## IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

Problem 32:Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.

Source:IMO shortlist 2011

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

At first we prove the following lemma.

Lemma:$ABC$ be a triangle where $AB=AC$.Let $P$ be a brocard point of $\bigtriangleup ABC$ such that $\angle ABP=\angle BAP=\angle CAP$.Then $BP$ is a median.

Proof:Let $BP$ meets $AC$ at $Q$.After some angle chasing we get $\angle APQ=\angle A,\angle CPQ=\angle B$.
Again in $\bigtriangleup ACP$ and $\bigtriangleup BCP$
$\angle CAP=\angle BCP,\angle ACP=\angle C-\angle BCP=\angle B-\angle ABP=\angle CBP$
So $\bigtriangleup ACP \sim \bigtriangleup BCP$ and $\dfrac{AP}{CP}=\dfrac{AC}{BC}$
So $\dfrac{AQ}{BQ}=\dfrac{AP.sin APQ}{CP.sin CPQ}=\dfrac{AC.sin A}{BC.sin B}=\dfrac{AC.BC}{BC.AC}=1$
SO $Q$ is the midpoint of $AC$.(proved)

Main proof:By our lemma,$P$ lies on $BD$.In $\bigtriangleup AEC$ and $\bigtriangleup ABE$
$AC=BC,AE=BE$ and $\angle CAE=\angle BAE$.
So $\bigtriangleup ACE \cong \bigtriangleup ABE$
$\angle ABI=\angle ABE=\angle ACE=\angle DCE=\angle DBE=\angle \angle PBI$.So $BI$ bisects $\angle ABP$
Again $\angle PAF=\angle AFP-\angle ABP=\angle AEB-\angle ADP-\angle DAP$
$=180^{\circ}-\dfrac{\angle BEC}{2}-\angle ADP-\angle DAP$
$=90^{\circ}+\dfrac{180^{\circ}-\angle BDC}{2}-\angle ADP-\angle DAP=90^{\circ}+\dfrac{\angle ADB}{2}-\angle ADP-\angle DAP$
$=\dfrac{\angle DAP+\angle APD}{2}-\angle DAP=\dfrac{\angle APD-\angle DAP}{2}=\dfrac{\angle BAP}{2}$
So $\angle PAF=\angle BAF$.
So $I$ is the incenter of $\bigtriangleup ABP$
$\angle API=\angle BPI$
$\angle API=\angle API=\dfrac{\angle APB}{2}=90^{\circ}-\dfrac{\angle A}{2}=\angle C$
Again $\angle APC=180^{\circ}-\angle C$
$\angle APC+\angle API=\angle C+180-\angle C=180^{\circ}$
So $C,P,I$ collinear
So $P=CI \cap BD$
Again $K=P=CI \cap BD$
So $P\equiv K$
This implies $I$ is the incenter of $\bigtriangleup KAB$.

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: IMO Marathon

Problem $$\boxed{33}$$
Consider the circle $$\omega$$ with diameter $$BC$$ in $$\triangle ABC$$. $$AP$$ and $$AQ$$ are tangents to $$\omega$$ from $$A$$ where $$P$$ and $$Q$$ are the points of tangency, respectively. If $$H$$ is the orthocenter of $$\triangle ABC$$, prove that the points $$P,Q,H$$ are collinear.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

Let $\omega$ intersect $AB,AC$ at $S,T$. Then $CS\perp AB$ and $BT\perp AC$. So $H=CS\cap BT$. Now by Brokard's theorem, $H$ lies on the polar of $A$ wrt $\omega$. Again $PQ$ is the polar of $A$.So $H$ lies on $PQ$.

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: IMO Marathon

My solution is exactly the same. Brokard+Polar. Post a non-geometry problem.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

Problem $34$
Find all functions $f:N\rightarrow N$ for which for all natural $n$
$f(n+1)>\dfrac{f(n)+f(f(n))}{2}$
Last edited by asif e elahi on Sun Aug 31, 2014 6:21 pm, edited 1 time in total.

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: IMO Marathon

Is it $$f(f(n))$$ on the right side of your inequality?
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

Nirjhor wrote:Is it $$f(f(n))$$ on the right side of your inequality?
Yes.I have edited my post.

Fm Jakaria
Posts: 79
Joined: Thu Feb 28, 2013 11:49 pm

### Re: IMO Marathon

My solution to problem 34 is as follows:

The solution set is
S=$\left \{ \forall r\in\mathbb{N}\cup\left \{ +\infty\ \right \}|\;f_r(n)= \begin{cases} & n, \ {if }\ n<r \\ & n+1, \ { if }\ n\geqslant r \end{cases} \right \}$

We start by defining
$a_k$:= min{f(i)|$i\in\mathbb{N},i\geqslant k$}. For $\forall k \in\mathbb{N}, a_k$ is well-defined because $\mathbb{N}$ is well-ordered.
Also $S_k$:= {j|$j\in\mathbb{N}\ , \ j\geqslant k\ , \ f(j) = a_k$}. This must be non-empty; for $\forall k\in\mathbb{N}$, again.

We introduce the following lemma.
Lemma: Any solution f is strictly-increasing.
Proof: We’ll inductively show that for all n $\in\mathbb{N}$; for any m>n, f(m) > f(n) holds. But before that, we observe that for $\forall n\in\mathbb{N},\ f(n+1)> \frac{f(f(n))+f(n)}{2} \geqslant$ min{f(n), f(f(n))} …….(1)
So if for some i>1, i$\in S_1$;using (1), f(i) > f(i-1) or f(i)> f(f(i-1)), where f(i)=$a_1$; a contradiction.
So for i>1; f(i)>$a_1$. But as $S_1 \neq$ {}, $S_1$={1}; so $a_1$=f(1). This is the base step.
Now assume that the hypothesis is true for n=1,2,…,k. Suppose for some i>k+1; i$\in S_{k+1}$. Then using (1) again, f(i) > f(i-1) or f(i)> f(f(i-1)). The first is impossible as (i-1)$\geqslant$k+1, while f(i)=$a_{k+1}$. If the second is true , we must have f(i-1) < k+1; for the same reason. But (i-1) > k, so the inductive hypothesis implies that
1$\leqslant$f(1) < f(2) < ……<f(k) < f(i-1) < k+1.
This is clearly a contradiction because (k+1) distinct integers can’t exist in the interval [1,k+1).

So for $\forall$ i> k+1; f(i)> $a_{k+1}$. We now simply note that as $S_{k+1} \neq${}, $S_{k+1}$={k+1} and so $a_{k+1}$=f(k+1). This completes the induction.

With the lemma; we have $\forall n \in\mathbb{N}, f(n) \geqslant$ n. Further, if $\exists t\in\mathbb{N} : \exists c\in\mathbb{N}_0, f(t) \geqslant t+c$; then for $\forall j\geqslant t, f(j) \geqslant j+c.$
Suppose there exists such t$\in\mathbb{N}$ so that we can choose c=2. We now define
$b_i := f(t+i) – f(t+i-1)$, for $\forall i\in\mathbb{N}.$ By the lemma, $\forall i\in\mathbb{N}, b_i\in\mathbb{N}.$
Now for $\forall s \geqslant t, f(s) \geqslant$ s+2. The lemma gives that f(f(s)) $\geqslant f(s+2).$
Then for $\forall i\in\mathbb{N}; \ f(t+i) > \frac{f(t+i-1)+f(f(t+i-1))}{2} \geqslant \frac{f(t+i-1)+f(t+i+1)} {2}$
$\Rightarrow b_i$= f(t+i) – f(t+i-1) > f(t+i+1)- f(t+i) = $b_{i+1}$.
This gives $b_1>b_2>b_3>…….;$ an infinite strictly decreasing sequence of positive integers, clearly a contradiction.
Now we are only left with the set S from which f can come. Our final claim is that: any f in S is a solution.
When r=+$\infty$, f(n)=n for $\forall n \in\mathbb{N}$; is obviously a solution.
Let r< +$\infty$. For n $\geqslant r, f(f(n))=f(n+1)=n+2.$
$\Rightarrow n+2 = f(n+1) > n+1.5 = \frac{(n+2)+(n+1)}{2} = \frac{f(f(n))+f(n)}{2}.$

If n<r-1 or n=r-1 for some n$\in\mathbb{N}$, then respectively f(n+1) = n+1 or f(n+1) = n+2; in both cases greater than
$\frac{f(f(n))+f(n)}{2} = \frac{n+n}{2} = n.$
So we have established our claim.
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.

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: IMO Marathon

At first,we prove the following lemma.

Lemma: For all $n \in N$
$f(1)<f(2)<..........<f(n)<f(i)$ for all $i>n$
Proof: We induct on n.For the proof of base case,let $f(a)$ be the smallest element of the range of $f$.If $a>1$,then $a-1$ is positive.
$f(a)>\dfrac{f(a-1)+f(f(a-1))}{2} \geq min(f(a-1),f(f(a-1)) )$
But $f(a) \leq f(b)$ for all $b$.A contradiction.So $a=1$.
Now we assume that this is true for $n=k-1$.Now we will prove this for $n=k$.
Let $f(p)$ be the smallest element of the set $\left \{ f(k),f(k+1)........... \right \}$ .It's enough to prove $p=k$. We consider 2 cases.
$f(f(p-1)) \geq f(p-1)$:
$f(p)>\dfrac{f(p-1)+f(f(p-1))}{2} \geq f(p-1)$
As $f(p)$ is the smallest element of the set ${f(k),f(k+1),....................}$,we must have $f(p-1) \in \left \{ f(k),f(k+1)........... \right \}$
So $p-1 \leq k-1$
or $p \leq k$.This implies $p=k$.
$f(p-1)>f(f(p-1))$:
$f(p)>\dfrac{f(p-1)+f(f(p-1))}{2} \geq f(f(p-1))$
So $f(f(p-1)) \in {f(1),f(2)....f(k-1)}$.
$f(p-1) \leq k-1 \leq f(k-1)$
This implies $p-1 \leq k-1 \rightarrow p \leq k$.So $p=k$.
So our induction is complete

Main proof:From the lemma ,we can deduce that,$f$ is strictly increasing.Let there exist $m \in N$ s.t $f(m) \geq m+2$.
Then $f(m+k) \geq m+k+2$
$f(m+k+1)>\dfrac{f(m+k)+f(f(m+k))}{2} \geq \dfrac{f(m+k)+f(m+k+2)}{2}$
So $f(m+k+1)-f(m+k)>f(m+k+2)-f(m+k+1)$
So the function $f(n+1)-f(n)$ is strictly decreasing for $n \geq m$.But $f(n+1)-f(n)$ is positive for all $n$.So this is not possible.This implies $f(n) \leq n+1$
So for all $n$, $f(n)=n$ or $n+1$.Let $d$ be the smallest integer for which $f(d)=d+1$.Then $f(n)=n+1$ for all $n \geq d$ and $f(n)=n$ for all $n<d$.It's easy to see that this satisfies the given conmdition.

So the solutions are $f(n)=n$ for all $n \in N$
$f(n)=n$ for $n<d$ and $f(n)=n+1$ for $n\geq d$ where $d$ is any positive integer