ELMO 16 P6 lol

Yeah just ELMO P16. [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 = -7.530466480201327, xmax = 34.54580644948567, ymin = -22.3245606569796, ymax = 5.462172829703162; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen qqqqcc = rgb(0,0,0.8); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffqqff = rgb(1,0,1); /* draw figures */ draw((4.785225501676106,2.852100238498341)--(1.1,-5.82), linewidth(0.8) + ffqqtt); draw((1.1,-5.82)--(15.78,-5.56), linewidth(0.8) + ffqqtt); draw((4.785225501676106,2.852100238498341)--(15.78,-5.56), linewidth(0.8) + ffqqtt); draw(circle((8.406630284612802,-3.805894531215208), 7.579146848782504), linewidth(0.8) + green); draw(circle((6.170850655949693,-2.4002145455492294), 3.3294525882863786), linewidth(0.8) + green); draw((3.106599446102232,-1.098055161834919)--(6.229809915671388,-5.729145055989472), linewidth(0.8) + qqqqcc); draw((8.193984268203156,0.244059442640558)--(6.229809915671388,-5.729145055989472), linewidth(0.8) + qqqqcc); draw(circle((7.129914391470361,-6.03561667532645), 1.7466245607836695), linewidth(0.8) + rvwvcq); draw(circle((7.046122272796629,-1.3045847440557998), 4.731773901387759), linewidth(0.8) + green); draw(circle((10.910838957196441,-20.36749130439421), 14.718944377865363), linewidth(0.8) + green); draw((5.407856420687314,-5.743702815437419)--(7.129914391470361,-6.03561667532645), linewidth(0.8) + rvwvcq); draw((7.129914391470361,-6.03561667532645)--(8.84055530107375,-5.682905696302509), linewidth(0.8) + rvwvcq); draw((4.785225501676106,2.852100238498341)--(5.407856420687314,-5.743702815437419), linewidth(0.8) + rvwvcq); draw((4.785225501676106,2.852100238498341)--(8.84055530107375,-5.682905696302509), linewidth(0.8) + rvwvcq); draw((5.782528260879021,-10.916278172373117)--(11.2808740069574,-10.818896027006609), linewidth(0.8) + rvwvcq); draw(circle((7.004658792545392,-5.560826049601096), 5.332970386704358), linewidth(0.8) + ffqqff); draw((4.785225501676106,2.852100238498341)--(7.099096861561913,-10.892960199881458), linewidth(0.8) + rvwvcq); draw((4.785225501676106,2.852100238498341)--(10.910838957196441,-20.36749130439421), linewidth(0.8) + rvwvcq); draw((5.407856420687314,-5.743702815437419)--(5.782528260879021,-10.916278172373117), linewidth(0.8) + rvwvcq); draw((6.229809915671388,-5.729145055989472)--(7.575452137233042,-7.724460502752023), linewidth(0.8) + qqqqcc); draw((8.84055530107375,-5.682905696302509)--(11.2808740069574,-10.818896027006609), linewidth(0.8) + rvwvcq); draw((1.1,-5.82)--(-2.6356950345134758,-14.610865520991041), linewidth(0.8) + ffqqtt); draw((15.78,-5.56)--(19.854769023212206,-8.67760513844231), linewidth(0.8) + ffqqtt); draw((3.106599446102232,-1.098055161834919)--(8.193984268203156,0.244059442640558), linewidth(0.8) + qqqqcc); /* dots and labels */ dot((4.785225501676106,2.852100238498341),dotstyle); label("$A$", (4.901159612660738,3.1124588868764977), NE * labelscalefactor); dot((1.1,-5.82),dotstyle); label("$B$", (1.2126551675258397,-5.5486959954031825), NE * labelscalefactor); dot((15.78,-5.56),dotstyle); label("$C$", (15.884706182617993,-5.27547344391171), NE * labelscalefactor); dot((6.170850655949693,-2.4002145455492294),linewidth(4pt) + dotstyle); label("$I$", (6.267272370118108,-2.18805861205807), NE * labelscalefactor); dot((3.106599446102232,-1.098055161834919),linewidth(4pt) + dotstyle); label("F", (3.2071797934135997,-0.8765903648990017), NE * labelscalefactor); dot((8.193984268203156,0.244059442640558),linewidth(4pt) + dotstyle); label("$E$", (8.316441506304162,0.462200137409214), NE * labelscalefactor); dot((6.229809915671388,-5.729145055989472),linewidth(4pt) + dotstyle); label("D", (6.349239135565551,-5.5213737402540355), NE * labelscalefactor); dot((6.6843766457076805,-4.346772847900878),linewidth(4pt) + dotstyle); label("$X$", (6.786395217951909,-4.127938727647525), NE * labelscalefactor); dot((7.575452137233042,-7.724460502752023),linewidth(4pt) + dotstyle); label("$Y$", (7.688029637873773,-7.515898366141785), NE * labelscalefactor); dot((7.129914391470361,-6.03561667532645),linewidth(4pt) + dotstyle); label("$J$", (7.250873555487415,-5.821918546894655), NE * labelscalefactor); dot((5.407856420687314,-5.743702815437419),linewidth(4pt) + dotstyle); label("S", (5.5295714810911285,-5.5213737402540355), NE * labelscalefactor); dot((8.84055530107375,-5.682905696302509),linewidth(4pt) + dotstyle); label("T", (8.944853374734553,-5.466729229955741), NE * labelscalefactor); dot((10.910838957196441,-20.36749130439421),linewidth(4pt) + dotstyle); label("$M$", (11.021344766069754,-20.13878024504782), NE * labelscalefactor); dot((7.986291436780996,-5.942015735126961),linewidth(4pt) + dotstyle); label("$O$", (8.097863465110983,-5.712629526298066), NE * labelscalefactor); dot((5.782528260879021,-10.916278172373117),linewidth(4pt) + dotstyle); label("$P$", (5.884760798030045,-10.685279963442866), NE * labelscalefactor); dot((11.2808740069574,-10.818896027006609),linewidth(4pt) + dotstyle); label("$Q$", (11.376534083008671,-10.603313197995424), NE * labelscalefactor); dot((7.099096861561913,-10.892960199881458),linewidth(4pt) + dotstyle); label("$V$", (7.19622904518912,-10.685279963442866), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Firstly we'll prove $\gamma$ tangent to $\odot(ABC)$. We claim that $X$ and $Y$ are the incentre and the $A$ -excentre of $\Delta AST$ (this is just egmo problem 11.12 or sharygin 2013 problem on page 210 . Proof on page 290).This infact implies that $AS,AT$ isogonal by definition. Now be the line tangent to $\odot(ABC)$ at $A$ . We claim that this is also tangent to $\Delta AST$. For this just note that $\angle(l,AB)=\angle ACB$ and $\angle (l,AS)=\angle (l,AB)+\angle BAS=\angle ACB+\angle BAS=\angle ACB +\angle CAT =\angle ATB\equiv \angle ATS \implies l$ tangent to $\odot(AST)$ done with the first part. Now by the sawayama thebaults theorem for the triangle $\Delta AST$ with $C$ as the point lying on $ST$ and $I_A-X-I$ collinear where $X$ is the incentre of $\Delta AST$ and $I_A, I$ are the incentre and the $A$-excentre of $\Delta ABC$ have that $\gamma$ tangent to incircle and the $A$ - excircle. Thus we are done with the problem. $\blacksquare$

If one even does not know that ELMO problem;no worries: Just consider the internal tangents of mixtillinear incircle and mixtillinear excircle.By homothety prove that one of these is parallel to $BC$.Now invert this line about $A$ with $\sqrt bc$ inversion.

Just an extension of ELMO 2016 P6. Sharygin 2020 CR P14 wrote: A non-isosceles triangle is given. Prove that one of the circles touching internally its incircle and circumcircle and externally one of its excircles passes through a vertex of the triangle. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -7.477777777777773, xmax = 10.242222222222233, ymin = -3.31, ymax = 5.943333333333335; /* image dimensions */ pen ffcctt = rgb(1,0.8,0.2); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen yqqqqq = rgb(0.5019607843137255,0,0); pen qqttzz = rgb(0,0.2,0.6); /* draw figures */ draw(circle((-0.96,2.19), 1.5001991637326682), linewidth(0.4) + ffcctt); draw((-1.88,5.29)--(-2.82,0.71), linewidth(0.4) + wrwrwr); draw((-1.88,5.29)--(2.76,0.59), linewidth(0.4) + wrwrwr); draw((2.76,0.59)--(-2.82,0.71), linewidth(0.4) + wrwrwr); draw((-1.88,5.29)--(xmin, 4.872340425531915*xmin + 14.45), linewidth(0.4) + wrwrwr); /* ray */ draw((-1.88,5.29)--(xmax, -1.0129310344827587*xmax + 3.385689655172414), linewidth(0.4) + wrwrwr); /* ray */ draw((xmin, -3.3695652173913047*xmin-1.044782608695653)--(xmax, -3.3695652173913047*xmax-1.044782608695653), linewidth(0.4) + wrwrwr); /* line */ draw(circle((0.8317949493422048,-3.847569938000908), 4.478112270662846), linewidth(0.4) + green); draw((-2.82,0.71)--(-0.7098623278034928,1.3471448002074213), linewidth(0.4) + wrwrwr); draw((2.76,0.59)--(-0.2087501540789476,-0.3413853503861555), linewidth(0.4) + wrwrwr); draw(circle((-0.4593062409412202,0.5028797249106329), 0.8806599059964217), linewidth(0.4) + yqqqqq); draw((0.8965607303825336,0.6300739627874724)--(-3.6453098708955975,-3.311190647555148), linewidth(0.4) + wrwrwr); draw((-2.4391148451251157,2.565802137581883)--(-0.2087501540789476,-0.3413853503861555), linewidth(0.4) + wrwrwr); draw((0.0829335281227166,3.3016837107377652)--(-0.9844444444444377,0.69), linewidth(0.4) + wrwrwr); draw(circle((0.01012061147211442,2.5156084334533215), 3.357052947219786), linewidth(0.4) + red); draw(circle((-0.358242041949058,3.1341620198172904), 2.638822593447623), linewidth(0.4) + qqttzz); /* dots and labels */ dot((-0.96,2.19),dotstyle); label("$I$", (-0.9044444444444376,2.3166666666666678), NE * labelscalefactor); dot((-0.9844444444444377,0.69),dotstyle); label("$D$", (-1.077777777777771,0.3833333333333342), NE * labelscalefactor); dot((-1.88,5.29),dotstyle); label("$A$", (-2.1711111111111046,5.436666666666668), NE * labelscalefactor); dot((-2.82,0.71),dotstyle); label("$B$", (-3.1444444444444386,0.5566666666666675), NE * labelscalefactor); dot((2.76,0.59),dotstyle); label("$C$", (3.00222222222223,0.5166666666666675), NE * labelscalefactor); dot((-2.4391148451251157,2.565802137581883),dotstyle); label("$F$", (-2.384444444444438,2.7033333333333345), NE * labelscalefactor); dot((0.0829335281227166,3.3016837107377652),dotstyle); label("$E$", (0.13555555555556267,3.4366666666666683), NE * labelscalefactor); dot((0.8317949493422048,-3.847569938000908),dotstyle); label("$H$", (-7.477777777777773,5.943333333333335), NE * labelscalefactor); dot((0.8965607303825336,0.6300739627874724),dotstyle); label("$M$", (1.0022222222222297,0.3966666666666675), NE * labelscalefactor); dot((-0.7098623278034928,1.3471448002074213),dotstyle); label("$P$", (-0.9311111111111043,1.4766666666666677), NE * labelscalefactor); dot((-0.2087501540789476,-0.3413853503861555),dotstyle); label("$Q$", (-0.07777777777777073,-0.576666666666666), NE * labelscalefactor); dot((-0.4593062409412202,0.5028797249106329),linewidth(4pt) + dotstyle); label("$O$", (-0.41111111111110415,0.61), NE * labelscalefactor); dot((-3.6453098708955975,-3.311190647555148),dotstyle); label("$N$", (-3.904444444444439,-3.11), NE * labelscalefactor); dot((4.036069616003411,-0.7025705162103524),dotstyle); dot((0.4146822322652519,0.6110740811952683),linewidth(4pt) + dotstyle); label("$Y$", (0.45555555555556276,0.7433333333333343), NE * labelscalefactor); dot((-1.3224208725853333,0.6777939972598994),linewidth(4pt) + dotstyle); label("$X$", (-1.5711111111111045,0.4766666666666675), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] This is same as showing that a Circle passing through the Vertex $A$ of $\triangle ABC$ and tangent to both it's incircle and circumcircle internally is externally tangent to the $A-\text{Excircle.}$ Let $AK$ be the bisector of $\angle A$ where $K\in BC$. Let $\{P,Q\}\in AK$ such that $BP\perp AK$ and $CQ\perp AK$. Let $\odot(PQ)\cap BC=\{X,Y\}$ such that $B,X,Y,C$ are in this order. We claim that $\odot(AXY)$ is internally tangent to both $\odot(ABC)$ and $\odot(I)$. Notice that $$\frac{BX\cdot BY}{CX\cdot CY}=\frac{BP^2}{CQ^2}=\frac{AB^2}{AC^2}$$So by Steiner's Ratio Theorem we get that $\{AX,AY\}$ are isogonal WRT $\triangle ABC$. Let $\{AX,AY\}\cap\odot(ABC)=\{X',Y'\}$. So, $\angle AXY=\angle B+\angle BAX=\angle CY'A+\angle CAY'=180^\circ-\angle ACY'=\angle AX'Y'\implies X'Y'\|BC$. So, by a Homothety $\mathcal H$ at $A$ we get that $\odot(AXY),\odot(ABC)$ are internally tangent at $A$. Let $DEF$ be the Contanct Triangle of $\triangle ABC$ and $O$ be the circumcenter of $\odot(PQ)$. So, $OX=OY$ and $\angle XAO=\angle YAO\implies O\in\odot(AXY)$. Now it's well known that $\overline{Q-D-F}$ (Refer to Lemma 1.45 from Euclidean Geometry in Mathematics Olympiad) $B\in\text{ Polar of } Q$ by La Hire's Theorem WRT $\odot(I)$ and as $BP\perp AQ$, so $BP$ is the polar of $Q$. So, $\odot(PQ)\perp \odot(I)$ (by Lemma 9.27 from Euclidean Geomtry in Mathematical Olympiads). So, Inversion around $\odot(PQ)$ fixes $\odot(I)$ and swaps $\{\odot(AXY),BC\}$. But as $\odot(I)$ and $BC$ are tangent to other, hence, $\odot(AXY)$ and $\odot(I)$ must also be tangent to each other. Hence, we prove our claim. Now we will prove that $\odot(AXY)$ is tangent to the $A-\text{Excircle}(\odot(I_A))$ as well. Let $\{M,N\}$ be the tangency points made by $\odot(I_A)$ with $BC,AB$ respectively. Similarly refer to Lemma 1.45 from Euclidean Geometry in Mathematics Olympiad book to get that $\overline{M-Q-N}$. So, By La Hire's Theorem we get that $B\in \text{ Polar of } Q$ WRT $\odot(I_A)$ and as $BP\perp AQ$. Hence, $BP$ is the Polar of $Q$ WRT $\odot(I_A)$. Hence, by Lemma 9.27 from Euclidean Geometry in Mathematics Olympiad we get that $\odot(PQ)\perp\odot(I_A)$. Now again Inversion around $\odot(PQ)$ fixes $\odot(I_A)$ and swaps $\{BC,\odot(AXY)\}$ but as $BC,\odot(I_A)$ are tangent hence, $\odot(AXY), \odot(I_A)$ are also tangent to each other externally. $\blacksquare$

This solution is nearly identical to above solutions.

Attachments:
Problem 14.pdf (52kb)

After ELMO 2016/6 , I did a lot of unnecessary things to prove that the circle is tangent to the excircle . Let $AK$ be the angle bisector of $\angle BAC$ with $K$ on $BC$ . Let $D,E,F$ be the touchpoints of incircle with sides $BC,AC,AB$ respectively . Let $DE\cap AK= X ; DF\cap AK= Y$ Let $\omega_1$ be the circle with diameter $XY$. Let $\omega_1\cap BC = S ,T $ . We claim that one of the circle which is tangent to incircle ,circumcircle and excircle as described in the question is $\odot (AST)$ . This circle indeed passes through $A$ . So , the only thing to be proved is $\odot (AST)$ is tangent to circumcircle, incircle and excircle [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -1., xmax = 6., ymin = -1., ymax = 6.; /* image dimensions */ pen qqwuqq = rgb(0.,0.392156862745,0.); pen xdxdff = rgb(0.490196078431,0.490196078431,1.); pen qqffff = rgb(0.,1.,1.); pen ffdxqq = rgb(1.,0.843137254902,0.); pen bfffqq = rgb(0.749019607843,1.,0.); pair A = (2.52,2.42), C = (0.682887099703,5.21560658741), D = (-0.775182532809,1.84365628704), G = (-0.000564511004626,3.63504773574), H = (2.06529064264,4.18884817754), I = (1.17328212699,1.73139822855), J = (1.42845604051,2.43439277373), K = (1.95810238597,0.458640228594), L = (1.69327921324,1.44651650116), M = (2.69686346899,1.64361922649), O = (1.62386921699,1.70543828535), P = (1.25175936133,3.09352866053), Q = (1.81148283025,3.49817830179); draw(C--D--(5.7271885701,1.46903129608)--cycle, linewidth(1.2) + blue); /* draw figures */ draw(circle(A, 3.34520552433), qqwuqq); draw(C--D, linewidth(1.2) + blue); draw(D--(5.7271885701,1.46903129608), linewidth(1.2) + blue); draw((5.7271885701,1.46903129608)--C, linewidth(1.2) + blue); draw(circle(P, 1.364389237), linewidth(0.4) + qqffff); draw(circle(L, 1.02275649238), linewidth(0.4) + ffdxqq); draw(circle(Q, 2.05506404748), linewidth(0.4) + green); label("$A$",(0.68,5.46),SE*labelscalefactor,blue); label("$B$",(-0.78,2.08),SE*labelscalefactor,blue); label("$C$",(5.72,1.7),SE*labelscalefactor,blue); label("$I$",(1.32,3.46),SE*labelscalefactor,qqffff); label("$F$",(-0.22,3.96),SE*labelscalefactor,qqffff); label("$E$",(2.12,4.48),SE*labelscalefactor,qqffff); label("$Y$",(1.96,0.7),SE*labelscalefactor,ffdxqq); label("$X$",(1.42,2.68),SE*labelscalefactor,ffdxqq); label("$T$",(0.72,2.),SE*labelscalefactor,green); label("$S$",(2.7,1.88),SE*labelscalefactor,green); label("$V$",(1.78,1.48),SE*labelscalefactor,blue); label("$D$",(1.18,1.98),SE*labelscalefactor,qqffff); label("$K$",(1.62,1.94),SE*labelscalefactor,blue); draw(G--K, linewidth(0.4) + blue); draw(I--H, linewidth(0.4) + blue); draw(C--K, linewidth(0.4) + red); label("$L$",(1.82,3.74),SE*labelscalefactor,green); label("$O$",(2.62,2.72),SE*labelscalefactor,qqwuqq); draw(C--A, linewidth(0.4) + blue); draw(C--(0.718971820001,1.75757267968), linewidth(0.4) + linetype("2 2") + blue); draw(C--M, linewidth(0.4) + linetype("2 2") + blue); /* dots and labels */ dot(A,blue); dot(C,xdxdff); dot(D,blue); dot((5.7271885701,1.46903129608),blue); dot(G,qqffff); dot(H,qqffff); dot(I,qqffff); dot(J,ffdxqq); dot(K,ffdxqq); dot(L,blue); dot(M,green); dot((0.718971820001,1.75757267968),green); dot(O,blue); dot(P,qqffff); dot(Q,bfffqq); dot(A,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] WLOG $AB<AC$ and $T$ is between $B$ and $D$. Let $O$ be the circumcenter of triangle $ABC$. Let $I$ be the incenter of triangle $ABC$. $AI$ cuts $BC$ at $K$. We see that $\angle FAI= \angle IAC$ and $\angle AFY= \angle AIC(=90^{\circ}+ \tfrac 12 \angle ABC)$ so $\triangle AIC \sim \triangle AFY$ implies $\tfrac{AI}{AF}= \tfrac{AC}{AY}$ or $AI \cdot AY = AC \cdot AF= AC \cdot AE$. Hence, $I,Y,E,C,D$ are concyclic. Similarly, we have $\triangle ABI \sim \triangle AEX$ implies $\tfrac{AI}{AE}= \tfrac{AB}{AX}$, which follows that $B,F,I,D,X$ are concyclic. Also, note that $\tfrac{AI}{AE}= \tfrac{AI}{AF}$ so $\tfrac{AB}{AX}= \tfrac{AC}{AY}$ or $\tfrac{AX}{AY}= \tfrac{AB}{AC}$. Since $AB<AC$ so $X$ is inside triangle $ABC$. Since $BFIX$ is cyclic AND $IF \perp AB$ so $BX \perp AI$. Similarly, $CY \perp AI$. Hence, $BX \parallel CY$ implies $\tfrac{KX}{KY}= \tfrac{KB}{KC}= \tfrac{AB}{AC}$. Thus, we have $\tfrac{AX}{AY}= \tfrac{KX}{KY} \left( = \tfrac{AB}{AC} \right)$ implies $(AK,XY)=-1$. Note that $\angle XSY=\angle XTY=90^{\circ}$ so we follows that $TX,SX$ are bisectors of $\angle ATS, \angle AST$, repsectively. Therefore, $X$ is the incenter of triangle $ATS$. This means $\angle TAX= \angle SAX$ implies $\angle TAB= \angle SAC$. Similar we also obtain $Y$ is the $A$-excenter of $\triangle ATS$. Let $L$ be the circumcenter of triangle $AST$. If we let $H$ be the orthocenter of triangle $ABC$. It is well-known property that $\angle HAC= \angle OAB$. This property is also applied for $\triangle AT S$, which means $\angle HAS= \angle LAT$. We already have $\angle TAB= \angle SAC$ so $\angle HAS+\angle SAC= \angle LAT+ \angle TAB$ or $\angle HAC= \angle LAB$. Thus, $\angle LAB= \angle OAB(= \angle HAC)$. Since $L,O$ are on the same side wrt $AB$ so $A,L,O$ are collinear, which means $(L)$ and $(O)$ are tangent to each other. $AI$ cuts $(L)$ the second time at $V$ Since $AI$ bisects $\angle TAS$ so $V$ is the midpoint of arc $TS$ of $(L)$. Since $X,Y$ is the incenter, $A$-excenter of $\triangle AST$ respectively so it is well-known that $V$ is the midpoint of $XY$ or $V$ is the center of $(XTYS)$. Note that $(AK,XY)=-1$ so from Newton's identity we have $VT^2=VS^2=VX^2=VK \cdot VA$. Consider an inversion about $V$ with radius $VX$ then since $V \in (L)$ so the inversion will send $(L)$ to a line $\ell$. Note that under this inversion, it sends $T$ to $T$ and $S$ to $S$ so $\ell$ is $BC$. We have $\angle IDE=\tfrac 12 \angle C = \angle AYD$ since $\triangle AFY \sim \triangle AIE$. This means $\triangle IDX \sim \triangle IYD$ implies $ID^2=IX \cdot IY$. Since $ID$ is the radius of $(I)$ so we obtain that $(I)$ and $(V, \tfrac{XY}{2})$ are orthogonal circles. This means that the image of $(I)$ under inversion about $V$ radius $VX$ is itself. Note that $BC$ is tangent to $(I)$ or under inversion, the image of $(I)$ is tangent to image of $(L)$. This follows that $(L)$ is tangent to $(I)$. Let the A-excircle be $\omega_A$ . Let $O'$ be the reflection of $O$ over $A$ . Let $\Omega$ be the circle with centre $O'$ and radius $O'A$ . As $O,A,O'$ are collinear and $\odot (ABC)$ is tangent to $\odot (AST)$ , even $\Omega$ is tangent to $\odot (AST)$ . Now we apply casey's theorem on the circles (1) Point circle ${A}$ ;(2) $\Omega$ (3) Incircle of $\triangle ABC$ (4)$\omega_a$ $t_{12}=0 , t_{13}=s-a , t_{14}= s$ . $t_{23}$= length of transverse common tangent of incircle and $\Omega$. $t_{23}=\sqrt {O'I^2-(R+r)^2}$. Now in $\triangle OIO' $, by Appolonius Theorem , $O'I^2= 2AI^2+2OA^2 - OI^2 = 2(s-a)^2+2r^2+2R^2-R^2+2Rr= 2(s-a)^2+(R+r)^2+r^2 $ . $\implies t_{23}= \sqrt {2(s-a)^2+r^2}$ . $t_{24}$=length of direct common tangent of excircle and $\Omega$ . $t_{24}=\sqrt {O'I_a^2-(R-r_a)^2}$ Similarly applying Appolonius Theorem in $\triangle O'I_aO$, we get that : $O'I_a^2= 2AI_a^2+2OA^2-OI_a^2 = 2s^2+2r_a^2+2R^2-R^2-2Rr_a= 2s^2+(R-r_a)^2+r_a^2$ $\implies t_{24}= \sqrt { 2s^2+(R-r_a)^2+r_a^2 - (R-r_a)^2} = \sqrt {2s^2+r_a^2} $ If excircle was tangent to $\odot (AXY)$ , then Casey's theorem would be true . That is , $t_{12}\cdot t_{34} + t_{14}\cdot t_{23}=t_{13}\cdot t_{24}$ . $\Longleftrightarrow 0 + s\cdot \sqrt {2(s-a)^2+r^2} = (s-a) \cdot \sqrt {O'I_a^2-(R-r_a)^2} $ $\Longleftrightarrow \frac{s}{s-a}= \sqrt{\frac{2s^2+r_a^2}{2(s-a)^2+r^2}}$ But as $r_a= \frac{s}{s-a}\cdot r$ , $\Longleftrightarrow \frac{s}{s-a}= \sqrt{\frac{2s^2+r_a^2}{2(s-a)^2+r^2}}$= $\sqrt{\frac{\frac{2(s-a)^2\cdot s^2 +s^2\cdot r^2}{(s-a)^2}}{2(s-a)^2+r^2}} = \frac{s}{s-a}\cdot \sqrt{\frac{2s^2+r^2}{2s^2+r^2}} = \frac{s}{s-a}$ So, LHS=RHS . So , by converse of Casey's theorem on point circle $\odot(A) , \Omega,\Gamma$ and $\omega_A$ and the main circle $\odot (AST)$ , we get that $\omega_A$ is also tangent to $\odot (AST)$. Thus , $\odot (AST)$ is one circle which satisfies the tangency conditions given in the problem as well as passes through $A$ , thus completing our proof . $\blacksquare$ PS : I have copied many things directly from some Elmo's solution by aopser for this problem as I didn't have time to type because of boards

Problem 14.}A nonisosceles triangle is given. Prove that one of the circles touching internally its incircle and circumcircle and touching externally one of its excircles passes through a vertex of the triangle. Solution. Notations. We will be solving the problem in $\triangle ABC$. Let $I$ be the incenter of $\triangle ABC$ and let the foot of altitude from $B,C$ to $AI$ be $X,Y$ respectively. Let $\gamma$ be the circle with diameter $XY$ and let $\gamma\cap BC=S,T$. Also, $\omega, \Gamma, \lambda$ denotes the incircle, cimcumcircle, $A$-excircle of $\triangle ABC$. Let $M,N$ be the midpoint of $BC,XY$ respectively. Let $D,E,F$ be the contact point of $\omega$ with $BC,CA,AB$ respectively. Proof. So basically, the problem asks us to show the exsistance of a circle that is internally tangent to circumcircle and incircle and is externally tangent to the excircle and passes through one of the vertices of the triangle. Without loss of generality, let the excircle be $A$-excircle. We claim that $\odot(AST)$ is internally tangent to circumcircle and incircle and is externally tangent to the excircle and passes through one of the vertices of the triangle. To prove this, note that $IX\perp BX$ and let $B_0$ be the midpoint of $DF$, so that $B, B_0$ correspond in inversion about the incircle. Thus, if $X_0$ is the image of $X$ under inversion about the incircle, we should have that $\angle IXB=\angle IB_0X_0 = 90$ and so that $X_0$ lies on $DF$. Thus, $X_0 = Y$ so $X, Y$ are inverses under inversion about the incircle. Thus, inverting the diagram about $\omega$ fixes $\gamma$ and thus, $\omega$ and $\gamma$ are orthogonal to each other. Now, as $N$ is midpoint of $XY$ and $M$ is the midpoint of $BC$, we must have $MN\| BX\| CY\implies MN\perp AI$ and as $MN$ bisects the common tangent of $\lambda,\omega$, we must have $MN$ is the radical axis of $\lambda,\omega$. So, power of $N$ with respect to $\omega=$ power of $N$ with respect to $\lambda$ but as $N$ is the center of $\gamma$ which is orthogonal to $\omega$, the power of $N$ with respect to $\omega=NX^2=NY^2=$ power of $N$ with respect to $\lambda$ and hence, $\gamma$ and $\lambda$ are also orthogonal. Now, let $G$ be the foot of $AI$ on $BC$. We now claim that circle with diameter $AG$ and $\odot(XSYT)$ are orthogonal. It is enough to show $(A,G;X,Y)=-1$. Let external angle bisector of $\angle A$ meet $BC$ at $H$ and let $P_{\infty}$ is the point at infinity for $AH$. Project $A,G,P,Q$ to $BC$ from $P_{\infty}$. We get $(A,G;P,Q)=(H,G;B,C)=-1$ because $BX\|HA$ proving the claim. Thus, $A,G$ are inverses with respect to $\odot(XSYT)$. Now invert $\odot(AST)$ about $\odot(XSYT)$. Clearly $S, T$ remain fixed while $A$ goes to $G$ so $(AST)$ and line $BC$ are inverses. Thus, $\odot(AST)$ passes through $N$. Now, as $BC$ is tangent to $\omega,\lambda$, inverting about $\odot(XSYT)$, we get $BC\mapsto \odot(AST), \omega\mapsto \omega, \lambda\mapsto \lambda\implies (AST)$ is tangent to both $\omega$ and $\lambda$. Now, it is enough to show that $(AST)$ is tangent to $\Gamma$. But as $N\in\odot(AST)$ and $N$ is midpoint of arc $ST$ as $NS=NT$, we must have that $AN$ bisects $\angle SAT$ but $AN$ also bisects $\angle BAC\implies AS,AT$ are isogonal with respect to $\angle BAC$. Now, as $B,S,T,C$ are collinear, the orthocenter of $AST$ lies on $AH$. Thus, if $O'$ is the circumcenter of $\triangle AST$, then $AH,AO'$ are isogonal but $AH$ and $AO$ are isogonal too. This is possible if $A,O,O'$ are collinear and hence, $\odot(AST)$ is tangent to $\Gamma$ completeing the proof. \textbf{Feedback.} This problem was very similar to ELMO 2016 P6, however the new part was o show that this circle is also tangent to excircle.

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NJOY wrote: A non-isosceles triangle is given. Prove that one of the circles touching internally its incircle and circumcircle and externally one of its excircles passes through a vertex of the triangle. Just observe by memory (or surfing the net) that this problem is EGMO 2016/6. As mentioned above, it is equivalent to find such a circle passing through $A$. Construction : Without loss of generality let $AB < AC$ and let $\triangle DEF$ be the contact triangle with $D, E, F$ on $BC, AC, AB$ respectively. Line $AI$ intersects $DE, DF$ at $X, Y$ respectively. $S, T$ be points on $BC$ such that $BS < BT$ and $\angle XSY = \angle XTY = 90^\circ$. Then $\odot (\triangle AST)$ is the desired circle. We see that $\triangle AXB \sim \triangle AYC$ and similarly $\triangle BXQ \sim \angle CYQ$ where $Q = BC \cap AI$. Together using C.S.S.T property, we have that $(A, Q; X, Y) = -1$ and because $\angle XSY = \angle XTY = 90^\circ$, we get that $ST$ bisects angles $\angle AST$ and $\angle ATS$ using the following theorem : If any two of the following three conditions hold, then the third does too for collinear points $A, B, C, D$ and a point $X$ not lying on line $AB$ or line $CD$ whatever - $(A, C; B, D)$ is harmonic $XB \perp XD$ $XC$ is internal angle bisector of $\angle AXB$ Hence $X$ is incenter of $\triangle AST$. Now, similarly we have that $Y$ is excenter of $\triangle AST$ because $\angle XSY = \angle XTY = 90^\circ$. Now see that $AS, AT$ are isogonal, and so $\gamma = \odot (\triangle AST)$ is tangent to $\odot (\triangle ABC)$. Now, we know that midpoint of incenter and excenter lies on the circumcircle of reference triangle using Fact $5$ and so midpoint of $X$ and $Y$ = $Z$ lies on $\gamma$. But in fact $Z$ is center of circle $\odot (\square XSYT)$ because $\angle XSY = \angle XTY = 90^\circ$ and so, we would like to define an inversion $\Psi $ with circle of Inversion being $\Omega = \odot (\square XSYT)$. We see that $\gamma$ passes through center of circle $\Omega$ and cuts $\Omega$ at $S, T$, so in fact inversion under $\Psi$ will take $\gamma$ to $BC$. But trivial that $\triangle IXD \sim \triangle IDY$ using suitable angle chasing, so $ID^2 = IX \cdot IY$, but this means that $r^2=IX\cdot IY$ which means that incircle or $\Gamma$ is orthogonal to $\Omega$. But this means that $\Psi$ takes $\Gamma$ to $\Gamma$ and since $\Gamma$ is tangent to $BC$ at $D$, $\gamma$ is tangent to $\Gamma$. Now let $T_1, T_2$ be the tangency point of $A$-excircle ($A$-ex for short) with $BC, AB$. $T_1, T_2, Y$ are collinear by using some EGMO first lesson lemma I don't recall which number please forgive me. But $BT_1, BT_2$ are tangent to $A$-ex and so $Y \in Polar_{A-\mathrm{ex}}(B)$ and by La Hire, we have that $B \in Polar_{A-\mathrm{ex}}(Y)$. This is just $BX$ as $AY = I_AY \perp BX$. So $\Omega$ is orthogonal to $A$-ex and so inverting $A$-ex about $\Psi$ gives $A$-ex which is tangent to $BC$ at $T_1$ and so $\gamma$ is tangent to $A$-ex and we're done. Sorry for no diagram it's late night. And I might've blundered a few points but fine.

Let the triangle be $\Delta ABC$, and the circle is tangent to the $A$-excircle. We claim that it passes through $A$. Let the incircle, $A$-excircle, $A$-mixtilinear incircle, $A$-mixtilinear excircle be $\omega, \Omega, \omega', \Omega'$ respectively, which have radii $r,R,r',R'$ respectively. Take $\sqrt{bc}$ inversion. Then we need to show that an internal common tangent to $\omega'$ and $\Omega'$ is parallel to $BC$. We also have that $r \cdot R' = r' \cdot R$, so there exists a homothety centred at $A$ with ratio $\frac{r'}{r}=\frac{R'}{R}$ which takes $\{ \omega, \Omega \}$ to $\{ \omega', \Omega' \}$. This homothety also takes $BC$ to a line tangent to $\omega'$ and $\Omega'$, so we are done. $\square$