Special Problem Marathon

Discussion on Bangladesh National Math Camp
User avatar
Mursalin
Posts:68
Joined:Thu Aug 22, 2013 9:11 pm
Location:Dhaka, Bangladesh.
Special Problem Marathon

Unread post by Mursalin » Tue Jun 01, 2021 7:02 pm

Let \(G\) be the centroid of a right-angled triangle \(ABC\) with \(\angle BCA = 90^\circ\). Let \(P\) be the point on ray \(AG\) such that \(\angle CPA=\angle CAB\), and let \(Q\) be the point on ray \(BG\) such that \(\angle CQB =\angle ABC\). Prove that the circumcircles of triangles \(AQG\) and \(BPG\) meet at a point on side \(AB\).
This section is intentionally left blank.

User avatar
FuadAlAlam
Posts:30
Joined:Wed Sep 16, 2020 11:10 am
Location:Dhaka, Bangladesh
Contact:

Re: Special Problem Marathon

Unread post by FuadAlAlam » Tue Jun 01, 2021 8:42 pm

Mursalin wrote:
Tue Jun 01, 2021 7:02 pm
Let \(G\) be the centroid of a right-angled triangle \(ABC\) with \(\angle BCA = 90^\circ\). Let \(P\) be the point on ray \(AG\) such that \(\angle CPA=\angle CAB\), and let \(Q\) be the point on ray \(BG\) such that \(\angle CQB =\angle ABC\). Prove that the circumcircles of triangles \(AQG\) and \(BPG\) meet at a point on side \(AB\).
Let $M$ be the midpoint of $AB$ and $H$ be the feet of the altitude from $C$ to $AB$. We claim that $H$ is the desired point of intersection.
We have $\measuredangle CPG=\measuredangle CPA=\measuredangle BAC=\measuredangle MAC=\measuredangle ACM=\measuredangle ACG$. Thus, $AC$ is tangent to $(CPG)$ at $C$. Now, $AH \cdot AB=AC^2=AG \cdot AP$ which implies that $H$ lies on $(BPG)$. Similarly, $H$ lies on $(AQG)$, as desired.

User avatar
FuadAlAlam
Posts:30
Joined:Wed Sep 16, 2020 11:10 am
Location:Dhaka, Bangladesh
Contact:

Re: Special Problem Marathon

Unread post by FuadAlAlam » Tue Jun 01, 2021 8:49 pm

Problem 2:

There are $N$ boxes labelled $B_1, B_2, \ldots, B_N$ which are filled with balls of $N$ different colours $C_1, C_2, \ldots, C_N$. Further, it is known that for each colour we can partition the boxes into two sets, such that the total number of balls of that colour in both sets is the same.
Prove that one can choose a box $B_i$, such that at most half of the balls in box $B_i$ are of colour $C_i$.

User avatar
SMA_Nahian
Posts:5
Joined:Thu Dec 03, 2020 9:34 pm
Location:Demra, Dhaka, Bangladesh

Re: Special Problem Marathon

Unread post by SMA_Nahian » Tue Jun 01, 2021 10:13 pm

$\textbf{Solution (P2):}$

Let $D_i$ and $ M_i$ be the number of ball of color $C_i$ in box $B_i$ and not in box $B_i$ respectively. And $S_i$ be the number of ball in $B_i$.

For the sake of contradiction , lets assume that, $$D_i > \dfrac{1}{2} S_i \ \forall \ i \in \{1,2,3, \dots, N\}$$
$$\Rightarrow 2 \cdot D_i > S_i$$
Now,
$$\sum \limits_{i=1} ^N D_i + \sum \limits_{i=1} ^N M_i = \sum \limits_{i=1} ^N S_i < \sum \limits_{i=1} ^N 2\cdot D_i$$
$$\Rightarrow\sum \limits_{i=1} ^N M_i < \sum \limits_{i=1} ^N D_i$$
So, there will be at least a colour $C_j$ suct that $$M_j < D_j$$
So, for any partition of two sets of the boxes, the number of ball coloured $C_j$ is always bigger in the set containing box $B_j$ than the other set. And this contradicts our problem statement. $\blacksquare$
Last edited by SMA_Nahian on Tue Jun 01, 2021 10:43 pm, edited 1 time in total.

User avatar
SMA_Nahian
Posts:5
Joined:Thu Dec 03, 2020 9:34 pm
Location:Demra, Dhaka, Bangladesh

Re: Special Problem Marathon

Unread post by SMA_Nahian » Tue Jun 01, 2021 10:26 pm

$\textbf{Problem 3:}$

Let $a, b, c, d$ be positive real numbers such that $a + b + c + d = 1$. Prove that
,
$$6(a^3 + b^3 + c^3 + d^3) \geq (a^2 + b^2 + c^2 + d^2) + \dfrac{1}{8}$$

User avatar
FuadAlAlam
Posts:30
Joined:Wed Sep 16, 2020 11:10 am
Location:Dhaka, Bangladesh
Contact:

Re: Special Problem Marathon

Unread post by FuadAlAlam » Tue Jun 01, 2021 11:12 pm

Solution (P3):

By AM-GM, we have $$4(a^3+b^3+c^3+d^3) \geq (a^3+b^3+c^3+d^3)+(ab^2+bc^2+cd^2+da^2)+(ac^2+a^2c)+(bd^2+b^2d)+(ad^2+ba^2+cb^2+dc^2) \geq (a^2+b^2+c^2+d^2)(a+b+c+d)$$ and $$16(a^3+b^3+c^3+d^3) \geq (a^3+b^3+c^3+d^3)+3(ab^2+bc^2+cd^2+da^2)+3(ac^2+a^2c)+3(bd^2+b^2d)+3(ad^2+ba^2+cb^2+dc^2)+6(abc+bcd+cda+dab) \geq (a+b+c+d)^3$$

Therefore, $6(a^3+b^3+c^3+d^3) = 4(a^3+b^3+c^3+d^3)+2(a^3+b^3+c^3+d^3) \geq (a^2+b^2+c^2+d^2)(a+b+c+d)+ \frac{1}{8}(a+b+c+d)^3 = (a^2+b^2+c^2+d^2)+\frac{1}{8}$ and we are done.
Last edited by FuadAlAlam on Wed Jun 02, 2021 6:37 pm, edited 1 time in total.

User avatar
FuadAlAlam
Posts:30
Joined:Wed Sep 16, 2020 11:10 am
Location:Dhaka, Bangladesh
Contact:

Re: Special Problem Marathon

Unread post by FuadAlAlam » Tue Jun 01, 2021 11:16 pm

Problem 4:

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$$
for all $x, y \in \mathbb{Z}$.

User avatar
Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: Special Problem Marathon

Unread post by Mehrab4226 » Wed Jun 02, 2021 12:07 pm

FuadAlAlam wrote:
Tue Jun 01, 2021 11:16 pm
Problem 4:

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$$
for all $x, y \in \mathbb{Z}$.
Let,
$p(x,y)=f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$
Now, by $p(0,0)$ we get,
$f(0)+f(0)^3=f(0)^3+f(0)^3$
Or, $f(0)^3-f(0)=0$
Either, $f(0)=0$, or $f(0)=1$ or, $f(0)=-1$
Now, if $f(0)=0$ then by $p(x,0)$ we get,
$f(x)=f(x)^3$
$\rightarrow f(x)=0,1,-1$ [But the latter two are not acceptable]

Again if $f(0)-1$ then by $p(x,0)$ we get,
$f(x)+f(x)^2=1+f(x)^3$
Solving for $f(x)$ we get,
$f(x)=1,-1$ but $f(x)=-1$ is not acceptable.
Similarly for $f(0)=-1$,
$f(x)=-1,1$ [where the 2nd is not acceptable]

$\therefore f(x)=-1,0,1 $
Last edited by Mehrab4226 on Wed Jun 02, 2021 12:10 pm, edited 1 time in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

User avatar
Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: Special Problem Marathon

Unread post by Mehrab4226 » Wed Jun 02, 2021 12:09 pm

Problem: 05
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Source:
EGMO 2012(1)
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

User avatar
Anindya Biswas
Posts:264
Joined:Fri Oct 02, 2020 8:51 pm
Location:Magura, Bangladesh
Contact:

Re: Special Problem Marathon

Unread post by Anindya Biswas » Wed Jun 02, 2021 12:33 pm

Mehrab4226 wrote:
Wed Jun 02, 2021 12:07 pm
FuadAlAlam wrote:
Tue Jun 01, 2021 11:16 pm
Problem 4:

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$$
for all $x, y \in \mathbb{Z}$.
Let,
$p(x,y)=f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$
Now, by $p(0,0)$ we get,
$f(0)+f(0)^3=f(0)^3+f(0)^3$
Or, $f(0)^3-f(0)=0$
Either, $f(0)=0$, or $f(0)=1$ or, $f(0)=-1$
Now, if $f(0)=0$ then by $p(x,0)$ we get,
$f(x)=f(x)^3$
$\rightarrow f(x)=0,1,-1$ [But the latter two are not acceptable]

Again if $f(0)-1$ then by $p(x,0)$ we get,
$f(x)+f(x)^2=1+f(x)^3$
Solving for $f(x)$ we get,
$f(x)=1,-1$ but $f(x)=-1$ is not acceptable.
Similarly for $f(0)=-1$,
$f(x)=-1,1$ [where the 2nd is not acceptable]

$\therefore f(x)=-1,0,1 $
But for which $x$, $f(x)=0$ and for which $x$, $f(x)=\pm1$?
I think you showed that the range of the function is $\{-1,0,1\}$ but that doesn't mean $f(x)$ has the same value for every input.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Post Reply