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Projective training on circumscribds
Assassino9931   1
N 25 minutes ago by VicKmath7
Source: Bulgaria Balkan MO TST 2025
Let $ABCD$ be a circumscribed quadrilateral with incircle $k$ and no two opposite angles equal. Let $P$ be an arbitrary point on the diagonal $BD$, which is inside $k$. The segments $AP$ and $CP$ intersect $k$ at $K$ and $L$. The tangents to $k$ at $K$ and $L$ intersect at $S$. Prove that $S$ lies on the line $BD$.
1 reply
Assassino9931
Yesterday at 10:17 PM
VicKmath7
25 minutes ago
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Orthocenter config once again
Assassino9931   7
N 25 minutes ago by VicKmath7
Source: Bulgaria National Olympiad 2025, Day 2, Problem 4
Let \( ABC \) be an acute triangle with \( AB < AC \), midpoint $M$ of side $BC$, altitude \( AD \) (\( D \in BC \)), and orthocenter \( H \). A circle passes through points \( B \) and \( D \), is tangent to line \( AB \), and intersects the circumcircle of triangle \( ABC \) at a second point \( Q \). The circumcircle of triangle \( QDH \) intersects line \( BC \) at a second point \( P \). Prove that the lines \( MH \) and \( AP \) are perpendicular.
7 replies
Assassino9931
Tuesday at 1:53 PM
VicKmath7
25 minutes ago
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Angle EBA is equal to Angle DCB
WakeUp   6
N 40 minutes ago by Nari_Tom
Source: Baltic Way 2011
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
6 replies
WakeUp
Nov 6, 2011
Nari_Tom
40 minutes ago
J
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if xy+xz+yz+2xyz+1 prove that...
behdad.math.math   5
N 41 minutes ago by Sadigly
if xy+xz+yz+2xyz+1 prove that x+y+z>=3/2
5 replies
behdad.math.math
Sep 25, 2008
Sadigly
41 minutes ago
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no. of divisors of the form
S_14159   0
an hour ago
Source: idk
If $P=2^5 \cdot 3^6 \cdot 5^4 \cdot 7^3$ then number of positive integral divisors of $``P"$

(A) of form $(2 n+3), n \in \mathbb{N}$, is $=138$
(B) of form $(4 n+1), n \in \mathbb{W}$, is $=70$
(C) of form $(6 n+3), n \in \mathbb{W}$, is $=120$
(D) of form $(4 n+3), n \in \mathbb{W}$, is $=56$

(more than one option may be correct)
0 replies
S_14159
an hour ago
0 replies
J
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max n with n times n square are black
NicoN9   0
an hour ago
Source: Japan Junior MO Preliminary 2022 P5
Find the maximum positive integer $n$ such that for $45\times 45$ grid, no matter how you paint $2022$ unit squares black, there exists $n\times n$ square with all unit square painted black.
0 replies
NicoN9
an hour ago
0 replies
J
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BDE tangent to EF
NicoN9   0
an hour ago
Source: Japan Junior MO Preliminary 2022 P4
Let $ABC$ be a triangle with $AB=5$, $BC=7$, $CA=6$. Let $D, E$, and $F$ be points lying on sides $BC, CA, AB$, respectively. Given that $A, B, D, E$, and $B, C, E, F$ are cyclic respectively, and the circumcircle of $BDE$ are tangent to line $EF$, find the length of segment $AE$.
0 replies
NicoN9
an hour ago
0 replies
J
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Maximum number of m-tastic numbers
Tsukuyomi   30
N an hour ago by cursed_tangent1434
Source: IMO Shortlist 2017 N4
Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m-$tastic if there exists a number $c\in \{1,2,3,\ldots ,2017\}$ such that $\dfrac{10^t-1}{c\cdot m}$ is short, and such that $\dfrac{10^k-1}{c\cdot m}$ is not short for any $1\le k<t$. Let $S(m)$ be the set of $m-$tastic numbers. Consider $S(m)$ for $m=1,2,\ldots{}.$ What is the maximum number of elements in $S(m)$?
30 replies
Tsukuyomi
Jul 10, 2018
cursed_tangent1434
an hour ago
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interesting ineq
nikiiiita   6
N an hour ago by KhuongTrang
Source: Own
Given $a,b,c$ are positive real numbers satisfied $a^3+b^3+c^3=3$. Prove that:
$$\sqrt{2ab+5c^{2}+2a}+\sqrt{2bc+5a^{2}+2b}+\sqrt{2ac+5b^{2}+2c}\le3\sqrt{3\left(a+b+c\right)}$$
6 replies
nikiiiita
Jan 29, 2025
KhuongTrang
an hour ago
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Prove that x1=x2=....=x2025
Rohit-2006   7
N an hour ago by Fibonacci_math
Source: A mock
The real numbers $x_1,x_2,\cdots,x_{2025}$ satisfy,
$$x_1+x_2=2\bar{x_1}, x_2+x_3=2\bar{x_2},\cdots, x_{2025}+x_1=2\bar{x_{2025}}$$Where {$\bar{x_1},\cdots,\bar{x_{2025}}$} is a permutation of $x_1,x_2,\cdots,x_{2025}$. Prove that $x_1=x_2=\cdots=x_{2025}$
7 replies
Rohit-2006
Yesterday at 5:22 AM
Fibonacci_math
an hour ago
J
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Inspired by old results
sqing   1
N an hour ago by sqing
Source: Own
Let $a ,b,c \geq 0 $ and $a+b+c=1$. Prove that
$$\frac{1}{2}\leq \frac{ \left(1-a^{2}\right)^2+2\left(1-b^{2}\right) \left(1-c^{2}\right) }{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\leq 1$$$$1 \leq \frac{\left(1-a^{2}\right)^{2}+2\left(1-b^{2}\right) +\left(1-c^{2}\right)^{2}}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\leq \frac{3}{2}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
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Interesting Locus S
IvanGD   1
N Oct 5, 2021 by vanstraelen
Hello everyone.

This is my first actual AOPS post, even though I'm a long time IMO competitor and an avid AOPS user.

If you don't want to read the Problem History, you can scroll down to read the statement right away.

Problem History

The reason why I'm writing this post is to publish a very interesting geometry problem, which I've created with the help of a group of friends of mine. I started creating the problem around 2 years ago, when I heard about another geometry problem, and I developed the idea used in that problem into something much more, together with two of my friends. We reached some interesting conclusions, but couldn't prove any of it, so we stopped working on it after a month or so. I continued working on that problem alone for another two months maybe, and then I let it go.

About half a year ago, while I was on a math competition abroad, I told one of my teammates about the problem and he was interested in solving it together with me, so we resumed the investigation. He proposed one crucial idea that reshaped the problem into something totally different, and after I'd noticed one more important fact, we completely formed the problem. We still couldn't prove any of it, because it all seemed too difficult. After the competition, we stopped working on it.

Finally, around two weeks ago, I remembered the problem, and used Wolfram Mathematica to prove it. And guess what... Our hunch was indeed correct. A brand new geometry problem was born, and I was quite satisfied. That means that our solution to the problem was indeed correct, since we finally got a computerized proof. But to me, it's still not over. I want to know if there are any normal ways to prove this...Without using extreme analytic computations performed by a sophisticated mathematical program.

That is why I'm posting this problem as a challenge to everyone else who wants to try and solve it. Maybe someone here can pull it off? I don't know. I've shown this problem to a lot of successful IMO competitors, and none of them were close to solving it. I honestly think the problem is difficult, but I don't know how difficult it actually is. Anyhow, I'm going to post the problem statement text now. Good luck and best wishes to everyone!

Problem Statement

Three fixed points, $A$, $B$ and $C$, are given such that $\triangle ABC$ is a scalene triangle. A variable real parameter $\lambda$ is also given, and six points, $P$, $Q$, $R$, $K$, $L$ and $M$, are defined by the following six vector equalities:

$\overrightarrow{AP}=\lambda\overrightarrow{AB}$
$\overrightarrow{BQ}=\lambda\overrightarrow{BC}$
$\overrightarrow{CR}=\lambda\overrightarrow{CA}$
$\overrightarrow{BK}=\lambda\overrightarrow{BA}$
$\overrightarrow{CL}=\lambda\overrightarrow{CB}$
$\overrightarrow{AM}=\lambda\overrightarrow{AC}$

Let $X$ be the circumcenter of triangle $\triangle PQR$, and $Y$ the circumcenter of triangle $\triangle KLM$. The point $S$ is defined as the midpoint of $XY$.

For any fixed triple of points $A$, $B$ and $C$, find the locus of $S$, as $\lambda$ takes all possible real values.
1 reply
IvanGD
Nov 4, 2015
vanstraelen
Oct 5, 2021
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Interesting Locus S
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geometry
geometry unsolved
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IvanGD
1 post
IvanGD
#1 h  Nov 4, 2015, 6:34 PM • 2 Y
Y by GoJensenOrGoHome, Adventure10
Hello everyone.

This is my first actual AOPS post, even though I'm a long time IMO competitor and an avid AOPS user.

If you don't want to read the Problem History, you can scroll down to read the statement right away.

Problem History

The reason why I'm writing this post is to publish a very interesting geometry problem, which I've created with the help of a group of friends of mine. I started creating the problem around 2 years ago, when I heard about another geometry problem, and I developed the idea used in that problem into something much more, together with two of my friends. We reached some interesting conclusions, but couldn't prove any of it, so we stopped working on it after a month or so. I continued working on that problem alone for another two months maybe, and then I let it go.

About half a year ago, while I was on a math competition abroad, I told one of my teammates about the problem and he was interested in solving it together with me, so we resumed the investigation. He proposed one crucial idea that reshaped the problem into something totally different, and after I'd noticed one more important fact, we completely formed the problem. We still couldn't prove any of it, because it all seemed too difficult. After the competition, we stopped working on it.

Finally, around two weeks ago, I remembered the problem, and used Wolfram Mathematica to prove it. And guess what... Our hunch was indeed correct. A brand new geometry problem was born, and I was quite satisfied. That means that our solution to the problem was indeed correct, since we finally got a computerized proof. But to me, it's still not over. I want to know if there are any normal ways to prove this...Without using extreme analytic computations performed by a sophisticated mathematical program.

That is why I'm posting this problem as a challenge to everyone else who wants to try and solve it. Maybe someone here can pull it off? I don't know. I've shown this problem to a lot of successful IMO competitors, and none of them were close to solving it. I honestly think the problem is difficult, but I don't know how difficult it actually is. Anyhow, I'm going to post the problem statement text now. Good luck and best wishes to everyone!

Problem Statement

Three fixed points, $A$, $B$ and $C$, are given such that $\triangle ABC$ is a scalene triangle. A variable real parameter $\lambda$ is also given, and six points, $P$, $Q$, $R$, $K$, $L$ and $M$, are defined by the following six vector equalities:

$\overrightarrow{AP}=\lambda\overrightarrow{AB}$
$\overrightarrow{BQ}=\lambda\overrightarrow{BC}$
$\overrightarrow{CR}=\lambda\overrightarrow{CA}$
$\overrightarrow{BK}=\lambda\overrightarrow{BA}$
$\overrightarrow{CL}=\lambda\overrightarrow{CB}$
$\overrightarrow{AM}=\lambda\overrightarrow{AC}$

Let $X$ be the circumcenter of triangle $\triangle PQR$, and $Y$ the circumcenter of triangle $\triangle KLM$. The point $S$ is defined as the midpoint of $XY$.

For any fixed triple of points $A$, $B$ and $C$, find the locus of $S$, as $\lambda$ takes all possible real values.
This post has been edited 1 time. Last edited by IvanGD, Nov 4, 2015, 6:42 PM
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vanstraelen
8953 posts
vanstraelen
#2 h  Oct 5, 2021, 12:13 PM
Y by
Let $A(0,a),B(-b,0),C(c,0)$.

The locus of the points $S$ is the line segment $CD$ on $(a^{2}-3bc)x+a(b-c)y=bc(b-c)$,
with $C(\frac{c-b}{4},\frac{a^{2}+bc}{4a}\ )$ and $D(\frac{7(c-b)}{12},\frac{7a^{2}-9bc}{12a}\ )$.
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