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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
Can you construct the incenter of a triangle ABC?
PennyLane_31   3
N a few seconds ago by cj13609517288
Source: 2023 Girls in Mathematics Tournament- Level B, Problem 4
Given points $P$ and $Q$, Jaqueline has a ruler that allows tracing the line $PQ$. Jaqueline also has a special object that allows the construction of a circle of diameter $PQ$. Also, always when two circles (or a circle and a line, or two lines) intersect, she can mark the points of the intersection with a pencil and trace more lines and circles using these dispositives by the points marked. Initially, she has an acute scalene triangle $ABC$. Show that Jaqueline can construct the incenter of $ABC$.
3 replies
PennyLane_31
Oct 29, 2023
cj13609517288
a few seconds ago
Do not try to bash on beautiful geometry
ItzsleepyXD   3
N 5 minutes ago by FarrukhBurzu
Source: Own , Mock Thailand Mathematic Olympiad P9
Let $ABC$be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$
3 replies
ItzsleepyXD
Today at 9:30 AM
FarrukhBurzu
5 minutes ago
Cool functional equation
Rayanelba   2
N 7 minutes ago by ATM_
Source: Own
Find all functions $f:\mathbb{Z}_{>0}\to \mathbb{Z}_{>0}$ that verify the following equation for all $x,y\in \mathbb{Z}_{>0}$:
$max(f^{f(y)}(x),f^{f(y)}(y))|min(x,y)$
2 replies
Rayanelba
25 minutes ago
ATM_
7 minutes ago
1 line solution to Inequality
ItzsleepyXD   2
N 13 minutes ago by Vivaandax
Source: Own , Mock Thailand Mathematic Olympiad P8
Let $x_1,x_2,\dots,x_n$ be positive real integer such that $x_1^2+x_2^2+\cdots+x_n^2=2$ Prove that
$$\sum_{i=1}^{n}\frac{1}{x_i^3(x_{i-1}+x_{i+1})}\geqslant \left(\sum_{i=1}^{n}\frac{x_i}{x_{i-1}+x_{i+1}}\right)^3$$such that $x_{n+1}=x_1$ and $x_0=x_n$
2 replies
ItzsleepyXD
Today at 9:27 AM
Vivaandax
13 minutes ago
Generating Functions
greenplanet2050   7
N 3 hours ago by rchokler
So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)
7 replies
greenplanet2050
Yesterday at 10:42 PM
rchokler
3 hours ago
Algebraic Manipulation
Darealzolt   1
N 5 hours ago by Soupboy0
Find the number of pairs of real numbers $a, b, c$ that satisfy the equation $a^4 + b^4 + c^4 + 1 = 4abc$.
1 reply
Darealzolt
Today at 1:25 PM
Soupboy0
5 hours ago
BrUMO 2025 Team Round Problem 13
lpieleanu   1
N 5 hours ago by vanstraelen
Let $\omega$ be a circle, and let a line $\ell$ intersect $\omega$ at two points, $P$ and $Q.$ Circles $\omega_1$ and $\omega_2$ are internally tangent to $\omega$ at points $X$ and $Y,$ respectively, and both are tangent to $\ell$ at a common point $D.$ Similarly, circles $\omega_3$ and $\omega_4$ are externally tangent to $\omega$ at $X$ and $Y,$ respectively, and are tangent to $\ell$ at points $E$ and $F,$ respectively.

Given that the radius of $\omega$ is $13,$ the segment $\overline{PQ}$ has a length of $24,$ and $YD=YE,$ find the length of segment $\overline{YF}.$
1 reply
lpieleanu
Apr 27, 2025
vanstraelen
5 hours ago
Inequlities
sqing   33
N Today at 1:50 PM by sqing
Let $ a,b,c\geq 0 $ and $ a^2+ab+bc+ca=3 .$ Prove that$$\frac{1}{1+a^2}+ \frac{1}{1+b^2}+  \frac{1}{1+c^2} \geq \frac{3}{2}$$$$\frac{1}{1+a^2}+ \frac{1}{1+b^2}+ \frac{1}{1+c^2}-bc \geq -\frac{3}{2}$$
33 replies
sqing
Jul 19, 2024
sqing
Today at 1:50 PM
Very tasteful inequality
tom-nowy   1
N Today at 1:39 PM by sqing
Let $a,b,c \in (-1,1)$. Prove that $$(a+b+c)^2+3>(ab+bc+ca)^2+3(abc)^2.$$
1 reply
tom-nowy
Today at 10:47 AM
sqing
Today at 1:39 PM
Inequalities
sqing   8
N Today at 1:31 PM by sqing
Let $x\in(-1,1). $ Prove that
$$  \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{2+ x^2}  \geq  \dfrac{3}{2}$$$$ \dfrac{2}{\sqrt{1-x^2}} + \dfrac{1}{1+x^2} \geq 3$$
8 replies
sqing
Apr 26, 2025
sqing
Today at 1:31 PM
đề hsg toán
akquysimpgenyabikho   1
N Today at 12:16 PM by Lankou
làm ơn giúp tôi giải đề hsg

1 reply
akquysimpgenyabikho
Apr 27, 2025
Lankou
Today at 12:16 PM
Inequalities
sqing   2
N Today at 10:05 AM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
2 replies
sqing
Jul 12, 2024
sqing
Today at 10:05 AM
9 Physical or online
wimpykid   0
Today at 6:49 AM
Do you think the AoPS print books or the online books are better?

0 replies
wimpykid
Today at 6:49 AM
0 replies
Three variables inequality
Headhunter   6
N Today at 6:08 AM by lbh_qys
$\forall a\in R$ ,$~\forall b\in R$ ,$~\forall c \in R$
Prove that at least one of $(a-b)^{2}$, $(b-c)^{2}$, $(c-a)^{2}$ is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$.

I assume that all are greater than it, but can't go more.
6 replies
Headhunter
Apr 20, 2025
lbh_qys
Today at 6:08 AM
Parallel lines thru points on the altitudes of a triangle
Shu   14
N Dec 18, 2022 by levifb
Source: XVIII Cono Sur Mathematical Olympiad (2007)
Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, $CF$ where $D$, $E$, $F$ lie on $BC$, $AC$, $AB$, respectively. Let $M$ be the midpoint of $BC$. The circumcircle of triangle $AEF$ cuts the line $AM$ at $A$ and $X$. The line $AM$ cuts the line $CF$ at $Y$. Let $Z$ be the point of intersection of $AD$ and $BX$. Show that the lines $YZ$ and $BC$ are parallel.
14 replies
Shu
Aug 10, 2011
levifb
Dec 18, 2022
Parallel lines thru points on the altitudes of a triangle
G H J
Source: XVIII Cono Sur Mathematical Olympiad (2007)
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Shu
316 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, $CF$ where $D$, $E$, $F$ lie on $BC$, $AC$, $AB$, respectively. Let $M$ be the midpoint of $BC$. The circumcircle of triangle $AEF$ cuts the line $AM$ at $A$ and $X$. The line $AM$ cuts the line $CF$ at $Y$. Let $Z$ be the point of intersection of $AD$ and $BX$. Show that the lines $YZ$ and $BC$ are parallel.
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sunken rock
4388 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hint: take the Carnot circle $\odot BCH$ ($H$ - the orthocenter) and see that it passes through the reflection of $A$ in $M$, hence $X$ belongs to this circle as well. Subsequently, by symmetry about $BC$, we get $HZYX$ is cyclic, hence the two lines are parallel.

Best regards,
sunken rock
This post has been edited 1 time. Last edited by sunken rock, Aug 21, 2011, 3:01 PM
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genxium
118 posts
#3 • 2 Y
Y by Adventure10, Mango247
HXBC is cyclic, this is a very traditional problem, if not noticed that , maybe the Menelause theorem works, but the calculation must be tough
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yunxiu
571 posts
#4 • 5 Y
Y by mathisreal, NiltonCesar, primesarespecial, Adventure10, Mango247
Because $BDHF,MDHX$ are both cyclic, so $AF \cdot AB = AH \cdot AD =AX \cdot AM$, and $BMXF$ is cyclic, hence $\angle ZXY = \angle BXM = \angle BFM = \angle FBM = \angle DHC = \angle ZHY$, so $XYZH$ is cyclic, and we have $\angle XYZ = \angle AHX = \angle XMD$, so $YZ\parallel BC$.
Attachments:
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proglote
958 posts
#5 • 1 Y
Y by Adventure10
Let $P = AM \cap BE$, and note that \[{YZ \parallel BC \iff Y(Z,M;B,C) = -1 \iff B(Z, A; YB\cap AD, H) = - 1 \iff (X, A; Y, P) = -1,}\] which is true iff $P$ on the polar $\rho$ of $Y$ w.r.t. $(AEF).$ But the polar of $M$ is $EF$, hence $AA \cap EF  \in \rho.$ But also $AH \cap XF \in \rho$, hence by Pascal's theorem on $AAHEFX$ we get $P \in \rho$, as desired.
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jayme
9787 posts
#6 • 3 Y
Y by BALAMATDA, Adventure10, Mango247
Dear Mathlinkers,

1. T the second point of intersection of BX with the circle with diameter AH
2. the tangent to this circle at F goes through M
3. by using two time the Pascal theorem in two degenerated case, we are done.

Sincerely
Jean-Louis
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NiltonCesar
163 posts
#7 • 2 Y
Y by Adventure10, Mango247
sunken rock wrote:
Hint: take the Carnot circle $\odot BCH$ ($H$ - the orthocenter) and see that it passes through the reflection of $A$ in $M$, hence $X$ belongs to this circle as well. Subsequently, by symmetry about $BC$, we get $HZYX$ is cyclic, hence the two lines are parallel.

Best regards,
sunken rock

Is there any motivation to take the symmetric of $A$ in $M$?
Z K Y
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math_pi_rate
1218 posts
#8 • 2 Y
Y by Adventure10, Mango247
Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$, and let $BX \cap \odot (AEF)=K$. Then, by Pascal on $AHFKXA$, we get that the given statement is equivalent to proving that $FK$ is parallel to $BC$. But this directly follows from $$\angle AFK=\angle MXB=\angle ABC \Rightarrow FK \parallel BC$$where we use the fact that $\odot (AXB)$ is tangent to $BC$ (as $X$ is the $A$-Humpty point). Hence, done. $\blacksquare$
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omriya200
317 posts
#9 • 1 Y
Y by Adventure10
$\text {Lemma}$:$FM$ and $EM$ is tangent to the $\odot (AEF)$.

$\text {proove}$; Just from the nine point circle theorem, the midpoint of $AH$ and $M$ are diametrically opposite points and hence prof the lemma.

Then, we can easily observe that $BFMD$ and $HXDM$ are conclic and so $\angle AFX=\angle 180-BFX =\angle XMD$ and hence, $BXFM$ is conclic.

But $MFC=90-\angle B$ and $BFC=90$ and so $BFM=\angle B$ and $BXM=\angle B$ and $DHX=180-(\angle A+\angle C$ =$\angle B$ .
Hence, $HXZY$ is conclic and so $ZY \parallel DM$.
This post has been edited 3 times. Last edited by omriya200, Jan 7, 2019, 4:19 AM
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jayme
9787 posts
#10 • 3 Y
Y by Mango247, Mango247, Mango247
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 51...

Sincerely
Jean-Louis
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EulersTurban
386 posts
#11
Y by
[asy]
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import graph; size(20cm); 
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pen dotstyle = black; /* point style */ 
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draw((-4.087697920306253,9.765832300575541)--(-8.8,-2.42)--(7.48,-5.14)--cycle, linewidth(1) + zzttqq); 
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draw((7.48,-5.14)--(-4.087697920306253,9.765832300575541), linewidth(1) + zzttqq); 
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[/asy]


We denote with $x=\angle DAM$.

Miquel gives us two cyclic quads $BFXM$ and $CMXE$.
Thus we have that:
$$\angle MBX = \angle MFX = 90-\angle B + x = \angle FAX$$we have this because of the three tangents lemma.

The problem will be killed when we show that $YZHX$ is a cyclic quad.

By an easy angle chase we have that $\angle XZH = \angle B - x$ and we have that $\angle FYA = \angle B - x $ (look at the triangle $YFA$). thus the quad $XYZH$ is cylic.
Now we have that $\angle YZX=\angle YHX = \angle FAX = \angle MBX = 90-\angle B+x$.Thus we have that $ZY \parallel BC$
Z K Y
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yellowhanoi
49 posts
#12
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Very nice problem and solitions!
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dgreenb801
1896 posts
#13 • 2 Y
Y by math31415926535, BALAMATDA
See my solution on my Youtube channel here:

https://www.youtube.com/watch?v=CZkZxd-fC2o
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trying_to_solve_br
191 posts
#14
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Mehhhhhhhh
Let $H$ be the orthocenter, just remembering: if you know Humpty Dumpty stories, you're done.

Opinion:
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Hint 1:

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Hint 2:

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Hint 3:
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All of those hints are well known facts, note that $YZ || BC \Leftrightarrow YXHZ$ cyclic, and this is true iff $\angle XZH=\angle HYX$, and hence we just need to prove $(ABX)$ tangent to $BC$, by easy angle chase ($\angle XMD=\angle XHA$, and $\angle BAM=90-\angle HYX$, which is equal to $\angle 90-DZB$ iff the tangency is true).

Now just $ME^2=MB^2=MX.MA$ finishes the problem.
This post has been edited 2 times. Last edited by trying_to_solve_br, Jul 1, 2021, 1:17 AM
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levifb
61 posts
#15
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half line sol with proj:
let $K = BX \cap CF$. We know that $(A,X;E,F) \stackrel{H}{=} (B,K;X,Z) \stackrel{Y}{=} (B,C;M, YZ \cap BC) = -1 \Rightarrow YZ // BC$.
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