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Constant Angle Sum
i3435   6
N 24 minutes ago by bin_sherlo
Source: AMASCIWLOFRIAA1PD (mock oly geo contest) P3
Let $ABC$ be a triangle with circumcircle $\Omega$, $A$-angle bisector $l_A$, and $A$-median $m_A$. Suppose that $l_A$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$. A line $l$ parallel to $\overline{BC}$ meets $l_A$, $m_A$ at $G$, $N$ respectively, so that $G$ is between $A$ and $D$. The circle with diameter $\overline{AG}$ meets $\Omega$ again at $J$.

As $l$ varies, show that $\angle AMN + \angle DJG$ is constant.

MP8148
6 replies
i3435
May 11, 2021
bin_sherlo
24 minutes ago
NEPAL TST 2025 DAY 2
Tony_stark0094   8
N 40 minutes ago by cursed_tangent1434
Consider an acute triangle $\Delta ABC$. Let $D$ and $E$ be the feet of the altitudes from $A$ to $BC$ and from $B$ to $AC$ respectively.

Define $D_1$ and $D_2$ as the reflections of $D$ across lines $AB$ and $AC$, respectively. Let $\Gamma$ be the circumcircle of $\Delta AD_1D_2$. Denote by $P$ the second intersection of line $D_1B$ with $\Gamma$, and by $Q$ the intersection of ray $EB$ with $\Gamma$.

If $O$ is the circumcenter of $\Delta ABC$, prove that $O$, $D$, and $Q$ are collinear if and only if quadrilateral $BCQP$ can be inscribed within a circle.

$\textbf{Proposed by Kritesh Dhakal, Nepal.}$
8 replies
Tony_stark0094
Apr 12, 2025
cursed_tangent1434
40 minutes ago
Interesting inequalities
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq  0 $ and $  ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq  \frac{28}{9}$$
4 replies
1 viewing
sqing
3 hours ago
sqing
an hour ago
NEPAL TST DAY 2 PROBLEM 2
Tony_stark0094   6
N an hour ago by cursed_tangent1434
Kritesh manages traffic on a $45 \times 45$ grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the $45 \times 45$ grid.

Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove $k$ cars from the $45 \times 45$ grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.

What is the minimum value of $k$ that guarantees that Kritesh's job is possible?

$\textbf{Proposed by Shining Sun. USA}$
6 replies
Tony_stark0094
Apr 12, 2025
cursed_tangent1434
an hour ago
NEPAL TST DAY-2 PROBLEM 1
Tony_stark0094   9
N an hour ago by cursed_tangent1434
Let the sequence $\{a_n\}_{n \geq 1}$ be defined by
\[
a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}
\]Prove that
\[
a_n^{2025} >n^{2024}
\]for all positive integers $n \geq 2$.

$\textbf{Proposed by Prajit Adhikari, Nepal.}$
9 replies
Tony_stark0094
Apr 12, 2025
cursed_tangent1434
an hour ago
Inspired by Omerking
sqing   1
N an hour ago by lbh_qys
Source: Own
Let $ a,b,c>0 $ and $  ab+bc+ca\geq \dfrac{1}{3}. $ Prove that
$$  ka+ b+kc\geq \sqrt{\frac{4k-1}{3}}$$Where $ k\geq 1.$$$  4a+ b+4c\geq \sqrt{5}$$
1 reply
sqing
2 hours ago
lbh_qys
an hour ago
Weird Inequality Problem
Omerking   4
N 2 hours ago by sqing
Following inequality is given:
$$3\geq ab+bc+ca\geq \dfrac{1}{3}$$Find the range of values that can be taken by :
$1)a+b+c$
$2)abc$

Where $a,b,c$ are positive reals.
4 replies
Omerking
Yesterday at 8:56 AM
sqing
2 hours ago
A Projection Theorem
buratinogigle   2
N 2 hours ago by wh0nix
Source: VN Math Olympiad For High School Students P1 - 2025
In triangle $ABC$, prove that
\[ a = b\cos C + c\cos B. \]
2 replies
buratinogigle
6 hours ago
wh0nix
2 hours ago
Turbo's en route to visit each cell of the board
Lukaluce   18
N 3 hours ago by yyhloveu1314
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
18 replies
Lukaluce
Monday at 11:01 AM
yyhloveu1314
3 hours ago
Perhaps a classic with parameter
mihaig   1
N 3 hours ago by LLriyue
Find the largest positive constant $r$ such that
$$a^2+b^2+c^2+d^2+2\left(abcd\right)^r\geq6$$for all reals $a\geq1\geq b\geq c\geq d\geq0$ satisfying $a+b+c+d=4.$
1 reply
mihaig
Jan 7, 2025
LLriyue
3 hours ago
Connected graph with k edges
orl   26
N 4 hours ago by Maximilian113
Source: IMO 1991, Day 2, Problem 4, IMO ShortList 1991, Problem 10 (USA 5)
Suppose $ \,G\,$ is a connected graph with $ \,k\,$ edges. Prove that it is possible to label the edges $ 1,2,\ldots ,k\,$ in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.

Note: Graph-Definition. A graph consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices $ \,u,v\,$ belongs to at most one edge. The graph $ G$ is connected if for each pair of distinct vertices $ \,x,y\,$ there is some sequence of vertices $ \,x = v_{0},v_{1},v_{2},\cdots ,v_{m} = y\,$ such that each pair $ \,v_{i},v_{i + 1}\;(0\leq i < m)\,$ is joined by an edge of $ \,G$.
26 replies
orl
Nov 11, 2005
Maximilian113
4 hours ago
Path within S which does not meet itself
orl   5
N Apr 6, 2025 by atdaotlohbh
Source: IMO 1982, Day 2, Problem 6
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
5 replies
orl
Nov 11, 2005
atdaotlohbh
Apr 6, 2025
Path within S which does not meet itself
G H J
Source: IMO 1982, Day 2, Problem 6
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orl
3647 posts
#1 • 1 Y
Y by Adventure10
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
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SCP
1502 posts
#3 • 2 Y
Y by Adventure10, Mango247
There is sth wrong ith my idea:

The length of $X,Y>4*99-1$ is quite trivial in one direction, but if they ean the shortest distance, we just take points very close to the boundary and with angles close to $180$ degrees except in the corners, what is the mistake in those ideas?
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mavropnevma
15142 posts
#4 • 2 Y
Y by Adventure10, Mango247
See John Scholes' solution at http://mks.mff.cuni.cz/kalva/.
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Phorphyrion
395 posts
#6
Y by
The condition should be $A_0\neq A_n$. That's the statement appearing on the official IMO website.
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Lukariman
394 posts
#7
Y by
The solution is difficult to understand because it involves compact sets. Can anyone explain the properties of compact sets
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atdaotlohbh
184 posts
#8
Y by
The first one to post a solution!
Start walking from $A_0$ to $A_n$ along $L$. Paint in red all points on the boundary of $S$ that are within distance $\frac{1}{2}$ from the current point. Obviously at any point the red points are a finite union of red segments (with possibly length zero). Suppose such points $X$ and $Y$ do not exist.

The key observation: If $AB$ and $CD$ are opposite sides of squares, then we must completely paint $AB$ before we start painting $CD$ (or vice versa).
Proof: Suppose it is false and $AB$ was partly painted when we firstly painted a point on $CD$. Let $X$ be a red point on $AB$ such that on some side of it on $AB$ its close is not painted (More simply, an endpoint of a segment). Say $X$ was painted when passing $M$ on $L$. We must return to paint the neighbourhood of $X$, and as we always paint segments, we will point $X$ once again as well, say when passing through $N$. Notice that from triangle inequality $MN \leq MX+NX \leq \frac{1}{2}+\frac{1}{2}=1$. Also, to get from $M$ to a point which painted $CD$ a bit we must have travelled at least $99$, and to get back to $N$ we need $99$ again. Thus, the distance travelled between $M$ and $N$ is at least 198, and so $M$ and $N$ work as $X$ and $Y$ in the problem statement.

Now we finish by proving that the key observation is impossible. To do that, let the square be $ABCD$ and $AB$ painted before $CD$, $AD$ painted before $BC$. WLOG $B$ was painted before $D$, but $B$ is a part of $BC$, thus before $B$ was painted $AD$ was completely done already, and hence $D$ also, which is the desired contradiction.

Notice we never used the fact that $L$ doesn't intersect itself.
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