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Taiwan IMO Selection Camp

Geometry Difficulty 6.5 National Olympiad Prove it Taiwan

Let $O_1, O_2$ be circles with radius $R_1$ and $R_2$ , respectively. Let the two circles intersect each other at $A$ and $D$ . Let $L$ be a straight line passes through $D$ , and intersects $O_1$ and $O_2$ at $B$ and $C$ , respectively. Now, suppose that the distance between the center of $O_1$ and $O_2$ , and the line $L$ are chosen so that the area of $\triangle ABC$ is maximized. Find the length of $AD$ .

Solution

Observe that when $AD$ is a certain fixed value, $\angle B, \angle C$ (or their supplementary angles, hereafter the same) are fixed angles, and therefore $\angle A$ is a fixed angle. Since
$\begin{aligned} |\triangle ABC| &= \frac{1}{2} AB \cdot AC \sin \angle A \\ &= 2R_1 R_2 \sin \angle ADB \sin \angle ADC \sin \angle A \\ &= 2R_1 R_2 \sin^2 \angle ADB \sin \angle A. \end{aligned}$
the area attains its maximum value when $\angle ADB = 90^\circ$ . When $AD$ is allowed to vary, this value attains its maximum when $\angle A = 90^\circ$ . And when the two circles are orthogonal, that is, when $\angle O_1AO_2 = 90^\circ$ , we have $\angle A = 90^\circ$ . Let $AD = l$ at this time, then
$\begin{aligned} l^2 &= \sqrt{4R_1^2 - l^2} \sqrt{4R_2^2 - l^2} \\ \Rightarrow \quad l^4 &= (4R_1^2 - l^2)(4R_2^2 - l^2) = 16R_1^2 R_2^2 - 4(R_1^2 + R_2^2)l^2 + l^4, \end{aligned}$
therefore
$l = \frac{2}{\sqrt{\frac{1}{R_1^2} + \frac{1}{R_2^2}}}$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.