[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.5261061752813205cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.0787451085122726, xmax = 4.193010037555495, ymin = -2.1247406533848845, ymax = 1.052847504054537; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pair M = (-0.0071174265422121645,-0.9999746707989239), A = (-0.7071067811865475,0.7071067811865476), F = (0.564801013735846,-0.17966418222163313), B = (-0.9138814749768283,-0.40598109524234843), C = (0.9080099497325882,-0.4189486020821921), D = (-0.014234853084424329,-1.9999493415978478), G = (-0.717100446487059,-0.6969698341019219), J = (-0.42131422732250695,-0.4094869952462738), I = (-0.4183784647003867,0.002977853415996461), H = (-0.7121036138368031,0.005068473542312774); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.8) + linetype("0 3 4 3")); draw(circle(M, 1.), linewidth(0.8) + linetype("0 3 4 3") + ffdxqq); draw(A--(-0.9902969049784449,-0.8173326351612942), linewidth(0.4)); draw(A--C, linewidth(0.4)); draw(D--(0.,0.), linewidth(0.4)); draw(A--D, linewidth(0.4)); draw(A--G, linewidth(0.4)); draw((0.,0.)--G, linewidth(0.4)); draw(B--C, linewidth(0.4)); draw((0.,0.)--A, linewidth(0.4)); draw(G--B, linewidth(0.4)); draw(A--M, linewidth(0.4)); draw(J--I, linewidth(0.4)); draw(circle(I, 0.41247529633199), linewidth(0.8) + linetype("0 3 4 3") + ffdxqq); draw(H--(0.,0.), linewidth(0.4)); draw(M--F, linewidth(0.4)); draw(M--C, linewidth(0.4)); draw(M--(-0.9902969049784449,-0.8173326351612942), linewidth(0.4)); draw(M--B, linewidth(0.4)); /* dots and labels */ dot((0.,0.),linewidth(4.pt) + blue); label("$D'$", (0.01436785438337898,0.029042250464270024), NE * labelscalefactor,blue); dot(M,blue); label("$M$", (0.03711908224094041,-1.085767914556243), NE * labelscalefactor,blue); dot(A,blue); label("$A$", (-0.6909202092010254,0.7457059279774569), NE * labelscalefactor,blue); dot((-0.9902969049784449,-0.8173326351612942),linewidth(4.pt) + blue); label("$E$", (-1.0625235975411953,-0.8203369228846923), NE * labelscalefactor,blue); dot(F,linewidth(4.pt) + blue); label("$F$", (0.5793566795128212,-0.1491757010866283), NE * labelscalefactor,blue); dot(B,linewidth(4.pt) + blue); label("$B$", (-1.0094373992068855,-0.37289610835264964), NE * labelscalefactor,blue); dot(C,linewidth(4.pt) + blue); label("$C$", (0.9547519391625848,-0.4221904353773662), NE * labelscalefactor,blue); dot(D,linewidth(4.pt) + blue); label("$D$", (-7.996308549953056e-4,-1.9692739296915478), NE * labelscalefactor,blue); dot(G,linewidth(4.pt) + blue); label("$G$", (-0.7364226649161483,-0.8241287941942859), NE * labelscalefactor,blue); dot(J,linewidth(4.pt) + blue); label("$J$", (-0.3913623757431332,-0.4676928910924892), NE * labelscalefactor,blue); dot(I,linewidth(4.pt) + blue); label("$I$", (-0.3951542470527268,0.04041786439305077), NE * labelscalefactor,blue); dot(H,linewidth(4.pt) + blue); label("$H$", (-0.7819251206312711,-0.005084591322072211), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Here's the equivalence of the problem. Quote: $D'$ is the circumcenter of the circle $ABC$ and $M$ is the circumcenter of the circle $FD'E$ and $AB$ and $AC$ are tangent to circle $FD'E$ also D is the reflection of $D'$ across $M$.$G$ is the intersection of perpendicular from $A$ to line $BC$.prove that $BC$,$AG$ and $G,D'$ are concurrent Quote: Let $I$ be $ABC$ incenter and $J$ is the perpendicular from $I$ to the line $BC$.by homothety center from $A$ we get that $BC$ and $AD$ both pass through $J$. now we prove that $BC$ is parallel to $ID'$. it is trivial since both are perpendicular to $IJ$. now by homothety centered at $D'$ we get that $D'G$ also passes through $J$, and the proof is complete.$\blacksquare$

$\because AI$ bisects $\angle XAY\Rightarrow I$ is the midpoint of arc $XY$ not containing $A$. $\Rightarrow XY$ is parallel to $BC$. Let's use complex number to solve it, and take $\Gamma$ as the unit circle. WLOG, let $I=1\Rightarrow D=2$ Let $A=-1/a, X=b\Rightarrow Z=a, Y=1/b$ Let $E$ be the point on $AX$ such that $IE\perp AX$, and $E=(-1/a+b+b/a+1)/2$. $\because IE=ID'=1\Rightarrow ((-1/a+b+b/a-1)/2)\cdot((-a+1/b+a/b-1)/2)=1$ $\Rightarrow 2a+2/a=2b+2/b+b/a+a/b+ab+1/ab$ $\Rightarrow 2a^2b+2b=2ab^2+2a+a^2+b^2+a^2b^2+1$ Let $P=XY\cap D'Z\Rightarrow P=a^2(b^2+1)/b(a^2+1)$ $P\in AD\iff\frac{a^2(b^2+1)/b(a^2+1)-2}{-1/a-2}=\frac{(b^2+1)/b(a^2+1)-2}{-a-2}$ $\iff a^4b^2+a^4-2a^4b-2a^2b+2a^3b^2+2a^3=b^2+1-2a^2b-2b+2ab^2+2a$ $\iff a^4b^2+a^4-2a^4b-2a^2b+2a^3b^2+2a^3=-a^2-a^2b^2$ $\iff a^2b^2+a^2-2a^2b-2b+2ab^2+2a=-1-b^2$ $\iff 2a^2b+2b=2ab^2+2a+a^2+b^2+a^2b^2+1$ $\therefore AD, D'Z, XY$ concurrent at a point $P$.

Li4 wrote: Let $I$ be the incenter of triangle $ABC$ and let $D$ the foot of altitude from $I$ to $BC$. Suppose the reflection point $D’$ of $D$ with respect to $I$ satisfying $\overline{AD’} = \overline{ID’}$. Let $\Gamma$ be the circle centered at $D’$ that passing through $A$ and $I$, and let $X$, $Y\neq A$ be the intersection of $\Gamma$ and $AB$, $AC$, respectively. Suppose $Z$ is a point on $\Gamma$ so that $AZ$ is perpendicular to $BC$. Prove that $AD$, $D’Z$, $XY$ concurrent at a point. Let $\Delta DEF$ be the contact triangle of $\Delta ABC$. Clearly $IX=IY$ and $D'X=D'Y$. Hence, $\overline{ID}\perp\overline{XY}\parallel\overline{BC}$. Let $\overline{EF}\cap\overline{XY}=T$. Clearly $T$ is the Miquel point of $\{\overline{AB},\overline{AC},\overline{EF},\overline{XY}\}$. Hence, $\overline{IT}\perp\overline{XY}\perp\overline{OD}\implies I,T,D$ are collinear. Let $M$ be the midpoint of $BC$. It's well known that $A,T,M$ are collinear and $\overline{AD'}$ is the $A-$ nagel cevian, thus we have that $-1=(D,AD'\cap BC;M,\infty_{BC})\overset{T}{=}(D',AD'\cap BC;A,XT\cap AD'=K)$. Thus, $\overline{DK}$ is the reflection of $\overline{AD}$ over $\overline{DI}$. Now ,reflect $\overline{AD'},\overline{DI},\overline{DK}$ over $\overline{DI}$ to get the desired conclusion. $\quad\blacksquare$

I applied Cartesian coordinate during the test. Let $D’$ be the origin and $\Gamma$ be the unit circle, and else could be easily done by supposing $A=(a,b)$ and solving some linear equations.

Since $\angle XAI=\angle IAY$, $I$ is middle point of arc $XIY$ and thus $D’I\perp XY\implies XY\parallel BC$. So $\Delta AXY$ similar to $\Delta ABC$. AD’ pass through circumcenter of $\Delta AXY$ and Nagel point of $\Delta ABC$, then Nagel point (write Na) and circumcenter (D’) of $\Delta AXY$ collinear with $A$. Let $AD\cap XY=W,AD’\cap XY=V,AD’\cap (AXY)=\{A,U\}$, then the pair of point $(V,U)$ is reflection of $(W,Z)$ over $D’I$ An. and thus $D’,W,Z$ collinear.

HeHe, I got 7 points on this problem, but I didn't get 7 on P1, which everyone said it is easy. AOPS said that I can't post images here since I am a new user, so I decide not to post my solution here.

p1 is in the shortlist of IMO 2020?

Really similar solution to #3 We use complex numbers and let $(AXY)$ be the unit circle. Let $D'=0,I=1,D=2,A=a,X=x$. From all of above's solution we can know that $XY||BC\Rightarrow XY\perp AZ$ East to see that $Y=\overline{x}=\frac{1}{x}$ By EGMO Lemma ???, we have $AZ+XY=0$. Therefore $Z=-\dfrac{1}{a}$ The foot from $I$ to $AX=\dfrac{a+x+1-ax}{2}$ $d(I,AX)=1\Rightarrow (\dfrac{a+x+1-ax}{2}-1)\overline{\big(\dfrac{a+x+1-ax}{2}-1\big)}=1$ $\Leftrightarrow a^2x^2-2a^2x-2ax^2+a^2+x^2-2a-2x+1=0\ (\star)$ Let $P'=AD\cap D'Z$ Then by the intersection formula and some computation to get $P'=\dfrac{2}{(a-1)^2}$ Now it's sufficient to show that $P'\in XY\Leftrightarrow$ $$\dfrac{\frac{2}{(a-1)^2}-x}{\frac{2}{(a-1)^2}-\frac{1}{x}}=\dfrac{\frac{2a^2}{(a-1)^2}-\frac{1}{x}}{\frac{2a^2}{(a-1)^2}-x}\\\Leftrightarrow (2x-x^2(a-1)^2)(2a^2x-x^2(a-1)^2)=(2x-(a-1)^2)((2a^2x-(a-1)^2))$$After a little expanding (Don't need to expand $(a-1)^2$) we have $(2x+2a^2x-2a^2x^3-2x^3)(a-1)^2+(x^4-1)(a-1)^4=0\\\Leftrightarrow (2x+2a^2x)(1-x^2)(a-1)^2+(1-x^2)(-x^2-1)(a-1)^4=0\\\Leftrightarrow (2x+2a^2x)+(a-1)^2(-x^2-1)=0$ And now we expand the last equation and it is just $(\star)$, as desired.$\blacksquare$ Hope there's no typo.

since $AI $ is the bisector of $XAY$ then $XI=YI$ thus $D'I$ is perpendicular to $XY$ hence $AXY\sim ABC$ more it s known that $AD'$ is the $A-$ nagel cevian of the triangle $ABC$ . By homothety $AZ,AD,AD'$ are also resp. the altitude ,the $A-$ gergonne cevian and the $A-$ nagel cevian of the triangle $AXY$ : if $AD \cap XY=G,AD'\cap XY=N$ and $M$ is the midpoint of $ XY\implies NM=MG $ and $NG\parallel N'G'$ where $N'=D'N\cap (AXY) , G'D'G\cap (AXY)$ therefore $AG'$ is the isogonal of $AD$; since $AN'$ is diameter then $AG'$ is the altitude which means that $G'=Z$ .

Attachments:

Here it is another solution. As $AI$ is the angle bisector of $\angle BAC$ then $I$ is the midpoint of the minor arc $XY$. Let's consider $l$ the tagent to $\Gamma$ at $I$, we know that $XY \parallel l$ and because $l,BC \perp DI$ then $XY \parallel BC$. Now let's condsider $H$ the homotety centered at A, such that $H(X,Y)=B,C$. Let $O$ be the circuncenter of $\triangle ABC$. We have $OI^2=R(R-2r)=Pot_{\Gamma}O$ then $O \in l$. Now its easy to see that as $A,D',O$ are colinear, hence $O,D,H(Z)$ are colinear, ie, $H(D')H(Z),H(X)(Y),AD$ concur, ie, $XY,AD,ZD'$ concur.

Pretty similar to ISL 2005 G6, especially if you try projective. My solution is the same as that in #4. Let me just note that proving that the intersection of $FE$ and $XY$ lies on $DD'$ can be done by Simson line ($FE$).

Since $AI$ is the bisector of $\angle AXY$ and $I$ lies on $(AXY)$, it follows that $IX = IY$ and since $DX = DY$, we have $ID \perp XY \parallel BC$. So the homothety centered at $A$ taking $\triangle AXY$ to $\triangle ABC$ takes $D'$ to $O$, where $O$ is the circumcenter of $\triangle ABC$, so $A,D',O$ are collinear. Let $N = AO \cap BC$ and $M$ be the midpoint of $BC$. Notice that by the same homothety that we mentioned, it suffices to prove that $AK \perp BC$, where $K$ is the intersection of ray $OD$ and $(ABC)$. Since $DD' \perp BC$, $AK \perp BC$ is equivalent to proving $OD' = OD$, or that $OI \perp DD'$. Since $BD = NC$, $M$ is the midpoint of $DN$ and hence $DM = MN$. Also, since $OM \perp BC \perp DD'$, $OM \parallel DD'$ and hence, $OD' = ON \implies OI \parallel BC \implies OI \perp DD'$. $\square$

straightforward Note that $I$ and the midpoint $S$ of arc $BC$ in $(ABC)$ are homologous in the positive homothety sending $\Gamma$ to $\Delta ABC$, and $IS$ passes through $A$ which lies on both circles - this is enough to imply that $A$ is the exsimilicenter of $\Gamma$ and $(ABC)$, and hence $\Delta AXY \stackrel +\sim \Delta ABC$. It suffices to show that $R = XY \cap D'Z$ gets sent to $D$ under this homothety. Let $O$ be the circumcenter of $\Delta ABC$. It is well-known that $AD'O$ passes through the $A$-extouch point which is the reflection of $D$ across the midpoint of $BC$, so $\angle(AO) + \angle(OD) = 2 \angle(BC)$. However, we also have $\angle(AD') + \angle(ZD') = 2 \angle(XY)$. Hence $OD \parallel ZD'$, which is enough to conclude.