A solution: Let $T$ be a point on the circle $(IPQ)$, on the opposite side of $PQ$ from $I$, s.t. $\frac{TP}{TQ}=\frac{PI}{QC}$ (it's easy to argue that such a point exists, for continuity reasons). The triangles $TBI,TIC$ are now directly similar, and a simple angle chase will show that $\angle BTC=\pi-\angle BAC$, so $T$ belongs to the circle $(ABC)$. Since the triangles $TBI,TIC$ are directly similar, we have $\angle BTI=\angle ITC$, so $TI$ is the bisector of $\angle BTC$, thus cutting $(ABC)$ in the midpoint $M$ of the large arc $BC$ (I mean the arc which also contains $A$). Let $P',Q'$ be the points where $M_bT,M_cT$ cut $CI,BI$ respectively, where $M_b,M_c$ are the points where $BI,CI$ cut $(ABC)$, other than $B,C$. Now observe (angle chase) that the figures $TM_cBI,TM_bIC$ are directly similar, so $IP'\cdot IQ'=BP'\cdot CQ'\ (*)$. Then, observe that $TM$ is the median of $TM_bM_c$, so $P'Q'\|M_bM_c\ (**)$. $P',Q'$ are uniquely determined by $(*),(**)$ on $BI,CI$, and $P,Q$ also have these two properties, so $P=P',Q=Q'$, and $T$ is a homothetic center of $MM_cM_b$ and $IPQ$. The circles $(IPQ),(ABC)=(MM_cM_b)$ are then homothetic with center $T$, and since this point lies on both of them, as we have established above, the two circles must be tangent in $T$, as desired.

Why in the title you said <<rusian geometry>> ?This problem was posted again in Russia?

grobber wrote: Then, observe that $ TM$ is the median of $ TM_bM_c$, so $ P'Q'\|M_bM_c\ (**)$. How does the second statement follow from the first? I can't see...

hollandman wrote: grobber wrote: Then, observe that $ TM$ is the median of $ TM_bM_c$, so $ P'Q'\|M_bM_c\ (**)$. How does the second statement follow from the first? I can't see... notice these two figures are similar so TP'/TMc=TQ'/TMb, thus we get P'Q'//McMb

Well, I might be dumb, but could you please tell me why the two figures are similar? (Can you prove that $ TI$ bisects $ P'Q'$?)

hollandman wrote: Well, I might be dumb, but could you please tell me why the two figures are similar? (Can you prove that $ TI$ bisects $ P'Q'$?) hi,dear hollandman: the angles are correspondingly equal which can be proved by angle chasing combined with IT/BT=TC/IT so two quadrilaterals are similar

let the tangency be S,we can also get the following conclusion: 1$\frac{BS}{SC}=\frac{BI^{2}}{IC^{2}}$ 2IS bisects arc BAC 3triangle BSI and triangle ISC are similiar triangle BPS and triangle IQC are similiar triangle IPS and triangle CQS are similiar

Dear Mathlinkers, in order to make a link with another problem, what is the geometric nature of the point S of the last message? Sincerely Jean-Louis

Consider the inversion centred at $I$ that maintains the circle $(B,C,P,Q)$. We denote by $X'$ the invert image of $X$. We have to show that $P'Q'$ is tangent to $(A',B',C')$. Clearly, $I$ is the orthocentre of triangle $(A'B'C')$ and let $D,E,F$ be the feet of the altitudes of $A,B,C$ respectively. We have $EF//P'Q'$. By Nagel theorem we clearly deduce that $AT$ and $EF$ are orthogonal to each other. Clearly $P'Q'//EF$, so it suffices to show that $T$ belongs to $P'Q'$. Since $IBTC$ is a parallelogram, we need to show that $\displaystyle {\frac{IC'}{IQ'}+\frac{IB'}{IP'}=1}$. This is equivalent to the given relation.

Dear Mathlinkers, considering my conjecture and its proof, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=467843 now we can have a nice proof of the above result Sincerely Jean-Louis

A fairly standard and easy problem,consider $R$ on $BC$ such that $PM//IC$ and $QM//IB$(this point exists since the condition of the problem),now just observe that $IR$ is the radical axis of $BRM$ and $CQR$,now just from an easy angle chase we get that the circumcircles of $BRM$,$CQR$,$IPQ$ and $ABC$ concur at point $F$ lying on $IR$,so just observe that $FP$ intersects circumcircle of $ABC$ at the midpoint of arc $AB$,$FQ$ at the midpoint of $AC$ and $IF$ at the midpoint of $BC$ containing $A$,so from this it is an obvious angle chase.