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Product of opposite sides
nabodorbuco2   0
an hour ago
Source: Original
Let $ABCDEF$ a regular hexagon inscribed in a circle $\Omega$. Let $P_i$ be a point inside $\Omega$ and $P_e$ its polar reflection wrt $\Omega$. The rays $AP_i,BP_i,CP_i,DP_i,EP_i,FP_i$ meet $\Omega$ again at $A_i,B_i,C_i,D_i,E_i,F_i$. Call $Q_I$ the polygon formed by the vertices $A_i,B_i,C_i,D_i,E_i,F_i$. Similarly construct the polygon $Q_E$ using $P_e$ instead.

Show that $Q_I$ and $Q_E$ are congruent.
0 replies
nabodorbuco2
an hour ago
0 replies
In Cyclic Quadrilateral ABCD, find AB^2+BC^2-CD^2-AD^2
Darealzolt   1
N an hour ago by Beelzebub
Source: KTOM April 2025 P8
Given Cyclic Quadrilateral \(ABCD\) with an area of \(2025\), with \(\angle ABC = 45^{\circ}\). If \( 2AC^2 = AB^2+BC^2+CD^2+DA^2\), Hence find the value of \(AB^2+BC^2-CD^2-DA^2\).
1 reply
Darealzolt
Today at 4:10 AM
Beelzebub
an hour ago
Another FE
M11100111001Y1R   3
N an hour ago by AndreiVila
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
3 replies
M11100111001Y1R
Yesterday at 8:03 AM
AndreiVila
an hour ago
Iran TST Starter
M11100111001Y1R   4
N an hour ago by flower417477
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
4 replies
M11100111001Y1R
May 27, 2025
flower417477
an hour ago
Inequality with abc=1
tenplusten   10
N an hour ago by Adywastaken
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
10 replies
tenplusten
May 15, 2016
Adywastaken
an hour ago
A game of cutting
k.vasilev   11
N 2 hours ago by NicoN9
Source: All-Russian Olympiad 2019 grade 10 problem 2
Pasha and Vova play the following game, making moves in turn; Pasha moves first. Initially, they have a large piece of plasticine. By a move, Pasha cuts one of the existing pieces into three(of arbitrary sizes), and Vova merges two existing pieces into one. Pasha wins if at some point there appear to be $100$ pieces of equal weights. Can Vova prevent Pasha's win?
11 replies
k.vasilev
Apr 23, 2019
NicoN9
2 hours ago
Problem 10
SlovEcience   1
N 2 hours ago by lbh_qys
Let \( x, y, z \) be positive real numbers satisfying
\[ xy + yz + zx = 3xyz. \]Prove that
\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]
1 reply
SlovEcience
3 hours ago
lbh_qys
2 hours ago
Cup of Combinatorics
M11100111001Y1R   8
N 2 hours ago by sansgankrsngupta
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
8 replies
M11100111001Y1R
May 27, 2025
sansgankrsngupta
2 hours ago
2-var inequality
sqing   2
N 3 hours ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
2 replies
sqing
May 25, 2025
sqing
3 hours ago
2-var inequality
sqing   10
N 3 hours ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
10 replies
sqing
May 27, 2025
sqing
3 hours ago
Complex number
ronitdeb   1
N 4 hours ago by alexheinis
Let $z_1, ... ,z_5$ be vertices of regular pentagon inscribed in a circle whose radius is $2$ and center is at $6+i8$. Find all possible values of $z_1^2+z_2^2+...+z_5^2$
1 reply
ronitdeb
Yesterday at 6:13 PM
alexheinis
4 hours ago
Collinearity with orthocenter
liberator   182
N May 19, 2025 by CrazyInMath
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
182 replies
liberator
Jan 4, 2016
CrazyInMath
May 19, 2025
Collinearity with orthocenter
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2013 Problem 4
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liberator
95 posts
#1 • 26 Y
Y by Davi-8191, tenplusten, Arthur., Tawan, Wizard_32, Arshia.esl, OlympusHero, Instance, HWenslawski, donotoven, centslordm, jhu08, TheCollatzConjecture, mathlearner2357, tiendung2006, megarnie, RedFlame2112, rayfish, ImSh95, Lamboreghini, newinolympiadmath, Adventure10, Mango247, Rounak_iitr, farhad.fritl, ItsBesi
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
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High
876 posts
#2 • 9 Y
Y by Tawan, RedFlame2112, ImSh95, megarnie, Danielzh, lian_the_noob12, Adventure10, Mango247, Rounak_iitr
Just happened to stumble upon this problem recently; I guess I'll post a quick solution. Also, I think this problem needs to be added as the 2013 ISL G1 in the contest section.
Solution
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rkm0959
1721 posts
#3 • 6 Y
Y by Tawan, JUSTemom, RedFlame2112, ImSh95, Adventure10, Mango247
As $\angle BNC = \angle BMC = 90$, $B, N, M, C$ is cyclic. Let this circle be $\omega_3$. Let $Z(\not= W)  = \omega_1 \cap \omega_2$. Let $D = AH \cap BC$.
Now $BN$ is the radical axis of $\omega_1$ and $\omega_3$ and $CM$ is the radical axis of $\omega_2$ and $\omega_3$.
This implies that $A=BN \cap CM$ is the radical center of these three circles, so $A, Z, W$ are colinear.
Now by Power of a Point, we have $AN \cdot AB = AH \cdot AD= AZ \cdot AW = AM \cdot AC$, so we have $$\angle AEW = \angle AZH = \angle XZW = \angle WZY = \angle AZX = \angle AZY = 90$$by simple angle chasing - so $X, Y, Z, H$ are colinear.
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PROF65
2016 posts
#4 • 4 Y
Y by RedFlame2112, ImSh95, Adventure10, Mango247
Let $Z $ the second point of intersection of $\omega _1$ and $\omega _2$.
$(BCMN)$ are concyclic and the radical axes of its circumcircle , $\cal{C}$ $\ \text{and}\ \omega _1,$ $\cal{C}$ $  \text{and} \ \omega _2,\omega _1\ \text{and}\ \omega _2,$ are concurrent at $A$ . $\widehat{WZY}=\frac{\pi}{2},\widehat{WZX}=\frac{\pi}{2}$ thus $X,Y$ and $Z$ are colinear .if $H'$ is the intersection of $XY$ and $AH$ then $H'H_1ZW$ is cyclic where $H_1$ is the foot of $AH$.hence $AZ\cdot AW=AH_1\cdot AH'$ but $AZ\cdot AW=AN\cdot AB$ then $AH'\cdot AH_1=AB\cdot AN$ besides $BHH_1N$ is cyclic implies $AN\cdot AB=AH\cdot AH_1$ so $H=H'$ and the result follows.
R.HAS
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liberator
95 posts
#5 • 27 Y
Y by rkm0959, jam10307, vlohani, A_Math_Lover, Vfire, FadingMoonlight, Polynom_Efendi, zuss77, hara-huri, OlympusHero, sabrinamath, myh2910, mathlearner2357, prMoLeGend42, RedFlame2112, rayfish, ImSh95, megarnie, newinolympiadmath, Danielzh, Dimanas23, aqwxderf, Adventure10, Mango247, saltamonte, Math_.only., raffigm
The easiest way to do this is with Reim's theorem.
[asy]
/* IMO 2014 Question 4, free script by liberator, 12 August 2014 */
unitsize(2cm);
defaultpen(fontsize(10pt));
/* Initialize objects */
pair A = (-2.5, 2.5);
pair B = (-3.5, -0.5);
pair C = (0.5, -0.5);
pair H = orthocenter(A,B,C);
pair M = foot(B,C,A);
pair N = foot(C,A,B);
pair W = (-1.5, -0.5);
pair X = rotate(180, circumcenter(B,W,N))*W;
pair Y = rotate(180, circumcenter(C,W,M))*W;
pair P = intersectionpoint(H--Y, circumcircle(B,W,N)); 
/* Draw objects */
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(X--Y, rgb(0.4,0.6,0.8)+dashed+linewidth(1));
draw(A--H, rgb(0.4,0.6,0.8));
draw(B--X, rgb(0.4,0.6,0.8));
draw(C--Y, rgb(0.4,0.6,0.8));
draw(circumcircle(B,W,N), red);
draw(circumcircle(C,W,M), red);
draw(circumcircle(A,M,N), red);
/* Place dots on and label each point, label circles */
dot(A); label("$A$", A, dir(90));
dot(B); label("$B$", B, dir(200));
dot(C); label("$C$", C, dir(340));
dot(H); label("$H$", H, dir(-90));
dot(M); label("$M$", M, dir(0));
dot(N); label("$N$", N, dir(180));
dot(P); label("$P$", P, dir(-70));
dot(W); label("$W$", W, 2.5*dir(-80));
dot(X); label("$X$", X, dir(180));
dot(Y); label("$Y$", Y, dir(60));
label("$\omega_1$", B--W, 15*dir(-90));
label("$\omega_2$", C--Y, 10*dir(0));
[/asy]
Let $P$ be the second intersection of $\omega_1, \omega_2$. Then $P$ is the Miquel point of $\triangle MNW$ w.r.t $\triangle ABC$, so $AMPHN$ is cyclic. $BX \parallel AH$, hence $X, H, P$ are collinear by Reim's theorem; similarly, $H,P,Y$ are collinear, and the result follows.
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K6160
276 posts
#6 • 4 Y
Y by Mr.Mister, RedFlame2112, Adventure10, Mango247
Let $Z$ be the second intersection of $\omega_1$ and $\omega_2$. Let $A'$ be the foot of the $A$-altitude. We have that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ because $BCMN$ is cyclic. Also by power of a point, $AZ\cdot AW=AN\cdot AB = AH\cdot AA'$, so $A'WZH$ is cyclic. Therefore, $\angle XZW=\angle YZW=\angle HZW=90^{\circ}$ and the conclusion follows. $\boxed{}$
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Kezer
986 posts
#7 • 7 Y
Y by MSTang, S117, RedFlame2112, Adventure10, Mango247, Rounak_iitr, Math_.only.
Different angle chasing approach without the radical axis/Reim

Was too noob after drawing in the Miquel point to think of proving that it lies on $XH$, so I needed Evan's hint from his book to do so. :blush:
This post has been edited 2 times. Last edited by Kezer, Mar 29, 2016, 8:39 PM
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MSTang
6012 posts
#8 • 5 Y
Y by donotoven, RedFlame2112, Adventure10, Mango247, Natrium
Spiral sims!
This post has been edited 2 times. Last edited by MSTang, Jun 18, 2016, 3:03 AM
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Reynan
634 posts
#9 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
$AN\cdot AB=AM\cdot AC = AP\cdot AW\cdots (1)$

we have $\angle XPW=90=\angle WPX$ so $XPY$ collinear

let $D$ the perpendicular from $A$ to BC$
we have $NHDB$ and HDPW$ cylich from $(1)$
$\angle XHD=\angle DWP=180-\angle DHP$ so $XHP$ collinear
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the_executioner
533 posts
#10 • 4 Y
Y by Tawan, RedFlame2112, Adventure10, gulab_jamun
Let $Z$ be the second intersection point $\omega_1$ and $\omega_2$ . Now $\angle YZW$$+$$ \angle XZW $$=$$ 90^o+90^o=180^o$. Therefore $Y$, $Z$ and $X$ are collinear. So we are left to prove that $Z$, $H$ and $X$ are collinear. Now since $Z$ is the miquel point, we have $A$, $N$, $H$, $Z$, $M$ are concyclic. So$\angle AZM$$ +$$\angle MZW$ $= C+(A+B)=180^o$.
Therefore $A$, $Z$ and $W$ are collinear. Now, $\angle AZH$ $=90^o$ and thus $\angle HZW$$ =90^o$. But $\angle WZX$$ =90^o$, therefore $Z$, $H$ and $X$ are collinear and we are done.
This post has been edited 2 times. Last edited by the_executioner, Jul 1, 2016, 3:06 AM
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anantmudgal09
1980 posts
#11 • 4 Y
Y by Tawan, Aryan-23, RedFlame2112, Adventure10
A non standard way to do it...

Let $\ell_b,\ell_c$ be the lines perpendicular to line $BC$ at points $B,C$. Notice that as $W$ varies, $\angle WMY=\angle WNX=90^{\circ}$ and points $X,Y$ vary on lines $\ell_b,\ell_c$ respectively. Therefore, since this implies that $N$ and $NW$ are related projectively, i.e., the pencil and range have same cross ratio and essentially this is the same cross ratio for $X$ as it varies on the line $\ell_b$ we see that points $X$ and $W$ are projectively related to each other. Similarly, points $W$ and $Y$ are projectively related to each other. Thus, points $X$ and $Y$ are projectively related, i.e., there exist a projectivity mapping $X$ to $Y$. We prove that it is in fact a perspectivity. Thus, we only need to consider three positions of $W$ for which $X,Y,H$ are collinear. These are clearly true for $W=B,C,D$ where $D$ is the feet of altitude from $A$. Hence it holds for all $W$.
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Ghost_rider
35 posts
#12 • 2 Y
Y by RedFlame2112, Adventure10
Let $\omega_1 \cap \omega_2$ = $G$. $A$ has the same power with respect $\omega_1$ and $\omega_2$ so $A,G$ and $W$ are collinear.
Then $\angle NGA$ = $\angle ABW$ = $\angle NHA$, so $H,N,A$ and $G$ are concyclic ($\angle HGA$ = $90^\circ$).
Finally $X,H,G$ and $Y$ are collinear.
This post has been edited 1 time. Last edited by Ghost_rider, Jul 6, 2016, 2:43 AM
Reason: type
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jayme
9801 posts
#13 • 6 Y
Y by AlastorMoody, Kanep, Siddharth03, RedFlame2112, Adventure10, Mango247
Dear Mathlinkers,
just for history, this result comes from Mannhein in 1893... actually a rediscovery...

Sincerely
Jean-Louis
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Vexation
49 posts
#14 • 2 Y
Y by RedFlame2112, Adventure10
$K$ is the foot of the altitudes from $A$.

We need to check that $\frac{XB \cdot KC+YC \cdot KB}{BC} = {HK}$ ... (*)

From $NXB \sim NWC$ and $MYC \sim MWB$, (*) becomes $KH \cdot KA = KB \cdot KC$, which is true when $H$ is reflected in $BC$ by power of $K$ wrt the circumcircle.
This post has been edited 2 times. Last edited by Vexation, Aug 10, 2016, 6:44 AM
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CountofMC
838 posts
#15 • 4 Y
Y by RedFlame2112, Adventure10, Mango247, Rounak_iitr
Solution
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G
H
=
a