Tangential quadrilateral and 8 lengths

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popcorn1

1098 posts

popcorn1

#1 h Jul 20, 2021, 8:44 PM • 9 Y

Y by centslordm, Aritra12, HWenslawski, megarnie, deplasmanyollari, Rounak_iitr, cubres, Mathandski, ehuseyinyigit

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

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timon92

224 posts

timon92

#2 h Jul 20, 2021, 8:47 PM • 35 Y

Y by 62861, tapir1729, Kimchiks926, insertionsort, centslordm, TheMath_boy, tigerzhang, geometrylover123, arqady, DaRKst0Rm, Aritra12, khina, AllanTian, Offset, V880, pog, megarnie, erkhes_mng, pikapika007, opIst_w, Quidditch, Gnisrepfa2, Aryan-23, CyclicISLscelesTrapezoid, nguyenducmanh2705, ihatemath123, Tellocan, ljwn357, Polkishtrn_Kazakhstan, Sedro, arceuspokemon, Funcshun840, cubres, Mathandski, ehuseyinyigit

This problem was proposed by Burii and me

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adj0109

17 posts

adj0109

#3 h Jul 20, 2021, 8:57 PM • 6 Y

Y by Com10atorics, centslordm, Executioner230607, Elnuramrv, megarnie, cubres

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 $\frac{XB}{BC} = \frac{DY}{DC}$

the manipulation

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

This post has been edited 3 times. Last edited by adj0109, Jul 20, 2021, 9:19 PM
Reason: typos :'

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albind

28 posts

albind

#4 h Jul 20, 2021, 9:34 PM • 7 Y

Y by Leartia, Hamel, centslordm, Executioner230607, Elnuramrv, megarnie, cubres

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.

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dangerousliri

928 posts

dangerousliri

#5 h Jul 20, 2021, 9:56 PM • 3 Y

Y by centslordm, primesarespecial, cubres

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).

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albind

28 posts

albind

#6 h Jul 20, 2021, 10:07 PM • 4 Y

Y by centslordm, Leartia, Elnuramrv, cubres

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:
$\frac{DT}{DA}=\frac{sin\angle DAT}{sin\angle DTA}=\frac{sin\angle XAZ}{sin\angle AZC}=\frac{sin\angle ZAB}{sin\angle AZB}=\frac{BZ}{BA}$ .

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lahmacun

259 posts

lahmacun

#7 h Jul 20, 2021, 10:18 PM • 4 Y

Y by centslordm, CROWmatician, ZHEKSHEN, cubres

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 $VT=WZ$ , $TX=ZY$ , $XU=YS$ .

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MarkBcc168

1595 posts

MarkBcc168

#8 h Jul 20, 2021, 10:33 PM • 5 Y

Y by Kimchiks926, Elnuramrv, Steff9, leonhard8128, cubres

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.

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childishgambino

8 posts

childishgambino

#9 h Jul 20, 2021, 11:54 PM • 14 Y

Y by albind, Kagebaka, samrocksnature, anyone__42, steppewolf, Executioner230607, edfearay123, hakN, megarnie, mc_mojsije, 507219, howcanicreateacccount, sabkx, cubres

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

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IndoMathXdZ

691 posts

IndoMathXdZ

#10 h Jul 21, 2021, 12:18 AM • 4 Y

Y by Richangles, Elnuramrv, arceuspokemon, cubres

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
$\frac{DY}{DC} = \frac{\sin \angle DCY}{\sin \angle DYC} = \frac{\sin \angle TCY}{\sin \angle AYC} = \frac{\sin \angle XCB}{\sin \angle AXC} = \frac{\sin \angle BCX}{\sin \angle BXC} = \frac{BX}{BC}$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -46.0708539290498, xmax = 62.48458359050935, ymin = -64.89645919767288, ymax = 21.777384895169188; /* image dimensions */ /* draw figures */ draw(circle((-2.14696,-5.31757), 7.228701099824228)); draw(circle((7.726218568658888,-33.565788515407505), 29.923928624811893),red); draw((-22.19604779208755,-33.250383728857265)--(30.937206770698506,-14.679509606431964), blue); draw((36.05020676055918,-23.91212614054014)--(-20.444529007458954,-43.65788866084243), blue); draw((-22.19604779208755,-33.250383728857265)--(-5.196144423610607,5.785301945153355)); draw((36.05020676055918,-23.91212614054014)--(-5.196144423610607,5.785301945153355)); draw((-12.886450048922157,-11.873407438065486)--(30.937206770698506,-14.679509606431964)); draw((-20.444529007458954,-43.65788866084243)--(7.89779473280283,-3.6423517815907456)); /* dots and labels */ dot((-2.14696,-5.31757),dotstyle); label("$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$", (36.62451863166149,-22.83833157229378), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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USJL

535 posts

USJL

#11 h Jul 21, 2021, 12:33 AM • 6 Y

Y by albind, TheMath_boy, Richangles, primesarespecial, sabkx, cubres

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...)

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hukilau17

283 posts

hukilau17

#12 h Jul 21, 2021, 1:39 AM • 2 Y

Y by IAmTheHazard, cubres

complex bash

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square_root_of_3

78 posts

square_root_of_3

#13 h Jul 21, 2021, 1:46 AM • 1 Y

Y by cubres

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.

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USJL

535 posts

USJL

#14 h Jul 21, 2021, 2:07 AM • 3 Y

Y by MassimilianoF, L567, cubres

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.

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VulcanForge

626 posts

VulcanForge

#15 h Jul 21, 2021, 3:28 AM • 10 Y

Y by Eyed, tree_3, Zivugumiri, oolite, Steff9, Illuzion, babygrogu, IAmTheHazard, cubres, RecursiveCo

Solved with Isaac Zhu, Jeffery Chen, Kevin Wu, Albert Wang, Nacho Cho.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.376164608838126, xmax = 8.801371249903822, ymin = -3.5496621121255836, ymax = 4.1243854587926085; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(1)); draw(circle((2.7769452165639037,0.17263506887001415), 2.782306166258649), linewidth(1)); draw((xmin, -16.0856379572264*xmin + 35.22882371093464)--(xmax, -16.0856379572264*xmax + 35.22882371093464), linewidth(1) + linetype("4 4")); /* line */ draw((xmin, -16.08563795722648*xmin + 61.315129460947006)--(xmax, -16.08563795722648*xmax + 61.315129460947006), linewidth(1) + linetype("4 4")); /* line */ draw((-0.4262175384204444,0.904620699488364)--(3.636555100008105,2.8188207107111056), linewidth(1)); draw((-0.4262175384204444,0.904620699488364)--(-2.120509992007347,0.10634451661904495), linewidth(1)); draw((-2.120509992007347,0.10634451661904495)--(-0.5145856140906708,-0.8574390041110372), linewidth(1)); draw((-0.5145856140906708,-0.8574390041110372)--(2.3502694893580354,-2.5767603967940733), linewidth(1)); draw((0.18909695235571686,1.1945298600474499)--(0.6847835074112292,0.7287465594962181), linewidth(1)); draw((0.6847835074112292,0.7287465594962181)--(3.957681843636246,-2.346707825674261), linewidth(1)); draw((2.0130261092774537,2.8480145186534602)--(0.9478178654136119,-0.31881231783415154), linewidth(1)); draw((0.9478178654136119,-0.31881231783415154)--(0.5514465847093561,-1.4972102660081708), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$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$", (0.7390919300353486,0.8373566510131465), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
$E,F,G,H$ 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.

This post has been edited 3 times. Last edited by VulcanForge, Jul 21, 2021, 6:05 PM

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GeronimoStilton

1521 posts

GeronimoStilton

#63 h Mar 19, 2022, 8:05 PM

Y by

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.

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Wizard0001

336 posts

Wizard0001

#64 h May 25, 2022, 7:48 AM

Y by

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.

This post has been edited 1 time. Last edited by Wizard0001, May 25, 2022, 7:49 AM

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#65 h Jun 1, 2022, 3:18 AM

Y by

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.

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BVKRB-

322 posts

BVKRB-

#66 h Jun 5, 2022, 7:56 AM • 1 Y

Y by Mango247

1 Year late :blush:

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 $\blacksquare$

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khanhnx

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khanhnx

#67 h Jun 30, 2022, 11:46 AM

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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 $DA + AX + DT + TX = DY + CD + CZ + YZ$

This post has been edited 2 times. Last edited by khanhnx, Jun 30, 2022, 11:47 AM

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JAnatolGT_00

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JAnatolGT_00

#68 h Jul 14, 2022, 2:49 PM • 1 Y

Y by PRMOisTheHardestExam

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 $|BX|+|DT|=|BZ|+|DY|.$ $|DY|-|DT|=\frac{|TZ|}{|AC|}(|DC|-|DA|)=\frac{|XY|}{|AC|}(|BC|-|BA|)=|BX|-|BZ|\text{ } \blacksquare$

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awesomeming327.

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awesomeming327.

#69 h Feb 11, 2023, 4:02 AM

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Diagram

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 post has been edited 3 times. Last edited by awesomeming327., Feb 11, 2023, 4:03 AM

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ihatemath123

3443 posts

ihatemath123

#70 h Nov 23, 2023, 6:03 PM • 1 Y

Y by IAmTheHazard

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.

This post has been edited 1 time. Last edited by ihatemath123, Nov 23, 2023, 6:08 PM

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IAmTheHazard

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IAmTheHazard

#71 h Dec 29, 2023, 3:02 AM

Y by

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.

This post has been edited 1 time. Last edited by IAmTheHazard, Dec 29, 2023, 3:11 AM

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jp62

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jp62

#72 h Feb 11, 2024, 10:27 AM

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Let $\Gamma$ meet $AB$ , $BC$ , $CD$ , $DA$ at $E$ , $F$ , $G$ , $H$ respectively.
IX=IY
XE=YH
Apply symmetry

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.

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AngeloChu

470 posts

AngeloChu

#73 h Apr 4, 2024, 3:50 PM

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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, $AD+DT+TX+XA=XM-a+ZN-d+a+d+TX=XM+ZN+TX=XM-d+ZN-c+d+c+YZ=CD+DY+YZ+ZC$

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sami1618

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sami1618

#74 h May 16, 2024, 9:37 PM

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

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cursed_tangent1434

595 posts

cursed_tangent1434

#75 h Mar 3, 2025, 3:11 PM

Y by

Honestly I found this problem a pretty nice one, it was good to have a slightly different sort of geometry problem for a change. We start off with the following key observation.

Claim : We have the equal pairs of line segments $TX=YZ$ .

Proof : Let $P$ be a point on the extension of ray $AI$ beyond $A$ . Observe that,
$\measuredangle XAT = \measuredangle XAP + \measuredangle PAT = \measuredangle BAI + \measuredangle ICT = \measuredangle IAD + \measuredangle BCI = \measuredangle IAY + \measuredangle ZAI = \measuredangle ZAY$ Thus, arcs $TX$ and $YZ$ subtend equal angles on $\Omega$ which is sufficient to imply the claim.

Now, using the above claim and the fact that by Pitot's Theorem on tangential quadrilateral $ABCD$ we have $AB+CD = BC +DA$ it suffices to show that,
$BX+DT=DY+BZ$ However note that by Power of a Point at points $B$ and $D$ we can conclude,
$BX = \frac{BC}{BA}\cdot BZ \text{ and }DT = \frac{DA}{DC}\cdot DY$ Thus the desired rewrites to,
$\begin{align*} BX+DT &= DY +BZ\\ \frac{BC}{BA}\cdot BZ + \frac{DA}{DC}\cdot DY &= DY+BZ\\ \frac{(BC-BA)BZ}{BA} &= \frac{(DC-DA)DY}{DC}\\ \frac{BZ}{BA} &= \frac{DY}{DC} \end{align*}$ But this is clear since $\measuredangle AZC = \measuredangle AYC = \measuredangle DYC$ and
$\measuredangle BAZ = \measuredangle XAZ = \measuredangle XCY + \measuredangle YCZ = \measuredangle XCY + \measuredangle TCX = \measuredangle TCY = \measuredangle DCY$
Thus applying the Law of Sines we observe,
$\frac{BZ}{BA}= \frac{\sin \angle BAZ }{\sin \angle BZA} = \frac{\sin \angle TAD}{\sin \angle CYD}=\frac{DY}{DC}$ which finishes the proof.

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hgomamogh

39 posts

hgomamogh

#76 h Mar 12, 2025, 2:32 AM • 1 Y

Y by ihatemath123

Pretty neat problem.

Let $E$ be the tangent of the incircle of $\Gamma$ to $AB$ , and define $F$ , $G$ , and $H$ to be the tangents to $BC$ , $CD$ , and $DA$ . Additionally, let $L$ be the midpoint of the arc $AC$ (not containing $I$ ) of circle $(ACI)$ . Then, $L$ is the antipode of $I$ on $(ACI)$ .

We first claim that the angle bisector of $\angle XAY$ passes through $L$ . Observe that this bisector is clearly perpendicular to the external bisector of this angle, i.e. $AI$ . The perpendicular to $AI$ through $A$ , however, is clearly $AL$ , which completes the proof of the claim. Similarly, the angle bisector of $\angle TCZ$ passes through $L$ .

We will now proceed to the main part of the proof, in which we will show that $\begin{align*} AD + DT + TX + XA - (CD + DY + YZ + ZC) = 0. \end{align*}$
First, we claim that $TX - YZ = 0$ . To see this, observe that by our previous claim, $LX = LY$ and $LZ = LT$ . Hence, $XYZT$ is an isosceles trapezoid, and thus $TX = YZ$ .

Next, we claim that $AD + XA - (CD + ZC) = EX - FZ$ . This follows by a simple length chase. $\begin{align*} &~AD + XA - (CD + ZC)\\ = &~AH + HD + XA - (CG + GD + ZC)\\ = &~AH + AX - (GC + CZ)\\ = &~AE + AX - (FC + CZ)\\ = &~EX - FZ, \end{align*}$
as desired.

Finally, we will show that $DT - DY = FZ - EX$ . Observe that $I$ is clearly the arc midpoint of arc $TIZ$ (since $L$ , $I$ 's antipode, is the midpoint of arc $TZ$ not containing $I$ ). Hence, $IT = TZ$ . Furthermore, $IG = IF$ , and $\angle IGT = \angle IFZ$ . Hence, we conclude that triangles $\triangle IGT$ and $\triangle IFZ$ are congruent. Therefore, $GT = FZ$ . Hence, $\begin{align*} DT = GT - DG = FZ - DG. \end{align*}$
Similarly, we obtain $DY = EX - HD$ .

Finally, $\begin{align*} DT - DY =&~(FZ - DG) - (EX - HD)\\ =&~(FZ - EX) - (DG - HD)\\ =&~FZ - EX. \end{align*}$
This completes the proof of the claim. Finally, adding up our equalities, we obtain $\begin{align*} AD + DT + TX + XA - (CD + DY + YZ + ZC) = EX - FZ + FZ - EX = 0, \end{align*}$
which completes the proof.

This post has been edited 2 times. Last edited by hgomamogh, Mar 12, 2025, 3:32 AM

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Ilikeminecraft

351 posts

Ilikeminecraft

#77 h Mar 13, 2025, 3:14 PM

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Observe that $AI$ bisects $\angle DAB,$ and thus is an external angle bisector of $\angle XAY.$ Thus, $I$ is the midpoint of arc $XY.$ Similarly, $I$ is the midpoint of arc $ZT,$ which implies that $TX = ZY.$
It suffices to prove that $DT + AX + AD = DC + YD + CZ.$ Add $AD + BC = AB + CD$ to opposite sides, and we get $AX + DT + AB = YD + CZ + BC,$ or $XB + DT = DY + BZ.$

Note that by POP, we get:
$\begin{align*} BA\cdot BX & = BC\cdot BZ \\ \implies BA\cdot BX - BA\cdot BZ = BA(BX-BZ)& = BC\cdot BZ - BZ \cdot BA \\ & = BZ(BC-BA) \end{align*}$ while
$\begin{align*} DA\cdot DY & = DC\cdot DT \\ \implies DA\cdot DY - AD\cdot DT = AD(DY-DT)& = DC\cdot DT - AD \cdot DT \\ & = DT(DC-AD) \\ & = DT(BC-BA) \end{align*}$ where the last step is by Pitot’s.
Thus, it suffices to show that $\frac{DT}{AD} = \frac{BZ}{BA}.$ Note that $\frac{DT}{AD} = \frac{DY}{CD}$ by POP.
However, observe that $\angle AZB = \angle AYC$ while $\angle BAZ = 180- \angle XAZ = 180-\angle TCY,$ where the last step follows from the fact that $TX = YZ.$ Simple LOS finishes.

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