BdMO Online Forum

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

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 $\bigtriangleup KAB$.

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.

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 $PQ$.

My solution is exactly the same. Brokard+Polar. Post a non-geometry problem.

Problem $34$
Find all functions $f:N\rightarrow N$ for which for all natural $n$
$f(n+1)>\dfrac{f(n)+f(f(n))}{2}$

Is it \(f(f(n))\) on the right side of your inequality?

Nirjhor wrote:Is it \(f(f(n))\) on the right side of your inequality?

Yes.I have edited my post.

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.

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