2019 Serbia MO Day 1 P3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

XbenX

590 posts

XbenX

#1 h Apr 7, 2019, 10:04 AM • 2 Y

Y by Adventure10, Mango247

Let $k$ be the circle inscribed in convex quadrilateral $ABCD$ . Lines $AD$ and $BC$ meet at $P$ ,and circumcircles of $\triangle PAB$ and $\triangle PCD$ meet in $X$ . Prove that tangents from $X$ to $k$ form equal angles with lines $AX$ and $CX$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rmtf1111

698 posts

rmtf1111

#2 h Apr 7, 2019, 10:35 AM • 3 Y

Y by danepale, Adventure10, Mango247

not only it trivially follows from desargues, but it was already posted before https://artofproblemsolving.com/community/q1h1435358p8912832

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MarkBcc168

1595 posts

MarkBcc168

#3 h Apr 7, 2019, 10:52 AM • 2 Y

Y by Adventure10, Mango247

As @above has said, it's trivial by Desargues Involution Theorem.

Let the two tangents be $XK_1, XK_2$ . Then there exists involution swapping $(XA, XC)$ , $(XB, XD)$ , $(XK_1,XK_2)$ . However, as $\triangle XAD\stackrel{+}{\sim} \triangle XBC$ , we get $XB, XD$ are isogonal w.r.t $\angle AXC$ . Hence this involution is isogonality w.r.t. $\angle AXC$ or $XK_1,XK_2$ are isogonal w.r.t $\angle AXC$ so we are done.

EDIT : I found a quick (but not synthetic) solution.

Let $I$ be the incenter. Then it suffices to show that $XI$ bisects $\angle AXC$ . Invert around $X$ with power $XA\cdot XC$ $=XB\cdot XD$ , followed by reflection across the angle bisector of $\angle AXC$ . Clearly it maps $A\to C$ , $B\to D$ . Now by some long angle bashing, one can show that this inversion fixes $I$ so we are done.

This post has been edited 1 time. Last edited by MarkBcc168, Apr 7, 2019, 11:03 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

IMO2021

34 posts

IMO2021

#4 h Apr 7, 2019, 11:02 AM • 5 Y

Y by me9hanics, MilosMilicevonAoPS, khan.academy, Adventure10, Mango247

First of all, what is the point in disrespecting the contest that way? I respect those advanced techniques, but somewhere they can't be used without proof. Then, who cares if it was posted before? AoPS is not the place where we spend all time of our lives...

This post has been edited 1 time. Last edited by IMO2021, Apr 7, 2019, 11:11 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rmtf1111

698 posts

rmtf1111

#5 h Apr 7, 2019, 11:06 AM • 17 Y

Y by MathKnight16, math_pi_rate, Pluto1708, GianDR, tenplusten, khan.academy, danepale, mira74, AlastorMoody, DanDumitrescu, Kayak, Andrei246, Kamran011, KST2003, Quidditch, Adventure10, Mango247

IMO2021 wrote:

AoPS is not the place where we spend all time of our lives...

speak for yourself

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

me9hanics

375 posts

me9hanics

#7 h Apr 7, 2019, 11:17 AM • 4 Y

Y by danepale, HECAM-CA-CEBEPA, Adventure10, Mango247

IMO2021 wrote:

First of all, what is the point in disrespecting the contest that way? I respect those advanced techniques, but somewhere they can't be used without proof. Then, who cares if it was posted before? AoPS is not the place where we spend all time of our lives...

disagree, one problem can change the whole selection of the team for the IMO, and usually other national olympiads (other than China, USA, Russia etc.) are spreaded with problems that can be solved by older methods or directly by older ShortList problems, look at Hungary TST1, these problems are always solved by students who just read problems and solutions but spend no time in developing their problem-solving technique; when they face IMO problems they fail to live up to their expected results whereas students who finished below them in the team selection tests would've done a better job. and this is VERY common

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

me9hanics

375 posts

me9hanics

#8 h Apr 7, 2019, 11:20 AM • 4 Y

Y by danepale, ThE-dArK-lOrD, Adventure10, Mango247

also don't get why you can't "disrespect" a contest, do I have to respect it anyways if it's rigged and badly designed? totally disagree they'll just keep on designing it similarly and not the correct team will go to the IMO from their nation

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

IMO2021

34 posts

IMO2021

#9 h Apr 7, 2019, 11:37 AM • 3 Y

Y by MilosMilicevonAoPS, Adventure10, Mango247

I agree with you about the structure, but it turned out that nobody had seen this problem before. Also, if the jury knew it had been posted, they would simply select another problem. Also, there are two main contests for IMO selection, so that adventage of solving problem before would be a bit reduced. I totally agree it can make changes, but this time it didn't.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

math_pi_rate

1218 posts

math_pi_rate

#10 h Apr 7, 2019, 11:48 AM • 2 Y

Y by Adventure10, Mango247

XbenX wrote:

Let $k$ be the circle inscribed in convex quadrilateral $ABCD$ . Lines $AD$ and $BC$ meet at $P$ ,and circumcircles of $\triangle PAB$ and $\triangle PCD$ meet in $X$ . Prove that tangents from $X$ to $k$ form equal angles with lines $AX$ and $CX$ .

Btw I recently used this problem in my solution to Taiwan TST3 2014 P3.

MarkBcc168 wrote:

Now by some long angle bashing, one can show that this inversion fixes $I$ so we are done.

Hey pls share this angle chase part if you've got some time. I tried to prove this thing without DDIT, but unfortunately failed (I also wrote about this in my remarks to Taiwan TST3 2014 P3).

This post has been edited 1 time. Last edited by math_pi_rate, Apr 7, 2019, 11:52 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MilosMilicevonAoPS

3 posts

MilosMilicevonAoPS

#11 h Apr 7, 2019, 12:06 PM • 6 Y

Y by XbenX, IMO2021, tenplusten, p_square, gvole, Adventure10

On the actual olympiad, there were actually 5 completely different solutions, only one of which involved dual Desargues. The rest were pretty nice,(mostly) elementary solutions. I think this is a big plus for this problem, as it allowed many different approaches. The national PSC probably didn't know this theorem, as such overkill theorems aren't a focal point in our olympiad practice (and the national PSC has a lot of previous IMO contestants, so you can't say it's a matter of incompetence). Lastly, I would like to state that I think that if the PSC doesn't know (or doesn't notice an application of) a theorem, and a student does, that student deserves full marks (take, for example last years IMO problem 3, which was a "well known" problem, yet a committee of 100 team leaders didn't recognize it (as they wouldn't pick an actual well known problem))

This post has been edited 1 time. Last edited by MilosMilicevonAoPS, Apr 7, 2019, 1:18 PM
Reason: typo

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

FISHMJ25

293 posts

FISHMJ25

#15 h Apr 7, 2019, 5:40 PM • 2 Y

Y by Adventure10, Mango247

We will complex bash this. Let $k$ be unit circle and its center be $I$ .Let $x,y,z,t,$ be points where $k$ touches sides $AD,AB,BC,CD$ respectively.Now we have $a=\frac{2xy}{x+y}$ $b=\frac{2zy}{z+y}$ $c=\frac{2zt}{z+t}$ $d=\frac{2tx}{x+t}$ . Rename $X$ to $l$ .So $l$ is the center of spiral similarity that sends $BA$ to $CD$ so $l$ is given by formula $l=\frac{bd-ac}{b+d-a-c}$ (well known but can be derived from similarity of $\Delta ABX$ and $\Delta DXC$ ) . Now lets compute $l$ . We have
$l=\frac{\frac{2zy}{z+y}\frac{2tx}{x+t}-\frac{2xy}{x+y}\frac{2tx}{x+t}}{\frac{2zy}{z+y}+\frac{2tx}{x+t}-\frac{2xy}{x+y}-\frac{2zt}{z+t}}$
$l=\frac{2xyzt(t-y)(x-z)}{tx(y^2t+z^2x)+yz(x^2z+t^2y)-xy(t^2y+z^2x)-tz(x^2z+y^2t)+x^2t^2(y+z)+y^2z^2(x+t)-x^2y^2(t+z)-t^2z^2(y+x)}$
(here some factors cancel $\Sigma x^2yzt$ is in all four products)
Now we notice
$tx(y^2t+z^2x)+yz(x^2z+t^2y)-xy(t^2y+z^2x)-tz(x^2z+y^2t)=0$
and $x^2t^2(y+z)+y^2z^2(x+t)-x^2y^2(t+z)-t^2z^2(y+x)=(t-y)(x-z)(xyz+xtz+yzt+xyt)$ (from here on $xyz+xtz+yzt+xyt=\Sigma xyz$ )
So we have $l=\frac{2xyzt}{\Sigma xyz}$ . Next we notice that problem statement is equivalent to showing that $XI$ is angle bisector of angle $AXB$ . We show this by showing that $\Delta AXI$ is similar to $\Delta IXC$ or equivalently $\frac{0-a}{0-l}=\frac{c-0}{c-l}$ .
LHS equals $\frac{\Sigma xyz}{(x+y)yt}$ and $c-l=\frac {2z^2t^2(x+y)}{(\Sigma xyz)(z+t)}$ so $\frac{c-0}{c-l}=\frac{\Sigma xyz}{(x+y)yt}$ .
And we are done.

This post has been edited 1 time. Last edited by FISHMJ25, Apr 7, 2019, 6:08 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

tenplusten

1000 posts

tenplusten

#17 h Apr 8, 2019, 7:52 AM • 1 Y

Y by Adventure10

Let $O$ be incenter,and let incircle touch $CB,BA,AD,DC$ at $L,M,N,K$ .Let $B',A',D',C'$ be the midpoints of $LM,MN,NK,KL$ .Perform an inversion which fixes incircle.
Let's denote inverse of $T$ with $T'$ .
So $P'$ is midpoint of $LN$ ,and $X'=\odot(P'B'A')\cap \odot(P'C'D')$ .
We will show that $\triangle OXB\sim \triangle DXO$ (this clearly finishes the problem) which is equivalent to show that $\frac{OX}{XD}=\frac{XB}{OX}=\frac{OB}{OD}$ .But we have the following length information from distance formula of inversion:
$\frac{OX}{XD}=\frac{OD'}{X'D'}$
$\frac{XB}{OX}=\frac{XB'}{OB'}$
$\frac{OB}{OD}=\frac{OD'}{OB'}$
So if we prove $OB'=X'D'$ , $OD'=X'B'$ we are done.So let's show just one of equalities;other one can be proven similarly.
Let $\angle LKM=\alpha$ , $\angle NLK=\beta$ , $\angle X'P'D'=\phi$ , $\angle MLN=x$ , then $\angle C'X'B'=\angle C'LB'$ .So by Sine Law in triangles $X'B'C,B'C'L,X'B'P'$ we get the following equalities:
$\frac{X'B'}{sin(\phi-\alpha-\beta)}=\frac{LM}{2sin \alpha}=\frac{MN}{2sin x}=\frac{B'P'}{sin x}$
$\frac{X'B'}{sin(\beta+\phi-\alpha)}=\frac{B'P'}{sin x}$
So $sin(\phi-\alpha-\beta)=sin(\beta+\phi-\alpha)$ which means $\phi=90+\alpha$ .
Then $X'D'=\frac{P'D'\cdot sin(\angle LNK)}{cos \alpha}=\frac{P'B'\cdot sin\angle MLN}{cos \alpha}=OB'$ .so done

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

liekkas

370 posts

liekkas

#18 h Apr 8, 2019, 2:30 PM • 5 Y

Y by Adventure10, Mango247, Mango247, Mango247, Mysteriouxxx

Rewrite the problem condition as following: In triangle $ABC$ , $I$ is the incenter, $E,F$ lie on $AB,AC$ such that $EF$ is tangent to $(I)$ , $(AEF)$ intersects $(ABC)$ at $A,J$ , prove that $\angle BJI=\angle FJI$ .
Let $JI$ , $AI$ intersect $(AEF)$ , $(ABC)$ at $Q,P,N,M$ . By Reim's theorem, $PM \parallel NQ$ .
East to notice that $\Delta NEF \sim \Delta MCB$ , and $I$ is the $A$ -excentre of $\Delta AEF$ . Thus $\frac{NQ}{MP}=\frac{NI}{MI}=\frac{NE}{BM}=\frac{EF}{BC}$ .
Combined with $E,Q,N,F$ and $B,M,P,C$ concyclic, respectively, we have $NEFQ \sim MBCP$ , thus $\angle EJI=\angle EJQ=\angle EFQ=\angle CBP=\angle CJI$ .
But $\Delta JEB \sim \Delta JFC$ , so $\angle BJE=\angle CJF$ . Adding these two equality yields the conclusion.

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mela_20-15

125 posts

mela_20-15

#19 h Apr 17, 2019, 9:46 AM • 2 Y

Y by Adventure10, Mango247

This is indeed trivial by desargues involution but you can also do some really simple stuff.
Let $PI$ intersect $(PAB),(PCD)$ at midpoints of arcs $AB,CD$ : $M,N$ .You easily get that $MA=MI=MB$ , $NC=ND=NI$
and by similarities you get that $MI/NI=XM/XN$ thus $XI$ bisects $MXN$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fcomoreira

18 posts

fcomoreira

#20 h Mar 16, 2020, 8:50 PM

Y by

A very similar problem is on Brazilian MO 2016, which can be done through an inversion with center X swapping $(A,C)$ and $(B,D)$ .
https://artofproblemsolving.com/community/c6h1343440p7304630

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

guptaamitu1

656 posts

guptaamitu1

#21 h Aug 5, 2022, 11:12 AM • 1 Y

Y by Mango247

MarkBcc168 wrote:

As @above has said, it's trivial by Desargues Involution Theorem.

Let the two tangents be $XK_1, XK_2$ . Then there exists involution swapping $(XA, XC)$ , $(XB, XD)$ , $(XK_1,XK_2)$ . However, as $\triangle XAD\stackrel{+}{\sim} \triangle XBC$ , we get $XB, XD$ are isogonal w.r.t $\angle AXC$ . Hence this involution is isogonality w.r.t. $\angle AXC$ or $XK_1,XK_2$ are isogonal w.r.t $\angle AXC$ so we are done.

EDIT : I found a quick (but not synthetic) solution.

Let $I$ be the incenter. Then it suffices to show that $XI$ bisects $\angle AXC$ . Invert around $X$ with power $XA\cdot XC$ $=XB\cdot XD$ , followed by reflection across the angle bisector of $\angle AXC$ . Clearly it maps $A\to C$ , $B\to D$ . Now by some long angle bashing, one can show that this inversion fixes $I$ so we are done.

Here's a proof of why the inversion fixes $I$ . Look at this Blog of p_square. It is proven that the inversion is same as clawson schmidt conjugation. It was further shown the inversion is same as isogonal conjugation (provided isogonal conjugate exists). Since clearly $I$ is isogonal conjugate of itself, so $I$ is fixed.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ihategeo_1969

217 posts

ihategeo_1969

#22 h Nov 20, 2024, 6:13 PM

Y by

Solving this instead.

Quote:

A quadrilateral $ABCD$ has an circle inscribed in it with center $I$ . Let $P$ be a point outside $ABCD$ such that $\angle APC$ and $\angle BPD$ have the same angle bisector. Prove that $I$ lies on this angle bisector.

Let $Y$ and $Z$ be the points of tangency from the two tangents from $P$ to the incircle. Now apply DDIT from $P$ to $ABCD$ and so we get that $(\overline{PA},\overline{PC})$ ; $(\overline{PB},\overline{PD})$ ; $(\overline{PY},\overline{PZ})$ are pairs under some involution. The problem statement tells us this involution is a reflection and we are done.

This post has been edited 2 times. Last edited by ihategeo_1969, Nov 20, 2024, 6:15 PM

Z K Y