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f(f(n))=2n+2
Jackson0423   0
22 minutes ago
Source: 2013 KMO Second Round

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function satisfying the following conditions for all \( n \in \mathbb{N} \):
\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]Find the value of \( f(2013) \).
0 replies
Jackson0423
22 minutes ago
0 replies
Tangents involving a centroid with an isosceles triangle result
pithon_with_an_i   0
22 minutes ago
Source: Revenge JOM 2025 Problem 5, Revenge JOMSL 2025 G5, Own
A triangle $ABC$ has centroid $G$. A line parallel to $BC$ passing through $G$ intersects the circumcircle of $ABC$ at a point $D$. Let lines $AD$ and $BC$ intersect at $E$. Suppose a point $P$ is chosen on $BC$ such that the tangent of the circumcircle of $DEP$ at $D$, the tangent of the circumcircle of $ABC$ at $A$ and $BC$ concur. Prove that $GP = PD$.

Remark 1
Remark 2
0 replies
pithon_with_an_i
22 minutes ago
0 replies
Thailand MO 2025 P2
Kaimiaku   2
N 27 minutes ago by carefully
A school sent students to compete in an academic olympiad in $11$ differents subjects, each consist of $5$ students. Given that for any $2$ different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least $4$ different subjects.
2 replies
Kaimiaku
Today at 7:38 AM
carefully
27 minutes ago
Aime 2005a #15
4everwise   22
N 29 minutes ago by Ilikeminecraft
Source: Aime 2005a #15
Triangle $ABC$ has $BC=20$. The incircle of the triangle evenly trisects the median $AD$. If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n$.
22 replies
4everwise
Nov 10, 2005
Ilikeminecraft
29 minutes ago
Problem 2 (First Day)
Valentin Vornicu   84
N 44 minutes ago by cj13609517288
Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab+bc+ca = 0$ we have the following relations

\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]
84 replies
Valentin Vornicu
Jul 12, 2004
cj13609517288
44 minutes ago
Maximum number of divisor in Binom Coeff.
nataliaonline75   0
an hour ago
Let $k \geq 2$, determine the maximal number of divisors from $n \choose k $ may have in the range $n-k+ 1,...,n$ , as $n$ runs through integers $\geq k$.
0 replies
nataliaonline75
an hour ago
0 replies
Nice geometry...
Sadigly   1
N an hour ago by aaravdodhia
Source: Azerbaijan Senior NMO 2020
Let $ABC$ be a scalene triangle, and let $I$ be its incenter. A point $D$ is chosen on line $BC$, such that the circumcircle of triangle $BID$ intersects $AB$ at $E\neq B$, and the circumcircle of triangle $CID$ intersects $AC$ at $F\neq C$. Circumcircle of triangle $EDF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. Lines $FD$ and $IC$ intersect at $Q$, and lines $ED$ and $BI$ intersect at $P$. Prove that $EN\parallel MF\parallel PQ$.
1 reply
Sadigly
Sunday at 10:17 PM
aaravdodhia
an hour ago
Inspired by lbh_qys.
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +b+1}+ \frac{b}{b^2+a +b+1}  \leq  \frac{1}{2} $$$$ \frac{a}{a^2+ab+a+b+1}+ \frac{b}{b^2+ab+a+b+1} \leq   \sqrt 2-1  $$$$\frac{a}{a^2+ab+a+1}+ \frac{b}{b^2+ab+b+1} \leq  \frac{2(2\sqrt 2-1)}{7} $$$$\frac{a}{a^2+ab+b+1}+ \frac{b}{b^2+ab+a+1} \leq  \frac{2(2\sqrt 2-1)}{7} $$
5 replies
sqing
Today at 3:45 AM
sqing
an hour ago
AM=CN in Russia
mathuz   25
N an hour ago by Ilikeminecraft
Source: AllRussian-2014, Grade 11, day1, P4
Given a triangle $ABC$ with $AB>BC$, $ \Omega $ is circumcircle. Let $M$, $N$ are lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. $K(.)=MN\cap AC$ and $P$ is incenter of the triangle $AMK$, $Q$ is K-excenter of the triangle $CNK$ (opposite to $K$ and tangents to $CN$). If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.

M. Kungodjin
25 replies
mathuz
Apr 29, 2014
Ilikeminecraft
an hour ago
IMO 2010 Problem 6
mavropnevma   41
N an hour ago by pi271828
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
41 replies
mavropnevma
Jul 8, 2010
pi271828
an hour ago
Existence of a circle tangent to four lines
egxa   3
N Apr 21, 2025 by mathuz
Source: All Russian 2025 10.2
Inside triangle \(ABC\), point \(P\) is marked. Point \(Q\) is on segment \(AB\), and point \(R\) is on segment \(AC\) such that the circumcircles of triangles \(BPQ\) and \(CPR\) are tangent to line \(AP\). Lines are drawn through points \(B\) and \(C\) passing through the center of the circumcircle of triangle \(BPC\), and through points \(Q\) and \(R\) passing through the center of the circumcircle of triangle \(PQR\). Prove that there exists a circle tangent to all four drawn lines.
3 replies
egxa
Apr 18, 2025
mathuz
Apr 21, 2025
Existence of a circle tangent to four lines
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Source: All Russian 2025 10.2
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egxa
210 posts
#1 • 1 Y
Y by buratinogigle
Inside triangle \(ABC\), point \(P\) is marked. Point \(Q\) is on segment \(AB\), and point \(R\) is on segment \(AC\) such that the circumcircles of triangles \(BPQ\) and \(CPR\) are tangent to line \(AP\). Lines are drawn through points \(B\) and \(C\) passing through the center of the circumcircle of triangle \(BPC\), and through points \(Q\) and \(R\) passing through the center of the circumcircle of triangle \(PQR\). Prove that there exists a circle tangent to all four drawn lines.
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MathLuis
1526 posts
#2
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Let $O_1,O_2,O_3$ be the centers of $(BPC), (PQR), (BQRC)$ respectively (last circle exists from invert at $(A, AP)$ so dw) we claim that the quadrilateral asked for has as incenter $O_3$, we already have two angle bisectors so all we need is to check that the angle bisectors are in fact perpendicular bisectors of the sides of $BQRC$. See that:
\[ \measuredangle O_2RC=\measuredangle PRC-\measuredangle PRO_2=\measuredangle APC+90+\measuredangle RQP=\measuredangle APC+90+\measuredangle RQB+\measuredangle BQP=\measuredangle BPC+90+\measuredangle ACB=\measuredangle BCO_1+\measuredangle ACB=\measuredangle RCO_1 \]And this shows that the perpendicular bisector of $RC$ becomes an angle bisector so you do the same to get that so does $BQ$ and from here we are obviously done :cool:.
This post has been edited 3 times. Last edited by MathLuis, Apr 18, 2025, 11:32 PM
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mathlearner-sd
17 posts
#3
Y by
stop the sweat
...
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mathuz
1524 posts
#4
Y by
Probably, a similar solution to the above:

We have $AP^2 = AQ\cdot AB=AR\cdot AC$. So the inversion centered at $A$ with radius $AP$ maps $(PQR)$ to $(BPC)$, and vice versa. In other words, $A$ is the exsimilicenter of $(PQR)$ and $(BPC)$. Furthermore, the points $B, Q, R, C$ are lie on a circle, assume centered at $O$.

Let the centers of $(BPC)$ and $(PQR)$ are $O_1$ and $O_2$, respectively. Suppose $AR$ intersects $(BPC)$ again at $R'$ (other than $C$). Then $O_2R \parallel O_1R'$ and $O_1CR'$ is an isosceles triangle, which implies $\angle O_2RC = \angle O_1CR$.
A similar argument finishes the proof now; $O$ turns out to be the incenter for the quadrilateral.
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