First part is easy as it has only angle chasing. And showing angle in the same segment.

Dear Mathlinkers, for the second part http://www.oei.es/oim/revistaoim/numero52/DuranSolucElem_a_probelem.pdf Sincerely Jean-Louis

This was the easiest P6 ever on RMO

I agree but my first part just got half of it wrong.

Giveaway This was stupidly way too easy for RMO.

Any one #P4

Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh*

it is one of the easiest question in the paper.

Vrangr wrote: Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh* I think this is so because EGMO is such a complete book; it's hard to create a geometry problem that is not doable by methods taught in EGMO.

I'm sure it is possible without lifting a problem directly from EGMO.

MathematicalPhysicist wrote: Vrangr wrote: Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh* I think this is so because EGMO is such a complete book; it's hard to create a geometry problem that is not doable by methods taught in EGMO. I'd make a large distinction between 'problems solvable by techniques in EGMO' and 'restatements of EGMO problems'.

In fact it is an old well known lemma It states that there is a spiral similarity at $B$ that maps $I$ to $F$ and $C$ to $Y$, with $\widehat{CYI}=90$... $\widehat{BYX}=\widehat{BYF}=\widehat{BCI} $ Hence $BXYC$ and $IDCY$, the conclusion follows and we are done.$\blacksquare$

This is just too well-known

Posting my solution from contest just for completeness sake (and also cause I have been using this link in the answers of some other questions too ): We have $$\angle YEC=\angle AEF=90^{\circ}-\frac{A}{2}=180^{\circ}-\left(90^{\circ}+\frac{A}{2} \right)=180^{\circ}-\angle BIC=\angle YIC$$which means that $CIEY$ is cyclic. Similarly, we get that $BIFX$ is also cyclic. Thus, $$\angle CYI=\angle CEI=90^{\circ}=\angle BFI=\angle BXI$$This means that $X$ and $Y$ lie on the circle with $BC$ as diameter, proving Part (i). As for Part (ii), notice that, as $D$ also lies on $\odot (CIE)$, and cause $ID=IE$, i.e. $I$ is the midpoint of minor arc $DE$ in $\odot (CDIEY)$, we get that $YI$ bisects $\angle DYE$, i.e. $\angle DYX$. Similarly, $XI$ bisects $\angle DXY$, giving that $I$ is the incenter of $\triangle DXY$. $\blacksquare$ REMARK: Part (ii) can also be proved in another manner. Let $H$ be the orthocenter of $\triangle BIC$. Then $I$ is the orthocenter of $\triangle BHC$, and $DXY$ is its orthic triangle, and so by a well known property, we get that $I$ is the incenter of $\triangle DXY$.

The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school

Wizard_32 wrote: The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school Or probably no one wanted to waste time on this problem

https://artofproblemsolving.com/community/c6h1712858p11053650 Part 1 of the problem was on FBH regional 2012 for grade 9

Seems really easy right now:

Wizard_32 wrote: The part 2 can also be proved using congruency, which I am astonished no one in the posts above has pointed out. Probably because congruency is not used outside school I can't even remember the last time I used congruent triangles :p

INMO 1989 P6

Nordic Math Contest 2015 P1 @below Let's hope for a well-known Lemma this year too...I am getting very scared

I am sorry for unnecessary bump, but is this problem a joke? Vrangr wrote: Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh* Also, could someone please tell which lemma is being referred to?

Delta0001 wrote: I am sorry for unnecessary bump, but is this problem a joke? Vrangr wrote: Just like last year, this year's second geometry problem was just a restatement of a lemma/problem in EGMO. *sigh* Also, could someone please tell which lemma is being referred to? I don't think that it's a Lemma, a non-tricky angle chasing

@Above, it's easy and I do know the solution, but I wanted to see if another problem from Chapter-4 appeared in the RMO

Finally solved it, phew! Angle-chased diagram attached for the sake of completeness, as well as reference. A bit too detailed solution compared to other solutions.. Let $\angle A=2\alpha, \angle B=2\beta, \angle C=2\gamma$. I guess the rest of angle chase is clear from figure. (i) $\angle XFB=\angle XIB\implies BIFX$ is cyclic$\implies \angle BFI=\angle BXI\implies\angle BXI=90^{\circ}$. Similarly, $\angle YEC=\angle YIC\implies YEIC$ is cyclic$\implies \angle IEC=\angle IYC\implies \angle IYC=90^{\circ}$ So, $\angle BXC=\angle BYC\implies BXYC$ is cyclic, as desired. (ii)As proved in (i), $BXYC$ ic cyclic $\implies \angle XCB=\angle XYB\implies \angle XYB=\gamma$. Similarly, $\angle YBC=\angle YXC\implies \angle YXC=\beta$. Also, as proved in (i), $\angle BXC=90^{\circ}=\angle BDI\implies BDIX$ is cyclic $\implies\angle DXI=\angle DBI=\beta$. Similarly, $\angle BYC=90^{\circ}=\angle CDI\implies CDIY$ is cyclic$\implies\angle IYD=\angle ICD=\gamma$. So, $IX$ bisects $\angle DXY$ and $IY$ bisects $\angle DYX$. Thus $I$ is also the incenter of $\triangle DXY$, as desired $\blacksquare$.

Trivial by angle chasing: Redefine $K=EF\cap BI$. Note that $\angle{EKI}=\angle{FKI}=180-\angle{B}/2-(90+\angle{A}/2)=\angle{C}/2$, meaning $EKCI$ is cyclic. Thus, since $\angle{IEC}=90$, we have $\angle{IKC}=90$ as well. Now let $M$ be the midpoint of $BC$ and $N$ be the midpoint of $AC$. Note that since $\triangle{KBC}$ is right, we have $KM=BM=CM$, so $\angle{KMB}=180-2\cdot (\angle{B}/2)=180-\angle{B}$, meaning $K, M, N$ are collinear.

By Iran lemma, $\angle BYC=\angle CXB=90$ and so, $BCXY$ is cyclic. Also, as $DIEYC, DIFXB$ are cyclic with $IF=ID=IE,$ by fact 5 we get $I$ is the incenter of $DXY$. $\blacksquare$

Ig a different solution than others, here we go [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.764483846876923, xmax = 5.176303477974182, ymin = -1.5715650474047884, ymax = 9.629055475231542; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffqqff = rgb(1,0,1); /* draw figures */ draw((-6.207733319596109,6.131918499823255)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); draw((-8.523569086328425,2.153812191539418)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); draw((-6.207733319596109,6.131918499823255)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); draw(circle((-5.8337845690589685,3.624772291402155), 1.5845331914338108), linewidth(0.8) + wvvxds); draw((-7.602495703688477,4.362399043957692)--(-3.3158178513850256,5.001770541894226), linewidth(0.8)); draw((-3.3158178513850256,5.001770541894226)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); draw((-7.602495703688477,4.362399043957692)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); draw(circle((-5.026798754007371,2.7813134577590994), 3.552627280775174), linewidth(0.8) + wvvxds); draw((-8.097069626094108,4.5686571203275985)--(-7.602495703688477,4.362399043957692), linewidth(0.8) + wrwrwr); draw((-8.097069626094108,4.5686571203275985)--(-2.611734153564187,5.386812176199431), linewidth(0.8)); draw((-2.611734153564187,5.386812176199431)--(-3.3158178513850256,5.001770541894226), linewidth(0.8)); draw((-2.611734153564187,5.386812176199431)--(-5.17857829741542,-0.7680700896066753), linewidth(0.8)); draw((-5.17857829741542,-0.7680700896066753)--(-8.097069626094108,4.5686571203275985), linewidth(0.8)); draw((-6.207733319596109,6.131918499823255)--(-5.17857829741542,-0.7680700896066753), linewidth(0.8)); draw((-7.602495703688477,4.362399043957692)--(-5.901480869656607,2.041685861152432), linewidth(0.8)); draw((-3.3158178513850256,5.001770541894226)--(-5.901480869656607,2.041685861152432), linewidth(0.8)); draw((-7.602495703688477,4.362399043957692)--(-8.523569086328425,2.153812191539418), linewidth(0.8)); draw((-3.3158178513850256,5.001770541894226)--(-1.5963622702582592,1.8575894042284895), linewidth(0.8)); draw(circle((-5.0599656782933415,2.0057007978839554), 3.4667687481403786), linewidth(0.8) + linetype("4 4") + ffqqff); /* dots and labels */ dot((-6.207733319596109,6.131918499823255),linewidth(4pt) + dotstyle); label("$A$", (-6.1588223129470405,6.229740513121389), NE * labelscalefactor); dot((-8.523569086328425,2.153812191539418),linewidth(4pt) + dotstyle); label("$B$", (-8.469867377115445,2.2557212228847003), NE * labelscalefactor); dot((-1.5963622702582592,1.8575894042284895),linewidth(4pt) + dotstyle); label("$C$", (-1.5489599362724982,1.9500274313280317), NE * labelscalefactor); dot((-5.8337845690589685,3.624772291402155),linewidth(4pt) + dotstyle); label("$I$", (-5.779762011416773,3.7230514223567086), NE * labelscalefactor); dot((-5.901480869656607,2.041685861152432),linewidth(4pt) + dotstyle); label("$D$", (-5.853128521390373,2.1334437062620326), NE * labelscalefactor); dot((-4.756624001282189,4.786869768373093),linewidth(4pt) + dotstyle); label("$E$", (-4.7037198651373044,4.884687830272049), NE * labelscalefactor); dot((-7.203178462372859,4.421958467718602),linewidth(4pt) + dotstyle); label("$F$", (-7.149270197590643,4.517855280404047), NE * labelscalefactor); dot((-3.3158178513850256,5.001770541894226),linewidth(4pt) + dotstyle); label("$X$", (-3.260845168989835,5.10478736019285), NE * labelscalefactor); dot((-7.602495703688477,4.362399043957692),linewidth(4pt) + dotstyle); label("$Y$", (-7.552786002445444,4.456716522092712), NE * labelscalefactor); dot((-8.097069626094108,4.5686571203275985),linewidth(4pt) + dotstyle); label("$Y_1$", (-8.054123820598377,4.664588300351247), NE * labelscalefactor); dot((-2.611734153564187,5.386812176199431),linewidth(4pt) + dotstyle); label("$X_1$", (-2.5638633242406335,5.483847661723119), NE * labelscalefactor); dot((-5.17857829741542,-0.7680700896066753),linewidth(4pt) + dotstyle); label("$Z_1$", (-5.131691173316638,-0.6667114243970499), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Notice that $XY\parallel X_{1}Y_{1}$ as $\Delta X_{1}Y_{1}Z_{1}$ has its orthocenter as $I\Rightarrow AI\perp X_{1}Y_{1}$ and $AI\perp \overline{YFEX}$. Then $\exists$ a homothety sending $XY\leftrightarrow X_{1}Y_{1}$, hence $(BY_{1}X_{1}C)\leftrightarrow (BYXC)$ and we complete our first part. Now for the second part, by Iran lemma or EGMO lemma 1.45 we know that $\angle CXB = 90^{\circ}. \angle XYC = \angle XBC = \angle DYC $( because $\square YIDB$ is cyclic). Hence $CY$ bisects $\angle DYX$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(40cm); 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 = -10.222, xmax = 13.736, ymin = -4.304, ymax = 10.216; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-4.216,8.17)--(-8.044,2.032)--(9.49,0.118)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.216,8.17)--(-8.044,2.032), linewidth(2) + rvwvcq); draw((-8.044,2.032)--(9.49,0.118), linewidth(2) + rvwvcq); draw((9.49,0.118)--(-4.216,8.17), linewidth(2) + rvwvcq); draw((xmin, -0.3313292275453181*xmin + 3.2623143694050687)--(xmax, -0.3313292275453181*xmax + 3.2623143694050687), linewidth(2)); /* line */ draw((xmin, 0.48579311234662764*xmin + 5.9397197957162735)--(xmax, 0.48579311234662764*xmax + 5.9397197957162735), linewidth(2)); /* line */ draw(circle((-3.2766273734056255,4.3479567859894), 2.819610910551786), linewidth(2)); draw((-5.669096776867536,5.840034478471021)--(-1.8483924116576476,6.7790780460139635), linewidth(2)); draw((-5.669096776867536,5.840034478471021)--(-9.342,5.134), linewidth(2)); draw((-1.8483924116576476,6.7790780460139635)--(7.862,9.6), linewidth(2)); draw((-8.044,2.032)--(-7.004925504045343,5.58325092567291), linewidth(2)); draw((9.49,0.118)--(7.047687174717885,9.363437683167886), linewidth(2)); draw((-4.216,8.17)--(-3.276627373405626,4.3479567859894), linewidth(2)); draw((-3.276627373405626,4.3479567859894)--(-3.5825967298465926,1.5449961298577837), linewidth(2)); draw((-3.276627373405626,4.3479567859894)--(-5.669096776867536,5.840034478471021), linewidth(2)); draw((-3.276627373405626,4.3479567859894)--(-1.8483924116576476,6.7790780460139635), linewidth(2)); /* dots and labels */ dot((-4.216,8.17),dotstyle); label("$A$", (-4.128,8.39), NE * labelscalefactor); dot((-8.044,2.032),dotstyle); label("$B$", (-7.956,2.252), NE * labelscalefactor); dot((9.49,0.118),dotstyle); label("$C$", (9.578,0.338), NE * labelscalefactor); dot((-3.276627373405626,4.3479567859894),dotstyle); label("$I$", (-3.182,4.562), NE * labelscalefactor); dot((-3.5825967298465926,1.5449961298577837),dotstyle); label("$D$", (-3.49,1.768), NE * labelscalefactor); dot((-5.669096776867536,5.840034478471021),dotstyle); label("$F$", (-5.58,6.058), NE * labelscalefactor); dot((-1.8483924116576476,6.7790780460139635),dotstyle); label("$E$", (-1.752,7.004), NE * labelscalefactor); dot((-9.342,5.134),dotstyle); label("$H$", (-9.254,5.354), NE * labelscalefactor); dot((7.862,9.6),dotstyle); label("$I_{1}$", (7.95,9.534), NE * labelscalefactor); dot((-7.004925504045343,5.58325092567291),dotstyle); label("$X$", (-6.922,5.794), NE * labelscalefactor); dot((7.047687174717885,9.363437683167886),dotstyle); label("$Y$", (7.136,9.578), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I) denote $\angle{AIF}=A, \angle{IAE}=A , \angle{IBD}=\angle{IBE}=B , \angle{ICD}=\angle{ICE}=C$ now we notice that $\angle{BIC}=180^{\circ}-(B+C) \implies \angle{BIX}=B+C$ now in $\triangle{AIB}=180^{\circ}-(B+A)$ , so we get $\angle{IXB}=90^{\circ}-A$ , also we have $\angle{AFE}=90^{\circ}-A=\angle{BFX}$ , hence $X,F,I,B$ are concyclic points, hence $\angle{BXI}=90^{\circ}$ similarly, we can prove that points $E,I,C,Y$ are concyclic points , hence $\angle{IYC}=90^{\circ}$ hence we have points $B,X,Y,C$ concyclic $\blacksquare$ II) we notice that $\angle{XBI}=A$ and we also quadrilateral $XBDI$ to be cyclic quadrilateral hence we have $\angle{XDI}=A$ and similarly $\angle{XDY}=A$ , by using the fact that quadrilateral $XBYC$ is also cyclic , it is not hard to observe that $\angle{IXD}=\angle{IXY}$ hence we have $I$ as the incenter of triangle $DXY$ $\blacksquare$

Amazing problem , we have to prove that BXFID is cyclic with BI diameter and CDIEY is cyclic with CI diameter, if we think about this configuration as its also come in 1989 INMO as P6 DI is radical axis and BC is radical axis of BCXY so D is radical center........

Part 1: This is just Iran Lemma. $\angle BXC=\angle CYB=90^{\circ}$. Part 2: Note that points $I,D,C,X,E$ are concyclic, So $\angle ICD=\angle IXD$ and $\angle ICE=\angle IXE$ $\implies$ $XI$ is the angle bisector of $\angle YXD$. Similarly chase angles for the other side and we are done.

Let $P= AB \cap CX$, $Q= AC \cap BY$. As $I$ is the incentre of $\Delta ABC$, $\angle ACI= \angle BCI = c$ (say). $\angle CBI = \angle ABI=b$ (say). $\therefore \angle A=2a$, $\angle B=2b$, $\angle C=2c$. Then, by angle chasing in $\Delta BQC$, we get that $\angle BQC= 90^{\circ}+a-c$. As $AE$ and $AF$ are tangents to the incircle of $\Delta ABC$, we get that $\angle AFE= \angle AEF= 90^{\circ}-a$. $\angle AEF = \angle YEC$ (vertically opposite). Thus, by angle chasing in $\Delta YEQ$, we get that $\angle EYB= \angle EYQ= c \Rightarrow BCYX$ is cyclic. Part (i) proved. Then, we have that $\angle CBI= \angle CXY=b$, as $BCYX$ is cyclic. Also, $\angle EYI = \angle ECI= c \Rightarrow EICY$ is cyclic $\Rightarrow \angle IEC = \angle IYC = 90^{\circ}$. Then, $\angle IYC + \angle IDC = 180^{\circ} \Rightarrow IDCY$ is cyclic $\Rightarrow \angle DCI = \angle DYI=c$. Thus, $YI$ bisects $\angle XYD$. Similarly, $BDIX$ is cyclic and $XI$ bisects $\angle DXY$. Then $I$ is the incentre of $\Delta DYX$. Proved