Special Problem Marathon
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 \(AB\).
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- FuadAlAlam
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Re: Special Problem Marathon
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.Mursalin wrote: ↑Tue Jun 01, 2021 7:02 pmLet \(G\) 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 \(AB\).
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.
- FuadAlAlam
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Re: Special Problem Marathon
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 $C_i$.
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 $C_i$.
- SMA_Nahian
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Re: Special Problem Marathon
$\textbf{Solution (P2):}$
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. $\blacksquare$
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.
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Re: Special Problem Marathon
$\textbf{Problem 3:}$
Let $a, b, c, d$ be positive real numbers such that $a + b + c + d = 1$. Prove that
,
$$6(a^3 + b^3 + c^3 + d^3) \geq (a^2 + b^2 + c^2 + d^2) + \dfrac{1}{8}$$
Let $a, b, c, d$ be positive real numbers such that $a + b + c + d = 1$. Prove that
,
$$6(a^3 + b^3 + c^3 + d^3) \geq (a^2 + b^2 + c^2 + d^2) + \dfrac{1}{8}$$
- FuadAlAlam
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Re: Special Problem Marathon
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.
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.
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Re: Special Problem Marathon
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 $x, y \in \mathbb{Z}$.
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 $x, y \in \mathbb{Z}$.
- Mehrab4226
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Re: Special Problem Marathon
FuadAlAlam wrote: ↑Tue Jun 01, 2021 11:16 pmProblem 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 $x, y \in \mathbb{Z}$.
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.
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- Mehrab4226
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Re: Special Problem Marathon
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:
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:
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
-Henri Poincaré
- Anindya Biswas
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Re: Special Problem Marathon
But for which $x$, $f(x)=0$ and for which $x$, $f(x)=\pm1$?Mehrab4226 wrote: ↑Wed Jun 02, 2021 12:07 pmFuadAlAlam wrote: ↑Tue Jun 01, 2021 11:16 pmProblem 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 $x, y \in \mathbb{Z}$.
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."
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— John von Neumann