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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Interesting inequalities
sqing   1
N 3 minutes ago by sqing
Source: Own
Let $ a,b,c\geq  0 $ and $  a^2+b^2+c^2+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq  \frac{28}{9}$$
1 reply
1 viewing
sqing
4 minutes ago
sqing
3 minutes ago
Turbo's en route to visit each cell of the board
Lukaluce   17
N 10 minutes ago by Gato_combinatorio
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
17 replies
Lukaluce
Monday at 11:01 AM
Gato_combinatorio
10 minutes ago
Connected graph with k edges
orl   26
N 11 minutes 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
11 minutes ago
3 var inquality
sqing   2
N 21 minutes ago by sqing
Source: Own
Let $ a,b,c> 0 $ and $  4(a+b) +3c-ab  \geq10$ . Prove that
$$a^2+b^2+c^2+kabc\geq k+3$$Where $0\leq k \leq 1. $
$$a^2+b^2+c^2+abc\geq 4$$
2 replies
sqing
an hour ago
sqing
21 minutes ago
Expected number of flips
Bread10   8
N 3 hours ago by mathprodigy2011
An unfair coin has a $\frac{4}{7}$ probability of coming up heads and $\frac{3}{7}$ probability of coming up tails. The expected number of flips necessary to first see the sequence $HHTHTHHT$ in that consecutive order can be written as $\frac{m}{n}$ for relatively prime positive integers $m$, $n$. Find the number of factors of $n$.

$\textbf{(A)}~40\qquad\textbf{(B)}~42\qquad\textbf{(C)}~44\qquad\textbf{(D)}~45\qquad\textbf{(E)}~48$
8 replies
Bread10
Yesterday at 8:58 PM
mathprodigy2011
3 hours ago
Help with Competitive Geometry?
REACHAW   3
N 5 hours ago by REACHAW
Hi everyone,
I'm struggling a lot with geometry. I've found algebra, number theory, and even calculus to be relatively intuitive. However, when I took geometry, I found it very challenging. I stumbled my way through the class and can do the basic 'textbook' geometry problems, but still struggle a lot with geometry in competitive math. I find myself consistently skipping the geometry problems during contests (even the easier/first ones).

It's difficult for me to see the solution path. I can do the simpler textbook tasks (eg. find congruent triangles) but not more complex ones (eg. draw these two lines to form similar triangles).

Do you have any advice, resources, or techniques I should try?
3 replies
REACHAW
Monday at 11:51 PM
REACHAW
5 hours ago
fractional part
Ecrin_eren   3
N Yesterday at 9:26 PM by rchokler
{x^2}+{x}=0.64

How many positive real values of x satisfy this equation?
3 replies
Ecrin_eren
Apr 13, 2025
rchokler
Yesterday at 9:26 PM
Angle oriented geometry
Problems_eater   0
Yesterday at 9:03 PM
Let $A, B, C,D$ be four distinct points in the plane.
Which of the following statements, expressed using oriented angles, are always true?

1.If lines $AB$ and $CD$ are distinct and parallel, then
the oriented angle $ABC$ is equal to the oriented angle DCB.

2.If $B$ lies on the segment $AC$, then
the oriented angle $DBA$ plus the oriented angle $DBC $equals $180°$.

3.If the oriented angle$ ABC$ plus the oriented angle $BCD$ equals 0°, then
lines $AB $and $CD$ are parallel.

4.If the oriented angle $ABC$ plus the oriented angle $BCD$ equals $180°$, then
lines $AB$ and $CD$are parallel.
0 replies
Problems_eater
Yesterday at 9:03 PM
0 replies
how many quadrilaterals ?
Ecrin_eren   6
N Yesterday at 5:31 PM by mathprodigy2011
"All the diagonals of an 11-gon are drawn. How many quadrilaterals can be formed using these diagonals as sides? (The vertices of the quadrilaterals are selected from the vertices of the 11-gon.)"
6 replies
Ecrin_eren
Apr 13, 2025
mathprodigy2011
Yesterday at 5:31 PM
Plane geometry problem with inequalities
ReticulatedPython   3
N Yesterday at 2:48 PM by vanstraelen
Let $A$ and $B$ be points on a plane such that $AB=1.$ Let $P$ be a point on that plane such that $$\frac{AP^2+BP^2}{(AP)(BP)}=3.$$Prove that $$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$
Source: Own
3 replies
ReticulatedPython
Apr 10, 2025
vanstraelen
Yesterday at 2:48 PM
Inequalities
sqing   1
N Yesterday at 1:55 PM by sqing
Let $   a,b    $ be reals such that $  a^2-ab+b^2 =3$ . Prove that
$$  \frac{13}{ 10 }> \frac{1}{ a^2+1 }+ \frac{1}{ b^2+1 } \geq \frac{1}{ 2 }$$$$   \frac{6}{ 5 }>\frac{1}{ a^4+1 }+ \frac{1}{ b^4+1 } \geq   \frac{1}{ 5 }$$$$  \frac{1}{ a^6+1 }+ \frac{1}{ b^6+1 } \geq   \frac{1}{ 14 }$$
1 reply
sqing
Yesterday at 8:59 AM
sqing
Yesterday at 1:55 PM
idk12345678 Math Contest
idk12345678   21
N Yesterday at 1:25 PM by idk12345678
Welcome to the 1st idk12345678 Math Contest.
You have 4 hours. You do not have to prove your answers.
Post \signup username to sign up. Post your answers in a hide tag and I will tell you your score.*


The contest is attached to the post

Clarifications

*I mightve done them wrong feel free to ask about an answer
21 replies
idk12345678
Apr 10, 2025
idk12345678
Yesterday at 1:25 PM
purple comet math competition question
AVY2024   4
N Yesterday at 1:02 PM by K1mchi_
Given that (1 + tan 1)(1 + tan 2). . .(1 + tan 45) = 2n, find n
4 replies
AVY2024
Yesterday at 11:00 AM
K1mchi_
Yesterday at 1:02 PM
Inequalities
sqing   25
N Yesterday at 12:06 PM by sqing
Let $ a,b,c,d>0 $ and $(a+c)(b+d)=ac+\frac{3}{2}bd.$ Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{20-\sqrt{10}}{3}$$Let $ a,b,c,d>0 $ and $(a+c)(b+d)=ac+\frac{4}{3}bd.$ Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{21-\sqrt{6}}{3}$$
25 replies
sqing
Dec 3, 2024
sqing
Yesterday at 12:06 PM
Perpendicularity implies special intersection
Kezer   3
N Jan 17, 2021 by Mikeglicker
Source: Germany VAIMO 2015 - #5
Let $ABC$ be an acute triangle with the circumcircle $k$ and incenter $I$. The perpendicular through $I$ in $CI$ intersects segment $[BC]$ in $U$ and $k$ in $V$. In particular $V$ and $A$ are on different sides of $BC$. The parallel line through $U$ to $AI$ intersects $AV$ in $X$.
Prove: If $XI$ and $AI$ are perpendicular to each other, then $XI$ intersects segment $[AC]$ in its midpoint $M$.

(Notation: $[\cdot]$ denotes the line segment.)
3 replies
Kezer
Jul 11, 2015
Mikeglicker
Jan 17, 2021
Perpendicularity implies special intersection
G H J
Source: Germany VAIMO 2015 - #5
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Kezer
986 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an acute triangle with the circumcircle $k$ and incenter $I$. The perpendicular through $I$ in $CI$ intersects segment $[BC]$ in $U$ and $k$ in $V$. In particular $V$ and $A$ are on different sides of $BC$. The parallel line through $U$ to $AI$ intersects $AV$ in $X$.
Prove: If $XI$ and $AI$ are perpendicular to each other, then $XI$ intersects segment $[AC]$ in its midpoint $M$.

(Notation: $[\cdot]$ denotes the line segment.)
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huynguyen
535 posts
#2 • 2 Y
Y by Adventure10, Mango247
Are you sure it's true?
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TelvCohl
2312 posts
#4 • 5 Y
Y by Kezer, AlastorMoody, enhanced, Adventure10, Mango247
My solution :

Let $ T \equiv XU \cap AB $ and $ Y \equiv IX \cap AB $ .

From $ \angle BTU=\angle BAI=\tfrac{1}{2} \angle BAC=\angle BIC-90^{\circ}=\angle BIU \Longrightarrow B, I, T, U $ are concyclic ,
so $ I $ is the midpoint of arc $ TU $ in $ \odot (BTU) \Longrightarrow IX $ is the perpendicular bisector of $ TU \Longrightarrow XT=XU $ ,

hence $ \frac{VA}{VX}=\frac{AI}{XU}=\frac{AI}{XT}=\frac{YA}{YT} \Longrightarrow YV \parallel TU \parallel AI \Longrightarrow \angle BYV=\angle BIV \Longrightarrow V \in \odot (BIY) $ .

From $ \angle VBI=\angle VYI=\angle AIX=90^{\circ} \Longrightarrow V $ is the midpoint of arc $ AC $ (contain $ B $) in $ \odot (ABC) $ ,
so $\angle VMC=90^{\circ} \Longrightarrow M \in \odot (CIV) \Longrightarrow \angle VIM=180^{\circ}-\angle ACV=\angle VBA=180^{\circ}-\angle YIV $ ,
hence we get $ Y, I, M $ are collinear $ \Longrightarrow M \in XI  \Longrightarrow YI \equiv XI $ passes through the midpoint $ M $ of $ CA $ .

Q.E.D
____________________________________________________________
P.S. For more properties in this configuration you can see 2014 IMO Shortlist G7 :)
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Mikeglicker
258 posts
#6
Y by
sorry to bump such an old question but I think this statement is just true without ever being true because there is no diagram satisfying this.
We would like to prove that if I is incener of ABC and M is middle of AC then there cant be MI perpendicular to AI.
assume it does then let Mi meet AB at D and let CI meet AB at E and the perpendicular throught I to AC meet AB at F. Now if we project the harmonic bundle A,M,C andpoint at infiniti through point E to line AB we get that A,D,E,F is also a harmonic bundle. Now it is well known that the harmonic conjugate of AED must satisfy that angles FID=EID but we claim that F and E both lies on the same side of D on line AB (the side closer to A) which makes the angles equality impossible.
First because CAB is a sharp angle we get AFI=180 - CAB>90>90 - IAD=ADI so F is on segment AD.
let a,b,c be the angles of the triangle then AEI=180-a-c/2 and ADI = 90-a/2 and by the inequality a+c<180 it is easy to verify that AEI>ADI so E is on segment AD as well.

Can you please tell me wether this explanation is actually right, if someone can find mistakes I have made or verify that my solution is indeed right I will appreciate it.
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