Let $O_1$, $O_2$ be two circles with radii $R_1$ and $R_2$, and suppose the circles meet at $A$ and $D$. Draw a line $L$ through $D$ intersecting $O_1$, $O_2$ at $B$ and $C$. Now allow the distance between the centers as well as the choice of $L$ to vary. Find the length of $AD$ when the area of $ABC$ is maximized.
Problem
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Tags: geometry, trigonometry, geometric transformation
djmathman
19.07.2014 04:07
Let $\angle ADB=\theta$. Note that $\angle AO_1B=\angle AO_2C=2\theta$, so $\triangle AO_1B\sim\triangle AO_2C$ since both are isosceles. Spiral similarity gives $\triangle ABC\sim\triangle AO_1O_2$. Hence the area of $\triangle ABC$ can be written as \begin{align*}[AO_1O_2]\left(\dfrac{AB}{AO_1}\right)^2=\left(\dfrac12 R_1 R_2\sin\angle O_1AO_2\right)\left(\dfrac{AB}{R_1}\right)^2=2R_1R_2\sin\angle O_1AO_2\sin^2\theta,\end{align*} where the last equality is a result of the Extended Law of Sines. Maximizing $[ABC]$ is thus equivalent to maximizing both $\sin\theta$ and $\sin\angle O_1AO_2$. The former is maximized by letting $\theta=\tfrac{\pi}2$ and the latter is maximized by letting $AO_1\perp AO_2$. This is the angle equality we care about.
Since $AO_1=O_1D=R_1$ and $AO_2=O_2D=R_2$, $AO_1DO_2$ is a kite, and so $AD\perp O_1O_2$. Let $X=AD\cap O_1O_2$. Then by some simple area calculations we get the desired answer: \[AX=\dfrac{AO_1\cdot AO_2}{O_1O_2}=\dfrac{R_1R_2}{\sqrt{R_1^2+R_2^2}}\implies AD=\boxed{\dfrac{2R_1R_2}{\sqrt{R_1^2+R_2^2}}}.\]
EDIT: Needed to make minor correction in the solution. Fortunately, this only affected the expression to maximize and not how to maximize it.
fclvbfm934
16.08.2014 08:25
Suppose we have two different $B, C$, say $B_1, C_1$ and $B_2, C_2$. It is evident that $AB_1C_1 \sim AB_2C_2$, and since we are trying to maximize area, we maximize $AB$, which clearly happens when $AB$ is a diameter of $O_1$. This would lead us to $AC$ being the diameter of $O_2$. Hence, we have a triangle $ABC$ with $AB = 2R_1$ and $AC = 2R_2$; clearly, maximizing the area of $ABC$ is achieved by $\angle BAC = 90^{\circ}$, so we have $BC = 2\sqrt{R_1^2 + R_2^2}$. We want to find the length of the altitude from $A$ to $BC$, which should clearly be \[\frac{2R_1R_2}{\sqrt{R_1^2 + R_2^2}}\]
Kezer
10.09.2016 23:56
How troll. After realising how the area can be maximised, the problem just crumbles. Note $[ABC]=\tfrac12 \sin{\angle CAB} \cdot |AB| \cdot |CA| \leq \tfrac12 \cdot 1 \cdot (2R_1) \cdot (2R_2)$. Equality happens iff $\angle CAB=90^{\circ}$, $|AB|=2R_1$ and $|CA|=2R_2$. Then by Thales $AD \perp BC$. Now by the Pythagoras on $\triangle ABC$ we get $|BC|=2 \sqrt{R_1^2+R_2^2}$. So \[ 2R_1R_2 = [ABC]=\frac12 |BC| \cdot |AD| \iff |AD| = \frac{2R_1R_2}{\sqrt{R_1^2+R_2^2}}. \]And that's it.