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

G
Topic
First Poster
Last Poster
3 var inequality
JARP091   7
N 41 minutes ago by MathsII-enjoy
Source: Own
Let \( x, y, z \in \mathbb{R}^+ \). Prove that
\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]without using the Rearrangement Inequality or Chebyshev's Inequality.
7 replies
JARP091
Yesterday at 8:54 AM
MathsII-enjoy
41 minutes ago
System of Equations
Math-wiz   18
N an hour ago by justaguy_69
Source: RMO 2019 Paper 2 P3
Find all triples of non-negative real numbers $(a,b,c)$ which satisfy the following set of equations
$$a^2+ab=c$$$$b^2+bc=a$$$$c^2+ca=b$$
18 replies
Math-wiz
Nov 10, 2019
justaguy_69
an hour ago
Geometry
MathsII-enjoy   5
N an hour ago by MathsII-enjoy
Given triangle $ABC$ inscribed in $(O)$ with $M$ being the midpoint of $BC$. The tangents at $B, C$ of $(O)$ intersect at $D$. Let $N$ be the projection of $O$ onto $AD$. On the perpendicular bisector of $BC$, take a point $K$ that is not on $(O)$ and different from M. Circle $(KBC)$ intersects $AK$ at $F$. Lines $NF$ and $AM$ intersect at $E$. Prove that $AEF$ is an isosceles triangle.
5 replies
MathsII-enjoy
May 15, 2025
MathsII-enjoy
an hour ago
Self-evident inequality trick
Lukaluce   18
N an hour ago by sqing
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
18 replies
Lukaluce
May 18, 2025
sqing
an hour ago
Cauchy FE
InftyByond   1
N 2 hours ago by maromex
For the Cauchy FE, is x^nf(x)=f(x^(n+1)) enouh to go from rationals to reals? thats what wikipedia says but idk really
1 reply
InftyByond
Mar 1, 2025
maromex
2 hours ago
Train yourself on folklore NT FE ideas
Assassino9931   3
N 2 hours ago by MathLuis
Source: Bulgaria Spring Mathematical Competition 2025 9.4
Determine all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that $f(a) + 2ab + 2f(b)$ divides $f(a)^2 + 4f(b)^2$ for any positive integers $a$ and $b$.
3 replies
Assassino9931
Mar 30, 2025
MathLuis
2 hours ago
IMO ShortList 1998, number theory problem 2
orl   16
N 2 hours ago by littlefox_amc
Source: IMO ShortList 1998, number theory problem 2
Determine all pairs $(a,b)$ of real numbers such that $a \lfloor bn \rfloor =b \lfloor an \rfloor $ for all positive integers $n$. (Note that $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.)
16 replies
orl
Oct 22, 2004
littlefox_amc
2 hours ago
IMO Shortlist 2013, Algebra #2
lyukhson   27
N 3 hours ago by ezpotd
Source: IMO Shortlist 2013, Algebra #2
Prove that in any set of $2000$ distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d $, such that \[ \left| \frac{a-b}{c-d} - 1 \right|< \frac{1}{100000}. \]
27 replies
lyukhson
Jul 9, 2014
ezpotd
3 hours ago
center of (XYZ) lies on a fixed circle
VicKmath7   10
N 3 hours ago by Blackbeam999
Source: All-Russian 2022 11.3
An acute-angled triangle $ABC$ is fixed on a plane with largest side $BC$. Let $PQ$ be an arbitrary diameter of its circumscribed circle, and the point $P$ lies on the smaller arc $AB$, and the point $Q$ is on the smaller arc $AC$. Points $X, Y, Z$ are feet of perpendiculars dropped from point $P$ to the line $AB$, from point $Q$ to the line $AC$ and from point $A$ to line $PQ$. Prove that the center of the circumscribed circle of triangle $XYZ$ lies on a fixed circle.
10 replies
VicKmath7
Apr 19, 2022
Blackbeam999
3 hours ago
Hard Functional Equation in the Complex Numbers
yaybanana   3
N 5 hours ago by Burmf
Source: Own
Find all functions $f:\mathbb {C}\rightarrow \mathbb {C}$, s.t :

$f(xf(y)) + f(x^2+y) = f(x+y)x + f(f(y))$

for all $x,y \in \mathbb{C}$
3 replies
yaybanana
Apr 9, 2025
Burmf
5 hours ago
Incenter and concurrency
jenishmalla   7
N Apr 20, 2025 by brute12
Source: 2025 Nepal ptst p3 of 4
Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
7 replies
jenishmalla
Mar 15, 2025
brute12
Apr 20, 2025
Incenter and concurrency
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 Nepal ptst p3 of 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jenishmalla
5 posts
#1
Y by
Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
This post has been edited 2 times. Last edited by jenishmalla, Mar 15, 2025, 3:00 PM
Reason: formatting
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1008 posts
#2
Y by
Main idea for nuke
This post has been edited 2 times. Last edited by AshAuktober, Mar 15, 2025, 3:00 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Thapakazi
67 posts
#3 • 1 Y
Y by ABYSSGYAT
My problem. We present two solutions :love:

Solution 1: Let lines $D'Y$ and $DX$ intersect at $T$. Define $I$ as the incenter of $\triangle{ABC}$ and let $Z = DD' \cap AT$. First, observe that $D'$ is the orthocenter of $\triangle{TAD}.$ Since $IZ \perp AT$, it follows that $Z$ lies in a circle with diameter $AI$. Consequently, $AZFIE$ is a cyclic pentagon.

Now consider the radical axes of the circles $(AZFE), (EFXD)$, and $(AZXD)$. This implies that $AZ$, $EF$, and $DX$ are concurrent at $T$, meaning $EF$ and $DX$ intersect at $T$. However, since $YD'$ and $DX$ also intersect at $T$, it follows that $DX$, $D'Y$, and $EF$ are concurrent at $T$, as desired.

Solution 2: Observe that $FYED$ is a harmonic quadrilateral. Brianchon's Theorem on hexagon $YD'FXDE$, we see that $YX$ and $D'D$ concur on line $FE$. Lastly, by Pascal’s Theorem on hexagon $D'YYXDD$, the desired result follows.
This post has been edited 1 time. Last edited by Thapakazi, Mar 15, 2025, 5:02 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Captainscrubz
78 posts
#4
Y by
Let $G=D'Y\cap XD$
For the complete quadrilateral $D'XDY$ Let $J$ be the center of spiral similarity that sends $D'Y\rightarrow XD$
$\implies$ that $AG\parallel BC$ as $D'XDY$ is cylcic
Let $GE$ meet the incircle again at $F'$
$\implies GE\cdot GF'=GX\cdot GD=GJ\cdot GA$
$\therefore F\equiv F'$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Double07
93 posts
#5
Y by
This can also be easily bashed with complex numbers:

Take the incircle of $\Delta ABC$ to be the unit circle.

Then $a=\frac{2ef}{e+f}, b=\frac{2df}{d+f}, c=\frac{2de}{d+e}, d'=-d$.

$A, D', X$ are collinear $\implies -d'x=\frac{a-d'}{\overline{a}-\overline{d'}}\implies dx=\frac{\frac{2ef}{e+f}+d}{\frac{2}{e+f}+\frac{1}{d}}=d\cdot\frac{2ef+de+df}{2d+e+f}\implies x=\frac{2ef+de+df}{2d+e+f}$.

$A, D, Y$ are collinear $\implies -dy=\frac{a-d}{\overline{a}-\overline{d}}\implies -dy=d\cdot\frac{2ef-de-df}{2d-e-f}\implies y=\frac{2ef-de-df}{e+f-2d}$.

Let $P=EF\cap DX\implies p=\frac{ef\cdot\frac{2d^2+2de+2df+2ef}{2d+e+f}-\frac{d(2ef+de+df)}{2d+e+f}(e+f)}{ef-\frac{d(2ef+de+df)}{2d+e+f}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}$.

Let $Q=EF\cap D'Y\implies q=\frac{ef\cdot\frac{2d^2-2de-2df+2ef}{e+f-2d}+\frac{d(2ef-de-df)}{e+f-2d}(e+f)}{ef+\frac{d(2ef-de-df)}{e+f-2d}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}=p$, so $P=Q$ and we're done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidenkim119
34 posts
#6
Y by
Trivial by Six-point-line!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihategeo_1969
238 posts
#7
Y by
All pole-polars are w.r.t incircle. It is well known midpoint of $\overline{BC}$ say $M$ is pole of $\overline{DX}$ (and $A$ is pole of $\overline{EF}$); hence we just need to prove pole of $\overline{YD'}$ lies on $\overline{AM}$ by La Hire. Now $\overline{AM}$, $\overline{DD'}$, $\overline{EF}$ concurrence is well known and since $(YD;EF)=(D'X;EF)=-1$, we have the concurrency point also lies on $\overline{YX}$. Now Pascal at $YYDD'D'X$ finishes.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
brute12
1 post
#9
Y by
So, like one little solution which i noticed during exam time :blush:
Define:
\[
P = EF \cap AD, \quad Q = EF \cap AD'.
\]
We claim that lines $DX$, $D'Y$, and $EF$ are concurrent.

Note that the cross-ratios satisfy:
\[
(A, Q; D', X) = (A, P; Y, D) = -1.
\]
This follows because $P$ and $Q$ are defined by the intersections of $AD$ and $AD'$ with $EF$, and $(A, P; Y, D)$ and $(A, Q; D', X)$ are harmonic divisions.

Moreover, since $D'$ is the reflection of $D$ across the incenter $I$ of triangle $ABC$, it follows that $D'$ and $Y$ lie on the same side of $EF$, and likewise for $D$ and $X$.

Now, by the **Prism Lemma**, which states that if $(A, P; Y, D) = -1$ and $(A, Q; D', X) = -1$ and if $D'$ and $X$ lie on the same side of $EF$, then the lines $DX$, $D'Y$, and $EF$ concur, the result follows.


REMARK: : we haven't used the property that D is the intouch point of the circle with side BC, and D' is the antipode of D, It holds for general point on the incircle, this property of D opens the path for the nice synthetic solution by radical axis, only required property was that EDFD' harmonic!
This post has been edited 1 time. Last edited by brute12, Apr 20, 2025, 4:13 PM
Reason: just for adding nice remark
Z K Y
N Quick Reply
G
H
=
a