This problem was proposed by Burii and me

WLOG assume $\angle C > \angle A$ Claim 1 $TX = YZ$ proof. Angle chase to get $\angle TAX = YCZ$ The problem is now reduced to proving $$AD + DT + XA = CD + DY + ZC$$Which is equivalent to proving $DT + XB = DY + ZB$ by using $AB + CD = AD + BC$. Consider the power of the point $B$ and $D$ wrt $\Omega$. Then after some manipulation, it is sufficient to prove $\frac{XB}{BC} = \frac{DY}{DC}$

This can be shown by angle chasing and law of sine on $\triangle BXC$ and $\triangle DCY$

After proving $XT=YZ$, and using Pitot's Theorem twice, it suffices proving $\frac{DT}{DA}=\frac{BZ}{BA}$, which can be proven by law of sines on $\bigtriangleup ATD$ and $\bigtriangleup BAZ$. Sorry to the proposers, this is a good length chasing problem, but personally I don't think it's IMO type.

popcorn1 wrote: Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\] This problem is a good one but it might be just a bit slightly harder for easy problem. It is very near to medium problem but not a medium problem (if you understand what I mean).

Pretty much same as #3. After $XT=YX$, we need to prove that $AD+DT+XA=CD+DY+ZC$. $AD-CD+DT+XA=DY+ZC$. Use Pitot, $AD-CD=AB-BC$, and we have that, $AB+DT+XA=BC+DY+ZC$. $BX+DT=BZ+DY$. $BX-BZ=DY-DT$. Note that, $BX-BY=\frac{(BC-BA)BZ}{BA}$ and that $DY-DT=\frac{(DC-DA)DT}{DA}$. Since, $AD-CD=AB-BC$, suffices to show that $\frac{DT}{DA}=\frac{BZ}{BA}$. To show this we proceed as follows: $\frac{DT}{DA}=\frac{sin\angle DAT}{sin\angle DTA}=\frac{sin\angle XAZ}{sin\angle AZC}=\frac{sin\angle ZAB}{sin\angle AZB}=\frac{BZ}{BA}$.

Inverting around $\Gamma$, $A',C',X',Y',Z',T'$ are all collinear. Let $S,V,W,U$ be the tangencies of $\Gamma$ with $AD, DC, CB, BA$. Note that $D',C',T'\in (IV)$; $X',A',B'\in (IU)$; $Z',C',B'\in (IW)$; $A',D',Y'\in (IS)$. $$\measuredangle IZ'T'=\measuredangle IZ'C'=\measuredangle IWC'=\measuredangle C'VI=\measuredangle C'T'I=\measuredangle Z'T'I$$Hence, $IZ'=IT'$, equivalently $IZ=IT$. Similarly, $IY=IX$ Now, \[A D+D T+T X+X A=C D+D Y+Y Z+Z C\]is equivalent to \[D V+A U+D T+T X+X A=CW+DS+D Y+Y Z+Z C\]or \[VT+TX+XU=WZ+ZY+YS\]. Which is true since $VT=WZ$, $TX=ZY$, $XU=YS$.

First, we claim that: Claim: $XYZT$ is an isosceles trapezoid. Proof: Observe that $AI$ is the angle bisector of $\angle BAD \equiv \angle XAY$, so $IX=IY$. Similarly, $IZ=IT$, so $XY\parallel ZT$. This gives the claim. $\blacksquare$ In particular, the claim implies that $XT=YZ$, removing two terms in the equation to be proved. Now, here is the critical observation, if we let the $\Gamma$ touches $AB, AD, CB, CD$ at $P,Q,R,S$, respectively, then $\triangle IPX\cong\triangle IQY$ (as $IP=IQ$ and $IX=IY$). Thus, $PX=QY$. Similarly, $RZ=ST$. Therefore \begin{align*} PX + ST &= QY + RZ \\ (AX + AP) + (DT + DS) &= (DY + QD) + (RC + CZ) \\ (AX + DT) + (AP + DS) &= (DY + CZ) + (QD + RC) \\ (AX + DT) + (AQ + QD) &= (DY + CZ) + (DS + SC) \\ AX + DT + AD &= DY + CZ + DC, \end{align*}done.

Dear mathlinkers, Upon seeing this problem (and this year's test, generally) I had to write a word or two about this. It is inexplicable how this problem can make the test. First, there's no creativity required to solve it. The entire construction is given and you don't have to add any points or lines to make progress. The problem is a tiny bit of angle chasing followed by calculations that can be performed almost mindlessly. I'm not saying that this is easier than 2020 #1 or 2018 #1 but those problems at least encouraged finding the missing ingredient in the diagram and sadly this does not. Best wishes, Donald Glover

I think it is pretty decent for a P4. Claim 01. $XY \parallel ZT$, which implies $XT = YZ$. Proof. Note that $\measuredangle IXY = \measuredangle IAY = \measuredangle BAI = \measuredangle XAI = \measuredangle XYI$. Therefore, $I$ is the midpoint of arc $XY$. Similar argument gives us $I$ is the midpoint of arc $ZT$, and therefore $XY \parallel ZT$. It suffices to prove that \[ AD + DT + XA = CD + DY + ZC \]However, by Pitot, $AB - BC = AD - CD$, and therefore it suffices to prove that \[ BX + DT = DY + BZ \Leftrightarrow DT - DY = BZ - BX \ (\star) \]By power of point, we have $DT \cdot DC = DA \cdot DY \implies DT = \frac{DA}{DC} \cdot DY$ and $BZ \cdot BC = BX \cdot BA \implies BZ = \frac{BA}{BC} \cdot BX$. Substituting this to $(\star)$, and notice that $DA - DC = BA - BC$ by Pitot, it suffices to prove that \[ \frac{DY}{DC} = \frac{BX}{BC} \]The easiest way to do this is by Law of Sine, which is \[ \frac{DY}{DC} = \frac{\sin \angle DCY}{\sin \angle DYC} = \frac{\sin \angle TCY}{\sin \angle AYC} = \frac{\sin \angle XCB}{\sin \angle AXC} = \frac{\sin \angle BCX}{\sin \angle BXC} = \frac{BX}{BC} \] [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10 cm); 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 = -46.0708539290498, xmax = 62.48458359050935, ymin = -64.89645919767288, ymax = 21.777384895169188; /* image dimensions */ /* draw figures */ draw(circle((-2.14696,-5.31757), 7.228701099824228)); draw(circle((7.726218568658888,-33.565788515407505), 29.923928624811893),red); draw((-22.19604779208755,-33.250383728857265)--(30.937206770698506,-14.679509606431964), blue); draw((36.05020676055918,-23.91212614054014)--(-20.444529007458954,-43.65788866084243), blue); draw((-22.19604779208755,-33.250383728857265)--(-5.196144423610607,5.785301945153355)); draw((36.05020676055918,-23.91212614054014)--(-5.196144423610607,5.785301945153355)); draw((-12.886450048922157,-11.873407438065486)--(30.937206770698506,-14.679509606431964)); draw((-20.444529007458954,-43.65788866084243)--(7.89779473280283,-3.6423517815907456)); /* dots and labels */ dot((-2.14696,-5.31757),dotstyle); label("$I$", (-1.597225730591665,-3.940591794674114), NE * labelscalefactor); dot((-12.886450048922157,-11.873407438065486),dotstyle); label("$A$", (-14.253845905941057,-10.476652168888434), NE * labelscalefactor); dot((-5.196144423610607,5.785301945153355),dotstyle); label("$B$", (-4.581079379689495,7.142293187689297), NE * labelscalefactor); dot((7.89779473280283,-3.6423517815907456),dotstyle); label("$C$", (8.491041368739094,-2.235532566618205), NE * labelscalefactor); dot((1.4190856331366912,-12.789414855299016), dotstyle); label("$D$", (1.9549809945247991,-12.613358320925707), NE * labelscalefactor); dot((-22.19604779208755,-33.250383728857265), dotstyle); label("$X$", (-25.93167166024852,-34.574069057596624), NE * labelscalefactor); dot((-20.444529007458954,-43.65788866084243), dotstyle); label("$T$", (-23.92661243219262,-44.5886009639414), NE * labelscalefactor); dot((30.937206770698506,-14.679509606431964), dotstyle); label("$Y$", (31.509340947493783,-13.602594086990935), NE * labelscalefactor); dot((36.05020676055918,-23.91212614054014), dotstyle); label("$Z$", (36.62451863166149,-22.83833157229378), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

dangerousliri wrote: popcorn1 wrote: Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\] This problem is a good one but it might be just a bit slightly harder for easy problem. It is very near to medium problem but not a medium problem (if you understand what I mean). I stand with this too. I find this as a surprisingly hard P4 geo. I don't particularly like how the problem camouflaged itself though---2020 P1 was better in this regard (I know it was easy, but hey, look at P1 this year...)

USJL wrote: I don't particularly like how the problem camouflaged itself though.. I actually really like it, the statement looks really attractive. It tastes like an IMO problem.

square_root_of_3 wrote: USJL wrote: I don't particularly like how the problem camouflaged itself though.. I actually really like it, the statement looks really attractive. It tastes like an IMO problem. Yeah it's pretty surprising that one could put the angle chasing and congruence together and somehow make it into a statement about quadrilateral having equal perimeter. Unpacking it makes it a bit underwhelming though.

Solved with Isaac Zhu, Jeffery Chen, Kevin Wu, Albert Wang, Nacho Cho. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -3.376164608838126, xmax = 8.801371249903822, ymin = -3.5496621121255836, ymax = 4.1243854587926085; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(1)); draw(circle((2.7769452165639037,0.17263506887001415), 2.782306166258649), linewidth(1)); draw((xmin, -16.0856379572264*xmin + 35.22882371093464)--(xmax, -16.0856379572264*xmax + 35.22882371093464), linewidth(1) + linetype("4 4")); /* line */ draw((xmin, -16.08563795722648*xmin + 61.315129460947006)--(xmax, -16.08563795722648*xmax + 61.315129460947006), linewidth(1) + linetype("4 4")); /* line */ draw((-0.4262175384204444,0.904620699488364)--(3.636555100008105,2.8188207107111056), linewidth(1)); draw((-0.4262175384204444,0.904620699488364)--(-2.120509992007347,0.10634451661904495), linewidth(1)); draw((-2.120509992007347,0.10634451661904495)--(-0.5145856140906708,-0.8574390041110372), linewidth(1)); draw((-0.5145856140906708,-0.8574390041110372)--(2.3502694893580354,-2.5767603967940733), linewidth(1)); draw((0.18909695235571686,1.1945298600474499)--(0.6847835074112292,0.7287465594962181), linewidth(1)); draw((0.6847835074112292,0.7287465594962181)--(3.957681843636246,-2.346707825674261), linewidth(1)); draw((2.0130261092774537,2.8480145186534602)--(0.9478178654136119,-0.31881231783415154), linewidth(1)); draw((0.9478178654136119,-0.31881231783415154)--(0.5514465847093561,-1.4972102660081708), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$I$", (0.05321584022310287,0.125598444604208), NE * labelscalefactor); dot((0.18909695235571686,1.1945298600474499),dotstyle); label("$A$", (0.2343906563999225,1.3291168663502313), dir(120) * labelscalefactor); dot((0.5514465847093561,-1.4972102660081708),dotstyle); label("$C$", (0.5967402887535618,-1.3626232597053909), dir(0) * labelscalefactor); dot((1.152383829468616,0.28935513536869273),linewidth(4pt) + dotstyle); label("$D$", (1.2049700287757419,0.39736066886943905), NE * labelscalefactor); dot((-2.120509992007347,0.10634451661904495),linewidth(4pt) + dotstyle); label("$B$", (-2.0691177207053557,0.21618585269261834), NE * labelscalefactor); dot((3.636555100008105,2.8188207107111056),linewidth(4pt) + dotstyle); label("$X$", (3.6896532220578395,2.920867037046585), NE * labelscalefactor); dot((2.0130261092774537,2.8480145186534602),linewidth(4pt) + dotstyle); label("$T$", (2.059079876466463,2.9467491536432733), NE * labelscalefactor); dot((2.3502694893580354,-2.5767603967940733),linewidth(4pt) + dotstyle); label("$Z$", (2.408488450521758,-2.4755542733630036), NE * labelscalefactor); dot((3.957681843636246,-2.346707825674261),linewidth(4pt) + dotstyle); label("$Y$", (4.013179679516446,-2.242615223992806), NE * labelscalefactor); dot((-0.4262175384204444,0.904620699488364),linewidth(4pt) + dotstyle); label("$E$", (-0.3738390836222577,1.005590408891623), dir(120) * labelscalefactor); dot((-0.5145856140906708,-0.8574390041110372),linewidth(4pt) + dotstyle); label("$F$", (-0.4644264917106675,-0.7543935196832069), NE * labelscalefactor); dot((0.9478178654136119,-0.31881231783415154),linewidth(4pt) + dotstyle); label("$G$", (0.9979130960022338,-0.21086907115274478), dir(0) * labelscalefactor); dot((0.6847835074112292,0.7287465594962181),linewidth(4pt) + dotstyle); label("$H$", (0.7390919300353486,0.8373566510131465), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $E,F,G,H$ are the tangency points as shown above. Note that $AI$ is an angle bisector of $\overline{AX}, \overline{AY},$ so $IX = IY$. Similarly, $IT = IZ$ so $TXYZ$ is a isosceles trapezoid. In particular $TX=YZ$, and it remains to show the quantities \begin{align*} TD + DA + AX = TG + EX \\ ZC + CD + DY = ZF + HY \end{align*}are equal. But this follows by symmetry: the lengths of the tangents to $\Gamma$ from $Y$ and $X$ are equal, and so are the tangents from $T$ and $Z$. The end.

Notice that $\triangle ITX=\triangle IZY, \triangle IMX=\triangle IQY, \triangle IPT=\triangle INZ$ Thus, $DA+DT+TX+XA=CD+DY+YZ+ZC \iff DA+DT+XA=CD+DY+ZC$ $DA+TP-DP+XM-AM=CD+YQ-DQ+ZN-NC \iff DA-DP-AM=CD-DQ-NC$. This is trivial as $RHS=LHS=0$.

Solved with Alex Zhao. Note since \(\overline{AI}\) bisects \(\angle XAY\) that \(IX=IY\), and analogously \(IT=IZ\). It follows that \(TX=YZ\). Now if \(E\), \(F\), \(G\), \(H\) are the feet from \(I\) to \(\overline{AB}\), \(\overline{BC}\), \(\overline{CD}\), \(\overline{DA}\), note the congruences \(\triangle IEX\cong\triangle IHY\) and \(\triangle IGT\cong\triangle IFZ\), from which we obtain \begin{align*} (AD-CD)+(XA-CZ)+(DT-DY) &=(AB-CB)+(XA-CZ)+(DT-DY)\\ &=(BX-BZ)+(DT-DY)\\ &=(EX-FZ)+(GY-HY)=0. \end{align*} [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair I,EE,F,G,H,A,B,C,D,O,X,Z,Y,T; I=origin; EE=dir(150); F=dir(45); G=dir(-70); H=dir(245); A=2/(1/H+1/EE); B=2/(1/EE+1/F); C=2/(1/F+1/G); D=2/(1/G+1/H); O=circumcenter(A,I,C); X=2*foot(O,A,B)-A; Z=2*foot(O,C,B)-C; Y=2*foot(O,A,D)-A; T=2*foot(O,C,D)-C; draw(EE--I--F,gray+dashed+linewidth(.3)); draw(H--I--G,gray+dashed+linewidth(.3)); draw(unitcircle); draw(circumcircle(A,I,C),dashed+linewidth(.4)); draw(X--B--Z); draw(A--Y); draw(C--T); clip( (-100,-3.5)--(100,-3.5)--(100,100)--(-100,100)--cycle); dot("\(E\)",EE,EE,gray); dot("\(F\)",F,F,gray); dot("\(G\)",G,SE,gray); dot("\(H\)",H,SW,gray); dot("\(A\)",A,NW); dot("\(B\)",B,N); dot("\(C\)",C,NE); dot("\(D\)",D,S); dot("\(I\)",I,N); dot("\(X\)",X,W); dot("\(Z\)",Z,E); dot("\(Y\)",Y,E); dot("\(T\)",T,W); [/asy][/asy]

I am extremely grateful to SatisfiedMagma who allowed me to use his Latex, for storage purposes as we had the same solution

A digram for above! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -11.082564640442863, xmax = 32.127932663797246, ymin = -8.306215585080063, ymax = 20.058957491936464; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); pen ttffqq = rgb(0.2,1,0); /* draw figures */ draw(circle((0,0), 4), linewidth(0.4) + ffxfqq); draw(circle((8.514218322195653,7.076662743701791), 11.071181925436376), linewidth(0.4) + blue); draw((-2.1440077320691855,4.081118285293722)--(3.769581411268206,3.864759987347469), linewidth(0.4) + ttffqq); draw((3.769581411268206,3.864759987347469)--(4.199660919624339,-3.1192022831620996), linewidth(0.4) + ttffqq); draw((4.199660919624339,-3.1192022831620996)--(-8.656519374133946,-6.151538640655174), linewidth(0.4) + ttffqq); draw((-8.656519374133946,-6.151538640655174)--(-2.1440077320691855,4.081118285293722), linewidth(0.4) + ttffqq); draw((2.9812613768336007,16.666107809339263)--(-2.1440077320691855,4.081118285293722), linewidth(0.4) + ttffqq); draw((0,0)--(-2.1440077320691855,4.081118285293722), linewidth(0.4) + ttffqq); draw((6.715004399887755,18.000668350066267)--(18.925047974899115,3.3102725397399757), linewidth(0.4) + red); draw((13.03183605295998,10.400633454143106)--(12.514221798481419,10.401298606223266), linewidth(0.4) + red); draw((13.03183605295998,10.400633454143106)--(13.125830576305455,10.909642283582976), linewidth(0.4) + red); draw((18.925047974899115,3.3102725397399757)--(4.199660919624339,-3.1192022831620996), linewidth(0.4) + ttffqq); draw((2.9812613768336007,16.666107809339263)--(16.930253971139187,-0.11648769138718018), linewidth(0.4) + red); draw((10.16756753955294,8.019973068216027)--(9.649953285074378,8.020638220296187), linewidth(0.4) + red); draw((10.16756753955294,8.019973068216027)--(10.261562062898411,8.528981897655894), linewidth(0.4) + red); draw((16.930253971139187,-0.11648769138718018)--(18.925047974899115,3.3102725397399757), linewidth(0.4) + ttffqq); draw((2.9812613768336007,16.666107809339263)--(6.715004399887755,18.000668350066267), linewidth(0.4) + ttffqq); draw((3.769581411268206,3.864759987347469)--(29.77557862599852,2.9132881895284855), linewidth(0.4) + ttffqq); draw((-2.1440077320691855,4.081118285293722)--(6.715004399887755,18.000668350066267), linewidth(0.4) + ttffqq); draw((4.199660919624339,-3.1192022831620996)--(29.77557862599852,2.9132881895284855), linewidth(0.4) + ttffqq); draw((3.769581411268206,3.864759987347469)--(2.9812613768336007,16.666107809339263), linewidth(0.4) + ttffqq); /* dots and labels */ dot((0.1462486698437907,3.997325521716854),dotstyle); label("$E$", (0.31652360546096975,4.418348038254454), NE * labelscalefactor); dot((3.769581411268206,3.864759987347469),dotstyle); label("$D$", (3.9394896371048236,4.285800500511385), NE * labelscalefactor); dot((4.199660919624339,-3.1192022831620996),dotstyle); label("$C$", (4.381314762915049,-2.6950364872901895), NE * labelscalefactor); dot((-8.656519374133946,-6.151538640655174),dotstyle); label("$B$", (-10.4,-6.58309759442018), NE * labelscalefactor); dot((-2.1440077320691855,4.081118285293722),linewidth(4pt) + dotstyle); label("$A$", (-3.8,4.19743547534934), NE * labelscalefactor); dot((0,0),dotstyle); label("$I$", (0.18397606771790193,0.4419219059624171), NE * labelscalefactor); dot((18.925047974899115,3.3102725397399757),linewidth(4pt) + dotstyle); label("$Y$", (19.580099090786824,3.755610349539114), NE * labelscalefactor); dot((2.9812613768336007,16.666107809339263),linewidth(4pt) + dotstyle); label("$T$", (3.1442044106464166,17.010364123845903), NE * labelscalefactor); dot((6.715004399887755,18.000668350066267),linewidth(4pt) + dotstyle); label("$X$", (6.899717980033338,18.33583950127658), NE * labelscalefactor); dot((16.930253971139187,-0.11648769138718018),linewidth(4pt) + dotstyle); label("$Z$", (16.97333084850649,-1.2370135721164426), NE * labelscalefactor); dot((29.77557862599852,2.9132881895284855),linewidth(4pt) + dotstyle); label("$K$", (29.962989547327137,3.2696027111478654), NE * labelscalefactor); dot((0.9182670419408749,-3.89317166840677),linewidth(4pt) + dotstyle); label("$F$", (0.7141662186901732,-5.213439704408479), NE * labelscalefactor); dot((-3.3745233312354657,2.1476946447173293),linewidth(4pt) + dotstyle); label("$G$", (-4.455187753289471,2.2092224092033224), NE * labelscalefactor); dot((3.9924370627734307,0.24585829209742088),linewidth(4pt) + dotstyle); label("$H$", (4.160402200009936,0.6186519562865076), NE * labelscalefactor); dot((8.514218322195653,7.07666274370179),linewidth(4pt) + dotstyle); label("$Q$", (8.711200995855265,7.422758893763992), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

popcorn1 wrote: Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\] Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland You can prove \(XYZT\) an isoceles trap and then pitot twice ftw..

Denote $AD$ tangents to the circle $I$ at $M$, $CD$ tangents to the same circle at $N$; $XB$ tangents at $F$ and $ZB$ tangents at $J$. We can get that $AD=AM+MD;CD=DN+CN$.Since $AM=AF,XA=XF-AM;ZC=ZJ-CN$ Same reason, we can get that $DT=TN-DM;DY=YM-DM$ We can find that $AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM$. Connect $IM,IT,IT,IZ,IX,IN,IJ,IF$ separately, we can create two pairs of congruent triangles. In $\triangle{XIF},\triangle{YIM}$, since $\widehat{AI}=\widehat{AI},\angle{FXI}=\angle{MYI}$ After getting that $\angle{IFX}=\angle{IMY};\angle{FXI}=\angle{MYI};IF=IM$, we can find that $\triangle{IFX}\cong \triangle{IMY}$. Getting that $YM=XF$, same reason, we can get that $ZJ=TN$. Now the only thing left is that we have to prove $TX=YZ$. Since $\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}$ we can subtract and get that $\widehat{XT}=\widehat{YZ}$,means $XT=YZ$ and we are done

Two pairs of congruent triangles and equal arcs

Beautiful.......

Let the foot from $I$ to line $AB$ be $P$ and the foot from $I$ to line $AD$ be $Q$. Let $r$ denote the radius of $\Gamma$. As $\angle IYA = \angle IXA$, we must have $\triangle IYQ\cong \triangle IXP$, meaning that $IY = IX$. Analogously, we have $IT=IZ$. Hence we must have $TX=YZ$, so the question reduces to demonstrating \[YD+DC+CZ = XA+AD+DT.\]Let the foot from $I$ to line $DC$ be $R$ and the foot from $I$ to line $BC$ be $S$. Observe that $YD+DC+CZ$ is equal to \[YQ - QD + DC + ZS - SC = YQ+ZS.\]We analogously have $XA+AD+DT=XP+TR$. As we ought to have $YQ+ZS = XP + TR$, we are done.

Let $\Gamma$ touch $AB,BC,CD,DA$ at $H,G,F,E$. Then $\angle HXI=\angle AXI=\angle AYI=\angle EYI \implies \triangle HXI \cong \triangle EYI\implies HX=EY (1), IX=IY (2)$ Similarly $\triangle IZG\cong \triangle ITF \implies FT=GZ (3), IT=IZ (4)$. Now using $(2),(4)$, we have that $\triangle ITX \cong \triangle IZY \implies TX=ZY (5)$. Now $(1)+(3)+(5)$ gives $HX+FT+TX=EY+GZ+ZY \implies (AX+AE)+(DE+DT)+TX=(DY+DF)+(CF+CZ)+ZY$ $ \implies AX+AD+DT+TX=DY+CD+CZ+YZ$ as desired.

Claim $: XT = YZ$. Proof $:$ Note that $\angle IXY = \angle IAY = \angle IAB = \angle IYX$ so $IX = IY$. with same approach $IZ = IT$. Now Let $\Gamma$ meet $AB,BC,CA,AD$ at $S,R,Q,P$. Note that $T,Z$ and $X,Y$ are equidistant from $I$ so they have same power w.r.t $\Gamma$ so $TQ = ZR$ and $XS = YP$. Now we want to prove $AP+PD+TQ-DQ+TX+XS-AS = CQ+QD + YP - DP + YZ + ZR - CR$ which is true. we're Done.

1 Year late Same length chase solution as a lot of ppl here, gotta say the synthetic solution is beautiful Claim: $TX=YZ$ Proof: Notice that $\angle IAB = \angle IAD \implies \angle IYX = \angle IXY \implies IX = IY$ and similarly $IZ=IT$ which implies that $XYZT$ is an isosceles trapezoid because the perp bisector of $XY$ and $ZT$ pass through $I$ which implies $TX=YZ \ \square$ It remains to prove that $$AD+DT+XA=CD+DY+ZC$$By pitot's theorem we know that $$AD+DT+XA=CD+DY+ZC \iff AB+XA+DT=BC+ZC+DY \iff BX+DT=BZ+DY$$Now using PoP and Pitot's theorem again we get that this condition is equivalent to $$BX-BZ=DY-DT \iff BZ\cdot(\frac{BC}{AB}-1)=DT\cdot(\frac{DC}{AD}-1) \iff BZ\cdot(\frac{BC-AB}{AB})=DT\cdot(\frac{DC-DA}{DA}) \iff \frac{BZ}{AB}=\frac{DT}{DA} \iff \dfrac{\sin \angle BAZ}{\sin \angle BZA} = \dfrac{\sin \angle DAT}{\sin \angle DTA} $$as $BC-AB=DC-DA$ This is direct as $\angle BAZ=\angle DAT$ and $\angle BZA=180^{\circ}-\angle DTA$ which finishes the proof $\blacksquare$

We have $$\angle{YCT} = \angle{YDT} - \angle{CYD} = \angle{ADC} + \angle{AIC} - 180^{\circ}$$$$\angle{ZAX} = \angle{ABZ} + \angle{AZC} = \angle{ABC} - \angle{AIC} + 180^{\circ}$$But $$\angle{ADC} + 2 \angle{AIC} - \angle{ABC} - 360^{\circ} = \angle{ADC} - 2 (\angle{IAC} + \angle{ICA}) - \angle{ABC} = \angle{ADC} + 2 (\angle{CAD} + \angle{ACD})$$$$- \angle{DAB} - \angle{DCB} - \angle{ABC} = 2 \angle{ADC} + 2 (\angle{CAD} + \angle{ACD}) - 360^{\circ} = 0$$then $\angle{YCT} = \angle{ZAX}$. So $YT = XZ,$ hence $TX = YZ$. It's easy to see that $\triangle ABC$ $\sim$ $\triangle ZBX,$ then we have $$\dfrac{AC}{ZX} = \dfrac{BX}{BC} = \dfrac{BZ}{AB} = \dfrac{BX - BZ}{BC - AB}$$Similarly, we have $\dfrac{AC}{YT} = \dfrac{DY - DT}{CD - DA}$. So $$\dfrac{BX - BZ}{BC - AB} = \dfrac{AC}{ZX} = \dfrac{AC}{YT} = \dfrac{DY - DT}{CD - DA}$$or $BX - BZ = DY - DT$. From this, we have $$DY - DT = BX - BZ = AB + AX - BC - CZ = DA - CD + AX - CZ$$Therefore, $DA + AX + DT = DY + CD + CZ$ or $DA + AX + DT + TX = DY + CD + CZ + YZ$

Claim. $|TX|=|YZ|.$ Proof. Note that $I$ is the midpoint of both arcs $XAY,TCZ$ therefore $XY\parallel TZ\text{ } \Box$ Considering $|AD|-|CD|=|AB|-|CB|$ we reduce the desired equality to proving $|BX|+|DT|=|BZ|+|DY|.$ $$|DY|-|DT|=\frac{|TZ|}{|AC|}(|DC|-|DA|)=\frac{|XY|}{|AC|}(|BC|-|BA|)=|BX|-|BZ|\text{ } \blacksquare$$

Since $\angle IAD=\angle IAB$, we have arc $XI$, which is inscribed by $\angle IAB$, is equal to arc $YI$, which is inscribed by $\angle IAD$. Similarly, $TI=IZ$. Therefore, $XT$ and $ZY$ are reflections across the diameter from $I$ of $\Omega$. This implies that $TX=ZY$. Since $AD-DC=AB-BC$, we just need to show that $XB+TD=BZ+DY$. This rearranges to $XB-ZB=DY-DT$. However, $XB\cdot AB=ZB\cdot CB$ and $DY\cdot DA=DT\cdot DC$. Additionally, $\angle DCY=\angle XCZ=\angle XAZ$ is a supplement to $\angle BAZ$, and $\angle DYC=\angle AZB$. Therefore, by the law of sines, $\tfrac{BZ}{BA}=\tfrac{DY}{DC}.$ We have \begin{align*} XB-ZB &= ZB\cdot \left(\frac{BC}{BA}-1\right)\\ &=\frac{BZ}{BA}\cdot (BC-BA) \\ &=\frac{DY}{DC}\cdot (DC-DA) \\ &=DY-DT\end{align*}as desired.

This is my new favorite geometry problem The below solution is very simple with virtually no length chasing. Let $I'$ be the antipode of $I$ WRT $\Omega$. The angle bisector of $\angle XAY$ is perpendicular to $\overline{AI}$, so the angle bisector of $\angle XAY$ passes through $I'$. Similarly, the angle bisector of $\angle TCZ$ passes through $I'$. Since arcs $XY$ and $TZ$ both share a midpoint of $I'$, it follows that $XT = YZ$, so it suffices to show that \[ XA + AD + DT = ZC + CD + DY.\]Letting $f(P)$ denote the length of the tangent from $P$ to $\Gamma$, we can rewrite the LHS and RHS of the above equation to \[ f(X) + f(T) = f(Y) + f(Z).\]Because $I$ is the midpoint of arcs $XY$ and $TZ$, $f(X) = f(Y)$ and $f(T) = f(Z)$, so the above equation is true.

I would not be surprised to hear that this problem came from an alien civilization. It's certainly not geometry. The internal bisectors of $\angle XAY$ and $\angle TCZ$ intersect at the antipode of $I$ with respect to $(AIC)$. Thus, arcs $XY$ and $TZ$ not containing $A,C$ share a midpoint, so $XYZT$ is an isosceles trapezoid and thus $XT=YZ$. Furthermore, since $AD-CD=AS-AR=AP-AQ$, it suffices to show that $DT+PX=DY+QZ$. Adding $BP=BQ$ to both sides of this equation, it suffices to show that $DT+BX=DY+BZ$. Note that $\triangle DTY \sim \triangle DAC$, with similarity ratio $TY/AC$, and $\triangle BAC \sim \triangle BZX$, with similarity ratio $ZX/AC$. But since $XYZT$ is an isosceles trapezoid, we have $TY/AC=ZX/AC$, so it suffices to show that $DA+BC=DC+BA$, which is just Pitot. $\blacksquare$ Remark: I think the solution above, as well as in #15, are sufficiently mindblowing to put this problem on the IMO, especially given that most people (myself included) apparently did something slightly more complicated.

Let $\Gamma$ meet $AB$, $BC$, $CD$, $DA$ at $E$, $F$, $G$, $H$ respectively.

Therefore we can compute \begin{align*} (AD+DT&+TX+XA)-(CD+DY+YZ+ZC)\\&=(AH+HD-DG+GT+TX+XE-EA)\\&\phantom{=\quad}-(CG+GD-DH+HY+YZ+ZF-FC)\\ &=(AH-EA)+(HD-DH)+(GD-DG)\\&\phantom{=\quad}+(GT-ZF)+(TX-YZ)+(XE-HY)+(FC-CG)\\ &=0+0+0+0+0+0+0=0 \end{align*}as desired.

let $AB$ be tangent to $\Gamma$ at $M$, $BC$ be tangent to $\Gamma$ at $N$, $CD$ be tangent to $\Gamma$ at $O$, $AD$ be tangent to $\Gamma$ at $P$ let $AM=AP=a$, $BM=BN=b$, $CN=CO=c$, $DO=DP=d$ we have $AXI=AYI=MXI=PYI$, $MI=PI$, and $XMI=YPI$ therefore, $MIX$ and $PIY$ are congruent and $MX=PY$ similarly, $NZ=OT$ we can also prove that $TAX=ZCY$ via angle chasing, and $TX=ZY$ therefore, $AD+DT+TX+XA=XM-a+ZN-d+a+d+TX=XM+ZN+TX=XM-d+ZN-c+d+c+YZ=CD+DY+YZ+ZC$

The key observation is that $IX=IY$ and $IZ=IT$. After that everything is just manipulation.