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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 \(AB\).

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.

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 $C_i$.

$\textbf{Solution (P2):}$

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. $\blacksquare$

$\textbf{Problem 3:}$

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

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.

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 $x, y \in \mathbb{Z}$.

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 $x, y \in \mathbb{Z}$.

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:

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.