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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Lemma on tangency involving a parallelogram with orthocenter
Gimbrint   1
N 8 minutes ago by Captainscrubz
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$ and orthocenter $H$. Let $M$ be the midpoint of $AC$. Line $BH$ intersects $\omega$ again at $L\neq B$, and line $ML$ intersects $\omega$ again at $P\neq L$. Points $D$ and $E$ lie on $AB$ and $BC$ respectively, such that $BEHD$ is a parallelogram.

Prove that $BP$ is tangent to the circumcircle of triangle $BDE$.
1 reply
+1 w
Gimbrint
2 hours ago
Captainscrubz
8 minutes ago
Random walk
EthanWYX2009   1
N 11 minutes ago by XAN4
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
1 reply
EthanWYX2009
2 hours ago
XAN4
11 minutes ago
2025 KMO Inequality
Jackson0423   1
N 15 minutes ago by Quantum-Phantom
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
1 reply
Jackson0423
Yesterday at 4:32 PM
Quantum-Phantom
15 minutes ago
Simple Geometry
AbdulWaheed   2
N 23 minutes ago by Sadigly
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 $ABC$.
2 replies
AbdulWaheed
Today at 5:15 AM
Sadigly
23 minutes ago
Inequalities
sqing   18
N 5 hours ago by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
18 replies
sqing
May 13, 2025
sqing
5 hours ago
helpppppppp me
stupid_boiii   1
N 5 hours ago by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
5 hours ago
Algebra Polynomials
Foxellar   2
N Today at 5:43 AM by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
Today at 4:52 AM
Foxellar
Today at 5:43 AM
geometry
luckvoltia.112   0
Today at 2:29 AM
Let \( \triangle ABC \) be an acute triangle with \( AB < AC \), and its vertices lie on the circle \( (O) \). Let \( AD \) be the altitude from vertex \( A \). Let \( E \) and \( F \) be the feet of the perpendiculars from \( D \) to the lines \( AB \) and \( AC \), respectively. Let \( EF \) intersect the circle \( (O) \) again at points \( P \) and \( Q \) such that \( E \) lies between \( Q \) and \( F \). Let the lines \( AD \) and \( EF \) intersect at point \( G \). Let \( I \) be the midpoint of segment \( AD \). Let \( AO \) intersect line \( BC \) at point \( K \).
a) Prove that \( AP = AQ = AD \).
b) Prove that line \( OI \) is parallel to line \( KG \).
c)Let \( H \) be the orthocenter of triangle \( ABC \), and let \( M \) be the midpoint of segment \( BC \). $S$ is the center (HBC). Let point \( T \) lie on line \( DS \) such that ray \( KD \) is the angle bisector of \( \angle GKT \). Prove that lines \( AD \) and \( MT \) intersect at a point lying on circle \( (O) \).
0 replies
luckvoltia.112
Today at 2:29 AM
0 replies
Great Geometry with Squares on sides of triangles
SomeonecoolLovesMaths   2
N Today at 1:50 AM by happypi31415
Three squares are drawn on the sides of triangle \(ABC\) (i.e., the square on \(AB\) has \(AB\) as one of its sides and lies outside \(ABC\)). Show that the lines drawn from the vertices \(A\), \(B\), and \(C\) to the centers of the opposite squares are concurrent.

IMAGE
2 replies
SomeonecoolLovesMaths
Yesterday at 9:44 PM
happypi31415
Today at 1:50 AM
A suspcious assumption
NamelyOrange   2
N Today at 1:30 AM by maromex
Let $a,b,c,d$ be positive integers. Maximize $\max(a,b,c,d)$ if $a+b+c+d=a^2-b^2+c^2-d^2=2012$.
2 replies
NamelyOrange
Yesterday at 1:53 PM
maromex
Today at 1:30 AM
n is divisible by 5
spiralman   1
N Yesterday at 8:42 PM by KSH31415
n is an integer. There are n integers such that they are larger or equal to 1, and less or equal to 6. Sum of them is larger or equal to 4n, while sum of their square is less or equal to 22n. Prove n is divisible by 5.
1 reply
spiralman
Wednesday at 7:38 PM
KSH31415
Yesterday at 8:42 PM
Monochromatic Triangle
FireBreathers   1
N Yesterday at 8:08 PM by KSH31415
We are given in points in a plane and we connect some of them so that 10n^2 + 1 segments are drawn. We color these segments in 2 colors. Prove that we can find a monochromatic triangle.
1 reply
FireBreathers
Yesterday at 2:28 PM
KSH31415
Yesterday at 8:08 PM
how difficult are these problems
rajukaju   1
N Yesterday at 7:28 PM by Shan3t
I can solve only the first 4 problems of the last general round of the HMMT competition: https://hmmt-archive.s3.amazonaws.com/tournaments/2024/nov/gen/problems.pdf

As a prediction, would this mean I am good enough to qualify for AIME? How does the difficulty compare?

1 reply
rajukaju
Yesterday at 6:43 PM
Shan3t
Yesterday at 7:28 PM
Maximum value of function (with two variables)
Saucepan_man02   1
N Yesterday at 1:39 PM by Saucepan_man02
If $f(\theta) = \min(|2x-7|+|x-4|+|x-2 -\sin \theta|)$, where $x, \theta \in \mathbb R$, then maximum value of $f(\theta)$.
1 reply
Saucepan_man02
Yesterday at 1:25 PM
Saucepan_man02
Yesterday at 1:39 PM
Angle bisectors
bgn   3
N Aug 11, 2020 by Arefe
Source: Iran MO 3rd round 2016 finals - Geometry P3
Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively.
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$
3 replies
bgn
Sep 5, 2016
Arefe
Aug 11, 2020
Angle bisectors
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran MO 3rd round 2016 finals - Geometry P3
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bgn
178 posts
#1 • 2 Y
Y by doxuanlong15052000, Adventure10
Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively.
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$
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drmzjoseph
445 posts
#2 • 2 Y
Y by Adventure10, Mango247
Nice problem

General problem

Given triangle $\triangle ABC$ and let $D$ be the foot of angle bisectors of $A$. If $X \in AD, BX \cap AC \equiv E, CX \cap AB \equiv F$
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$

Proof

$Y \equiv EF \cap BC, R \equiv AH \cap EF$ let $W,Z$ the proyections of $A,D$ at $EF$ respectively. Denote $P$ a point at $ZD$ such that $\odot (APD)$ is tangent to $AZ$, and $Q \equiv EF \cap AP$

Is clear $\angle AYF =\angle ADP =\angle PAZ \Rightarrow AZ^2=ZY.ZQ=ZP.ZD \Rightarrow YQPD$ is cyclical $\Rightarrow \angle ZYD=\angle APZ \Rightarrow \triangle AZP \cup Q \sim \triangle HWA \cup R \Rightarrow \frac{AR}{RH}=\frac{PQ}{QA}$ and $\angle ARN=\angle MQA$

Denote $K'$ and $L'$ points at $EF$ such that $AN \parallel PK'$ and $AM \parallel PL' \Rightarrow AMN \cup H \sim PK'L' \cup A \Rightarrow \triangle HMN \sim \triangle AK'L' \Rightarrow K' \equiv K$ and $L \equiv L'$ is evident that $KP=PL \Rightarrow DK=DL$
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nguyenhaan2209
111 posts
#3 • 3 Y
Y by top1csp2020, Adventure10, Mango247
First we see that K,L can be construct uniquely by let the symmetric of H wrt EF be H' and draw two parallel line from A to H'M, H'N we have KL. The same idea as IMO 2018 G6, we will reconstruct it. Let the altitude of A,D and BC meet EF at U,V,W and P lies on DU such that UPA~UAD then UP.UD=UA^2 and UDA=UWA=UAP so UW.UQ=UD.UP gives us WQPD cyclic. Cause AVH=AUP=180-AWD and APU=UAD=VWH=VAR so AUP ~ HVA and PUQ=RVA=90 then AR/RH=PQ/QA + ARN=KQA. Re-construct K,L as PK//AN, PL//AM so AMN-H~PKL-A so AKL~HMN. (Q.E.D)
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Arefe
65 posts
#4
Y by
If we get $ T , P , Y $ foot of $ A , D , Y $ with $ EF$ and $Q , X$ intersection of $AH , AD$ with $EF$ .
Get $N$ the point on $TE$ and $L' , K'$ reflection of $L , K$ about $T$ .
$HN||AL' , HM||AK' \rightarrow \frac{NQ}{NT+TL}=\frac{NQ}{ML}=\frac{HQ}{HA}=\frac{MQ}{NK}=\frac{2NT-NQ}{NT+TK}$
$\rightarrow NT+TL=\frac{AH}{HQ}.NQ , NT+TK=\frac{AH}{HQ}.(2NT-NQ) \rightarrow |TL-TK|=2.\frac{AH}{HQ}.|NQ-NT| = 2.\frac{AH}{HQ}.QT$
so the problem is equal $\leftrightarrow PL=PK \leftrightarrow |TL-TK|=2TP \leftrightarrow TP=\frac{AH}{HQ}.QT $
If $D'$ intersection of $EF , BC$ we can see that $\angle{DAD'}=90$ so $ATHD' , PDD'A$ are cyclic .
$\frac{AH}{HQ}.QT=AH.\frac{QT}{QA}.\frac{AQ}{QH}=AH . \sin{BD'F} . \frac{\sin{AD'F}}{\sin{BD'F}}.\frac{AD'}{D'H} = AH . \sin{AD'F} . \frac{AD}{AH} = \sin{AD'F} . AD = TP $ and we're done $\blacksquare$
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