Obvious, that $W$ lies on bisector $CI$ of $\angle C$. $AI=AM \cos{\frac{\angle A}{2}}$ $\angle AWC=\angle B, \angle WAI=\frac{\angle A+ \angle C}{2}=90-\frac{\angle B}{2}$ $IW=\frac{AI \sin{(90-\frac{\angle B}{2}})}{\sin{\angle B}}= \frac{ AM \cos{\frac{\angle A}{2}}}{2 \sin{\frac{\angle B}{2}}}$ $\angle IMC= 90+\frac{\angle A}{2}, \angle MIC=\frac{\angle B}{2}$ $CI=\frac{MC \sin{\angle IMC}}{\sin{\angle MIC}}=\frac{ AM \cos{\frac{\angle A}{2}}}{\sin{\frac{\angle B}{2}}}=2IW$

RagvaloD wrote: Obvious, that $W$ lies on bisector $CI$ of $\angle C$. $AI=AM \cos{\frac{\angle A}{2}}$ $\angle AWC=\angle B, \angle WAI=\frac{\angle A+ \angle C}{2}=90-\frac{\angle B}{2}$ $IW=\frac{AI \sin{(90-\frac{\angle B}{2}})}{\sin{\angle B}}= \frac{ AM \cos{\frac{\angle A}{2}}}{2 \sin{\frac{\angle B}{2}}}$ $\angle IMC= 90+\frac{\angle A}{2}, \angle MIC=\frac{\angle B}{2}$ $CI=\frac{MC \sin{\angle IMC}}{\sin{\angle MIC}}=\frac{ AM \cos{\frac{\angle A}{2}}}{\sin{\frac{\angle B}{2}}}=2IW$ Good,but P'tolemy may work better at the end...

Stats- 21 students solved it in the contest.

Draw \(WL \perp AI\) to intersect \(AC\) at \(N.\) By the in-excenter lemma, \(AW = IW\), so, \(L\) is the midpoint of \(AI\) and also \(LN || IM\). \(=> N\) is the midpoint of \(AM\) \(=> CI : IW = 2 : 1\). \(\boxed{Q.E.D}\)

Answer: $\boxed{2}$. Let $a, b, c$ be sides of $ABC$; as usual. By Fact 5, $WI=WA=WB$; hence $$\frac{CI}{IW}=\frac{\left(\frac{(s-c)}{\cos \tfrac{C}{2}}\right)}{\left(\frac{c}{2\cos \tfrac{C}{2}}\right)}=\frac{2(s-c)}{c}.$$Observe that $$\angle AIM=90^{\circ} \implies \frac{b}{2}=\frac{(s-c)}{\cos^2 \tfrac{A}{2}} \implies c=\tfrac{a+b}{3},$$as desired.

Notice that $M$ is the touchpoint of the $A$-mixtilinear incircle on $\overline{AC}$. By $\sqrt{bc}$ inversion, we know that $\tfrac{b}{2}=\tfrac{2bc}{a+b+c}\Longrightarrow c=\tfrac{a+b}{3}$. Then $$\frac{CI}{CW}=\frac{CI\cdot CD}{ab}=\frac{CI^2\cdot \frac{a+b+c}{a+b}}{ab}=\frac{a+b-c}{a+b}=\frac{2}{3}\Longrightarrow CI:IW=2$$

Let $P$ and $N$ be the $B$-intouch and extouch point respectively, and let $S$ be the midpoint of arc $BC$(not containing $A$). By homothety, it can easily be shown that $IM \parallel BN \Rightarrow AI \bot BN$. Now, $\angle AIM = 90^{\circ} \Rightarrow MI$ is tangent to $(AIP)$, as $\angle API = 90^{\circ}$ $\Rightarrow MI^2 = MP \cdot MA = MN \cdot MC$, as $P$ and $N$ are isotomic points on $AC$ $\Rightarrow MI$ is tangent to $(MNC)$ And, $MI \bot AI \Rightarrow$ The center of $(INC)$ lies on $AI$ Also, The center of $(INC)$ lies on the perpendicular bisector of $IC \Rightarrow S$ is the center of $(INC)$ But, The circle passing through $I$ and $C$, centered at $S$ also passes through $B \Rightarrow BINC$ lie on a circle centered at S. $\Rightarrow SI$ is the perpendicular bisector of $BN \Rightarrow AB = AN \Rightarrow c = s-c$ Now, By Ptolemy's theorem on $BWAC$, we get $BW \cdot AC + AW \cdot BC = CW \cdot AB$ Using, $AW = BW = IW$ and $CW = CI+AW$, we get $IW(AC+BC-AB) = CI \cdot AB \Rightarrow IW \times 2(s-c) = CI \cdot c \Rightarrow CI:IW = 2:1$

Let $T_A$ be the $A$-mixtilinear intouch point and $V$ the midpoint of the arc $\widehat{AC}$ not containing $B$. The condition $\angle AIM=90^\circ$ implies that the $A$-mixtilinear incircle touches $AC$ at its midpoint. Because $MT_A$ bisects $\widehat{AVC}$, it follows that $T_AV$ is the perpendicular bisector of $\overline{AC}$, thus $VK\perp KT_A$, where $K=\overline{AI}\cap (ABC),\ K\neq A$. We also have $VK\perp IC$ being $VK$ the perpendicular bisector of $\overline{IC}$, thus $KT_A\parallel CW$, i.e. $CWT_AK$ is an isosceles trapezoid. It's known that $T_AIMC$ is cyclic, then $\angle T_AIC=90^\circ\therefore T_AIJK$ is a rectangle ($J=\overline{VK}\cap \overline{IC}$). By symmetry, $\bigtriangleup WT_AI\cong \bigtriangleup CKJ$ so $WI=CJ=JI$ therefore $CI:IW=2$.

Easy for #2 anantmudgal09 wrote: Let $I$ be the incenter of a triangle $ABC$, $M$ be the midpoint of $AC$, and $W$ be the midpoint of arc $AB$ of the circumcircle not containing $C$. It is known that $\angle AIM = 90^\circ$. Find the ratio $CI:IW$. Answer: $\boxed{CI : IW = 2:1}$. $ $ Proof. Since $\angle AIM = 90^{\circ} \implies N$ is the center of $\odot(AIM)$ where $N$ is the mid-point of $\overline{AM}$. Now it is well known that $\angle AIC = 90^{\circ} + \frac{1}{2} \angle B \implies \angle CIM = \frac{1}{2} \angle B$. So $\angle MIC = \angle IBC \implies \overline{MI} $ is tangent to $\odot(IBC) \implies \triangle CMI \sim \triangle CIB$. Now let $W_{c}$ denote the mid-point of $\widehat{AC}$ not containing $B$. Then the points $W , N , W_{c}$ are collinear. (The line determined by these points is the perpendicular bisector of $AI$). Now, $$ \angle CWW_{c} = \angle CWN = \frac{1}{2} \angle B \implies \triangle CMI \sim \triangle CNW \implies \frac{CM}{CI} = \frac{CN}{CW} \implies \frac{CI}{CW} = \frac{CM}{CN} = \frac{2}{3} \iff CI : IW = 2:1$$