Ny Oppgave: Zeroes are written in all cells of a $5 \times 5$ board. We can take an arbitrary cell and increase by 1 the number in this cell and all cells having a common side with it. Is it possible to obtain the number 2012 in all cells simultaneously?

Ny Oppgave: Let $CSTB$ be a square and let $A$ be an arbitrary point on $ST$. Let $H$ be the orthocenter of $\triangle{ABC}$ and $E, F$ be the altitudes from $B$ and $C$ to $AC$ and $AB$. If $P = TF \cap SE$ show that $PH$ passes through the midpoint of $ST$.
https://latex.artofproblemsolving.com ...

Solution: Let $P = NM \cap BC$, then let $Y = EX \cap BC$, by the Prism lemma we know that $(B, C; Y, P) = -1$, thus applying the Prism lemma again we obtain that $A, D$ and $P$ are collinear. Consequently by Brocard we know that $EF = \text{Polar}_{(BCX)}(P)$ which implies that $XP$ is tangent to ...

Solution: Notice,

Construction: Obviously by considering $\mathbb{D}$ to be the collection of strings starting with $B$ except for $BB\ldots B$ the grid cannot be filled. Thus $\mathbb{D}$ can have a size of $2^{k - 1} - 1$.

Bound: Assume that $|\mathbb{D}| \geq 2^{k - 1}$, then let us pair all ...

Ny Oppgave: In a triangle $\triangle ABC$, $D$ and $E$ are respectively on $AB$ and $AC$ such that $DE\parallel BC$. Let $P$ be $BE \cap CD$, $M$ be the second intersection of $(APD)$ and $(BCD)$, and $N$ be the second intersection of $(APE)$ and $(BCE)$. Let $\omega$ be the circle passing through ...

Solution: Let $K$ be the midpoint of $BC$, then it is well known that $KE$ and $KF$ are both tangent to $(AEF)$, thus taking homothety at $A$ we obtain that $DP \parallel HK$, however it is also well known that $HK$ passes through the antipode of $A$ in $(ABC)$ (let it be $A'$). It is also well ...

Problem: Find all triples of positive integers $(a,b,n)$ where $n\ge 2$, and both $a^n+b$ and $b^n+a$ are powers of $2$.

Solution: Let $n = xy$, then,
\[
x^2 + x + 1 \mid (xy)^2 + xy + 1
\]
notice that $x^2 + x + 1 \mid x^4 + x^2 + 1$, then we obtain,
\[
x^2 + x + 1 \mid x^2 (x^2 - y^2) + x^2 - xy
\]
next we obtain,
\[
x^2 + x + 1 \mid x(x^2 - y^2) + x - y = (x - y)(x(x + y) + 1)
\\ = (x - y)(x^2 + xy + 1 ...

Ny Oppgave: ( AoPS ) Notice that the Anchor Point Conjugation Theorem can be reformulated as,

https://latex.artofproblemsolving.com/7/f/a/7fa395c9aa5005401508aac09d5b4466b159195c.png

Theorem: Fix a point \(F \in (ABC)\) and an arbitrary point \(D\) in the plane. Let \(\omega\) be a circle ...

It is trivial to see that,
\[
\begin{cases}
AD^2 = \frac{b c \left(-12 a^2+12 b^2+25 b c+12 c^2\right)}{(3 b+4 c)^2}\\ BE^2 = \frac{a c \left(2 a^2+5 a c-2 b^2+2 c^2\right)}{(a+2 c)^2}\\
CF^2 = \frac{a b \left(6 a^2+13 a b+6 b^2-6 c^2\right)}{(2 a+3 b)^2}\\
\frac{T}{S} = 1-\frac{np}{ab}-\frac{qr ...

Ny Oppgave: Vis at for alle $a, b, c \in \mathbb{R}^{+}$ har vi,
\[
\sum_{\text{cyc}} \frac{\sqrt{b + c}}{a} \geq \frac{4(a + b + c)}{\sqrt{(a + b)(b + c)(c + a)}}
\]

Satisfying inequality :mrgreen:

Proof: Notice,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + \sqrt{2a + 2c}} \right)^3 = \sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac{a + c}{2}} } \right)^3
\]
however by AM-GM,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac ...

Ny Oppgave: Bevise at det eksiterer en permutasjon av de naturlige tallene så summen av de første n er delelig på n.

Vurdere fargeleggingen i $4$ farger slik at,
\[
\begin{array}{|c|c|c|c}
\hline
1 & 2 & 1 & \cdots \\
\hline
3 & 4 & 3 & \cdots \\
\hline
1 & 2 & 1 & \cdots \\
\hline
\vdots & \vdots & \vdots & \ddots
\end{array}
\]
La grafens noder være ruter med $1$ og kanter gjennom $(u, v)$ hvis en domino med ...

Ny Oppgave: Bevis at det eksisterer et tall som er større enn $2^\binom{2025}{1012} + 1$, slik at summen av primtall mindre enn tallet er relativt primt med det.