Approach by vladimir92: Observe that $AP$ is symmedian and let it intersect $BC$ at $E$ and the circumcircle of $\triangle{ABC}$ at $G$ and denote $N\equiv (AO)\cap (O_1O_2)$. An easy angle chasing gives that: $\angle{NAO_1}=\angle{PCG}$ and $\angle{NAO_2}=\angle{PCB}$. Therefor, $\frac{NO_1}{NO_2}=\frac{AO_1}{AO_2}.\frac{sin(\angle{PCG})}{sin(\angle{PBG})}$, since $AO_1$ is the circumraduis of $\odot(APC)$ we have $2AO_1=\frac{PC}{sin(\angle{PAC})}$ similary $2AO_2=\frac{PB}{sin(\angle{PAB})}$. Then $\frac{AO_1}{AO_2}=\frac{PC}{PB}.\frac{GB}{GC}$ It follow that $\frac{NO_1}{NO_2}=\frac{EC}{EB}.\left(\frac{BG}{CG}\right)^2=\left(\frac{AC}{CG}.\frac{AB}{BG}\right)^2$ because $AE$ is symmedian. Another angle chasing gives that $\triangle{ABG}\sim \triangle{AMC}$ and $\triangle{ACG}\sim\triangle{AMB}$ from which follow immediatly that $\frac{NO_1}{NO_2}=1$ meaning that $AO$ passe trough the midpoint of $O_1O_2$.

Let the tangents of $\odot(ABC) \equiv (O)$ through $B,C$ intersect at $D.$ Then $P \in AD.$ Let $N,L$ be the orthogonal projections of $M$ on $AB,AC$ and let $(O')$ be the circumcircle of $ANML.$ $C'$ denotes the orthogonal projection of $C$ onto $AB$ and $U$ denotes the midpoint of $DM.$ From $\angle BCD=\angle BAC,$ we have $\frac{_{AC}}{^{DC}}=\frac{_{CC'}}{^{DM}}=\frac{_{MN}}{^{MU}}$ $\Longrightarrow$ $\triangle ACD \sim \triangle NMU$ by SAS criterion. Thus, $\angle UNM=\angle DAC=\angle NLM$ $\Longrightarrow$ $UN$ is tangent to $(O')$ through $N.$ Likewise, $UL$ is tangent to $(O')$ through $L$ $\Longrightarrow$ $OM \equiv MU$ is the M-symmedian of $\triangle MNL.$ If $E$ is the midpoint of $NL,$ then $\angle LME= \angle NMO$ yields $AO \parallel ME.$ Now, since $O_1O_2 \perp AP \perp NL,$ $OO_1 \parallel MN$ and $OO_2 \parallel ML,$ it follows that $\triangle MNL$ and $\triangle OO_1O_2$ are homothetic with corresponding cevians $ME,OA.$ Therefore, ray $OA$ is the O-median of $\triangle OO_1O_2.$

$\frac{O_{1}N}{\sin\angle O_{1}ON}=\frac{ON}{\sin\angle NO_{1}O}\Rightarrow O_{1}N=\frac{ON\cdot\sin\angle C}{\sin\angle BAP}$ $~$ $~$ Similarly $~$ $O_{2}N=\frac{ON\cdot\sin\angle B}{\sin\angle PAC}$ $~$ $(1)$ $\frac{BA}{\sin\angle AMB}=\frac{BM}{\sin\angle BAM}$ $~$ and $~$ $\frac{CA}{\sin\angle AMC}=\frac{CM}{\sin\angle CAM}\Rightarrow \frac{\sin\angle B}{\sin\angle C}=\frac{\sin\angle PAC}{\sin\angle BAP}$ $~$ $(2)$ $\therefore O_{1}N=O_{2}N$

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Let the perpendicular line from B cut $(O),(O_1)$ at second points $E,I$,respectively.Denote the points F,J in the same way. Obviously $O_1,O_2$ are midpoints of $AI,AJ$.Hence it's suffice to prove that $AE$ pass through the midpoint of $IJ$. Also,$AD$ is the A-symmedian of triangle $ABC$ and $D \in (O)$.So $A,D,B,C$ form a harmonic quadilateral.It's equivalent with $(ED,EA,EI,EF)=-1$.Because $IJ \perp AD \perp ED$ so $IJ \| ED$.From those we deduce that $EA$ pass through the midpoint of $IJ$ (Q.E.D) Image not found

Let $AM$ intersect circumcircle of triangle $ABC$ in $N$. It's easy to prove, that $\triangle BNC \sim \triangle O_1OO_2$ and $\angle AOO_2=\angle ANC$ which proves that $AO$ is median in triangle $OO_1O_2$.

This is another approach. Let lines $CA$ , $BA$ and $AP$ meet $(O_1)$ , $(O_2)$ and $(O)$ at $B_1$ , $C_1$ and $S$ respectively. Easy to see that $\triangle{PC_1C}\sim\triangle{PBB_1}$ and $\triangle{PC_1B}\sim \triangle{SCB} \sim \triangle{PCB_1}$, So, $\frac{BB_1}{CC_1}=\frac{PB}{PC_1}=\frac{SB}{SC}=\frac{AB}{AC}$ because quadrilateral $CABS$ is harmonic, thus $(B_1C_1)\parallel(BC)$. Now let $C_2$ and $B_2$ be the orthogonale projections of $O_1$ and $O_2$ into $CA$ and $BA$ respectively, So $(B_2C_2)\parallel(M_bM_c)$ where $M_b$ and $M_c$ are midpoints of $AC$ and $AB$ respectively. Denote $X\equiv(O_1C_2)\cap(O_2B_2)$ and ${Y\equiv(AO)\cap(O_2C_2}$ So $\angle{C_2XA}=\angle{C_2B_2A}=\angle{AM_cM_b}=\angle{AOM_b}=\angle{C_2YA}$, Then $X\equiv Y$, or again $A\in OX$ and since $O_2OO_1X$ is a parallelogram, we deduce that $AO$ bissect segement $O_1O_2$.

Let $AP$ intersect the circumcircle of $ABC$ at $D$, let $AM$ intersect the circumcircle of $ABC$ at $E$. I will show that $\triangle O_1OO_2 \sim \triangle BEC$, from which we can use the fact that $\angle O_1OA = \angle ACB = \angle BEM$ to conclude that $OA$ intersects $O_1O_2$ at its midpoint. Note that we have a spiral similarity making $\triangle DOC \sim \triangle O_2PC$, so $\frac {DP} {OO_2} = \frac {DC} {OC}$. Similarly, we can get $\frac {DP} {OO_1} = \frac {DB} {OB}$. Combining these two equations and using $OB = OC$, we obtain $\frac {DC} {DB} = \frac {OO_1} {OO_2}$. But we also have $DC = EB$ and $DB = EC$ since $AE$ and $AD$ are isogonal lines. In addition, $\angle BEC = 180 - \angle BAC = \angle O_1OO_2$, so by SAS similarity, $\triangle O_1OO_2 \sim \triangle BEC$, as desired.

We just need to prove that the angle between the $A-$median and $A-$attitude of $\Delta AO_1O_2$ is equal to the angle between $AO$ and $AP$ , or the angle between $AM$ and $A-$attitude of $\Delta ABC$ . Consider the isogonal conjugate of $P$ w.r.t. $\Delta ABC$ , $Q$ , which is on segment $AM$ . We have $ \angle QBC = \angle AO_1O_2 $ , $ \angle QCB = \angle AO_2O_1 $ so $ \Delta AO_1O_2 $ ~ $ \Delta QBC $ . Since $AO_1 > AO_2 $ , we can see that the $A-$median of $\Delta AO_1O_2 $ and $AO$ are of the same side of $AP$ . At the same time ,the angle between the $A-$median and $A-$attitude of $\Delta AO_1O_2$ $~=$ the angle between the $A-$median and $A-$attitude of $\Delta QBC$ $~=$ the angle between $AM$ and $A-$attitude of $\Delta ABC$ , done .

My solution: Let $AO\cap O_1O_2=S,AP\cap O_1O_2=L$ and $R,N$ are midpoints of $AB,AC$ Note that $OO_1,OO_2$ are perpendicular bisectors of $AB,AC\longrightarrow \angle AOO_1=\angle C,\angle AOO_2=\angle B$ so in $\triangle O_1OO_2$: $\frac{SO_2}{SO_1}=\frac{\sin B}{\sin C}.\frac{\sin OO_2O_1}{\sin OO_1O_2}$(1) but observe that in cyclic quadrilaterals $ALNO_2,ALRO_1$: $\angle O_1O_2O=\angle CAP,\angle O_2O_1O=\angle PAB\longrightarrow \frac{\sin OO_2O_1}{\sin OO_1O_2}=\frac{\sin PAC}{\sin PAB}=\frac{\sin MAB}{\sin MAC}=\frac{\sin C}{\sin B}$(2) combining (1),(2) we get the result. DONE

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My solution: $O$ is the center of $(ABC)$ $t$ is the line passing through $A$ which is perpendicular to $AO$ $x$ is the line passing through $O$ which is perpendicular to $AP$ $\Rightarrow x\parallel{O_1O_2}$ According to the subject: $AP$ is the symmedian of $\triangle{ABC}$ $\Rightarrow A(BCPt) = -1 \Rightarrow O(O_1O_2xA) = -1$ (orthogonal harmonic pencil) $\Rightarrow OA$ bisects $O_1O_2$ (because $Ox\parallel{O_1O_2}$) Q.E.D

Let $AO$ cut $O_1O_2$ at $K$, and let $X, Y$ be the midpoints of $\overline{AB}, \overline{AC}$, respectively. Consider the inversion $\mathcal{T} : X \mapsto X'$, composed of an inversion with pole $A$ and radius $r = \sqrt{bc}$ combined with a reflection in the $A$-angle bisector. It is easy to see that $B' \equiv C$ and $C' \equiv B.$ Then since $P$ lies on the $A$-symmedian in $\triangle ABC$, it follows that $P'$ lies on the $A$-median in $\triangle AB'C'.$ Furthermore note that $X'$ is the reflection of $A$ in $B'$, and $Y'$ is the reflection of $A$ in $C'.$ Then because $\angle AXO = \angle AYO = 90^{\circ}$, it follows under inversion that $\angle AO'X' = \angle AO'Y' = 90^{\circ} \implies O'$ is the projection of $A$ onto $X'Y'.$ Hence, $O'$ is the reflection of $A$ in $B'C'.$ Similarly, we find that $O_1'$ is the reflection of $A$ in $B'P'$ and $O_2'$ is the reflection of $A$ in $C'P'.$ Meanwhile, $K'$ is the second intersection of $AO'$ and $\odot (AO_1'O_2').$ From the inversive distance formula, we have \[K'O_1' = KO_1 \cdot \frac{r^2}{AK \cdot AO_1} \quad \text{and} \quad K'O_2' = KO_2 \cdot \frac{r^2}{AK \cdot AO_2} \implies \frac{K'O_1'}{K'O_2'} = \frac{KO_1}{KO_2} \cdot \frac{AO_2}{AO_1}.\] Furthermore, \[AO_1 \cdot AO_1' = AO_2 \cdot AO_2' = r^2 \implies \frac{AO_2}{AO_1} = \frac{AO_1'}{AO_2'}.\] Therefore, in order to show that $K$ is the midpoint of $\overline{O_1O_2}$, we need only show that $\tfrac{K'O_1'}{K'O_2'} = \tfrac{AO_1'}{AO_2'}$, i.e. show that quadrilateral $AO_1'K'O_2'$ is harmonic. To see this, let $O_1^*, O_2^*$ be the projections of $A$ onto $B'P', C'P'$, respectively, and let $K^*$ be the midpoint of $\overline{AK'}.$ By considering the homothety with center $A$ and ratio $1 / 2$, it is sufficient to prove that quadrilateral $AO_1^*K^*O_2^*$ is harmonic. Because $AO_1^* \perp P'O_1^*$ and $AO_2^* \perp P'O_2^*$, it is clear that $A, O_1^*, O_2^*, P'$ are inscribed in the circle $\omega$ of diameter $\overline{AP'}.$ Hence, $K^*$ also lies on $\omega \implies P'K^* \perp AK^* \implies P'K^* \parallel B'C'.$ Then if $P_{\infty}$ denotes a point at infinity on line $B'C'$ and $M^*$ is the midpoint of $\overline{B'C'}$, the division $\left(B', C'; M^*, P_{\infty}\right)$ is harmonic. By taking perspective at $P'$ onto $\omega$, we obtain the desired result. $\square$

use that in triangel $O2O1O$if $AO$ is median $\angle O_1O_2O=\angle CAP,\angle O_2O_1O=\angle PAB\longrightarrow \frac{\sin OO_2O_1}{\sin OO_1O_2}=\frac{\sin PAC}{\sin PAB}=\frac{\sin MAB}{\sin MAC}=\frac{\sin C}{\sin B}$

Let $M,N$ be the midpoints of $AB,BC$ respectively and let $l$ be a line perpendicular to the $A-$symmedian passing through $A$. Let $OM \cap l= X$ and $ON \cap l=Y$ Then obviously $\triangle OXY$ is homothetic to $\triangle OO_1O_2$ with $O$ the centre of homothety. Thus we are left to prove that $A$ is the midpoint of $XY$. This is easy trigo.

my solution = we have $OO_1\perp AB$ and $OO_2\perp AC$ and $O_1O_2\perp AP$ also angle chasing we get $\angle O_2O_1O=\angle BAP$ and $\angle O_1O_2O=\angle PAC$. let $AO\cap O_1O_2=D$. so by sine law in triangle $DO_1O$ and triangle $DO_2O$ we get $\frac{O_1D}{O_2D}=\frac{sin C.sinBAP}{sinB.sinPAC}=1$ as by sine law in triangle $ABP$ and $ACP$ along with using $AM=BM$ we get $\frac{sinC}{sinB}=\frac{sinPAC}{sinBAP}$ so we are done.

We will bash this problem. Observe that showing \[ \frac{\cos{\angle{MAB}}}{\cos{\angle{MAC}}} = \frac{\sin{C}}{\sin{B}} = \frac{AB}{AC} \]is sufficient to conclude. Now, observe that \[ \cos{\angle{MAB}} = \frac{MA^2 + AB^2 - MB^2}{2(MA)(AB)}, \]so the problem transforms into \[ \frac{MA^2 + AB^2 - MB^2}{MA^2 + AC^2 - MC^2} = \frac{AB^2}{AC^2}. \]This is equivalent to showing that \[ \frac{AB^2}{AC^2} = \frac{MA^2 - MB^2}{MA^2 - MC^2} = \frac{BO_1^2 - O_1A^2}{CO_2^2 - AO_2^2} = \frac{\operatorname{Pow}(B, (APC))}{\operatorname{Pow}(C, (APB))} \]Extend $BA$ to hit $(APC)$ at $B'$ and $CA$ to hit $(APB)$ at $C'$. We're obviously done if we can show that \[ \frac{PC}{PB'} = \frac{AC}{AB} \]since then by the spiral similarity $\bigtriangleup{CPC'} \sim \bigtriangleup{PB'B}$ we will have that $\frac{BB'}{CC'} = \frac{AB}{AC}$, which will of course conclude by the powers. But observe that \[ \frac{PC}{PB'} = \frac{\sin{\angle{PB'C}}}{\sin{\angle{PCB'}}} = \frac{\sin{\angle{PAC}}}{\sin{\angle{PAB}}} = \frac{AC}{AB} \]and we may conclude.

"You won't last more than 10 minutes" ahh problem. Reflect $A$ over $O,O_1,O_2$ and let them be $A',A_1,A_2$ respectively, then trivially due to all diameters we got that the triples $(A_1,B,A'), (A_2, C, A'), (A_1, P, A_2)$ are colinear and from ratio Lemma all we need to get is: \[\frac{\sin \angle AA'A_2}{\sin \angle AA'A_1}=\frac{AC}{AB}=\frac{\sin \angle PAC}{\sin \angle PAB}=\frac{\sin \angle PA_2A'}{\sin \angle PA_1A'}=\frac{A_1A'}{A'A_2} \]Giving that $AA'$ bisects $A_1A_2$ so by homothety at $A$ we are done .