Let ${A'}=BC\cap PQ$ P,Q,X,Y are concyclic if and only if $A'P*A'Q=AX*AY$ By Pop we have: $A'P*A'Q=A'B*A'C$ so $A'X*A'Y=A'B*A'C$ Suppose that there exist 2 points on $BC$: $M,N$ for which $MX*MY=MB*MC=NX*NY$=>$M,N$ are on the radical axis of $(ABC)$ and $(AXY)$ so A-M-N are collinear false.So $A'$ is the unique point on BC for which $A'X*A'Y=A'B*A'C$ Using barycentric coordinates:$A'=(0,\dfrac{b(a-b)(p-c)}{c(a-c)(p-b)+b(a-b)(p-c)},\dfrac{1}{1+\dfrac{b(a-b)(p-c)}{c(a-c)(p-b)}})$(this can be proved by intersecting $OI$ and $BC$) $A,D,L$ are collinear if and only if AD is the A-symedian: By Steiner this means:$\dfrac{p-b}{p-c}=\dfrac{c^2}{b^2}$<=>$b(a-b)=c(c-a)$ So A,D,L are collinear<=>$A'=(0,\dfrac{p-b}{c-b},\dfrac{c-p}{c-b})$<=> ${A'}=AA\cap BC$ we have $\triangle{A'YA}\sim\triangle{A'AX}$(a.a.)so $A'A^2=A'X*A'Y$ but $A'A^2=A'B*A'C$=>$A'X*A'Y=A'B*A'C$

Lemma 1: Let $ABC$ be a triangle with $O$ be its circumcenter, $H$ be its orthocenter and $X$ be the intersection of tangent from $A$ with $BC$. Let $AO\cap BC=T$,$AH\cap BC=S$ then $XB.XC=XS.XT$. Proof: $\angle TAC=\angle SAB\Longrightarrow \angle XAS=\angle XTA\Longrightarrow XB.XC=XA^2=XS.XT$. obviously the point $X$ is unique. Lemma 2: Let $ABC$ be a triangle with intouch triangle $DEF$ and $O,I$ be its circumcenter and incenter. let $D^{\star}$ be the reflection of $D$ WRT $EF$ then $AD^{\star},OI,BC$ are concurrent. Proof: see 2012 China TST -Quiz1- Day 1- P2. Back to the original problem: Let $K$ be the center of $A$ appolonian circle (the intersection of tangent from $A$ to $\odot(O)$ with the line $BC$). First assume that $A,D,L$ are concurrent then since $AL$ is polar of $K$ WRT $(O)$ we deduce that $OX\perp AL (\clubsuit)$ and $(K,D;B,C)=-1\Longrightarrow AD$ is polar of $K$ WRT $(I)$ hence $KI\perp AD$ combining this with $\clubsuit$ we get that $O,I,K$ are collinear so from first lemma $\Longrightarrow KQ.KP=KB.KC=KY.KX\Longrightarrow PQXY$ is cyclic. Now assume that $PQXY$ is cyclic qudrilateral. Let $U=OI\cap BC$ then $UQ.UP=UB.UC=UY.UX$ hence from first lemma we deduce that $U\equiv K\Longrightarrow KA$ is tangent to $(O)$. Let $A^{\star}$ be the reflection of $A$ WRT $EF$.($E,F$ are the tangency points of $(I)$ with $AC,AB$ respectively) From the second lemma the reflection of $D$ WRT $EF$ lies on $AK$ so $EF$ is internal angle bisector of $AKB$; Hence $A^{\star}$ lies on $BC$ so $AEA^{\star}F$ is rhombus so from thales's theorem: $\left.\begin{array}{lll} \frac{CE}{b}=\frac{A^{\star}E}{c}=\frac{AF}{c}\\ \\\frac{BF}{c}=\frac{A^{\star}F}{b}=\frac{AE}{b} \end{array}\right\}\Longrightarrow \frac{BD}{CD}=\frac{c^2}{b^2}\Longrightarrow AD$ is symmedian line of $ABC$. 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We will show that both conditions are equivalent to $a(b+c)=b^2+c^2$. (computations are omitted for brevity) Note that $AL$ is a symmedian, so $D$ lies on it iff $\frac{c^2}{b^2}=\frac{s-b}{s-c}$, which is equivalent to $a(b+c)=b^2+c^2$ after simplification. Let $AR$ and $AS$ be the angle bisectors of $\angle A$ and let $M$ be the midpoint of $RS$. Since $(B,C;R,S)=-1$, $AM$ is tangent to $(ABC)$. Because $(B,C;X,Y)=-1$ as well, we have that $MR^2=MB\cdot MC=MX\cdot MY$. By Power of a Point, the point $M'$ on $BC$ that lies on $PQ$ is the unique point that satisfies $M'X\cdot M'Y=M'P\cdot M'Q=M'B\cdot M'C$, so $M$ lies on $PQ$. The poles of lines $AM$, $OM$, and $BM$ with respect to $\odot O$ all lie on the polar of $M$, so $A$, $L$, and the point at infinity perpendicular to $OI$ are collinear. Hence, $PQXY$ is cyclic iff $AL\perp OI$. Consider the function $f(P)=p(P,\odot O)=p(P,\odot I)$, and let $E$ be the intersection of $AL$ with $BC$. Note that since $AE\perp OI$, $f(A)=f(E)$. Now, we can compute \[f(A)=(s-a)^2\quad f(E)=\frac{(b-c)^2(b^2+c^2-ab-ac)^2+4a^2b^2c^2}{4(b^2+c^2)^2}\]and verify that $f(A)=f(E)$ iff $a(b+c)=b^2+c^2$, as desired.

(all steps are reversible) Let $AR$ and $AS$ be the angle bisectors of $\angle A$ and let $M$ be the midpoint of $RS$. Since $(B,C;R,S)=-1$, $AM$ is tangent to $(ABC)$. Because $(B,C;X,Y)=-1$ as well, we have that $MR^2=MB\cdot MC=MX\cdot MY$. By Power of a Point, the point $M'$ on $BC$ that lies on $PQ$ is the unique point that satisfies $M'X\cdot M'Y=M'P\cdot M'Q=M'B\cdot M'C$, so $M$ lies on $PQ$. Alternative finish using this: andria wrote: Lemma: Let $ABC$ be a triangle with intouch triangle $DEF$ and $O,I$ be its circumcenter and incenter. let $D*$ be the reflection of $D$ WRT $EF$ then $AD*,OI,BC$ are concurrent. Proof: see 2012 China TST -Quiz1- Day 1- P2. Let $E$ and $F$ be the other intouch points and let $D'$ be the reflection of $D$ over $EF$. By that lemma, $D'$ lies on $AM$, so $MD$ reflects to $MA$ over $EF$. Hence, $M$ lies on $EF$, so $(B,C;D,M)=-1$, from which we are done.

ABCDE wrote: (all steps are reversible) Let $AR$ and $AS$ be the angle bisectors of $\angle A$ and let $M$ be the midpoint of $RS$. Since $(B,C;R,S)=-1$, $AM$ is tangent to $(ABC)$. Because $(B,C;X,Y)=-1$ as well, we have that $MR^2=MB\cdot MC=MX\cdot MY$. By Power of a Point, the point $M'$ on $BC$ that lies on $PQ$ is the unique point that satisfies $M'X\cdot M'Y=M'P\cdot M'Q=M'B\cdot M'C$, so $M$ lies on $PQ$. Alternative finish using this: andria wrote: Lemma: Let $ABC$ be a triangle with intouch triangle $DEF$ and $O,I$ be its circumcenter and incenter. let $D*$ be the reflection of $D$ WRT $EF$ then $AD*,OI,BC$ are concurrent. Proof: see 2012 China TST -Quiz1- Day 1- P2. Let $E$ and $F$ be the other intouch points and let $D'$ be the reflection of $D$ over $EF$. By that lemma, $D'$ lies on $AM$, so $MD$ reflects to $MA$ over $EF$. Hence, $M$ lies on $EF$, so $(B,C;D,M)=-1$, from which we are done. What's the mean of $(B,C;X,Y)=-1$?

Note that for all $T_1,T_2\in BC-\{ B,C\}$ that $AT_1$ and $AT_2$ are isogonal conjugate, $P,Q,T_1,T_2$ are concyclic iff $A,D,L$ are collinear

lwwwww wrote: ABCDE wrote: (all steps are reversible) Let $AR$ and $AS$ be the angle bisectors of $\angle A$ and let $M$ be the midpoint of $RS$. Since $(B,C;R,S)=-1$, $AM$ is tangent to $(ABC)$. Because $(B,C;X,Y)=-1$ as well, we have that $MR^2=MB\cdot MC=MX\cdot MY$. By Power of a Point, the point $M'$ on $BC$ that lies on $PQ$ is the unique point that satisfies $M'X\cdot M'Y=M'P\cdot M'Q=M'B\cdot M'C$, so $M$ lies on $PQ$. Alternative finish using this: andria wrote: Lemma: Let $ABC$ be a triangle with intouch triangle $DEF$ and $O,I$ be its circumcenter and incenter. let $D*$ be the reflection of $D$ WRT $EF$ then $AD*,OI,BC$ are concurrent. Proof: see 2012 China TST -Quiz1- Day 1- P2. Let $E$ and $F$ be the other intouch points and let $D'$ be the reflection of $D$ over $EF$. By that lemma, $D'$ lies on $AM$, so $MD$ reflects to $MA$ over $EF$. Hence, $M$ lies on $EF$, so $(B,C;D,M)=-1$, from which we are done. What's the mean of $(B,C;X,Y)=-1$? (B,C;X,Y)是调和点列

Let $E,F$ be intouch points of incircle with $AC,AB$ resp. Let $A'$ be intersection of $AD$ and $\odot(\Delta ABC),$ and let $U$ be intersection of tangent at $A$ and $BC.$ The "if" part. Since $AA'$ is symmedian in $ABC$ it follows that $ABA'C$ is harmonic and thus $UA'$ is also tangent to $\odot(\Delta ABC)$ hence $AA'$ is pole of $U$ hence $UO \perp AD...(*)$ Intersecting the harmonic bundle $(A,A';B,C)$ with line $BC$ we deduce $(U,D;B,C)=-1.$ Clearly this implies that $U$ lies on $EF.$ Now since $U$ is on pole of $A$ WRT incircle it follows that $A$ is on pole of $U$ WRT incircle $\implies$ $UI \perp AD...(**)$ Hence from $(*),(**)$ we get that $U,I,O$ are collinear. Hence $U,P,Q$ are collinear. Now since $AY,AX$ are isogonals it follows that $\odot(\Delta AYX)$ is tangent to $\odot(\Delta ABC) \implies$ $UP \cdot UQ=UA^2=UX \cdot UY \implies$ $P,Q,X,Y$ are concyclic.

The result being clear for an isosceles triangle we may assume that it is scalene. Let the tangent to $(ABC)$ at $A$ meet $BC$ at $T$ and the line $OI$ meet $BC$ at $T'$. We claim that both conditions are equivalent to the fact that $a(b+c)=b^2+c^2$ where $a,b,c$ are the side lengths of the triangle $ABC$ (denoted in the usual way). Indeed, $A,D,L$ are collinear is equivalent to $$\frac{s-b}{s-c}=\frac{a+c-b}{a+b-c}=\frac{c^2}{b^2} \Longrightarrow a(b^2-c^2)=(b-c)(b^2+c^2) \Longrightarrow a(b+c)=b^2+c^2.$$ If the points $P,Q,X,Y$ are concyclic then by power of a point, $$T'B\cdot T'C=T'P\cdot T'Q=T'X\cdot T'Y$$and since $AX,AY$ are isogonal lines in angle $A$, we conclude that $T'=T$. Thus, $T,I,O$ are collinear. Apply inversion at $A$ of radius $\sqrt{AB\cdot AC}$ followed by reflection in the bisector of angle $BAC$. Let $A'$ be the point such that $ABCA'$ is an isosceles trapezoid with $AA' \parallel BC$. Let $A_1$ be the reflection of $A$ in $BC$ and let $I_A$ be the $A$ excenter of triangle $ABC$. The condition $T,I,O$ are collinear becomes $I_A,A_1,A',A$ lie on a circle. Let $M$ be the midpoint of $BC$ then $MA=MI_A$. Note that $MA^2=MI_A^2$ is equivalent to $$I_AB^2+I_AC^2=b^2+c^2$$hence, we have $$b^2+c^2=CD^2+BD^2+2r_a^2=(s-b)^2+(s-c)^2+\frac{2s(s-b)(s-c)}{s-a} \Longrightarrow b^2+c^2=\left((s-b)+(s-c)\right)^2+\frac{a(a+b-c)(a+c-b)}{b+c-a}.$$Therefore, we have $$b^2+c^2=a\cdot \left(\frac{a(b+c-a)+a^2-(b-c)^2}{b+c-a}\right).$$Simplifying the last equation gives $$a^2(b+c)+2abc-(b+c)(b^2+c^2)=0.$$Considering this as a quadratic equation in variable $a$, since the product of both the roots is $-(b^2+c^2)<0$, it can have at most one positive real root. Clearly, $a=\frac{b^2+c^2}{b+c}$ satisfies this quadratic equation and so it is the only positive real root. Both conditions are equivalent to $a(b+c)=b^2+c^2$ as desired.

The conclusion is obvious if $AB=AC$; otherwise, define $T\equiv AA\cap BC$; it's easy to check that $TX\cdot TY=TB\cdot TC$, and accordingly $PQ$ passes through $T$. This is evidently equivalent to $AL \perp OI$, or by a well-known lemma, $AL$ is parallel to the line $\ell$ between the feet of the external angle bisectors. If $D' \equiv AL\cap BC$, we have $$\frac{CD'}{CA}=\frac{BC-AB}{AC-AB}$$and thus $$\frac{BD'}{CD'}=\frac{AB(AC-BC)}{AC(BC-AB)}=\frac{AB^2}{AC^2} \Longrightarrow BC(AB+AC)=AB^2+AC^2\Longleftarrow \frac{AB^2}{AC^2}=\frac{BC-AC+AB}{BC-AB+AC}$$as desired.

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$ \bullet $ $ \qquad $ If $ \color{blue} A, D, L $ are collinear. Let $ T $ be the intersection of $ BC $ with the polar of $ A $ WRT $ \odot (I). $ Clearly, $ \overline{ADL} $ is the polar of $ T $ WRT $ \odot (I), $ so $ \overline{ADL} $ $ \perp $ $ IT. $ On the other hand, as $ BC $ is the polar of $ L $ WRT $ \odot (O), $ so notice $ (B,C;D,T) $ $ = $ $ -1 $ we get $ \overline{ADL} $ is the polar of $ T $ WRT $ \odot (O), $ hence $ \overline{ADL} $ $ \perp $ $ OT $ $ \Longrightarrow $ $ T $ lies on $ IO $ and $ AT $ is tangent to $ \odot (O) $ at $ A. $ Since $ AX $ is the diameter of $ \odot (AXY), $ so $ AT $ is the radical axis of $ \odot (O) $ and $ \odot (AXY) $ $ \Longrightarrow $ $ TP $ $ \cdot $ $ TQ $ $ = $ $ TB $ $ \cdot $ $ TC $ $ = $ $ TX $ $ \cdot $ $ TY $ $ \Longrightarrow $ $ P, $ $ Q, $ $ X, $ $ Y $ are concyclic. $ \bullet $ $ \qquad $ If $ \color{blue} P, Q, X, Y $ are concyclic. Since $ AX $ is the diameter of $ \odot (AXY), $ so the tangent $ \tau_A $ of $ \odot (O) $ passing through $ A $ is the radical axis of $ \odot (O) $ and $ \odot (AXY) $ $ \Longrightarrow $ $ \tau_A, $ $ BC, $ $ OI $ are concurrent at $ S. $ Since $ AL $ is the polar of $ S $ WRT $ \odot (O) $ $ \Longrightarrow $ $ \overline{IOS} $ $ \perp $ $ AL, $ so note that $ A(B,C;L,S) $ $ = $ $ -1 $ we get $ AL $ is the polar of $ S $ WRT $ \odot (I), $ hence the tangency point $ D $ of $ \odot (I) $ with $ BC $ lies on $ AL. $

My solution: Assume $A,D,L$ collinear. Then consider the harmonic conjugate of $D$, say $D'$. Then as AD is a symmedian, $AD'$ is tangent to $(O)$. D' is the pole of $AD$ wrt $(I)$ and of $AL$ wrt $(O)$, so $D \in OI$. Also, $(AXY)$ is tangent to $(O)$ so by converse of radical axes, we get our result. Assume $PQXY$ is cyclic. Then by radical axes, we have $O, I, D'$ are collinear where the last point is the intersection of the tangent to $O$ at $A$ and $BC$. Also, we have $OD' \perp AL$ and contains $I$. So $\angle D'ID" = 90^\circ$ for the point $D"$ such that $AD"$ is a symmedian. Also the fact that BC is the pole of $D$ and the previous fact along with $OD' \perp AL$ directly give that $ D' \equiv D"$ which is the required relationship.

Too easy for a CMO.Here's my solution If $EA \cap \odot O=S \not =A,$ then $A,S,X,Y$ concyclic $\Longleftrightarrow AS \perp SX$ which is contradiction since $AX$ is not a diameter So $P,Q,X,Y$ concyclic $\Longleftrightarrow EA $ is a tangent let $M$ me the midpoint of $AF,\Longrightarrow OM \perp AF \Longrightarrow E'A$ is a tangent where $E'=BC \cap OM$. So $E'G$ is a tangent of $\odot I$(Because$ G,T_B,D,T_C$ is a harmonic quadrilateral), then $\triangle E'AF,\triangle E'GD \text{ isosceles},G,D $ lie on $AF \Longrightarrow \boxed{\text{ The middle of GD=M}} \Longrightarrow E'IO $ collinear $\Longrightarrow E=E'$.Done! Your sincerely CeuAzul

sketch during contest. maybe be coinciding with #6 or #14. if so, please delete it. maybe there be some mistake in it because this is written by memory. WLOG AB＞AC To prove these two conditions are equivalent to $a(b+c)=b^2+c^2$ collinear to this: AL is symmedian so $ \frac{BD}{CD}\mathrm{{=}}\frac{{c}^{2}}{{b}^{2}}\mathrm{\Longleftrightarrow}{b}^{2}{\mathrm{(}}{a}\mathrm{{+}}{c}\mathrm{{-}}{b}{\mathrm{)}}\mathrm{{=}}{c}^{2}{\mathrm{(}}{a}\mathrm{{+}}{b}\mathrm{{-}}{c}{\mathrm{)}}\mathrm{\Longleftrightarrow}{\mathrm{(}}{b}\mathrm{{-}}{c}{\mathrm{)(}}{ab}\mathrm{{+}}{ac}\mathrm{{-}}{b}^{2}\mathrm{{-}}{c}^{2}{\mathrm{)}}\mathrm{{=}}{0} $ OK cyclic to this OI intersects BC at R，so PQXY cyclic is equivalent to $ {RB}\mathrm{\times}{RC}\mathrm{{=}}{RX}\mathrm{\times}{RY} $(1) make M the midpoint of BC and let RM be x, the formula above is equivalent to $ {x}\mathrm{{=}}\frac{\frac{{a}^{2}}{4}\mathrm{{-}}{MX}\mathrm{\times}{MY}}{{MY}\mathrm{{-}}{MX}} $(2) make$ \mathrm{\frac{XC}{BX}}\mathrm{{=}}\mathrm{\alpha}\mathrm{{>}}{1}\hspace{0.33em}\mathrm{\frac{YC}{BY}}\mathrm{{=}}\mathrm{\beta}\mathrm{{<}}{1} $(3) since $ {MX}\mathrm{{=}}\mathrm{\frac{\mathrm{\alpha}{-}{1}}{{1}{+}\mathrm{\alpha}}}\mathrm{\times}\frac{a}{2}\hspace{0.33em}{MY}\mathrm{{=}}\mathrm{\frac{{1}{-}\mathrm{\beta}}{\mathrm{\beta}{+}{1}}}\mathrm{\times}\frac{a}{2} $(4) put (3)(4) into (2) we have (since AX,AY isogonal conjugate, so we have$ \mathit{\alpha}\mathit{\beta}\mathrm{{=}}\frac{{b}^{2}}{{c}^{2}} $) $ {x}\mathrm{{=}}\frac{a}{2}\mathrm{\times}\frac{{1}\mathrm{{+}}\mathit{\alpha}\mathit{\beta}}{{1}\mathrm{{-}}\mathit{\alpha}\mathit{\beta}}\mathrm{{=}}\frac{a}{2}\mathrm{\times}\frac{{b}^{2}\mathrm{{+}}{c}^{2}}{{c}^{2}\mathrm{{-}}{b}^{2}} $(5) and we have $ {x}\mathrm{{=}}{MR}\mathrm{{=}}\frac{OM}{{OM}\mathrm{{-}}{ID}}{DM}\mathrm{{=}}\frac{{R}\cos{A}\mathrm{\times}{\mathrm{(}}{c}\mathrm{{-}}{b}{\mathrm{)}}}{{2}{\mathrm{(}}{R}\cos{A}\mathrm{{-}}{r}{\mathrm{)}}}\mathrm{{=}}\frac{{\mathrm{(}}{c}\mathrm{{-}}{b}{\mathrm{)}}\cos{A}}{{2}{\mathrm{(}}{1}\mathrm{{-}}\cos{B}\mathrm{{-}}\cos{C}{\mathrm{)}}}\mathrm{{=}}\frac{{a}{\mathrm{(}}{c}\mathrm{{-}}{b}{\mathrm{)(}}\mathrm{{-}}{a}^{2}\mathrm{{+}}{b}^{2}\mathrm{{+}}{c}^{2}{\mathrm{)}}}{{2}{\mathrm{(}}{2}{abc}\mathrm{{-}}{cb}^{2}\mathrm{{-}}{bc}^{2}\mathrm{{-}}{ca}^{2}\mathrm{{-}}{ba}^{2}\mathrm{{+}}{b}^{3}\mathrm{{+}}{c}^{3}{\mathrm{)}}} $(6) so (5) is equivalent to  $ \frac{{c}^{2}\mathrm{{+}}{b}^{2}}{{c}^{2}\mathrm{{-}}{b}^{2}}\mathrm{{=}}\frac{{\mathrm{(}}{c}\mathrm{{-}}{b}{\mathrm{)(}}\mathrm{{-}}{a}^{2}\mathrm{{+}}{b}^{2}\mathrm{{+}}{c}^{2}{\mathrm{)}}}{{\mathrm{(}}{2}{abc}\mathrm{{-}}{cb}^{2}\mathrm{{-}}{bc}^{2}\mathrm{{-}}{ca}^{2}\mathrm{{-}}{ba}^{2}\mathrm{{+}}{b}^{3}\mathrm{{+}}{c}^{3}{\mathrm{)}}} $(7) full expend and make it be a polynomial of a we have $ {2}{abc}{\mathrm{(}}{ab}\mathrm{{+}}{ac}\mathrm{{-}}{b}^{2}\mathrm{{-}}{c}^{2}{\mathrm{)}}\mathrm{{=}}{0} $(8) and $ {ab}\mathrm{{+}}{ac}\mathrm{{=}}{b}^{2}\mathrm{{+}}{c}^{2} $ OK.

TelvCohl wrote: $ \bullet $ $ \qquad $ If $ \color{blue} A, D, L $ are collinear. Let $ T $ be the intersection of $ BC $ with the polar of $ A $ WRT $ \odot (I). $ Clearly, $ \overline{ADL} $ is the polar of $ T $ WRT $ \odot (I), $ so $ \overline{ADL} $ $ \perp $ $ IT. $ On the other hand, as $ BC $ is the polar of $ L $ WRT $ \odot (O), $ so notice $ (B,C;D,T) $ $ = $ $ -1 $ we get $ \overline{ADL} $ is the polar of $ T $ WRT $ \odot (O), $ hence $ \overline{ADL} $ $ \perp $ $ OT $ $ \Longrightarrow $ $ T $ lies on $ IO $ and $ AT $ is tangent to $ \odot (O) $ at $ A. $ Since $ AX $ is the diameter of $ \odot (AXY), $ so $ AT $ is the radical axis of $ \odot (O) $ and $ \odot (AXY) $ $ \Longrightarrow $ $ TP $ $ \cdot $ $ TQ $ $ = $ $ TB $ $ \cdot $ $ TC $ $ = $ $ TX $ $ \cdot $ $ TY $ $ \Longrightarrow $ $ P, $ $ Q, $ $ X, $ $ Y $ are concyclic. $ \bullet $ $ \qquad $ If $ \color{blue} P, Q, X, Y $ are concyclic. Since $ AX $ is the diameter of $ \odot (AXY), $ so the tangent $ \tau_A $ of $ \odot (O) $ passing through $ A $ is the radical axis of $ \odot (O) $ and $ \odot (AXY) $ $ \Longrightarrow $ $ \tau_A, $ $ BC, $ $ OI $ are concurrent at $ S. $ Since $ AL $ is the polar of $ S $ WRT $ \odot (O) $ $ \Longrightarrow $ $ \overline{IOS} $ $ \perp $ $ AL, $ so note that $ A(B,C;L,S) $ $ = $ $ -1 $ we get $ AL $ is the polar of $ S $ WRT $ \odot (I), $ hence the tangency point $ D $ of $ \odot (I) $ with $ BC $ lies on $ AL. $ Could you explain why $A(B,C;L,S) $ $ = $ $ -1 $ we get $ AL $ is the polar of $ S $ WRT $ \odot (I)$ ? Thanks!

Can I use Inversion to solve it?Help !

I wanna let T be the mid-point of inferior arc BC and then let it be the centre of inversion,and TB=TC=TI be the radius ,inversion then.But I cannot continue to finish it ,does any one can help me?

Solved with BVKRB. Part 1: If $\overline{A-D-L}$ Let $F$ be $A$-dumpty point and $T = OF \cap BC$ then $(T,D;B,C)=-1 \implies IT \perp AD$ hence $\overline{F-I-O}$ Note $F,O,X,Y$ is cyclic

by power of point $$TP.TQ=TB.TC=TF.TO=TY.TX$$we get $P,Q,X,Y$ cyclic [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.87858586650066cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.120249688786348, xmax = 27.758336177714312, ymin = -14.750468467629766, ymax = 6.080604828212421; /* image dimensions */ /* draw figures */ draw(circle((7.338775233469293,-1.273260927228962), 6.865571130988021), linewidth(0.4)); draw(circle((4.90042996788325,-1.7870737571762263), 2.9805630817448177), linewidth(0.4)); draw(circle((6.7699907490619,1.4182981458919826), 7.394754074111638), linewidth(0.4)); draw((2.9320681891566176,3.9914327941817325)--(6.810100606436214,-14.384391677649301), linewidth(0.4) + red); draw(circle((7.0744379199527545,-7.828826302439134), 6.560892592016091), linewidth(0.4)); 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Part 2: if $P,Q,X,Y$ is cyclic let $T'=PQ \cap BC$ then we get $T'P.T'Q=T'B.T'C=T'X.T'Y$ but we also have $TX.TY=TB.TC$ as both have same cross ratio $T=T' \implies \overline{F-I-O}$ Consider $G,H$ be touch point of incircle to side $AB$ and $AC$ with $U = HG \cap AF$ and $ N = AI \cap GH$ then we know $\angle AFI = 90$ so $U,N,I,F$ and $G,F,H,I$ are cyclic If $R=FI \cap GH$ and $K=AF \cap BC$ $$RF.RI = RU.RN=RG.RH \implies$$$$(R,U;G,H)\stackrel{D}{=}(AR\cap BC,K;B,C)=-1=(T,K;B,C)$$$R=T$ and so $\overline{T-G-H}$ which implies $K=D$ $\blacksquare$

Nice Problem! Let $f(X)= \frac{BX}{CX}$; then as $PQXY$ is cyclic so $f(P)f(Q)=f(X)f(Y)$, but $f(P)f(Q)=f(OI \cap BC = T)$ and $f(X) \cdot f(Y) = f(A)^2$ as $AX,AY$ are isogonal; therefore $f(A)^2=f(T)$, therefore $OI,AA,BC$ concur. Let $A',B',C'$ be the reflections of $A,B,C$ in $OI$ then by reflection $T \in \overline{B'C'}$, let $\overline{AD} \cap (ABC) = K$ then by DDIT on the degenerate quadrilateral $ABCD$ and the point $A'$ and projecting to the circumcircle we get if $A'D \cap (ABC) = U$ then $AU,B'C',BC,KA'$ concur this implies $A=U$ and therefore $K=A'$ and therefore $ABCA'$ is harmonic which implies $A,D,L$ are collinear, for the other direction assume $A,D,L$ are collinear, so let $T$ be the harmonic conjugate of $D$ in $BC$ then $T$ is pole of $AD$ in incircle and circumcircle so $T,O,I$ are collinear and so $f(X)f(Y)=f(A)^2=f(T)=f(P)f(Q)$, so done.

Let $E, F$ be the intouch points on $CA, AB$ respectively, $T = \overline{BC} \cap \overline{EF}$. If $\overline{A, D, L}$, then $\frac{BD}{CD} = \frac{c^2}{b^2}$. Since $AX, AY$ are isogonal, then $\frac{BX}{CX} \cdot \frac{BY}{CY} = \frac{c^2}{b^2}$. Let $G = \overline{AD} \cap (ABC)$. Because $T \in polar_{(O)}(L), \Rightarrow L \in polar_{(O)}(T)$, and since $(T, D; B,C) = -1), \Rightarrow D \in polar_{(O)}(T)$, so $\overline{DL} = polar_{(O)}(T)$, i.e. $\overline{TO} \perp \overline{AG}$. Let $H = \overline{AD} \cap (I)$, since $T \in polar_{(I)}(A), \Rightarrow A \in polar_{(I)}(T)$, i.e. $\overline{HD} = polar_{(I)}(T)$, so $\overline{TI} \perp \overline{AG}$, hence $\overline{T-I-O-P-Q}$. Since $T = \overline{PQ} \cap \overline{BC}$, it is well-known that $\frac{BP}{CP} \cdot \frac{BQ}{CQ} = \frac{BT}{CT}$. Therefore $$\frac{BP}{CP} \cdot \frac{BQ}{CQ} = \frac{BT}{CT} = \frac{BD}{CD} = \frac{c^2}{b^2} = \frac{BX}{CX} \cdot \frac{BY}{CY}$$which implies that $PQXY$ is cyclic, as desired. If $PQXY$ is cyclic, we have $$\frac{BP}{CP} \cdot \frac{BQ}{CQ} = \frac{BX}{CX} \cdot \frac{BY}{CY} = \frac{c^2}{b^2}$$Let $D' = \overline{AL} \cap (ABC), T' = \overline{PQ} \cap \overline{BC}, G' = \overline{AD'} \cap (ABC)$. Since $\frac{BD'}{CD'} = \frac{c^2}{b^2} = \frac{BT'}{CT'} \Rightarrow (T', D'; B,C) = -1$ and $T \in polar_{(O)}(L), \Rightarrow L \in polar_{(O)}(T)$, we have $\overline{AD'} = polar_{(O)}(T')$, and it is clear that $\overline{T'O}$ is the perpendicular bisector of $AG'$, so $IA = IG'$. We already have $\overline{AD} = polar_{(I)}(T)$, and $\overline{AD}$ is the perpendicular bisector of $AG$, so $IA = IG$. Hence, $IG = IG'$, so $G = G'$, $D = D'$, therefore $\overline{A-D-L}$.