Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.3 AMC 10/12 Find the answer

Given the Cartesian coordinate system $(xOy)$ , the parametric equations of the curve $C$ are $\begin{cases} x=3+5\cos \alpha \\ y=4+5\sin \alpha \end{cases}$ (where $\alpha$ is the parameter). Points $A$ and $B$ lie on the curve $C$ . Using the origin $O$ as the pole and the positive $x$ -axis as the polar axis, the polar coordinates of points $A$ and $B$ are $A(\rho\_1, \frac{\pi}{6})$ and $B(\rho\_2, \theta)$ , respectively, where $\theta \in [0, \pi]$ .

(1) Find the polar equation of the curve $C$ .

(2) Let $M$ be the center of the curve $C$ . Find the maximum area of $\triangle MAB$ and the polar coordinates of point $B$ in this case.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

(1) From the parametric equations of the curve $C$ , $\begin{cases} x=3+5\cos \alpha \\ y=4+5\sin \alpha \end{cases}$ , we obtain $(x-3)^2 + (y-4)^2 = 25$ , which can be rewritten as $x^2 + y^2 - 6x - 8y = 0$ . Thus, the polar equation of the curve $C$ is $\rho = 6\cos\theta + 8\sin\theta$ .

(2) To maximize the area of $\triangle MAB$ , it is clear that $\theta > \frac{\pi}{6}$ . The area of the triangle is given by $S_{\triangle MAB} = \frac{1}{2} \cdot 5 \cdot 5 \sin\left(2\left(\theta - \frac{\pi}{6}\right)\right) = \frac{25}{2} \sin\left(2\theta - \frac{\pi}{3}\right)$ .

The maximum area occurs when $2\theta - \frac{\pi}{3} = \frac{\pi}{2}$ , i.e., $\theta = \frac{5\pi}{12}$ . In this case, the maximum area is $S_{\triangle MAB_{\text{max}}} = \boxed{\frac{25}{2}}$ .

At this point, we have $\rho\_2 = \frac{7\sqrt{6}}{2}$ , so the polar coordinates of point $B$ are $\boxed{B\left(\frac{7\sqrt{6} + \sqrt{2}}{2}, \frac{5\pi}{12}\right)}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.