Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.6 AMC 10/12 Find the answer China

The straight line $\frac{x}{4} + \frac{y}{3} = 1$ intersects the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at two points $A$ and $B$ . There is a point $P$ on this ellipse such that the area of $\triangle PAB$ is equal to $3$ . There is/are ( ) such point/points $P$ .

Pick one

Solution

Suppose that there is a point $P(4\cos \alpha, 3\sin \alpha)$ on the ellipse. When $P$ and the origin $O$ are not on the same side of $AB$ , the distance from $P$ to $AB$ is
$\begin{aligned} & \frac{3(4\cos\alpha) + 4(3\sin\alpha) - 12}{5} \\ &= \frac{12}{5}(\cos\alpha + \sin\alpha - 1) \\ &\le \frac{12}{5}(\sqrt{2} - 1) \\ &< \frac{6}{5}. \end{aligned}$
But $AB = 5$ , so $\triangle PAB < \frac{1}{2} \times 5 \times \frac{6}{5} = 3$ .
Therefore, when the area of $\triangle PAB$ is equal to $3$ , points $P$ and $O$ are on the same side of $AB$ . There are two such points $P$ . Answer: B.

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