IMO Marathon
 asif e elahi
 Posts: 183
 Joined: Mon Aug 05, 2013 12:36 pm
 Location: Sylhet,Bangladesh
Re: IMO Marathon
Problem $35$
Every unit square of a $100 \times 100$ is coloured by one of $4$ colours red,green,blue or yellow so that in every line and column,there are $25$ squares of every colour.Prove that thre exists $2$ rows and columns so that their $4$ intersection points have $4$ different colours.
Every unit square of a $100 \times 100$ is coloured by one of $4$ colours red,green,blue or yellow so that in every line and column,there are $25$ squares of every colour.Prove that thre exists $2$ rows and columns so that their $4$ intersection points have $4$ different colours.
Last edited by asif e elahi on Wed Sep 10, 2014 2:21 pm, edited 1 time in total.

 Posts: 107
 Joined: Sun Dec 12, 2010 10:46 am
Re: IMO Marathon
I think there is a mistake in the sentence, instead of every line and column, it should be in all there are $25$ squares of each color.so that in every line and column,there are 25 squares of every colour
Re: IMO Marathon
Or the square can be $100 \times 100$.
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Nur Muhammad Shafiullah  Mahi
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Nur Muhammad Shafiullah  Mahi
 asif e elahi
 Posts: 183
 Joined: Mon Aug 05, 2013 12:36 pm
 Location: Sylhet,Bangladesh
Re: IMO Marathon
Mahi vai is right.The square is $100 \times 100$.
Re: IMO Marathon
I think this marathon should be revived. Also, post a solution to problem 35, Asif, if you have it.
Problem 36
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall x,y\in \mathbb{R}$,
$f(f(x)y^{2})=f(x^{2})+y^{2}f(y)2f(xy)$.
Problem 36
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall x,y\in \mathbb{R}$,
$f(f(x)y^{2})=f(x^{2})+y^{2}f(y)2f(xy)$.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$
 Fm Jakaria
 Posts: 79
 Joined: Thu Feb 28, 2013 11:49 pm
Re: IMO Marathon
Denote the main equation by (M). Suppose f(0)=c.
First we check for constant solutions:
(M) $\Rightarrow$ c=c+y$^2$c2c, taking y nonzero and y$\neq \pm\sqrt{2}$ implies c=0. Constant f=0 is a solution. Now assume f is nonconstant.
(M)x=y=0 $\Rightarrow$ f(c)=c…..(i).
(M) y = 0 $\Rightarrow$ f(f(x))=f(x$^2$)2c…..(1)
(M) x=0 $\Rightarrow$ f(cy$^2$) = y$^2$f(y)c ….(2)
We may interchange y by (y) to conclude y$^2$f(y)c= y$^2$f(y)c, so always
f(y)=f(y)….(ii)[including when y=0].
We now introduce our first lemma.
Lemma 1: If f(x)=c, then x=0.
Proof: Assume there exists nonzero z: f(z)=c. Then
(M)x=z $\Rightarrow$
f(z$^2$) + y$^2$f(y)2f(zy)= f(cy$^2$) $\rightarrow$(2) $\rightarrow$y$^2$f(y)c
$\Rightarrow$2f(zy) = f(z$^2$)+c, constant. But zy takes all values from R, so f is constant; contradiction.
Now (1) x=c $\Rightarrow$ f(c$^2$)2c = f(f(c)) = f(c) = f(c) = c(using (i) and (ii))
$\Rightarrow$f(c$^2$)=c. By lemma 1; we deduce that c=0.
Now (1), (2) turns to
f(f(x)) = f(x$^2$)…(1)
y$^2$f(y) = f(y$^2$) = f(y$^2$) ….(2)
Also lemma 1 turns to, f(x)=0 $\Leftrightarrow$ x=0.
Now our second lemma comes to hand.
Lemma 2: f(a)=f(b) $\Leftrightarrow$ a=b or a=b.
Proof: If f(a)=0, lemma (1) implies a=b=0. Assume f(a)$\neq$0.
(1)x=a,b $\Rightarrow$ f(a$^2$) = f(f(a)) = f(f(b)) = f(b$^2$)
Then(2)y=a,b $\Rightarrow$ a$^2$f(a)=f(a$^2$) = f(b$^2$) = b$^2$f(b)=b$^2$f(a)
$\Rightarrow$ a=b or a=b; as f(a) $\neq$0.
This suffices to conclude from (1) that, f(x)=x$^2$ or f(x)=x$^2$; $\forall$ x. Let f(1)=k$\neq$0.
(M)y=1 $\Rightarrow$ f(f(x)1) = f(x$^2$) – 2f(x) +k…(3)
We call some number x 'evil' if f(x) has opposite sign with k; where x is not 0 or $\pm$1. Our next lemma is stated now.
Lemma 3: Either there are no evil number, or else there are infinitely many.
Proof:
Note that if f(x)=x$^2$, (2) implies f(x$^2$)=x$^4$. If f(x)=x$^2$, (2) implies f(x$^2$)=x$^4$.
Also x$\neq 0, \pm 1$ implies x$^2$ is not 0 or $\pm$1. So if x is evil, so is x$^2$. We can inductively say that
x$^{{2}^{n}}$ are distinct evil numbers, for all nonnegative integers n; so the lemma is proved.
First let k=1, but there exists a: f(a)=a$^2$. Then RHS of (3) is 2(a$^2$1)$^2$, LHS is (a$^2$+1)$^2$ or (a$^2$+1)$^2$.Both equality can have at most finitely many solutions. Lemma 3 now implies a=0. So f(x)=x$^2$ for all x.
Now let k=1, but there exists a:f(a)=a$^2$. Now RHS of (3) is (a$^2$1)$^2$2, LHS is (a$^2$1)$^2$ or (a$^2$1)$^2$. Here the first case is impossible. In the second case, there are finitely many solutions. So again
a = 0. So f(x)=x$^2$ for all x.
It is easy to check that these last two solutions are indeed solutions.
First we check for constant solutions:
(M) $\Rightarrow$ c=c+y$^2$c2c, taking y nonzero and y$\neq \pm\sqrt{2}$ implies c=0. Constant f=0 is a solution. Now assume f is nonconstant.
(M)x=y=0 $\Rightarrow$ f(c)=c…..(i).
(M) y = 0 $\Rightarrow$ f(f(x))=f(x$^2$)2c…..(1)
(M) x=0 $\Rightarrow$ f(cy$^2$) = y$^2$f(y)c ….(2)
We may interchange y by (y) to conclude y$^2$f(y)c= y$^2$f(y)c, so always
f(y)=f(y)….(ii)[including when y=0].
We now introduce our first lemma.
Lemma 1: If f(x)=c, then x=0.
Proof: Assume there exists nonzero z: f(z)=c. Then
(M)x=z $\Rightarrow$
f(z$^2$) + y$^2$f(y)2f(zy)= f(cy$^2$) $\rightarrow$(2) $\rightarrow$y$^2$f(y)c
$\Rightarrow$2f(zy) = f(z$^2$)+c, constant. But zy takes all values from R, so f is constant; contradiction.
Now (1) x=c $\Rightarrow$ f(c$^2$)2c = f(f(c)) = f(c) = f(c) = c(using (i) and (ii))
$\Rightarrow$f(c$^2$)=c. By lemma 1; we deduce that c=0.
Now (1), (2) turns to
f(f(x)) = f(x$^2$)…(1)
y$^2$f(y) = f(y$^2$) = f(y$^2$) ….(2)
Also lemma 1 turns to, f(x)=0 $\Leftrightarrow$ x=0.
Now our second lemma comes to hand.
Lemma 2: f(a)=f(b) $\Leftrightarrow$ a=b or a=b.
Proof: If f(a)=0, lemma (1) implies a=b=0. Assume f(a)$\neq$0.
(1)x=a,b $\Rightarrow$ f(a$^2$) = f(f(a)) = f(f(b)) = f(b$^2$)
Then(2)y=a,b $\Rightarrow$ a$^2$f(a)=f(a$^2$) = f(b$^2$) = b$^2$f(b)=b$^2$f(a)
$\Rightarrow$ a=b or a=b; as f(a) $\neq$0.
This suffices to conclude from (1) that, f(x)=x$^2$ or f(x)=x$^2$; $\forall$ x. Let f(1)=k$\neq$0.
(M)y=1 $\Rightarrow$ f(f(x)1) = f(x$^2$) – 2f(x) +k…(3)
We call some number x 'evil' if f(x) has opposite sign with k; where x is not 0 or $\pm$1. Our next lemma is stated now.
Lemma 3: Either there are no evil number, or else there are infinitely many.
Proof:
Note that if f(x)=x$^2$, (2) implies f(x$^2$)=x$^4$. If f(x)=x$^2$, (2) implies f(x$^2$)=x$^4$.
Also x$\neq 0, \pm 1$ implies x$^2$ is not 0 or $\pm$1. So if x is evil, so is x$^2$. We can inductively say that
x$^{{2}^{n}}$ are distinct evil numbers, for all nonnegative integers n; so the lemma is proved.
First let k=1, but there exists a: f(a)=a$^2$. Then RHS of (3) is 2(a$^2$1)$^2$, LHS is (a$^2$+1)$^2$ or (a$^2$+1)$^2$.Both equality can have at most finitely many solutions. Lemma 3 now implies a=0. So f(x)=x$^2$ for all x.
Now let k=1, but there exists a:f(a)=a$^2$. Now RHS of (3) is (a$^2$1)$^2$2, LHS is (a$^2$1)$^2$ or (a$^2$1)$^2$. Here the first case is impossible. In the second case, there are finitely many solutions. So again
a = 0. So f(x)=x$^2$ for all x.
It is easy to check that these last two solutions are indeed solutions.
You cannot say if I fail to recite
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.
Re: IMO Marathon
@Jakaria, $f(x)=x^{2}$ is not a valid solution. Check it please.
My solution:
I am skipping obvious calculations here.
Let us denote the given statement by $P(x,y)$. Then,
$P(x,0)\Rightarrow f(f(x))=f(x^{2})2f(0)$
$P(1,1)\Rightarrow f(f(1)1)=f(1)+f(1)2f(1)=0$. Let $b=f(1)1$. Then $f(b)=0$. So,
$P(b,0)\Rightarrow b^{2}=3f(0)$. Let $f(0)=c$. Then $b^{2}=3c$. Now,
$P(0,y)\Rightarrow f(cy^{2})=y^{2}f(y)c$. Again,
$P(0,y)\Rightarrow f(cy^{2})=y^{2}f(y)c$. Thus, $f(y)=f(y) \forall y\in \mathbb{R}$. Now,
$P(b,b)\Rightarrow f(b^{2})=f(b^{2})$. But we already know that $f(b^{2})=f(b^{2})$. Thus,
$f(b^{2})=0\Rightarrow 3c=0\Rightarrow f(0)=c=0$. So $f(y^{2})=f(y^{2})=y^{2}f(y)$. Now,
$P(y,y)\Rightarrow f(f(y)y^{2})=0$.
Now if $\exists d\in \mathbb{R}$ such that $d\neq 0, f(d)=0$, then,
$P(0,d)\Rightarrow f(d^{2})=0=f(d^{2})$. So,
$P(d,y)\Rightarrow f(y^{2})=f(d^{2})+y^{2}f(y)2f(dy)=0$. From our previous results we conclude that $f(dy)=0\forall y\in \mathbb{R}$. Now set $y=\dfrac{x}{d}$. Here $d\neq 0$.
Then we get $f(x)=0\forall x\in \mathbb{R}$. So in this case $f$ is constant. If $f$ is nonconstant then this means a contradiction. So, when $f$ is nonconstant, $r=0$ whenever $f(r)=0$.
So because $f(f(y)y^{2})=0\forall y\in \mathbb{R}$, we have $f(y)y^{2}=0\forall y\in \mathbb{R}$.
So our solutions are:
$f(y)=0\forall y\in \mathbb{R}$
and
$f(y)=y^{2}\forall y\in \mathbb{R}$.
It is easily verified that these are indeed the desired solutions.
Problem $37$
Let $ABC$ be a triangle with $AB<AC$. Its incircle with centre $I$ touches the sides $BC,CA$ and $AB$ at the points $D,E$ and $F$ respectively. The angle bisector $AI$ intersects the lines $DE$ and $DF$ in the points $X$ and $Y$ respectively. Let $Z$ be the foot of the altitude through $A$ on $BC$. Prove that $D$ is the incentre of the triangle $XYZ$.
Source: Middle European Mathematical Olympiad $2014$ $T5$
My solution:
I am skipping obvious calculations here.
Let us denote the given statement by $P(x,y)$. Then,
$P(x,0)\Rightarrow f(f(x))=f(x^{2})2f(0)$
$P(1,1)\Rightarrow f(f(1)1)=f(1)+f(1)2f(1)=0$. Let $b=f(1)1$. Then $f(b)=0$. So,
$P(b,0)\Rightarrow b^{2}=3f(0)$. Let $f(0)=c$. Then $b^{2}=3c$. Now,
$P(0,y)\Rightarrow f(cy^{2})=y^{2}f(y)c$. Again,
$P(0,y)\Rightarrow f(cy^{2})=y^{2}f(y)c$. Thus, $f(y)=f(y) \forall y\in \mathbb{R}$. Now,
$P(b,b)\Rightarrow f(b^{2})=f(b^{2})$. But we already know that $f(b^{2})=f(b^{2})$. Thus,
$f(b^{2})=0\Rightarrow 3c=0\Rightarrow f(0)=c=0$. So $f(y^{2})=f(y^{2})=y^{2}f(y)$. Now,
$P(y,y)\Rightarrow f(f(y)y^{2})=0$.
Now if $\exists d\in \mathbb{R}$ such that $d\neq 0, f(d)=0$, then,
$P(0,d)\Rightarrow f(d^{2})=0=f(d^{2})$. So,
$P(d,y)\Rightarrow f(y^{2})=f(d^{2})+y^{2}f(y)2f(dy)=0$. From our previous results we conclude that $f(dy)=0\forall y\in \mathbb{R}$. Now set $y=\dfrac{x}{d}$. Here $d\neq 0$.
Then we get $f(x)=0\forall x\in \mathbb{R}$. So in this case $f$ is constant. If $f$ is nonconstant then this means a contradiction. So, when $f$ is nonconstant, $r=0$ whenever $f(r)=0$.
So because $f(f(y)y^{2})=0\forall y\in \mathbb{R}$, we have $f(y)y^{2}=0\forall y\in \mathbb{R}$.
So our solutions are:
$f(y)=0\forall y\in \mathbb{R}$
and
$f(y)=y^{2}\forall y\in \mathbb{R}$.
It is easily verified that these are indeed the desired solutions.
Problem $37$
Let $ABC$ be a triangle with $AB<AC$. Its incircle with centre $I$ touches the sides $BC,CA$ and $AB$ at the points $D,E$ and $F$ respectively. The angle bisector $AI$ intersects the lines $DE$ and $DF$ in the points $X$ and $Y$ respectively. Let $Z$ be the foot of the altitude through $A$ on $BC$. Prove that $D$ is the incentre of the triangle $XYZ$.
Source: Middle European Mathematical Olympiad $2014$ $T5$
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$
 Fm Jakaria
 Posts: 79
 Joined: Thu Feb 28, 2013 11:49 pm
Re: IMO Marathon
$Y$ lies outside $ABC$, $X$ lies inside.
As $AY$ is the perpendicular bisector of $EF, \angle YEF = \angle YFE = \angle DFE = 90(C/2)$. For this reason,
$\angle AYE = \angle AYF = ( \angle EYF)/2 = (1802\angle YEF)/2 = \angle C/2$.
Then $\angle DYE = \angle AYF+\angle AYE = \angle C,$ so $DYCE$ is cyclic. Then
$\angle EYC=\angle EDC=\angle DFE=90(\angle C/2)$.
Also, $\angle DYC = \angle AED = \angle AEF+\angle DEF = (90 (A/2))+(90(B/2)) = 90+(\angle C/2)$.
Now $\angle AYC = \angle AYE + \angle EYC = (\angle C/2) + (90 (C/2)) = 90= \angle AZC$, so $AZYC$ is cyclic.
So $\angle ZYC=180\angle ZAC=180(90C)= 90+C$.
Now $\angle ZYD=\angle ZYC\angle DYC =(90+C)(90+(C/2)) = C/2 = \angle AYF = \angle DYX$.
Similarly, we can prove $\angle ZXD=\angle DXY$. Actually, we make the pattern of angel chasing slightly different ; still the idea is to show first $BFXD$ is cyclic. Then to show $\angle AXB=90$; so $AXZB$ is cyclic. Using these together, angle chasing implies the desired result.
So $D$ is the incenter of $XYZ$.
As $AY$ is the perpendicular bisector of $EF, \angle YEF = \angle YFE = \angle DFE = 90(C/2)$. For this reason,
$\angle AYE = \angle AYF = ( \angle EYF)/2 = (1802\angle YEF)/2 = \angle C/2$.
Then $\angle DYE = \angle AYF+\angle AYE = \angle C,$ so $DYCE$ is cyclic. Then
$\angle EYC=\angle EDC=\angle DFE=90(\angle C/2)$.
Also, $\angle DYC = \angle AED = \angle AEF+\angle DEF = (90 (A/2))+(90(B/2)) = 90+(\angle C/2)$.
Now $\angle AYC = \angle AYE + \angle EYC = (\angle C/2) + (90 (C/2)) = 90= \angle AZC$, so $AZYC$ is cyclic.
So $\angle ZYC=180\angle ZAC=180(90C)= 90+C$.
Now $\angle ZYD=\angle ZYC\angle DYC =(90+C)(90+(C/2)) = C/2 = \angle AYF = \angle DYX$.
Similarly, we can prove $\angle ZXD=\angle DXY$. Actually, we make the pattern of angel chasing slightly different ; still the idea is to show first $BFXD$ is cyclic. Then to show $\angle AXB=90$; so $AXZB$ is cyclic. Using these together, angle chasing implies the desired result.
So $D$ is the incenter of $XYZ$.
Last edited by Phlembac Adib Hasan on Sat Sep 27, 2014 2:38 pm, edited 1 time in total.
Reason: Latexed
Reason: Latexed
You cannot say if I fail to recite
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
Re: IMO Marathon
@Jakaria, please write in latex. If you don't know the codes, at least write every mathematical expression between two dollar signs. But honestly I think learning a few codes like \angle, \Longrightarrow, \frac, \sum isn't hard. After all, you are a senior member now and this much is expected of you.
Problem 36: Another way to prove $f(0)=0$
Obviously $f(x)=0$ is the only constant solution. So we may assume $\exists a\in \mathbb R : f(a)\neq 0$. Let $f(0)=c$. Combining $P(0,0)$ and $P(c,0)$ we get $f(c)=c$ and $f(c^2)=c$. Now $P(c^2,0)$ yields $f(c^4)=c$.
$P(0,y)\Longrightarrow f(cy^2)=c+y^2f(y)2c$
$P(c^2,y)\Longrightarrow f(cy^2)=c+y^2f(y)2f(c^2y)$
Hence $f(c^2y)=c\forall y\in \mathbb R$ which is possible only if $c=f(0)=0$.
Problem 36: Another way to prove $f(0)=0$
Obviously $f(x)=0$ is the only constant solution. So we may assume $\exists a\in \mathbb R : f(a)\neq 0$. Let $f(0)=c$. Combining $P(0,0)$ and $P(c,0)$ we get $f(c)=c$ and $f(c^2)=c$. Now $P(c^2,0)$ yields $f(c^4)=c$.
$P(0,y)\Longrightarrow f(cy^2)=c+y^2f(y)2c$
$P(c^2,y)\Longrightarrow f(cy^2)=c+y^2f(y)2f(c^2y)$
Hence $f(c^2y)=c\forall y\in \mathbb R$ which is possible only if $c=f(0)=0$.
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Re: IMO Marathon
My solution to 37:
Let $P\in AC$ such that $AP=AB$. Let $BP\cap AI=X_0$. Let $EX_0$ meet $\odot DEF$ at $D_0\neq E$. Clearly $D_0$ is on the other side of $BP$ than $F$. Now, $AI$ is the perpendicular bisector of $EF$ and $BP$. Again, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$ i.e. $D_0BFX_0$ is cyclic. Thus, $\angle BDF=\angle BX_0F=90^{\circ}\angle FX_0A=\angle X_0FE=\angle X_0EF=\angle DEF$. So, $BD_0$ is tangent to $\odot DEF$. This means that $D\equiv D_0$. Thus $X\equiv X_0$. So we proved that $BX\perp AI$.
Now, $\angle DYE=\angle FYE=180^{\circ}\angle YFE\angle YEF$. But $Y$ is on the perpendicular bisector of $EF$. So, $\angle DYE=180^{\circ}2\angle YFE=180^{\circ}2\angle DFE=180^{\circ}2\angle DEC=\angle DCE$. Thus $DECY$ is cyclic, and so $\angle IYE=\dfrac{1}{2}\angle DYE=\dfrac{1}{2}\angle DCE=\angle ICE$ which means $IYCE$ is cyclic too. Thus $\angle IYC=\angle IEC=90^{\circ}$. So $CY\perp AI$ too.
Now, $AXZB$ and $AZYC$ both are cyclic. So, $\angle ZXY=90^{\circ}\angle BXZ=90^{\circ}\angle BAZ=\angle ABC$ and $\angle DXY=\angle AXE=90^{\circ}\angle DEF=90^{\circ}\angle BDF=\angle IBD=\dfrac{1}{2}\angle ABC$. This means $DX$ bisects $\angle ZXY$. Also, $\angle XZD=180^{\circ}\angle XZB=\angle XAB=\angle IAB=\angle IAC=\angle YAC=\angle YZC=\angle YZD$. Thus $DZ$ bisects $\angle XZY$.
All these results imply that $D$ is the incenter of $\triangle XYZ$.
Let $P\in AC$ such that $AP=AB$. Let $BP\cap AI=X_0$. Let $EX_0$ meet $\odot DEF$ at $D_0\neq E$. Clearly $D_0$ is on the other side of $BP$ than $F$. Now, $AI$ is the perpendicular bisector of $EF$ and $BP$. Again, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$ i.e. $D_0BFX_0$ is cyclic. Thus, $\angle BDF=\angle BX_0F=90^{\circ}\angle FX_0A=\angle X_0FE=\angle X_0EF=\angle DEF$. So, $BD_0$ is tangent to $\odot DEF$. This means that $D\equiv D_0$. Thus $X\equiv X_0$. So we proved that $BX\perp AI$.
Now, $\angle DYE=\angle FYE=180^{\circ}\angle YFE\angle YEF$. But $Y$ is on the perpendicular bisector of $EF$. So, $\angle DYE=180^{\circ}2\angle YFE=180^{\circ}2\angle DFE=180^{\circ}2\angle DEC=\angle DCE$. Thus $DECY$ is cyclic, and so $\angle IYE=\dfrac{1}{2}\angle DYE=\dfrac{1}{2}\angle DCE=\angle ICE$ which means $IYCE$ is cyclic too. Thus $\angle IYC=\angle IEC=90^{\circ}$. So $CY\perp AI$ too.
Now, $AXZB$ and $AZYC$ both are cyclic. So, $\angle ZXY=90^{\circ}\angle BXZ=90^{\circ}\angle BAZ=\angle ABC$ and $\angle DXY=\angle AXE=90^{\circ}\angle DEF=90^{\circ}\angle BDF=\angle IBD=\dfrac{1}{2}\angle ABC$. This means $DX$ bisects $\angle ZXY$. Also, $\angle XZD=180^{\circ}\angle XZB=\angle XAB=\angle IAB=\angle IAC=\angle YAC=\angle YZC=\angle YZD$. Thus $DZ$ bisects $\angle XZY$.
All these results imply that $D$ is the incenter of $\triangle XYZ$.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$