Concentric Circles, Tangent Circles

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Draw and label as in the above diagram.

First, notice that $\angle DPC=90^{\circ}$ because $CD$ is the diameter of circle $\Gamma_3$ and $P$ is on $\Gamma_3$ .

Also, we know that $\angle DO_1A=90^{\circ}$ .

Thus, $\triangle DO_1P\sim \triangle PO_1C$ .

Therefore the ratio of their sides are the same: $\dfrac{DO_1}{O_1P}=\dfrac{O_1P}{O_1C}$

Rearranging gives $(O_1P)^2=(DO_1)(O_1C)$ or $O_1P=\sqrt{DO_1\cdot O_1C}$

Now, we know that $DO_1=BO_1$ and $AO_1=CO_1$ because they are radii of their respective circles.

Thus, $O_1P=\sqrt{AO_1\cdot BO_1}$

Thus, $AP=AO_1-PO_1=AO_1-\sqrt{AO_1\cdot BO_1}$

Also, $AB=AO_1-BO_1$ .

Thus, we need to find the minimum of $\dfrac{AO_1-\sqrt{AO_1\cdot BO_1}}{AO_1-BO_1}$

By AM-GM, we have $-\sqrt{AO_1\cdot BO_1}\ge -\dfrac{AO_1+BO_1}{2}$ , so $\dfrac{AO_1-\sqrt{AO_1\cdot BO_1}}{AO_1-BO_1}\ge \dfrac{AO_1-\dfrac{AO_1+BO_1}{2}}{AO_1-BO_1}=\dfrac{\dfrac{AO_1-BO_1}{2}}{AO_1-BO_1}=\dfrac{1}{2}$

Thus, $m=\dfrac{1}{2}$ , so $\lfloor 1000m\rfloor = \boxed{500}$

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Note: Equality case is when $AO_1=BO_1$ . However, technically this can never be achieved, because when $AO_1=BO_1$ then $\dfrac{AP}{AB}=\dfrac{0}{0}$ which is not defined. Thus, we actually have $m > \dfrac{1}{2}$ . However, this does not affect the answer.

Let $O_1$ be $(0,0)$ in $Oxy$ Coordinate and let $O_1A=1$ , $O_1B=r(r<O_1A=1)$ .

We then have:

• $R_{O_2}=\frac{1+r}{2}$
• Coordinate of $O_2$ is $(\frac{1-r}{2},0)$
• $AB=1-r$
• $AP=O_1A-O_1P=\sqrt{{O_2P}^2-O_1O_2}=1-\sqrt{(\frac{1+r}{2})^2-(\frac{1-r}{2})^2}=1-\sqrt{r}$

That means $m=min(\frac{AP}{AB})=min(\frac{1-\sqrt{r}}{1-r})=min(\frac{1}{1+\sqrt{r}})=\frac{1}{2}$

when $r=1$

Call x=AP. $AP*(2R-AP)=\frac{R-r}{\sqrt{2}}*R*\sqrt{2}=(R-r)*R$ $\Rightarrow x^2-2Rx+(R-r)*R=0$ $AP=-\sqrt{rR}+R$ $S=\frac{AP}{AB}=\frac{R-\sqrt{rR}}{R-r}$ $S=\frac{\sqrt{R}(\sqrt{R}-\sqrt{r})}{R-r}$ $S=\frac{\sqrt{R}}{\sqrt{R}+\sqrt{r}}$ $Lim_{r \rightarrow R}{S}=\frac{1}{2}$ Because when r increases then S is decreases, but max(r)=R, at this r: S=1/2. $\lfloor 1000S \rfloor = \boxed{500}$