Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$ , and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$ . Draw $PC \perp x$ -axis and $PD \perp y$ -axis, with the feet of the perpendiculars being $C$ and $D$ , respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

4. 24.

Notice that
$\begin{array}{l} S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\ =2\left(x+\frac{9}{x}\right)+12 \geqslant 24 . \end{array}$

The equality holds if and only if $x=3$ . Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24.

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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.