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points on sides of a triangle, intersections, extensions, ratio of areas wanted
parmenides51   1
N 21 minutes ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1997 OMM P5
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
21 minutes ago
starting with intersecting circles, line passes through midpoint wanted
parmenides51   2
N 40 minutes ago by EmersonSoriano
Source: Peru Ibero TST 2014
Circles $C_1$ and $C_2$ intersect at different points $A$ and $B$. The straight lines tangents to $C_1$ that pass through $A$ and $B$ intersect at $T$. Let $M$ be a point on $C_1$ that is out of $C_2$. The $MT$ line intersects $C_1$ at $C$ again, the $MA$ line intersects again to $C_2$ in $K$ and the line $AC$ intersects again to the circumference $C_2$ in $L$. Prove that the $MC$ line passes through the midpoint of the $KL$ segment.
2 replies
1 viewing
parmenides51
Jul 23, 2019
EmersonSoriano
40 minutes ago
An inequality
Rushil   14
N 40 minutes ago by frost23
Source: Indian RMO 1994 Problem 8
If $a,b,c$ are positive real numbers such that $a+b+c = 1$, prove that \[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]
14 replies
Rushil
Oct 25, 2005
frost23
40 minutes ago
3 var inequality
SunnyEvan   6
N 41 minutes ago by JARP091
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
6 replies
SunnyEvan
May 17, 2025
JARP091
41 minutes ago
collinearity as a result of perpendicularity and equality
parmenides51   2
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1996 OMM P6
In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp  CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.
2 replies
parmenides51
Jul 28, 2018
FrancoGiosefAG
an hour ago
3 var inequality
JARP091   6
N an hour ago by JARP091
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.
6 replies
JARP091
Today at 8:54 AM
JARP091
an hour ago
Helplooo
Bet667   1
N an hour ago by Lil_flip38
Let $ABC$ be an acute angled triangle.And altitudes $AD$ and $BE$ intersects at point $H$.Let $F$ be a point on ray $AD$ such that $DH=DF$.Circumcircle of $AEF$ intersects line $BC$ at $K$ and $L$ so prove that $BK=BL$
1 reply
Bet667
2 hours ago
Lil_flip38
an hour ago
Cyclic sum of 1/(a+1/b+1)
v_Enhance   22
N 2 hours ago by Rayvhs
Source: ELMO Shortlist 2013: Problem A2, by David Stoner
Prove that for all positive reals $a,b,c$,
\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]Proposed by David Stoner
22 replies
v_Enhance
Jul 23, 2013
Rayvhs
2 hours ago
xf(x + xy) = xf(x) + f(x^2)f(y)
orl   14
N 2 hours ago by jasperE3
Source: MEMO 2008, Team, Problem 5
Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad  \forall  x,y \in \mathbb{R}.\]
14 replies
orl
Sep 10, 2008
jasperE3
2 hours ago
Beautiful Number Theory
tastymath75025   34
N 2 hours ago by Adywastaken
Source: 2022 ISL N8
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
34 replies
tastymath75025
Jul 9, 2023
Adywastaken
2 hours ago
Hard Functional Equation in the Complex Numbers
yaybanana   1
N 2 hours ago by jasperE3
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 $x,y \in \mathbb{C}$
1 reply
yaybanana
Apr 9, 2025
jasperE3
2 hours ago
Find all numbers
Rushil   11
N 2 hours ago by frost23
Source: Indian RMO 1994 Problem 3
Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.
11 replies
Rushil
Oct 25, 2005
frost23
2 hours ago
Do not try to bash on beautiful geometry
ItzsleepyXD   9
N May 2, 2025 by Captainscrubz
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 $MN\parallel DG$
9 replies
ItzsleepyXD
Apr 30, 2025
Captainscrubz
May 2, 2025
Do not try to bash on beautiful geometry
G H J
Source: Own , Mock Thailand Mathematic Olympiad P9
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ItzsleepyXD
147 posts
#1 • 1 Y
Y by Dasfailure
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 $MN\parallel DG$
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moony_
22 posts
#2
Y by
ddit for AEGF and point D: pairs (DE, DF), (DA, DG), (DB, DC) are in involution. Now move this lines to point A in parallel: (AK, AI), (AL, AD), (Ainf, Ainf) and we wanna proof that AL is bissector of angle BAC. Now we move this involution on line BC: (K, I), (L, D), (inf, inf).
CI/CD = CA/CF => CI/BK = BE/CF * CA/BA * CD/BD = 1 => CI = BK => this involution is symmetry by point M => D and L are symmetrical by M => BL/CL = BA/CA => BL - bissector of angle BAC => DG || AL || MN => DG || MN

soooo nice geo ^_^
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moony_
22 posts
#3
Y by
ItzsleepyXD wrote:
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 $MN\parallel DG$


another solution:
Let ABA'C - parallelogram
BD/DC = CA/BA = BA'/CA' => A'D - bissector of angle BA'C.
G, D, A' collinear cuz parallelogram bissector lemma
GD || bissector of angle BAC || MN => GD || MN

you can proove parallelogram bissector lemma by using pappus theorem for lines AB and AC and points E, B, inf and F, C, inf on them

it may be easier than first one... hahaha

upd: point names on pic are wrong, sry
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This post has been edited 2 times. Last edited by moony_, Apr 30, 2025, 12:38 PM
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FarrukhBurzu
5 posts
#4
Y by
Is $BE=CF$ true for D E F symmetrically ?
This post has been edited 1 time. Last edited by FarrukhBurzu, Apr 30, 2025, 7:59 PM
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Tkn
33 posts
#6
Y by
[asy]
size(9cm);
defaultpen(fontsize(11pt));
    
pair A = (-0.25,2);
pair B = (-1,0);
pair C = (2,0);

path C1 = circle(B,1);
path C2 = circle(C,1);

pair E1 = intersectionpoint(C1, A--B);
pair F = intersectionpoint(C2, A--C);

pair M = (E1+F)/2;
pair N1 = (B+C)/2;

pair P = 2N1-A;
pair Q = 2M-A;

pair R = extension(Q,F,B,P);
pair S1 = extension(E1,Q,C,P);
pair T = extension(B,F,C,E1);
pair D = extension(T,P,B,C);
pair U = extension(F,Q,C,E1);

draw(A--B--C--cycle, black);
draw(E1--Q--F, black);
draw(B--P--C, black);
draw(R--Q--S1, black+dashed);
draw(E1--F, black);
draw(B--F,blue);
draw(C--E1,blue);
draw(T--P, blue+dashed);
draw(M--N1, red+dashed);

        
dot(A);
dot(B);
dot(C);
dot(E1);
dot(F);
dot(P);
dot(Q);
dot(R);
dot(S1);
dot(M,red);
dot(N1,red);
dot(T,blue);
dot(D);

label("$A$", A, N, black);
label("$B$", B, SW, black);
label("$C$", C, E, black);
label("$E$", E1, NW, black);
label("$F$", F, NE, black);
label("$Q$", Q, 1.3W+0.5S, black);
label("$P$", P, S, black);
label("$R$", R, SW, black);
label("$S$", S1, SE, black);
label("$N$", N1, SW, black);
label("$M$", M, N, black);
label("$G$", T, S+0.5W, black);
label("$D$", D, NE, black);
[/asy]

Let $P$ and $Q$ denote the reflection of $A$ across $N$ and $M$, respectively.
Let $R=\overleftrightarrow{FQ}\cap\overleftrightarrow{BP}$, and $S=\overleftrightarrow{EQ}\cap \overleftrightarrow{CP}$
Note that $AM=MQ$, and $AN=NP$. It implies that $AEQF$, and $ABPC$ are parallelograms.
Since $BE=CF$, we have $QS=RP=RQ=PS$. This indicates that $PRQS$ is a rhombus. Furthermore, the line $\overleftrightarrow{PQ}$ must bisect $\angle{BPC}$.
By the given ratio condition of the point $D$:
$$\frac{BD}{CD}=\frac{AC}{AB}=\frac{BP}{CP},$$so, $D$ lies on the angle bisector of $\angle{BPC}$; that is $D\in \overleftrightarrow{PQ}$.
Next, it suffices to show that $G$ also lies on $\overleftrightarrow{PQ}$, and it could be proceed using Menelaus' theorem on $\triangle{ABF}$ and the line $\overleftrightarrow{EGC}$:
\begin{align*}
    1&=\frac{BE}{AE}\cdot \frac{AC}{CF}\cdot \frac{FG}{GB}\\
    &=\frac{AE}{BE}\cdot \frac{CF}{AC}\cdot \frac{GB}{FG}\\
    &=\frac{FQ}{QR}\cdot \frac{RP}{BP}\cdot \frac{GB}{GF}.
\end{align*}The converse of Menelaus' theorem also holds for $\triangle{FBR}$, so $G\in \overleftrightarrow{PQ}$.
The last step is to observe that $A$ is the homothetic center sending $\overline{MN}\mapsto \overline{QP}$. Therefore, $\overleftrightarrow{PQ}\parallel\overleftrightarrow{MN}$.
In conclusion $\overleftrightarrow{MN}\parallel\overleftrightarrow{GD}$ as desired.
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cj13609517288
1922 posts
#7
Y by
Swap the names of $E$ and $F$.

Move $E$ linearly along $AC$, then $F$ moves linearly along $AB$ and $N$ moves linearly. Assume WLOG $AB<AC$, then let the value of $E$ when $F=A$ be $P$. Then when $F=B$ and $F=A$ we get that $N$ lies on the line through $M$ and the midpoint of $AP$. A homothety of scale factor $2$ from $A$ takes this line through $D$ parallel to the $A$-angle bisector. A simple bary (or MMP) argument shows that $G$ moves on a line (since it's degree $1+1-1=1$). When $E=P$ we get $G=P$ which does lie on the desired line. When $E=A$ we get $G=DP\cap AB$ which does lie on the desired line. $\blacksquare$
This post has been edited 1 time. Last edited by cj13609517288, May 1, 2025, 3:55 PM
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DottedCaculator
7357 posts
#8
Y by
Beautiful problem with a beautiful solution!

Let $D=(0:c:b)$, $E=(t:c-t:0)$, and $F=(t:0:b-t)$. Then, $M=(0:1:1)$ and $N=((b+c)t:b(c-t):c(b-t))$, so $MN$ has direction $(b+c:-b:-c)$. In addition, $G=(t:c-t:b-t)$, so $DG$ also has direction $(t(b+c):(c-t)(b+c)-c(b+c-t):(b-t)(b+c)-b(b+c-t))=(t(b+c):-bt:-ct)=(b+c:-b:-c)$. Therefore, $MN$ and $DG$ are parallel.
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Assassino9931
1362 posts
#9
Y by
In this config it is also well known (follows through the midpoint of $BF$ or $CE$) that $MN$ is parallel to the angle bisector of $\angle BAC$.
In fact, the foot of this bisector is isotomic with $D$ by the angle bisector property.
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Ianis
415 posts
#10
Y by
Assassino9931 wrote:
In this config it is also well known (follows through the midpoint of $BF$ or $CE$) that $MN$ is parallel to the angle bisector of $\angle BAC$.
In fact, the foot of this bisector is isotomic with $D$ by the angle bisector property.

Indeed! And hence it suffices to prove that if $G'$ is the reflection of $G$ wart to $M$ then $G'$ lies on the angle bisector of $\angle BAC$, for which it suffices to prove that the distances from $G'$ to $CA$ and $AB$ are equal. Since $CF=BE$ it suffices to prove that $[G'CF]=[G'EB]$. Now, observe that$$[G'CF]=[G'CB]=[G'EB],$$because $BF\parallel BG\parallel CG'$ and $CE\parallel CG\parallel BF'$. Done.

Credits go to @magnusarg for this solution.
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Captainscrubz
78 posts
#11
Y by
Let $P$ be the arc midpoint of $BAC$
Let an arbitrary circle from $P$ and $A$ intersect $AB$ and $AC$ at $E'$ and $F'$ respectively.
As $P$ will be the center of spiral similarity that sends $E'F'\rightarrow BC$
$\implies \angle E'PB=\angle F'PC$ and $E'P=F'P, BP=PC \implies E'B=F'C$
$\therefore E'\equiv E$ and $F'\equiv F$
If $AB<AC$ then let $H=MN\cap AC$
see that as $P$ is the center of spiral similarity that sends $NF\rightarrow MC$ and a fixed point $\implies PHMC$ is cyclic$\implies H$ is a fixed point
and $\angle NHF=\angle NPF=\angle \frac{A}{2} \implies MN \parallel A\text{ angle bisector}$
Now after the introduction of point $A'$ such that $ABA'C$ is a parallelogram the problem reduces to-
Quote:
Let $G$ be a point on the angle bisector of $A$ and let $l_1$ and $l_2$ be lines from $B$ and $C$ such that $l_1$ and $l_2$ is parallel to $AC$ and $AB$. Let $E=CG\cap l_1$ and $F=BG\cap l_2$. Prove $BE=CF$.
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