O is the center of the regular hexagon. Then we obviously have $ABC\cong COA\cong EOC$. And therefore we have also obviously $ABM\cong AOM\cong CON$, as $\frac{AM}{AC} =\frac{CN}{CE}$. So we have $\angle{BMA} =\angle{AMO} =\angle{CNO}$ and $\angle{NOC} =\angle{ABM}$. Because of $\angle{AMO} =\angle{CNO}$ the quadrilateral $ONCM$ is cyclic. $\Rightarrow \angle{NOC} =\angle{NMC} =\angle{BMA}$. And as we also have $\angle{NOC} =\angle{ABM}$ we get $\angle{ABM} =\angle{BMA}$. $\Rightarrow AB=AM$. And as $AC=\sqrt{3} \cdot AB$ we get $r=\frac{AM}{AC} =\frac{AB}{\sqrt{3} \cdot AB} =\frac{1}{\sqrt{3}}$.

I think this problem can solve by using vector method. I am finding this solution.

Let $X$ be the intersection of $AC$ and $BE$. $X$ is the mid-point of $AC$. Since $B$, $M$, and $N$ are collinear, then by Menelaus Theorem, $\frac{CN}{NE}\cdot\frac{EB}{BX}\cdot\frac{XM}{MC}=1$. Let the sidelength of the hexagon be $1$. Then $AC=CE=\sqrt{3}$. $\frac{CN}{NE}=\frac{CN}{CE-CN}=\frac{\frac{CN}{CE}}{1-\frac{CN}{CE}}=\frac{r}{1-r}$ $\frac{EB}{BX}=\frac{2}{\frac{1}{2}}=4$ $\frac{XM}{MC}=\frac{AM-AX}{AC-AM}=\frac{\frac{AM}{AC}-\frac{AX}{AC}}{1-\frac{AM}{AC}}=\frac{r-\frac{1}{2}}{1-r}$ Substituting them into the first equation yields $\frac{r}{1-r}\cdot\frac{4}{1}\cdot\frac{r-\frac{1}{2}}{1-r}=1$ $3r^2=1$ $\therefore r=\frac{\sqrt{3}}{3}$

the solution in which you have not to think about anything... Let O (the center of the exagon) be the point $(0;0)$, and let $B$ be $(0;1)$. so: $A=(-\frac{\sqrt3}{2};\frac{1}{2})$ $C=(\frac{\sqrt3}{2};\frac{1}{2})$ $D=(\frac{\sqrt3}{2};-\frac{1}{2})$ $E=(0;-1)$ $F=(-\frac{\sqrt3}{2};-\frac{1}{2})$ and, $M=rC+(1-r)A=r(\frac{\sqrt3}{2};\frac{1}{2})+(1-r)(-\frac{\sqrt3}{2};\frac{1}{2})=(r\sqrt3-\frac{\sqrt3}{2};\frac{1}{2})$ $N=rE+(1-r)C=r(0;-1)+(1-r)(\frac{\sqrt3}{2};\frac{1}{2})=(\frac{\sqrt3}{2}-\frac{\sqrt3}{2}r;\frac{1}{2}-\frac{3}{2}r)$ for having B, M, N collinear it must be $\frac{y_M-y_B}{x_M-x_B}=\frac{y_N-y_B}{x_N-x_B}$ so, it must be $\frac{\frac{1}{2}-1}{r\sqrt3-\frac{\sqrt3}{2}}=\frac{\frac{1}{2}-\frac{3}{2}r-1}{\frac{\sqrt3}{2}-\frac{\sqrt3}{2}r}$ $\frac{1}{2r\sqrt3-\sqrt3}=\frac{3r+1}{\sqrt3-r\sqrt3}$ $1-r=(2r-1)(3r+1)$ $3r^2=1$ $r=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$ i used my brain VERY less than you..... bye bye

I note $x=m(\widehat {EBM})$. From the relation $r=\frac{AM}{AC}=\frac{CN}{CE}$ results $\frac{MA}{MC}=\frac{NC}{NE}$, i.e. $\frac{BA}{BC}\cdot \frac{\sin (60+x)}{\sin (60-x)}=\frac{BC}{BE}\cdot \frac{\sin (60-x)}{\sin x}$. Thus, $2\sin x\sin (60+x)=\sin^2(60-x)\Longrightarrow$ $2[\cos 60-\cos (60+2x)]=1-\cos (120-2x)\Longrightarrow \cos (60+2x)=0\Longrightarrow x=15^{\circ}.$ Therefore, $\frac{MA}{MC}=\frac{\sin 75}{\sin 45}=\frac{1+\sqrt 3}{2}$, i.e. $r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}.$

Nice solution, levi!

Who is Levi ??

Levi=Ph-An=Virgil Nicula

It is O.K. See my two solutions for the your nice and easy problem from the section "Geometry Own and Proposed Problems"

frengo wrote: the solution in which you have not to think about anything... (...) i used my brain VERY less than you..... Sometimes it's good to find solutions in which you do not have to think, mas all correct solutions are welcome and sometimes you have to think... Here's my solution. From the statement, we cleary have $ \overline{AC}=\overline{CE}$, therefore $ \overline{AM}=\overline{CN}$. We claim $ \overline{AM}=\overline{AB}$. Suppose $ \overline{AM}>\overline{AB}$. So we have $ \overline{CN}>\overline{CB}$, then: $ A\hat{B}M+C\hat{B}N>A\hat{M}B+C\hat{N}B\Leftrightarrow A\hat{B}M+C\hat{B}N>C\hat{M}N+C\hat{N}B\Leftrightarrow$ $ \Leftrightarrow A\hat{B}C>180^{\circ}-A\hat{C}E\Leftrightarrow A\hat{B}C>A\hat{F}E$ Suppose $ \overline{AM}<\overline{AB}$. So we have $ \overline{CN}<\overline{CB}$, then: $ A\hat{B}M+C\hat{B}N<A\hat{M}B+C\hat{N}B\Leftrightarrow A\hat{B}M+C\hat{B}N<C\hat{M}N+C\hat{N}B\Leftrightarrow$ $ \Leftrightarrow A\hat{B}C<180^{\circ}-A\hat{C}E\Leftrightarrow A\hat{B}C<A\hat{F}E$ In both situations we get a contradiction, so our claim is correct. Using Co-Sine's Law we get $ r=\frac{\sqrt{3}}{3}$.

$CM=EN \Rightarrow \angle NDE = \angle MBC$ $\Rightarrow \angle NBC + \angle NDC = 120^\circ \Rightarrow \angle BND = 120^\circ$ Since $BC=CD$ and $360^\circ - \angle BCD = 240^\circ = 2\cdot 120^\circ = 2 \cdot \angle BND$, $C$ is the center of the circle passing through $B,N,D$. So $CN=1$ and $CE=\sqrt 3$, $r=\dfrac {\sqrt 3}3$.

We assign the affixes $1,\epsilon,{\epsilon}^2,\cdots {\epsilon}^5$ to the points $B,C,D,E,F,A$ where ${%Error. "eplison" is a bad command. }=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, We denote the coordinates of an uppercase letter by a lowercase letter. Then $\frac{a-m}{a-c}=\frac{{\epsilon}^5-m}{{\epsilon}^5-\epsilon}=r \Rightarrow m={\epsilon}r+{\epsilon}^5(1-r)$.Similarly $n={\epsilon}^2r+\epsilon(1-r)$. $m-1={\epsilon}r+{\epsilon}^5(1-r)-1={\epsilon}r-{\epsilon}^2(1-r)-1=\frac{-1}{2}+\frac{\sqrt{3}(2r-1)}{2}i$. Similarly $n-1=\frac{-1-3r}{2}+\frac{\sqrt{3}(1-r)}{2}i$. Therefore $B,M,N$ collinear $\Leftrightarrow \text{Im}(\frac{m-1}{n-1})=0$ $\Leftrightarrow \sqrt{3}-\sqrt{3}r-(\sqrt{3}+3\sqrt{3}r)(2r-1)=0$ $\Leftrightarrow 6r^2-2=0$ $\Leftrightarrow r=\frac{1}{\sqrt{3}}$