1 + \cos\theta)^2 \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2} \int_{\frac{\pi}{6}}^{\pi} (1 + 2\cos\theta + \cos^2\theta) \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2} \int_{\frac{\pi}{6}}^{\pi} \left( 1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta \right) \,\mathrm{d}\theta\\[4mm] = &\,\frac{1}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{\frac{\pi}{6}}^{\pi} \end{align*}

Evaluate at the limits:

\begin{align*} \text{Upper limit } (\pi): \quad &\,\frac{1}{2} \left( \frac{3\pi}{2} + 0 + 0 \right) = \frac{3\pi}{4}\\[4mm] \text{Lower limit } \left(\frac{\pi}{6}\right): \quad &\,\frac{1}{2} \left( \frac{3\pi}{12} + 2\left(\frac{1}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \right) = \frac{\pi}{8} + \frac{1}{2} + \frac{\sqrt{3}}{16} \end{align*}

Area of this sector $= \frac{3\pi}{4} - \left( \frac{\pi}{8} + \frac{1}{2} + \frac{\sqrt{3}}{16} \right) = \frac{5\pi}{8} - \frac{1}{2} - \frac{\sqrt{3}}{16}$ .

Area of the Triangle: The triangle has a base along the initial line of length $k = \frac{\sqrt{3}}{2} + \frac{3}{4}$ , and a height $y_P$ at point $P$ . We can use the formula Area $= \frac{1}{2}ab\sin C$ , where $a = k\sec(\pi/6)$ (the distance $OP$ ), and $b = k$ , but it’s simpler to use the Cartesian area of a right triangle $\frac{1}{2}xy$ :

\begin{align*} x_P = &\,r\cos\left(\frac{\pi}{6}\right) = k = \frac{2\sqrt{3} + 3}{4}\\[4mm] y_P = &\,r\sin\left(\frac{\pi}{6}\right) = \left( 1 + \frac{\sqrt{3}}{2} \right)\left(\frac{1}{2}\right) = \frac{2 + \sqrt{3}}{4} \end{align*}

\begin{align*} \text{Area of Triangle} = &\,\frac{1}{2}x_Py_P\\[4mm] = &\,\frac{1}{2} \left( \frac{2\sqrt{3} + 3}{4} \right) \left( \frac{2 + \sqrt{3}}{4} \right)\\[4mm] = &\,\frac{1}{32} \left( 4\sqrt{3} + 6 + 6 + 3\sqrt{3} \right)\\[4mm] = &\,\frac{12 + 7\sqrt{3}}{32}\\[4mm] = &\,\frac{3}{8} + \frac{7\sqrt{3}}{32} \end{align*}

Total Area of $R$ :

\begin{align*} \text{Total Area} = &\,\left( \frac{5\pi}{8} - \frac{1}{2} - \frac{\sqrt{3}}{16} \right) + \left( \frac{3}{8} + \frac{7\sqrt{3}}{32} \right)\\[4mm] = &\,\frac{5\pi}{8} - \frac{4}{8} + \frac{3}{8} - \frac{2\sqrt{3}}{32} + \frac{7\sqrt{3}}{32}\\[4mm] = &\,\frac{5\pi}{8} + \frac{5\sqrt{3}}{32} - \frac{1}{8} \end{align*}

This is in the required form $p\pi + q\sqrt{3} + r$ , with $p = \frac{5}{8}$ , $q = \frac{5}{32}$ , and $r = -\frac{1}{8}$ .