IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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asif e elahi
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Re: IMO Marathon

Unread post by asif e elahi » Tue Sep 09, 2014 6:20 pm

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.
Last edited by asif e elahi on Wed Sep 10, 2014 2:21 pm, edited 1 time in total.

mutasimmim
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Re: IMO Marathon

Unread post by mutasimmim » Tue Sep 09, 2014 10:49 pm

so that in every line and column,there are 25 squares of every colour
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.

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*Mahi*
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Re: IMO Marathon

Unread post by *Mahi* » Tue Sep 09, 2014 11:28 pm

Or the square can be $100 \times 100$.
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asif e elahi
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Re: IMO Marathon

Unread post by asif e elahi » Wed Sep 10, 2014 2:22 pm

Mahi vai is right.The square is $100 \times 100$.

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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Thu Sep 25, 2014 11:38 pm

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 $\forall x,y\in \mathbb{R}$,
$f(f(x)-y^{2})=f(x^{2})+y^{2}f(y)-2f(xy)$.
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Fm Jakaria
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Re: IMO Marathon

Unread post by Fm Jakaria » Fri Sep 26, 2014 6:48 pm

Denote the main equation by (M). Suppose f(0)=c.
First we check for constant solutions:
(M) $\Rightarrow$ c=c+y$^2$c-2c, 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(c-y$^2$) = 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(c-y$^2$) $\rightarrow$(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.
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the umpteenth digit of PI,
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Fri Sep 26, 2014 10:33 pm

@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,
$P(0,y)\Rightarrow f(c-y^{2})=y^{2}f(y)-c$. Again,
$P(0,-y)\Rightarrow f(c-y^{2})=y^{2}f(-y)-c$. 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 non-constant then this means a contradiction. So, when $f$ is non-constant, $r=0$ 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 $2014$ $T-5$
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Fm Jakaria
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Re: IMO Marathon

Unread post by Fm Jakaria » Sat Sep 27, 2014 6:08 am

$Y$ 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,
$\angle AYE = \angle AYF = ( \angle EYF)/2 = (180-2\angle YEF)/2 = \angle C/2$.
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 $\angle ZYC=180-\angle ZAC=180-(90-C)= 90+C$.
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
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.

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Phlembac Adib Hasan
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Sat Sep 27, 2014 2:33 pm

@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 $f(c^4)=c$.
$P(0,y)\Longrightarrow f(c-y^2)=c+y^2f(y)-2c$
$P(c^2,y)\Longrightarrow f(c-y^2)=c+y^2f(y)-2f(c^2y)$
Hence $f(c^2y)=c\forall y\in \mathbb R$ which is possible only if $c=f(0)=0$.
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Sat Sep 27, 2014 6:49 pm

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 $\triangle XYZ$.
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