Let $O, S, T$ be the circumcenters of triangles $BCD$, $XAY$ and $ACD$. Claim: $A, C, O, D$ are concyclic. Proof. Note that \begin{align*} \angle (DA, AC)=360^{\circ}-(\angle DAB+\angle BAC) &=360^{\circ}-2\cdot(180^{\circ}-\angle DBC)=\angle DOC \end{align*}where $\angle DAB=\angle BAC=180^{\circ}-\angle DBC$ follows from the tangency, proving the claim. $\blacksquare$ Now remark that $\triangle ACB \sim \triangle ABD \sim \triangle AO_2O_1$, where $O_1, O_2$ are the centers of $\omega_1, \omega_2$. By looking at the spiral similarity pivoted at $A$, taking $CB$ to $BD$, we conclude that $Y$ goes to $X$, so, the quadrilaterals $ACYB, AO_2SO_1,$ and $ABXD$ are all similar, yielding $SO_1=SO_2$. Finally, note that $\triangle DO_1O \sim \triangle DBC \sim \triangle OO_2C$. Let $O'$ be the reflection of $O_2$ in $TO$, clearly, $O_1, O'$ are reflections in the perpendicular bisector of $DO$, thus, $TO_2=TO_1$ follows. $\blacksquare$

@above why did you change notation? Solution- We will show that $PO_1=PO_2$ and $QO_1=QO_2$ . Part I- Let $S,R,T$ be mid points of $AD,AC,AB$, then the quadrilaterals $O_1SAT$ and $O_2RAT$ are cylcic with diameter $O_1A$ and $O_2A$. Hence $\angle PO_1O_2=\angle SO_1T=180-\angle DAB=\angle ADB+\angle ABD= \angle ABC+\angle ACB=180-\angle BAC =\angle RO_2S=\angle PO_2O_1$ Hence $PO_1=PO_2$. Part II- From angle chase in part I we have $\angle DAB=\angle BAC \implies \angle XAB=\angle YAB$ Now let $M,N$ be midpoints of $XA,YA$, then as before we have quadrilaterals $MO_1AT$ and $NO_2AT$ are cyclic. So $\angle QO_1O_2=\angle MO_1T=\angle MAT=\angle XAB=\angle BAY=\angle TAN=\angle TO_2N=\angle O_1O_2N$ Hence $QO_1=QO_2$ Hence proved.

Ohh god .How come they put such a trivial geo on the paper . [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.606614403935124, xmax = 69.94950900043317, ymin = -33.53977373724896, ymax = 20.979432433038262; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); /* draw figures */ draw(circle((2.645133883320531,-2.6435670317322884), 8.85141709724894), linewidth(0.4) + linetype("4 4") + dtsfsf); draw(circle((28.168374622560695,-5.228199005326221), 20.374589572642932), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((9.5064509375227,2.948383693299657)--(8.246106084330277,-9.497521731975564), linewidth(0.4) + rvwvcq); draw((2.645133883320531,-2.6435670317322884)--(28.168374622560695,-5.228199005326221), linewidth(0.4) + rvwvcq); draw((21.30705756835853,-10.820149730358168)--(8.246106084330277,-9.497521731975564), linewidth(0.4) + rvwvcq); draw((-5.46224269888529,0.908619560181418)--(9.5064509375227,2.948383693299657), linewidth(0.4) + rvwvcq); draw((-5.46224269888529,0.908619560181418)--(3.1854287144896913,-11.478478809723208), linewidth(0.4) + rvwvcq); draw((8.246106084330277,-9.497521731975564)--(3.1854287144896913,-11.478478809723208), linewidth(0.4) + rvwvcq); draw((2.645133883320531,-2.6435670317322884)--(9.5064509375227,2.948383693299657), linewidth(0.4) + rvwvcq); draw((9.5064509375227,2.948383693299657)--(28.168374622560695,-5.228199005326221), linewidth(0.4) + rvwvcq); draw((2.645133883320531,-2.6435670317322884)--(3.1854287144896913,-11.478478809723208), linewidth(0.4) + rvwvcq); draw((28.168374622560695,-5.228199005326221)--(32.41148517830495,-25.156066755423865), linewidth(0.4) + rvwvcq); draw((9.5064509375227,2.948383693299657)--(43.96206250001865,7.643604422441301), linewidth(0.4) + rvwvcq); draw((43.96206250001865,7.643604422441301)--(32.41148517830495,-25.156066755423865), linewidth(0.4) + rvwvcq); draw((8.246106084330277,-9.497521731975564)--(32.41148517830495,-25.156066755423865), linewidth(0.4) + rvwvcq); draw((2.645133883320531,-2.6435670317322884)--(8.246106084330277,-9.497521731975564), linewidth(0.4) + rvwvcq); draw((8.246106084330277,-9.497521731975564)--(28.168374622560695,-5.228199005326221), linewidth(0.4) + rvwvcq); draw((-5.46224269888529,0.908619560181418)--(21.30705756835853,-10.820149730358168), linewidth(0.4) + rvwvcq); draw((9.5064509375227,2.948383693299657)--(3.1854287144896913,-11.478478809723208), linewidth(0.4) + rvwvcq); draw((8.246106084330277,-9.497521731975564)--(43.96206250001865,7.643604422441301), linewidth(0.4) + rvwvcq); draw((-5.46224269888529,0.908619560181418)--(8.246106084330277,-9.497521731975564), linewidth(0.4) + rvwvcq); draw((9.5064509375227,2.948383693299657)--(32.41148517830495,-25.156066755423865), linewidth(0.4) + rvwvcq); draw(circle((16.069768915035553,2.61138676965826), 14.416496286875642), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((21.30705756835853,-10.820149730358168)--(43.96206250001865,7.643604422441301), linewidth(0.4) + rvwvcq); draw(shift((17.522238898162183,16.954527853033788))*xscale(28.031367558997946)*yscale(28.031367558997946)*arc((0,0),1,214.91963350166375,340.6000159190857), linewidth(0.4) + wvvxds); draw(shift((12.887834679647847,-28.810213804795318))*xscale(19.862671508438734)*yscale(19.862671508438734)*arc((0,0),1,10.601120456923134,119.24035562207919), linewidth(0.4) + wvvxds); /* dots and labels */ dot((2.645133883320531,-2.6435670317322884),dotstyle); label("$O_1$", (2.24276623438307,-1.9647369286951997), NE * labelscalefactor); dot((28.168374622560695,-5.228199005326221),dotstyle); label("$O_2$", (28.403407936546525,-4.6987384180606355), NE * labelscalefactor); dot((9.5064509375227,2.948383693299657),linewidth(4pt) + dotstyle); label("B", (9.694260489712416,3.3960503053546747), NE * labelscalefactor); dot((8.246106084330277,-9.497521731975564),linewidth(4pt) + dotstyle); label("A", (8.461279425880942,-9.094583949981532), NE * labelscalefactor); dot((3.1854287144896913,-11.478478809723208),linewidth(4pt) + dotstyle); label("D", (3.4221394258740454,-11.024467354239487), NE * labelscalefactor); dot((32.41148517830495,-25.156066755423865),linewidth(4pt) + dotstyle); label("C", (32.63842985144594,-24.748082673407165), NE * labelscalefactor); dot((-5.46224269888529,0.908619560181418),linewidth(4pt) + dotstyle); label("$X$", (-5.262335893286775,1.3589511564157224), NE * labelscalefactor); dot((43.96206250001865,7.643604422441301),linewidth(4pt) + dotstyle); label("$Y$", (44.1641224046532,8.059935198978065), NE * labelscalefactor); dot((21.30705756835853,-10.820149730358168),linewidth(4pt) + dotstyle); label("$I$", (21.541600276962672,-10.381172886153502), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof : Let $\odot(XAY) \cap \odot(ADC)=\{I,A\}$. Now notice that $\angle IO_2B=90^\circ-\angle O_2BC=\angle CBD=\angle CYB \implies \{I,O_2,Y\}$ is a collinear triple. Similiarly we get that $\{I,O_1,X\}$ is a collinear triple. Now notice that $\angle IO_1O_2=\angle IO_1B-\angle O_2O_1B=2\angle IXB -\frac{\angle AO_1B}{2}=2\angle IXB-\angle AXB=\angle BXD-\angle AXB=\angle AXD=\angle ABD=\angle BYA=\angle O_1O_2A$. Now notice that $\angle AO_1O_2=\angle AXB$ . Similiarly $\angle AO_2O_1=\angle AYB \implies$ by spiral similiarity that $IAO_1O_2$ is cyclic. Along with the previous claim this implies $IAO_1O_2$ is an iscoseles trapezoid. From here the result follows trivially $\blacksquare$.

LOL

The key observation is that $A$ is the $B$ Dumpty point of $BCD.$ Then $CDAO_3$ are concyclic, where $O_3$ is the center of $(BCD).$ Then show that $O_3$ also lies on $XAY.$ Also $AO_1PO_2$ is cyclic by spiral similarity, and a simple angle chase shows $P$ is the arc midpoint. P.S. Humpty point in RMO and Dumpty point in INMO? Wow

Let me post a rigorious, neat, clean and clear proof for this Geo with three separate diagrams. INMO 2020 P1 wrote: Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$. Proposed by Prithwijit De [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.37, xmax = 11.37, ymin = -5.21, ymax = 8.09; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((-4.57,1.05), 3.959040287746514), linewidth(0.4) + red); draw(circle((0.71,0.97), 2.220720603768065), linewidth(0.4) + ttffqq); draw((-0.935921505623135,-0.5208193711269217)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.935921505623135,-0.5208193711269217)--(-5.63,4.83), linewidth(0.4)); draw((-5.63,4.83)--(-0.89,2.51), linewidth(0.4)); draw((-0.89,2.51)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.89,2.51)--(-0.935921505623135,-0.5208193711269217), linewidth(0.4)); draw((-5.63,4.83)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw(circle((-1.9421286846486794,6.362547428778127), 3.993631988559477), linewidth(0.4) + linetype("4 4") + ffdxqq); draw((-1.9421286846486794,6.362547428778127)--(-4.57,1.05), linewidth(2) + blue); draw((-1.9421286846486794,6.362547428778127)--(0.71,0.97), linewidth(2) + blue); draw((2.7563168338928214,0.10733702100596743)--(-7.50577002435651,-1.6061728791797008), linewidth(0.4)); draw((-7.50577002435651,-1.6061728791797008)--(-0.89,2.51), linewidth(0.4)); draw((-0.89,2.51)--(2.7563168338928214,0.10733702100596743), linewidth(0.4)); draw((-4.57,1.05)--(0.71,0.97), linewidth(2)); draw(circle((-1.9801363099243707,-3.1125906080471197), 5.727296248660026), linewidth(0.4) + linetype("4 4") + ffxfqq); draw((-1.9801363099243707,-3.1125906080471197)--(-4.57,1.05), linewidth(2) + blue); draw((-1.9801363099243707,-3.1125906080471197)--(0.71,0.97), linewidth(2) + blue); draw((-1.9421286846486794,6.362547428778127)--(-1.9801363099243707,-3.1125906080471197), linewidth(2)); draw((-5.63,4.83)--(-7.50577002435651,-1.6061728791797008), linewidth(0.4)); draw((0.4669852549326141,3.1773839343620836)--(2.7563168338928214,0.10733702100596743), linewidth(0.4)); /* dots and labels */ dot((-4.57,1.05),dotstyle); label("$O_1$", (-4.49,1.25), NE * labelscalefactor); dot((-0.89,2.51),dotstyle); label("$A$", (-0.81,2.71), NE * labelscalefactor); dot((0.71,0.97),dotstyle); label("$O_2$", (0.79,1.17), NE * labelscalefactor); dot((-0.935921505623135,-0.5208193711269217),dotstyle); label("$B$", (-0.95,-0.95), NE * labelscalefactor); dot((0.4669852549326141,3.1773839343620836),dotstyle); label("$C$", (0.55,3.37), NE * labelscalefactor); dot((-5.63,4.83),dotstyle); label("$D$", (-5.55,5.03), NE * labelscalefactor); label("$e$", (-3.95,9.35), NE * labelscalefactor,ffdxqq); dot((-1.9421286846486794,6.362547428778127),dotstyle); label("$P$", (-1.87,6.57), NE * labelscalefactor); dot((2.7563168338928214,0.10733702100596743),dotstyle); label("$Y$", (2.83,0.31), NE * labelscalefactor); dot((-7.50577002435651,-1.6061728791797008),dotstyle); label("$X$", (-7.83,-1.73), NE * labelscalefactor); dot((-1.9801363099243707,-3.1125906080471197),dotstyle); label("$Q$", (-1.69,-3.37), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Claim 1:-$\overline{X-B-Y}$. As $XD=XB$. Hence, $\angle DBX=90^\circ-\frac{\angle DXB}{2}$ , as $BC$ is tangent to $\odot(ABXD)$ we get that $\angle DBC=\angle DXB=\angle CYB$ and again as $YB=YC$. Hence, $\angle CBY=90^\circ-\frac{\angle DXB}{2}$. So, $$\angle DBX+\angle DBC+\angle CBY=180^\circ$$. So $\overline{X-B-Y}$. Claim 2:-$PO_1=PO_2$ [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.37, xmax = 11.37, ymin = -5.21, ymax = 8.09; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffdxqq = rgb(1,0.8431372549019608,0); /* draw figures */ draw(circle((-4.57,1.05), 3.959040287746514), linewidth(0.4) + red); draw(circle((0.71,0.97), 2.220720603768065), linewidth(0.4) + ttffqq); draw((-0.935921505623135,-0.5208193711269217)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.935921505623135,-0.5208193711269217)--(-5.63,4.83), linewidth(0.4)); draw((-5.63,4.83)--(-0.89,2.51), linewidth(0.4)); draw((-0.89,2.51)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.89,2.51)--(-0.935921505623135,-0.5208193711269217), linewidth(0.4)); draw((-5.63,4.83)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw(circle((-1.9421286846486794,6.362547428778127), 3.993631988559477), linewidth(0.8) + linetype("4 4") + ffdxqq); draw((-1.9421286846486794,6.362547428778127)--(-4.57,1.05), linewidth(2) + blue); draw((-1.9421286846486794,6.362547428778127)--(0.71,0.97), linewidth(2) + blue); /* dots and labels */ dot((-4.57,1.05),dotstyle); label("$O_1$", (-4.49,0.67), NE * labelscalefactor); dot((-0.89,2.51),dotstyle); label("$A$", (-0.81,2.71), NE * labelscalefactor); dot((0.71,0.97),dotstyle); label("$O_2$", (0.87,0.77), NE * labelscalefactor); dot((-0.935921505623135,-0.5208193711269217),dotstyle); label("$B$", (-0.85,-0.33), NE * labelscalefactor); dot((0.4669852549326141,3.1773839343620836),dotstyle); label("$C$", (0.55,3.37), NE * labelscalefactor); dot((-5.63,4.83),dotstyle); label("$D$", (-5.55,5.03), NE * labelscalefactor); label("$e$", (-3.95,9.35), NE * labelscalefactor,ffdxqq); dot((-1.9421286846486794,6.362547428778127),dotstyle); label("$P$", (-1.87,6.57), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Notice that \begin{align*}\angle PO_1O_2 &=\angle PO_1A+\angle AO_1O_2 &=\angle PO_1D+\angle O_2O_1B &\implies \angle PO_1O_2\frac{\angle DO_1B}{2} &=\angle DXB\end{align*} Similarly we get that $\angle PO_2O_2=\angle CYB=\angle DXB=\angle PO_1O_2\implies PO_1=PO_2$. Claim 3:-$QO_1=QO_2$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.37, xmax = 11.37, ymin = -5.21, ymax = 8.09; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((-4.57,1.05), 3.959040287746514), linewidth(0.4) + red); draw(circle((0.71,0.97), 2.220720603768065), linewidth(0.4) + ttffqq); draw((-0.935921505623135,-0.5208193711269217)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.935921505623135,-0.5208193711269217)--(-5.63,4.83), linewidth(0.4)); draw((-5.63,4.83)--(-0.89,2.51), linewidth(0.4)); draw((-0.89,2.51)--(0.4669852549326141,3.1773839343620836), linewidth(0.4)); draw((-0.89,2.51)--(-0.935921505623135,-0.5208193711269217), linewidth(0.4)); draw((2.7563168338928214,0.10733702100596743)--(-7.50577002435651,-1.6061728791797008), linewidth(0.4)); draw((-7.50577002435651,-1.6061728791797008)--(-0.89,2.51), linewidth(0.4)); draw((-0.89,2.51)--(2.7563168338928214,0.10733702100596743), linewidth(0.4)); draw(circle((-1.9801363099243707,-3.1125906080471197), 5.727296248660026), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-1.9801363099243707,-3.1125906080471197)--(-4.57,1.05), linewidth(2) + blue); draw((-1.9801363099243707,-3.1125906080471197)--(0.71,0.97), linewidth(2) + blue); /* dots and labels */ dot((-4.57,1.05),dotstyle); label("$O_1$", (-4.49,1.25), NE * labelscalefactor); dot((-0.89,2.51),dotstyle); label("$A$", (-0.81,2.71), NE * labelscalefactor); dot((0.71,0.97),dotstyle); label("$O_2$", (0.79,1.17), NE * labelscalefactor); dot((-0.935921505623135,-0.5208193711269217),dotstyle); label("$B$", (-0.95,-0.95), NE * labelscalefactor); dot((0.4669852549326141,3.1773839343620836),dotstyle); label("$C$", (0.55,3.37), NE * labelscalefactor); dot((-5.63,4.83),dotstyle); label("$D$", (-5.55,5.03), NE * labelscalefactor); dot((2.7563168338928214,0.10733702100596743),dotstyle); label("$Y$", (2.83,0.31), NE * labelscalefactor); dot((-7.50577002435651,-1.6061728791797008),dotstyle); label("$X$", (-7.83,-1.73), NE * labelscalefactor); dot((-1.9801363099243707,-3.1125906080471197),dotstyle); label("$Q$", (-1.69,-3.37), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] First notice that as $\{\triangle DXB,\triangle CYB\}$ are similar we get that $\angle XO_1B=\angle BO_2Y\implies \angle O_1BX=\angle O_2YB\implies O_2Y\|O_1B$ \begin{align*}\angle QO_1O_2 &=\angle AO_1Q-\angle AO_1O_2 &=\angle XDA-\angle BO_1O_2 &=\angle ABY-(180^\circ-\angle YO_2O_1) &=\angle QO_2O_1\end{align*} So we get that $QO_1=QO_2$. Combining the last two diagrams we get the 1st Diagram. Hence we get that $PO_1QO_2$ is a kite. Hence, $PQ$ is the perpendicular bisector of $O_1O_2$

Trivial geo. Why inmo setters, why?

TinTin028 wrote: Trivial geo. Why inmo setters, why? Trivial problems need not be bad problems. I found this to be a really good INMO problem (and non trivial). And I am pretty sure you would have asked the same question if this was a hard geo you couldn't solve

TinTin028 wrote: Trivial geo. Why inmo setters, why? I didn’t find it trivial at all. It was much harder than 2019 INMO problem 1.

Doing on paper isn't exactly trivial. Suitable problem

Wizard_32 wrote: TinTin028 wrote: Trivial geo. Why inmo setters, why? Trivial problems need not be bad problems. I found this to be a really good INMO problem (and non trivial). And I am pretty sure you would have asked the same question if this was a hard geo you couldn't solve When did I say it is bad? But if you have only one geometry it doesn't make sense to put it such that it can be done only by angle chasing. It's only an opinion so people's thoughts may differ.

Considering how the paper looked at first glance, this was a heck of a relief.

Here is a solution using only angle chasing This solution lacks details but is complete. Consider $l_1$ to be the perpendicular bisector of $AX$ and $l_2$ to be the perpendicular bisector of $AY$, $l_3$ to be the perpendicular bisector of $AD$ and $l_4$ of $AC$. $\angle {O_1BD}=\angle{O_2BC}$. Thus $\angle {BO_1D}=\angle {BO_2C}$ and $\angle {BAD}=\angle {BAC}$. Now, $l_3$, $l_4$ and $O_1O_2$ are respectively perpendicular to $AD$, $AC$ and $AB$. So, $\angle {BAD}=\angle {BAC}$ implies $l_3$ and $l_4$ are equally inclined to $O_1O_2$. Similarly $l_1$, $l_2$ and $O_1O_2$ are respectively perpendicular to $AX$, $AY$, $AB$. Since $\angle {BAY}=\angle {BAX}$ implies $l_1$ and $l_2$ are equally inclined to $O_1O_2$. Other configurations may be handled similarly.....

Wizard_32 wrote: The key observation is that $A$ is the $B$ Dumpty point of $BCD.$ Then $CDAO_3$ are concyclic, where $O_3$ is the center of $(BCD).$ Then show that $O_3$ also lies on $XAY.$ Also $AO_1PO_2$ is cyclic by spiral similarity, and a simple angle chase shows $P$ is the arc midpoint. P.S. Humpty point in RMO and Dumpty point in INMO? Wow Question setters love Humpty Dumpty I mean rhymes....

This is the simplest problem in this paper imo (and I do not like geometry) Due to obvious results $\angle BDA = \angle CBA = \alpha$, say, and $\angle ACB = \angle ABD = \beta$, say. This means $\Delta ABD \sim \Delta ACB$. It is easy to see that $\angle PO_1O_2 = \angle DAB$ and $\angle O_1O_2P = \angle BAC$, so $PO_1 = PO_2$. Now, consider $X$ and $Y$. First, is easy to see that $\Delta O_2AO_1 \sim \Delta DAB \sim \Delta BAC$. There is an obvious spiral similarity sending $\Delta ADB$ to $\Delta ABC$. Also, it sends $X$ to $Y$. This means $Q$ is the point where the angle bisector of $\angle O_1AO_2$ intersects the circumcircle of $\Delta O_1AO_2$. Therefore $Q$ also lies on the perpendicular bisector of $O_1O_2$, and we are done.

I somehow solved it with trigonometry.

Obviously, $\Delta ADB \sim \Delta ABC $. Thus, $\angle DAB $ $=$ $\angle BAC $ and $\angle XAB $ $=$ $\angle YAB $. \begin{align*} \angle PO_1O_2 =180^{\circ}-\angle DAB=180^{\circ}-\angle BAC=\angle PO_2O_1 \implies PO_1= PO_2 \end{align*}\begin{align*} \angle QO_1O_2=\angle XAB=\angle YAB =\angle QO_2O_1 \implies QO_1=QO_2 \qquad \blacksquare \end{align*}

What should be inmo's expected cutoff this year ???

Has anyone tried complex numbers on this?

AlastorMoody wrote: Obviously, $\Delta ADB \sim \Delta ABC $. Thus, $\angle DAB $ $=$ $\angle BAC $ and $\angle XAB $ $=$ $\angle YAB $. \begin{align*} \angle PO_1O_2 =180^{\circ}-\angle DAB=180^{\circ}-\angle BAC=\angle PO_2O_1 \implies PO_1= PO_2 \end{align*}\begin{align*} \angle QO_1O_2=\angle XAB=\angle YAB =\angle QO_2O_1 \implies QO_1=QO_2 \qquad \blacksquare \end{align*} can you please elaborate how QO1O2=XAB thanks

KS_789,I think. take $A_1A_2$ as midpoints of $AB$ and $AX$. Then, $O_1A_1A_2A$ is cyclic

Just Angle Chasing! May be similar to others above: Let $\odot(XAY)=K_2$ and $\odot(CAD)=K_1$ Let $ \angle XAD = \angle XAB = \alpha$ and $ \angle YAC = \angle YAB = \beta $ It is easy to see that $\triangle DAB \cong \triangle BAC$. So $\alpha = \beta = \gamma (let)$ $ \angle M_1O_1E = 180^\circ - 2 \gamma = \angle M_2O_2E$ So $K_1O_1 = K_1O_2$ Let $ \angle BXA = \angle ABC \ be \ \theta$ and $ \angle BYA = \angle ABD \ be \ \kappa$ $ \angle XBD = \angle YBC=\gamma$ $\angle R_1O_1E = 360^\circ -(\angle O_1R_1B + \angle R_1BE + \angle BEO_1)$ =$360^\circ -(90^\circ + \theta + \kappa + \gamma + 90^\circ)$ =$180^\circ - \theta + \kappa + \gamma$ $\angle R_2O_2E = 360^\circ -(\angle O_2R_2B + \angle R_2BE + \angle BEO_2)$ =$360^\circ -(90^\circ +\kappa + \gamma + \theta + 90^\circ)$ =$180^\circ - \theta + \kappa + \gamma$ $\implies \angle K_2O_1O_2 = \angle K_2O_2O_1 \implies K_2O_1 = K_2O_2$ Therefore $\triangle K_1O_1K_2 \cong \triangle K_1O_2K_2$ So $K_1K_2$ is the perpendicular bisector of $O_1O_2$ $\blacksquare$

PO1QO2 can be proved to be a kite and after that it’s an easy task!

Let $CA$ meet circle $\Gamma_1$ at $C^*$ and similarly $DA$ meet $\Gamma_2$ at $D^*$ Now by Tangent-Secant theorem in both the circles $\Gamma_1, \Gamma_2$ we get $\angle CC^*B=\angle CBA=\angle AXB, \angle DD^*B=\angle DBA=\angle AYB \implies \angle XAB =\angle BAY, \angle ABX+\angle ABY=180^\circ$ so $X\in BY$ Now just by some easy angle chase we can get $\angle O_1QO_2=\angle O_2QO \implies \boxed{O_1Q=O_2Q}$ Also observe $\triangle AO_1O_2 \sim \triangle AXY$ and also $\angle PO_1A=\angle PO_2A \implies \angle PO_1O_2=\angle PO_2O_1$ hence $\boxed{PO_1=PO_2}$ Hence $PO_1QO_2$ Forms a Kite. So diagonals $PQ \perp O_1O_2$ $\blacksquare$

Hi there! just made a video about it https://youtu.be/U0IyXX_3MFY

We can also solve this problem using another method (note that this is just a outline of solution, not the complete solution): Let $T$ be the intersection point of lines $XO_1$ , $YO_2$. One may prove that: $T$ lies on $\odot(ACD)$ $T$ lies on $\odot(AXY)$ quad. $O_1TAO_2$ is a isosceles trapeziod (which means that lines $AT$, $O_1O_2$ have the same perpendicular bisector). So the desired result follows.$\blacksquare$

Never thought that this day would come !

anantmudgal09 wrote: Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$. Proposed by Prithwijit De

YAY! AGAIN BACK FROM HOSPITAL :")

Can anyone tell why D, A, Y and C, A, X must be collinear?

We will prove $QO_1 = QO_2$ and $PO_1 = PO_2$. Let $QO_1$ meet $AX$ at $S$ and $O_1O_2$ meet $AB$ at $K$. we have $O_1SKA$ is cyclic so $\angle QO_1O_2 = \angle XAB = \angle CBY = \angle CAY = \angle BAY = \angle QO_2O_1$ so $QO_1 = QO_2$. with same approach for $P$ we have proved $QO_1 = QO_2$ and $PO_1 = PO_2$. we're Done.

Notice that \[ \angle QO_1O_2=\angle QO_1A-\angle O_2O_1A=\angle XDA-\angle BDA=\angle XAB \]By symmetry $\angle QO_2O_1=\angle YAB$. Notice that $\triangle ABD\sim\triangle ACD$. So $AB$ is the angle bisector of $\angle XAY$. Which means $\angle QO_2O_1=\angle QO_1O_2$. So $QO_1=QO_2$. In the same way we have \[ \angle PO_1O_2=\angle PO_1A+\angle AO_1O_2=\angle DBA+\angle ADB=\angle DBC \]By symmetry $\angle PO_2O_1=\angle DBC$. So $PO_1=PO_2$. So we are done.

I just kind of mocked inmo today, also this is probably the first inmo geo i have solved I remember when I was starting oly math, looking at the question paper and not even being able to draw the diagram My solution is basically AMAR04's solution only. (in the image, The things in boxes are things i have proven) I also thought that inversion around B might be good, because it turns ABCD into a parallelogram (this is what I have tried on left side of page) but inversion does not handle circumcircles well.

Just angle chasing Note that $\angle ACB=\angle ABD$ and $\angle BDA=\angle ABC$ and $\triangle ADB\sim \triangle ABC$. Now we just have to angle chase \begin{align*} \angle PO_1O_2=\angle PO_1A+\angle AO_1O_2&=\angle ABD+\angle BDA=\angle ABC+\angle ACB\\&=\frac{(\angle AO_2C+\angle AO_2B)}{2}=\angle PO_2O_1\implies \boxed{PO_1=PO_2}. \end{align*}\begin{align*} &\angle QO_1O_2=\angle QO_1A-\angle O_2O_1A=\angle XDA-\angle BDA=\angle XDB=\angle XAB\\ &\angle QO_2O_1=\angle QO_2A-\angle O_1O_2A=\angle YCA-\angle BCA=\angle YCB=\angle YAB\implies \boxed{QO_1=QO_2}. \end{align*}So $PO_1QO_2$ is a kite and we are done. 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