IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Fri Feb 01, 2013 8:06 pm

Tahmid Hasan wrote:Finally solved it! :D
Let $P(x,y) \Rightarrow f(x^2+y+f(y))=f(x^2)+\alpha.y$
If $\alpha=0$, then $f(x)=0,1$ both satisfy the equation, so a contradiction.
So $\alpha \neq 0$, then
$P(0,x) \Rightarrow f(x+f(x))=f(0)^2+\alpha.x$
since $f(x+f(x))-f(0)^2$ ranges over $\mathbb{R}$, so does $\alpha.x$ i.e $x$[$\alpha \in \mathbb{R}-\{0\}$]
So $f(x)$ is surjective.
Let $f(c)=0$.
$P(0,c) \Rightarrow f(c)=f(0)^2+\alpha.c \Rightarrow c=\frac{-f(0)^2}{\alpha}$.....$(1)$
$P(x,\frac{-f(x^2)}{\alpha}) \Rightarrow f(x^2-\frac{f(x)^2}{\alpha)}+f(\frac{-f(x)^2}{\alpha})=0$.....$(2)$
Let $f(c_1)=f(c_2)=0$
$P(0,c_1),P(0,c_2)$ implies $c_1=c_2$, so from $(1),(2)$ we conclude
$f(\frac{-f(x)^2}{\alpha})-f(\frac{f(x)^2}{\alpha})+x^2=-\frac{-f(0)^2}{\alpha}$.
$P(x,\frac{-f(y)^2}{\alpha}) \Rightarrow f(x^2-\frac{f(0)^2}{\alpha}-y^2)=f(x^2)-f(y)^2$. Let it be $P(x,y)$
If $\alpha >0$, plug $P'(\frac{f(0)}{\sqrt \alpha})$. If $\alpha <0$, plug $P'(\frac{f(0)}{\sqrt{- \alpha}},x)$
both will yield $f(\frac{f(0)^2}{\alpha})=0$
So $\frac{f(0)^2}{\alpha}=\frac{-f(0)^2}{\alpha} \Rightarrow f(0)=0$.
So $P(x,0) \Rightarrow f(x^2)=f(x)^2$.....$(3)$
So $P'(x,y) \Rightarrow f(x^2-y^2)=f(x^2)-f(y^2)$ which implies $f(x-y)=f(x)-f(y) \forall x \ge 0,y \ge 0$. Let it be $Q(x,y)$
$Q(x+y,y) \Rightarrow f(x+y)=f(x)+f(y) \forall x \ge 0,y \ge 0$. Let it be $R(x,y)$
Now $\forall x<0,f(-x)=f(0-x)=f(0)-f(x)=-f(x)$.
Let $y<0$, then $R(x,-y) \Rightarrow f(x-y)=f(x)-f(y)$
So $f(x-y)=f(x)-f(y) \forall x \ge 0, y \in \mathbb{R}$.
Then $Q(x,-y) \Rightarrow f(x+y)=f(x)+f(y)$
So $f(x+y)=f(x)+f(y) \forall x \ge 0,y \in \mathbb{R}$.
Let $x<0$, then take any $y<0$, so
$f(x+y)=-f(-x-y)=-(f(-x)+f(-y))=-(-f(x)-f(y))=f(x)+f(y)$. So $f(x)$ is additive $\forall x,y \in \mathbb{R}$.
I just came up with a different conclusion from here.
Since $f(x^2)=f(x)^2$, we conclude $f(x) \ge 0 \forall x \ge 0$.
So from this and the additive property we conclude $f(x)$ is continuous, so $f(x)=cx(c \neq 0),0$ for which the latter is a contracdiction.
$f(x^2)=f(x)^2 \Rightarrow cx^2=c^2x^2 \Rightarrow c=1$. So $f(x)=x$, which implies $\alpha=2$.
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Phlembac Adib Hasan
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Fri Feb 01, 2013 9:10 pm

Tahmid Hasan wrote: $P'(x,y) \Rightarrow f(x^2-y^2)=f(x^2)-f(y^2)$ which implies $f(x-y)=f(x)-f(y) \forall x \ge 0,y \ge 0$.
Sorry for my inconvenience. আমি আসলে বুঝি কম। এইটা কিভাবে হইল? ব্যাখ্যা করলে ভাল হয়।
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Fri Feb 01, 2013 9:45 pm

Phlembac Adib Hasan wrote:
Tahmid Hasan wrote: $P'(x,y) \Rightarrow f(x^2-y^2)=f(x^2)-f(y^2)$ which implies $f(x-y)=f(x)-f(y) \forall x \ge 0,y \ge 0$.
Sorry for my inconvenience. আমি আসলে বুঝি কম। এইটা কিভাবে হইল? ব্যাখ্যা করলে ভাল হয়।
কোন অংটুকু বোঝনি বলতে চাচ্ছ? লাল অংশটুকু কীভাবে আসল নাকি লাল অংশ থেকে পরের অংশ কীভাবে আসল? যাই হোক দুইটাই ব্যাখ্যা করে দেই।
Since $f(0)=0$, plugging this value in $f(x^2-\frac{f(0)^2}{\alpha}-y^2)=f(x^2)-f(y^2)$ results in the red part.
Since in the given function $x,y$ range over reals, $x^2,y^2$ range over non-negative reals.
That is why I concluded $f(x-y)=f(x)-f(y) \forall x \ge 0,y \ge 0$ from the red part.
বোঝাতে পারলাম কি? :?:
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Phlembac Adib Hasan
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Sat Feb 02, 2013 8:40 am

Tahmid Hasan wrote: $P(0,c) \Rightarrow f(c)=f(0)^2+\alpha.c \Rightarrow c=\frac{-f(0)^2}{\alpha}$.....$(1)$
$P(x,\frac{-f(x^2)}{\alpha}) \Rightarrow f(x^2-\frac{f(x)^2}{\alpha)}+f(\frac{-f(x)^2}{\alpha})=0$.....$(2)$
Let $f(c_1)=f(c_2)=0$
$P(0,c_1),P(0,c_2)$ implies $c_1=c_2$, so from $(1),(2)$ we conclude
$f(\frac{-f(x)^2}{\alpha})-f(\frac{f(x)^2}{\alpha})+x^2=-\frac{-f(0)^2}{\alpha}$.
$P(x,\frac{-f(y)^2}{\alpha}) \Rightarrow f(x^2-\frac{f(0)^2}{\alpha}-y^2)=f(x^2)-f(y)^2$. Let it be $P(x,y)$
শেষের লাইনটা কিভাবে আসলো? শেষের আগের লাইন থেকে সরাসরি কি এটা বলা যায়?
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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Sat Feb 02, 2013 9:14 am

Phlembac Adib Hasan wrote:
Tahmid Hasan wrote: $P(0,c) \Rightarrow f(c)=f(0)^2+\alpha.c \Rightarrow c=\frac{-f(0)^2}{\alpha}$.....$(1)$
$P(x,\frac{-f(x^2)}{\alpha}) \Rightarrow f(x^2-\frac{f(x)^2}{\alpha)}+f(\frac{-f(x)^2}{\alpha}))=0$.....$(2)$
Let $f(c_1)=f(c_2)=0$
$P(0,c_1),P(0,c_2)$ implies $c_1=c_2$, so from $(1),(2)$ we conclude
$f(\frac{-f(x)^2}{\alpha})-f(\frac{f(x)^2}{\alpha})+x^2=-\frac{-f(0)^2}{\alpha}$.
$P(x,\frac{-f(y)^2}{\alpha}) \Rightarrow f(x^2-\frac{f(0)^2}{\alpha}-y^2)=f(x^2)-f(y)^2$. Let it be $P(x,y)$
শেষের লাইনটা কিভাবে আসলো? শেষের আগের লাইন থেকে সরাসরি কি এটা বলা যায়?
$f(\frac{-f(x)^2}{\alpha})-f(\frac{f(x)^2}{\alpha})=-\frac{-f(0)^2}{\alpha}-x^2$....(*)
Plugging $P(x,\frac{-f(y)^2}{\alpha})$ gives the LHS of (*) inside the LHS, so I replaced it with RHS :?:
Is that what you wanted to know?
(2) সমীকরণে একটা ব্রাকেট কম দিয়ে দিয়েছিলাম, এইজন্যে বোধহয় বুঝতে সমস্যা হয়েছে :oops:
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FahimFerdous
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Re: IMO Marathon

Unread post by FahimFerdous » Sat Feb 02, 2013 11:39 pm

Problem $\boxed {28}$:

Trapezoid $ABCD$, with $AB$ parallel to $CD$, is inscribed in circle $w$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $w$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $AB$ intersect $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $BG$ bisects $\angle CBD$.
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Sun Feb 03, 2013 4:37 pm

FahimFerdous wrote:Problem $\boxed {28}$:

Trapezoid $ABCD$, with $AB$ parallel to $CD$, is inscribed in circle $w$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $w$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $AB$ intersect $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $BG$ bisects $\angle CBD$.

Well Here is my sexy proof of this cruxy problem .
Cru.PNG
Cru.PNG (27.82 KiB) Viewed 3232 times
Trapezoid $ABCD$ being inscribed in a circle imply its isoscales .
Let $w \cap QR=E , w \cap QS=F , DF \cap CE=H.$
Using pascel $R,H,S$ are colinear .

A simple angle chasing imply
$G,S,C,P$ and $B,H,S,F$ both are cyclic .

Now $\angle QRS=\angle QEC+\angle ECD=\angle QBC+\angle HCD$
again $\angle QPS=\angle QPG +\angle GPS=\angle ABQ+ \angle GCS$

So $\angle QRS+\angle QPS=\angle QBC+\angle HCD+\angle ABQ+ \angle GCS=180+\angle HCD$
Which imply $P,Q,R,S$ is cyclic $\Leftrightarrow G=H$

Note that $\angle HBS=\angle HFS=\angle DFQ=\angle DBQ=\angle DBG$ So

$P,Q,R,S$ is cyclic $\Leftrightarrow G=H $ $\Leftrightarrow$ $BQ$ bisect $\angle CBD$
$\frac{1}{0}$

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FahimFerdous
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Re: IMO Marathon

Unread post by FahimFerdous » Sun Feb 03, 2013 5:10 pm

Very nice solution, Nadim. Mine's quite dirty, but uses same concepts. :)

$\text{Problem }29$:

Fixed points $B$ and $C$ are on a fixed circle $w$ and point $A$ varies on this circle. We call the midpoint of arc $BC$ (not containing $A$) $D$ and the orthocenter of the triangle $ABC$, $H$. Line $DH$ intersects circle $w$ again in $K$. Tangent in $A$ to circumcircle of triangle $AKH$ intersects line $DH$ and circle $w$ again in $L$ and $M$ respectively. Prove that the value of $\frac{AL}{AM}$ is constant.
Last edited by *Mahi* on Tue Feb 05, 2013 7:17 pm, edited 2 times in total.
Reason: Edited once again
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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Sun Feb 03, 2013 5:45 pm

(unless I'm not being silly)

What is the difference between "circumcircle of triangle $ABC$" and $w$ ? :S
$\frac{1}{0}$

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Sun Feb 03, 2013 7:46 pm

FahimFerdous wrote:Problem $\boxed {28}$:

Trapezoid $ABCD$, with $AB$ parallel to $CD$, is inscribed in circle $w$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $w$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $AB$ intersect $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $BG$ bisects $\angle CBD$.
My solution: 'If' part: $\angle BQC=\angle BDC=\angle BRG,\angle CBQ=\angle RBG$.
So $\triangle BCQ \sim \triangle BGR$.
Hence $\frac{BC}{BG}=\frac{BQ}{BR} \Rightarrow \frac{BC}{BQ}=\frac{BG}{BR}$.....$(1)$
Since $\angle GBC=\angle QBR$ we conclude from this and $(1),\triangle BCG \sim \triangle BQR \Rightarrow \angle BCG=\angle BQR$.....$(2)$
Simlarly we can prove $\triangle BDQ \sim \triangle BGS \Rightarrow \triangle BDG \sim \triangle BQS \Rightarrow \angle BDG=\angle BQS$.....$(3)$
Now $\angle PCS=\angle PCB=\angle PAB=\angle PGS \Rightarrow PGCS$ is concyclic.
Again $\angle PDR=\angle PDB=\angle PAB=\angle AGR \Rightarrow PGRD$ is concyclic.
Now $\angle SPR=\angle SPG+\angle RPG=\angle SCG+\angle GDR=\angle BCG+\angle BDG=\angle BQR+\angle BQS=\angle SQR$
Hence $PQRS$ is concyclic.
'Only if' part: $CD \parallel RS \Rightarrow \omega,\odot BRS$ are homothetic with centre $B$ which implies they have a common tangent at $B$.
So $PQ,RS,BB$ are tangent at some point $X$ which is the radical centre of $\omega,\odot BRS,\odot PQRS$.
Now $\angle PBX=\angle PQB=\angle PAB=\angle PGX \Rightarrow PGBX$ is concyclic.
So $\angle QPG=\angle XBG \Rightarrow \angle QBA =\angle QAB \Rightarrow QA=QB$
Hence $Q$ lies on the perpendicular bisector of $AB$. Since $AB \parallel CD$, $Q$ lies on the perpendicular bisector of $CD$ too.
So $Q$ is the midpoint of minor arc $CD$ which implies $BG$ bisects $\angle CBD$.
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