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Collinear points
tenplusten   2
N 43 minutes ago by Blackbeam999
Let $A,B,C$ be three collinear points and $D,E,F$ three other collinear points. Let $G,H,I$ be the intersection of the lines $BE,CF$ $AD,CF$ and $AD,CE$,respectively. If $AI=HD$ and $CH=GF$.Prove that $BI=GE$



I hope you will use Pappus theorem in your solutions.
2 replies
tenplusten
Jun 20, 2016
Blackbeam999
43 minutes ago
Simple Geometry
AbdulWaheed   0
an hour ago
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
0 replies
AbdulWaheed
an hour ago
0 replies
Graph Theory
ABCD1728   0
an hour ago
Can anyone provide the PDF version of "Graphs: an introduction" by Radu Bumbacea (XYZ press), thanks!
0 replies
ABCD1728
an hour ago
0 replies
Number theory for people who love theory
Assassino9931   3
N 2 hours ago by NamelyOrange
Source: Bulgaria RMM TST 2019
Prove that there is no positive integer $n$ such that $2^n + 1$ divides $5^n-1$.
3 replies
Assassino9931
Jul 31, 2024
NamelyOrange
2 hours ago
Interesting inequalities
sqing   0
2 hours ago
Source: Own
Let $ a,b> 0 ,   a+b+a^2+b^2=2.$ Prove that
$$ab+ \frac{k}{a+b+ab} \geq \frac{3-k+(k-1)\sqrt{5}}{2}$$Where $ k\geq 2. $
$$ab+ \frac{2}{a+b+ab} \geq \frac{1+\sqrt{5}}{2}$$$$ab+ \frac{3}{a+b+ab} \geq  \sqrt{5} $$
0 replies
sqing
2 hours ago
0 replies
Stability of Additive Cauchy Equation
doanquangdang   1
N 3 hours ago by jasperE3
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
|f(x+y)-f(x)-f(y)-x y| \leq \varepsilon\left(|x|^p+|y|^p\right)
$$for some $\varepsilon>0,$ $p \in[0,1)$ and for all $x, y \in \mathbb{R}$, then there exists a unique solution $a: \mathbb{R} \rightarrow \mathbb{R}$ of the functional equation $a(x+y)=$ $a(x)+a(y)$ for all $x, y \in \mathbb{R}$ such that
$$
\left|f(x)-a(x)-\frac{1}{2} x^2\right| \leq \frac{2}{2-2^p} \varepsilon|x|^p
$$for all $x \in \mathbb{R}$.
1 reply
doanquangdang
Aug 16, 2024
jasperE3
3 hours ago
Polynomials with common roots and coefficients
VicKmath7   10
N 4 hours ago by math-olympiad-clown
Source: Balkan MO SL 2020 A3
Let $P(x), Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$.

Demetres Christofides, Cyprus
10 replies
VicKmath7
Sep 9, 2021
math-olympiad-clown
4 hours ago
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   1
N 4 hours ago by korncrazy
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
1 reply
OgnjenTesic
Yesterday at 4:06 PM
korncrazy
4 hours ago
Interesting inequalities
sqing   0
4 hours ago
Source: Own
Let $ a,b,c \geq  0 , a+b+c+abc = 4.$ Prove that
$$ a+ab^2+\frac{15}{4}ab^2c^3  \leq \frac{2(100+13\sqrt{13})}{27}$$$$ 2a+ab^2+ 4ab^2c^3\leq \frac{4(68+5\sqrt{10})}{27}$$$$ 3a+ab^2+ \frac{9}{2}ab^2c^3\leq \frac{2(172+7\sqrt{7})}{27}$$$$a+ab^2+ 3.75982ab^2c^3 \leq \frac{2(100+13\sqrt{13})}{27}$$$$ 2a+ab^2+ 4.21981ab^2c^3\leq \frac{4(68+5\sqrt{10})}{27}$$$$ 3a+ab^2+4.73626ab^2c^3\leq \frac{2(172+7\sqrt{7})}{27}$$
0 replies
sqing
4 hours ago
0 replies
Inspired by old results
sqing   0
4 hours ago
Source: Own
Let $ a,b,c \geq  0 , a+b+c+abc = 4.$ Prove that
$$ a+ab+2ab^2c^3  \leq \frac{25}{4}$$$$ 2a+ab+\frac{29}{10}ab^2c^3 \leq 9$$$$3a+ab+4ab^2c^3 \leq \frac{49}{4}$$$$ 2a+ab+2.9746371ab^2c^3 \leq 9$$$$3a+ab+4.062494ab^2c^3 \leq \frac{49}{4}$$
0 replies
sqing
4 hours ago
0 replies
Interesting inequalities
sqing   3
N 5 hours ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
3 replies
sqing
Yesterday at 12:44 PM
sqing
5 hours ago
Functional equation with powers
tapir1729   14
N 5 hours ago by Mathandski
Source: TSTST 2024, problem 6
Determine whether there exists a function $f: \mathbb{Z}_{> 0} \rightarrow \mathbb{Z}_{> 0}$ such that for all positive integers $m$ and $n$,
\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]Jaedon Whyte
14 replies
tapir1729
Jun 24, 2024
Mathandski
5 hours ago
Two circles and Three line concurrency
mofidy   0
Apr 10, 2025
Two circles $W_1$ and $W_2$ with equal radii intersect at P and Q. Points B and C are located on the circles$W_1$ and $W_2$ so that they are inside the circles $W_2$ and $W_1$, respectively. Also, points X and Y distinct from P are located on $W_1$ and $W_2$, respectively, so that:
$$\angle{CPQ} = \angle{CXQ}  \text{ and } \angle{BPQ} = \angle{BYQ}.$$The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
Thanks.
0 replies
mofidy
Apr 10, 2025
0 replies
Two circles and Three line concurrency
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mofidy
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Two circles $W_1$ and $W_2$ with equal radii intersect at P and Q. Points B and C are located on the circles$W_1$ and $W_2$ so that they are inside the circles $W_2$ and $W_1$, respectively. Also, points X and Y distinct from P are located on $W_1$ and $W_2$, respectively, so that:
$$\angle{CPQ} = \angle{CXQ}  \text{ and } \angle{BPQ} = \angle{BYQ}.$$The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
Thanks.
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