Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
Geometry Concurrence
KHOMNYO2   0
2 minutes ago
Given triangle $XYZ$ such that $XY \neq XZ$. Let excircle-$X$ be tangent with $YZ, ZX, XY$ at points $U, V, W$ respectively. Let $R$ and $S$ be points on the segment $XZ, XY$ respectively such that $RS$ is parallel to $YZ$. Lastly, let $\gamma$ be the circle that is externally tangent with the excircle-$X$ on point $T$. Prove that $VW, UT$, and $RS$ concur at a point.
0 replies
KHOMNYO2
2 minutes ago
0 replies
Sharygin 2025 CR P7
Gengar_in_Galar   5
N 31 minutes ago by Blackbeam999
Source: Sharygin 2025
Let $I$, $I_{a}$ be the incenter and the $A$-excenter of a triangle $ABC$; $E$, $F$ be the touching points of the incircle with $AC$, $AB$ respectively; $G$ be the common point of $BE$ and $CF$. The perpendicular to $BC$ from $G$ meets $AI$ at point $J$. Prove that $E$, $F$, $J$, $I_{a}$ are concyclic.
Proposed by:Y.Shcherbatov
5 replies
Gengar_in_Galar
Mar 10, 2025
Blackbeam999
31 minutes ago
helpppppppp me
stupid_boiii   1
N 4 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
4 hours ago
Algebra Polynomials
Foxellar   2
N 5 hours ago 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
6 hours ago
Foxellar
5 hours ago
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
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},
Vulch   3
N Yesterday at 11:58 AM by mohabstudent1
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},$ where $n$ is a natural number.What is the remainder when $n$ is divided by $13?$
3 replies
Vulch
Apr 9, 2025
mohabstudent1
Yesterday at 11:58 AM
Problem 1
SpectralS   146
N Apr 23, 2025 by YaoAOPS
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
146 replies
SpectralS
Jul 10, 2012
YaoAOPS
Apr 23, 2025
Problem 1
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SpectralS
5 posts
#1 • 17 Y
Y by buratinogigle, Mualpha7, Amir Hossein, iarnab_kundu, Davi-8191, math1181, AlastorMoody, A-Thought-Of-God, jhu08, megarnie, HWenslawski, ImSh95, Adventure10, Tastymooncake2, ItsBesi, Rounak_iitr, and 1 other user
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
FelixD
588 posts
#2 • 20 Y
Y by UraharaKisuke, buratinogigle, AndrewTom, biomathematics, karitoshi, Mathasocean, A-Thought-Of-God, translate, jhu08, yshk, ImSh95, Adventure10, Mango247, shafikbara48593762, Tastymooncake2, and 5 other users
If I'm not mistaken, it's easy to see that $\angle LFJ=\alpha/2$ and so $AFJL$ is cyclic. But $\angle JLA=90$ and so $\angle AFJ=90$. Thus, $AB=BS$ and hence $MS=AK$. Similarly, $MT=AL$, but $AK=AL$, so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AndreiAndronache
88 posts
#3 • 7 Y
Y by jhu08, ImSh95, Adventure10, Tastymooncake2, and 3 other users
Otherwise: $m(\angle{JFL})=m(\angle{JAL})=\frac{A}{2}\Rightarrow A,F,J,L$ are concyclic $\Rightarrow S,J,M,F$ are concyclic $\Rightarrow m(\angle{JST})=m(\angle{JTS})=\frac{A}{2}\Rightarrow JS=JT\Rightarrow SM=MT$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WakeUp
1347 posts
#4 • 9 Y
Y by buratinogigle, jhu08, metricpaper, ImSh95, Adventure10, Tastymooncake2, and 3 other users
Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $AJ$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Concyclicboy
49 posts
#5 • 2 Y
Y by jhu08, Adventure10
Yes, we only need that <MFB= <A/2 and then AFJL cyclic then <AFJ=<ALJ=90 then <SFJ=<SMJ=90 then SFMJ cyclic with <MSJ=<A/2. Similar to <JTM=<A/2
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Zhero
2043 posts
#6 • 7 Y
Y by buratinogigle, yugrey, jhu08, Adventure10, Tastymooncake2, and 2 other users
By a simple angle chase, $\angle FBM = \angle ABJ = 90 + \angle B/2$ and $\angle FMB = \angle AJB = \angle C / 2$, so $\triangle BFM \sim \triangle BAJ$. Thus, $\triangle BFA \sim \triangle BMJ$ (since $BF/BA = BM/BJ$ and $\angle FBA = \angle MBJ = 90 - \angle B / 2$.) Since $BM \perp MJ$, we have $BF \perp AF$. Since $\angle ABF = \angle FBS = 90 - \angle B/2$, we have $BS = BA$. Similarly, $CA = CT$, so $SM = (s-c) + c = s = (s-b) + b = MT$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
golem24
92 posts
#7 • 19 Y
Y by yugrey, amar_04, ETS1331, Aryan-23, jhu08, mistakesinsolutions, Adventure10, Mango247, MS_Kekas, Tastymooncake2, Funcshun840, and 8 other users
The solutions are smaller than the problem statement.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Kunihiko_Chikaya
14514 posts
#8 • 7 Y
Y by AceOfDiamonds, buratinogigle, jhu08, yshk, Adventure10, Mango247, Tastymooncake2
Here is the attached figure.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Orin
22 posts
#9 • 3 Y
Y by buratinogigle, Adventure10, Mango247
It is well-known that $\angle BJC=90^{\circ}-\frac {1}{2}\angle A$
now $\angle LJC=\angle MJC=90^{\circ}-\angle MCJ=90^{\circ}-(90^{\circ}- \frac {1}{2}\angle C)=\frac {1}{2} \angle C$
so $\angle FJL=\angle BJC+\angle LJC=90^{\circ} -\frac {1}{2} \angle A+\frac {1}{2} \angle C$....(1)
now $\angle MLJ=90^{\circ}- \angle LJC=90^{\circ}-\frac {1}{2} \angle C$....(2)
so from (1),(2) we get in $\triangle LJF,\angle LFJ=\frac {1}{2} \angle A$ which inplies $AFJL$ is cyclic.
since $\angle JLA=90^{\circ}$ we get $AFJ=90^{\circ}$
now $\angle FAB=\angle FAL- \angle A=\angle \frac {1}{2} \angle B$
and $\angle FSB=90^{\circ}-\angle FBS=90^{\circ}-\angle CBJ=\angle \frac {1}{2} \angle B$
hence $\angle FAB=\angle FSB$
so $\triangle FBS \cong \triangle FBA$[ASA] implying $AB=BS$
now $MS=MB+BS=BK+AB=AK$
simimlarly $MT=AL$
but $AK=AL$[tangent to the excircle from $A$]
hence $MS=MT$ which means $M$ is the midpoint of $ST$
P.S.:Edited only to correct a typo.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_genie
41 posts
#10 • 2 Y
Y by Adventure10, Mango247
Its quite easy to see that
$A,K,J,G$ is cyclic. (1)
$A,F,J,L$ is cyclic. (2)

Since $\angle AKJ = 90^{\circ}$, this implies that $\angle AGJ = 90^{\circ}$ from (1)
Similarly $\angle AFJ = 90^{\circ}$.
Hence we get
$A,F,K,J$ is cyclic (3)
$A,G,L,J$ is cyclic (4)

Thus from (1),(2),(3),(4) we get that all six points $A,G,F,J,K,L$ to be cyclic.
Now, note that from triangle ABS, $AB = BS = c$.
Similarly, $AC = CT = b$.
Thus $SM = SB + BM = c + (s-c) = s$
Similarly $TM = TC + CM = b + (s-b) = s$.
Hence proved.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sjaelee
485 posts
#11 • 2 Y
Y by Adventure10 and 1 other user
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Neerjhor
12 posts
#12 • 1 Y
Y by Adventure10
It was really easy. Mine is almost similar as everyone. I proved that $AFJL$ and $AGJK$ are cyclic, which concludes that $AB=BS$ and $AC=CT$. Then $BM=s-c$ and $CM=s-b$. Now, we add these and $MS=MT$. :-)

This was quite easy for IMO 1 I guess. :-)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi37
2079 posts
#13 • 1 Y
Y by Adventure10
Slightly different approach
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
guysfromKIROVOGRAD
3 posts
#14 • 2 Y
Y by Adventure10, Mango247
WakeUp wrote:
Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $A$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic.

Cool solution, bro. But ... too many letters as for me. Ты ещё не задолбался точечки отмечать? ( Use russian translator, bro)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AV-2
2 posts
#15 • 2 Y
Y by Adventure10, Mango247
Notice that $MK$ and $ML$ are parallel to the internal bisectors of $B$ and $C$ respectively. Then $MK \perp BJ$, $ML \perp CJ$, so $M$ is the orthocenter of $JFG$. Since $JM \perp BC$ from tangency, and $JM \perp FG$ from the orthocenter, $BC \parallel FG$. Now we can prove that $MF \parallel AG$ and $MG \parallel AF$ with any of the methods above, or by considering the excenters $I_b, I_c$, the fact tht $I_bI_c$ is antiparallel to $BC$ and a short angle chase. With this, $FMG$ is the medial triangle of $AST$ and we are done.
Z K Y
G
H
=
a