IMO Marathon

[*Mahi*]

My বিশ্রী solution:

I don't see why this is ugly . Is it for dirichlet ?

problem $\boxed {26}$
Let $I$ be the incenter of acute $\triangle ABC$. Let $\omega$ be a circle with center $I$ that lies inside $\triangle ABC$ . $D, E, F$ are the intersection points of circle $\omega$ with the perpendicular rays from $I$ to
sides $BC, CA, AB$ respectively. Prove that lines $AD, BE, CF$ are concurrent.

Source : Excalibur .

It was because of the completely constructive proof of the claim :/
And also, I found a proof of problem $26$ with inversion and some ugly stuff Can somebody please post a Euclidean/trigonometric solution?

[Nadim Ul Abrar]

I found a proof of problem $26$ with inversion and some ugly stuff

It will be great if you share your solution .

[*Mahi*]

It will be great if you share your solution .

Solution(outline mainly) $\boxed{26}$:

If we invert the whole thing WRT incircle $(I)$, Then the condition $AD, BE, CF$ is concurrent is similar to circle $A'ID'$, ,$B'IE'$, $C'IF'$ (where $A$ being the midpoint of $D''E''$, $D''$ and $E''$ being the touchpoints) having a common point other than $I$, or all three of them being co-axal circles, or that their centers lie on a line.
With a little help of homothety that can be reduced to this problem:
In an acute $\triangle ABC$, $X,Y,Z$ are points on $OA, OB, OC$ respectively with $OX=OY=OZ$, let $XX' \perp OX$ and $X' = XX' \cap BC$. Prove that $X', Y', Z'$ are collinear.
Proving this is quite easy, using Menelaus on $XX'$, $YY'$, $ZZ'$ and $\triangle ABC$, and noticing that, the ratio at which $XX'$ and $YY'$ divides $AB$, is reciprocal to each other.
And thus the claim is proved.

Now, please, Euclidean proof anybody?

[Nadim Ul Abrar]

You may see this .@ Mahi
http://www.math.ust.hk/excalibur/v17_n3.pdf
(page 4)

[*Mahi*]

Then we can move on to a new problem, can't we?
Also, can you people please post more non-geometry problem, for changing the taste?

[SANZEED]

Problem $\boxed {27}$:
Find all values for a real parameter $\alpha$ for which there exists exactly one function $f:\mathbb {R}\rightarrow \mathbb {R}$ satisfying
$f(x^2+y+f(y))=f(x)^2+\alpha \cdot y$

Vietnam-2005

[Tahmid Hasan]

problem $\boxed {26}$
Let $I$ be the incenter of acute $\triangle ABC$. Let $\omega$ be a circle with center $I$ that lies inside $\triangle ABC$ . $D, E, F$ are the intersection points of circle $\omega$ with the perpendicular rays from $I$ to
sides $BC, CA, AB$ respectively. Prove that lines $AD, BE, CF$ are concurrent.
Source : Excalibur .

Completely trig-bashed it!
Draw a directly homothetic triangle $A'B'C'$ of $\triangle ABC$ such that $(A,A'),(B,B'),(C,C')$ are corresponding points with homothetic centre $I$ and $\omega $ is the incircle of $\triangle A'B'C'$.
WLOG we assume the inradius of $\triangle ABC$ is $1$ and the radius of $\omega$ is $r$. Let the distance from $\omega$ to the incircle of $\triangle ABC$ be $d$.
Let $B'C' \cap AB=X,B'C' \cap AC=Y$.
$\frac{\sin BAD}{\sin CAD}=\frac{XD}{YD}.\frac{AY}{AX}$.....$(1)$
Now $XD=B'X+B'D=\frac{d}{\sin B}+x(s-b)$.
$YD=YC'+C'D=\frac{d}{\sin C}+x(s-c)$.
$AX=AB-BX=c-\frac{d}{\sin B}$.
$BX=AC-CY=b-\frac{d}{\sin C}$.
We put these values in $(1)$ and similarly calculate $\frac{\sin CBE}{\sin ABE},\frac{\sin ACF}{\sin BCF}$ then applying trig-Ceva will yield $AD,BE,CF$ are concurrent.
Horrible solution!!!

Problem \boxed{27}:
Find all values for a real parameter $\alpha$ for which there exists exactly one function $f: \mathbb {N} \rightarrow \mathbb {N}$ satisfying
$f(x^2+y+f(y))=f(x)^2+y$.

Sanzeed, could you please re-check your problem from the source? I've seen it in IMOMATH functional equation note and the problem states the function is $\mathbb{R} \mapsto \mathbb{R}$.

[SANZEED]

Sorry, Edited now. The source I used had a typo. Sorry for the inconvenience.

[Phlembac Adib Hasan]

Problem $\boxed {27}$:
Find all values for a real parameter $\alpha$ for which there exists exactly one function $f:\mathbb {R}\rightarrow \mathbb {R}$ satisfying
$f(x^2+y+f(y))=f(x)^2+\alpha \cdot y$

Vietnam-2005

At first define $\alpha=a$.
Notice that $f$ can't be a constant function. Because assuming $f(x)=b$ and $y=0$ give together $b=b^2\Longrightarrow b=0,1$, contradicting with the condition of $a$. So $f$ can't be constant and on the other hand $a\neq 0$.
Define $P(x,y)\Longrightarrow f(x^2+y+f(y))=f(x)^2+ay$
$P(0,y)\Longrightarrow f(y+f(y))=f(0)^2+ay.......(i)$
So $f$ is surjective.
$P(-x,y)\Longrightarrow f(x)=\pm f(-x).......(ii)$
From surjectivity, assume $f(c)=0$.
Suppose $\exists c,d\mid f(c)=f(d)=0$
From $(i)$,
\[\begin{array}{lr}f(c+f(c))=f(0)^2+ac \\ f(d+f(d))=f(0)^2+ad\end{array} \bigg\}c=d\]
From $(ii)$, $f(c)=\pm f(-c)=0\Longrightarrow c=-c\Longrightarrow c=0$
So $f(0)=0$. Now rewrite previous equations.
$f(y+f(y))=ay.......(iii)$
$P(x,0)\Longrightarrow f(x^2)=f(x)^2$
Now notice $f(1^2)=f(1)^2\Longrightarrow f(1)=0,1$. Since $1\neq 0$, $f(1)=1$.
So $f(-1)=\pm 1$. Suppose $f(-1)=1$.
$P(1,-1)\Longrightarrow a=0$ which is not possible. So $f(-1)=-1$.
So $P(1,-1)\Longrightarrow f(-1)=1-a \Longrightarrow a=2$.

Now it's time to check whether it's correct or not. It's easy to see $f(x)=x$ is a solution. So our task is to check if it is the only solution. This part is not very hard. I am tired of typing today, so I may post it tomorrow.

[Tahmid Hasan]

Finally solved it!
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 $\forall x,y \in \mathbb{R}$.
Since $f(x^2)=f(x)^2$, we get $f(1)=0,1$ but $f(0)=0$ so $f(1)=1$. Hence $f(-1)=-1$.
$P(1,-1) \Rightarrow \alpha=2$.
$P(0,f(y)) \Rightarrow f(f(y)+f(f(y)))=\alpha.f(y) \Rightarrow f(\alpha.y)=\alpha.f(y) \Rightarrow f(2y)=2f(y)$.....$(4)$
Now plug $x+y$ for $x$ in $(3)$, we get $f(2xy)=2f(x)f(y)$
Using $(4)$ we get $f(xy)=f(x)f(y)$, so $f(x)$ is multiplicative $\forall x,y \in \mathbb{R}$
The additive and multiplicative property combined yields $f(x)=x,0 \forall x \in \mathbb{R}$ in which the later is a contradiction.
So for $\alpha=2$ the unique solution is $f(x)=x$.
Very short solution Adib, I'm crying seeing your solution