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inequality
SunnyEvan   3
N a few seconds ago by SunnyEvan
Let $ a,b > 0 ,$ such that : $ a+b \geq \frac{3(a^4+b^4)}{a^2+b^2+1}\sqrt{\frac{\frac{1}{a}+\frac{1}{b}}{a+b}}.$
Prove that: $$ \frac{3(a^2+b^2)+4}{a^4+b^4} \geq 3(a+b)-1 $$
3 replies
SunnyEvan
Today at 6:53 AM
SunnyEvan
a few seconds ago
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tangent circles
m4thbl3nd3r   1
N an hour ago by Lil_flip38
Let $O,H,T$ be circumcenter, orthocenter and A-HM point of triangle $ABC$. Let $AH,AT$ intersect $(O)$ at $K,N$, respectively. Let $XYZ$ be the triangle formed by $TH,BC,KN$. Prove that $(XYZ)$ is tangent to $(O).$
1 reply
m4thbl3nd3r
2 hours ago
Lil_flip38
an hour ago
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inequality
SunnyEvan   8
N an hour ago by sqing
Let $ x,y \geq 0 ,$ such that : $ \frac{x^2}{x^3+y}+\frac{y^2}{x+y^3} \geq 1 .$
Prove that : $$ x^2+y^2-xy \leq x+y $$$$ (x+\frac{1}{2})^2+(x+\frac{1}{2})^2 \leq \frac{5}{2} $$$$ (x+1)^2+(y+1)^2 \leq 5 $$$$ (x+2)^2+(y+2)^2 \leq 13 $$
8 replies
SunnyEvan
Yesterday at 1:51 PM
sqing
an hour ago
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Why did all the old Japanese math Olympiad questions disappear?
parkjungmin   0
an hour ago
Why did all the old Japanese math Olympiad questions disappear?

Did the administrator delete it?
0 replies
parkjungmin
an hour ago
0 replies
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reals associated with 1024 points
bin_sherlo   2
N 2 hours ago by Tamam
Source: Türkiye 2025 JBMO TST P8
Pairwise distinct points $P_1,\dots,P_{1024}$, which lie on a circle, are marked by distinct reals $a_1,\dots,a_{1024}$. Let $P_i$ be $Q-$good for a $Q$ on the circle different than $P_1,\dots,P_{1024}$, if and only if $a_i$ is the greatest number on at least one of the two arcs $P_iQ$. Let the score of $Q$ be the number of $Q-$good points on the circle. Determine the greatest $k$ such that regardless of the values of $a_1,\dots,a_{1024}$, there exists a point $Q$ with score at least $k$.
2 replies
bin_sherlo
May 11, 2025
Tamam
2 hours ago
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Complex number
soruz   2
N 2 hours ago by alexheinis
$i)$ Determine $z \in \mathbb{C} $ such that $2|z| \ge |z^n-3|, \forall n \in \mathbb N^*.$
$ii)$ Determine $z \in \mathbb{C} $ such that $2|z| \ge |z^n+3|, \forall n \in \mathbb N^*.$
2 replies
soruz
Nov 28, 2024
alexheinis
2 hours ago
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Brilliant Problem
M11100111001Y1R   9
N 3 hours ago by The5
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[ \frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2 \]
9 replies
M11100111001Y1R
May 27, 2025
The5
3 hours ago
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Draw sqrt(2024)
shanelin-sigma   1
N 3 hours ago by CrazyInMath
Source: 2024/12/24 TCFMSG Mock p10
On a big plane, two points with length $1$ are given. Prove that one can only use straightedge (which draws a straight line passing two drawn points) and compass (which draws a circle with a chosen radius equal to the distance of two drawn points and centered at a drawn points) to construct a line and two points on it with length $\sqrt{2024}$ in only $10$ steps (Namely, the total number of circles and straight lines drawn is at most $10$.)
1 reply
shanelin-sigma
Dec 24, 2024
CrazyInMath
3 hours ago
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A beautiful Lemoine point problem
phonghatemath   3
N 3 hours ago by orengo42
Source: my teacher
Given triangle $ABC$ inscribed in a circle with center $O$. $P$ is any point not on (O). $AP, BP, CP$ intersect $(O)$ at $A', B', C'$. Let $L, L'$ be the Lemoine points of triangle $ABC, A'B'C'$ respectively. Prove that $P, L, L'$ are collinear.
3 replies
phonghatemath
Today at 5:01 AM
orengo42
3 hours ago
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Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   4
N 3 hours ago by Mathgloggers
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
4 replies
OgnjenTesic
May 22, 2025
Mathgloggers
3 hours ago
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Reflected point lies on radical axis
Mahdi_Mashayekhi   7
N Jun 1, 2025 by amogususususus
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
7 replies
Mahdi_Mashayekhi
Apr 19, 2025
amogususususus
Jun 1, 2025
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Reflected point lies on radical axis
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Source: Iran 2025 second round P4
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Mahdi_Mashayekhi
699 posts
Mahdi_Mashayekhi
#1 h  Apr 19, 2025, 11:01 AM • 2 Y
Y by GA34-261, Rounak_iitr
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Apr 19, 2025, 11:25 AM
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ItzsleepyXD
151 posts
ItzsleepyXD
#2 h  Apr 19, 2025, 11:26 AM • 1 Y
Y by GA34-261
Angle chasing :
$\angle BPX = \angle BPA + \angle APX = 180^{\circ} - \angle BAP - \angle ABO + 90^{\circ} + \angle ACB = \angle BAC + 90^{\circ} = 180^{\circ} - \angle O'BC $
and from $OX//BC$
Let $O'B \cap OX=E , O'B \cap OX = F$
so $\angle BEX + \angle BPX = 180^{\circ}$
thus $E,B,P,X$ concyclic .
in the same way $F,C,P,X$ concyclic.
so $O'$ is on the radical axis of $(BXP),(CXQ)$
implies that $X,Y,O'$ collinear . done $\square$
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gghx
1072 posts
gghx
#3 h  Apr 19, 2025, 11:35 AM • 2 Y
Y by blackidea, GA34-261
Define $Y'$ as the second intersection of $O'X$ and $(BO'C)$. We claim that $Y=Y'$. Note that $$\angle BY'X=\angle BCO'=\angle BCO=90^\circ-\angle PQO=\angle XPO,$$hence $BPYX'$ is concyclic. Similarly, $CXQY$ is concyclic, and we are done.
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Mahdi_Mashayekhi
699 posts
Mahdi_Mashayekhi
#4 h  Apr 19, 2025, 4:34 PM • 4 Y
Y by Parsia--, sami1618, GA34-261, Rounak_iitr
Note that $\angle XYC = \angle XQC =\angle XQO = 90 - \frac{\angle QXO}{2} = 90 - \angle QPO = 90 - (90 - \angle C + 90 - \angle B) = 90-\angle A$ and $\angle XYB = \angle XPO = 90 - \frac{\angle PXO}{2} = 90 - \angle PQO = 90 - (90 - \angle C + 90 - \angle B) = 90 - \angle A$ so $X$ lies on angle bisector of $BYC$.
Now note that $\angle BYC = 2(90 - \angle A) = 180 - 2\angle A = 180 - \angle BOC = 180 - \angle BO'C$ so $BYCO'$ is cyclic and since $BO'=CO'$ then $O'$ also lies on angle bisector of $BYC$ so $Y,X,O'$ are collinear as wanted.
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SimplisticFormulas
133 posts
SimplisticFormulas
#5 h  May 30, 2025, 12:34 PM • 1 Y
Y by GA34-261
ez
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Mahdi_Mashayekhi
699 posts
Mahdi_Mashayekhi
#6 h  May 31, 2025, 5:28 AM • 1 Y
Y by GA34-261
SimplisticFormulas wrote:
ez

Wrong solution. $O'B$ and $O'C$ aren't necessarily tangent.
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khanhnx
1619 posts
khanhnx
#7 h  May 31, 2025, 3:27 PM • 1 Y
Y by GA34-261
Let $U, V$ be second intersections of $O'B, O'C$ with $(BXP)$ and $(CXQ)$. We have
$$\angle{OXP} = 2\angle{OQP} = 2(90^{\circ} - \angle{OCB}) = 180^{\circ} - \angle{OBO'} = \angle{PBU}$$Then $U \in OX$. Similarly, we have $V \in OX$. But
$$\angle{XOB} = 90^{\circ} - \angle{OQP} = \angle{OCB} = \angle{OBC}$$so $OX \parallel BC$. Hence it's easy to see that $BCVU$ is isosceles trapezoid. From this, we have
$$\mathcal{P}_{O' / (BXP)} = \overline{O'B} \cdot \overline{O'U} = \overline{O'C} \cdot \overline{O'V} = \mathcal{P}_{O' / (CXQ)}$$This means $O'$ lies on radical axis of $(BXP)$ and $(CXQ)$ or $O' \in XY$
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amogususususus
371 posts
amogususususus
#8 h  Jun 1, 2025, 9:58 AM
Y by
When you don't use ur brain to solve geo belike :

Let $E$ be the second intersection of $BC$ with $(XBP)$, define $F$ similarly, let $M$ be the midpoint of segment $BC$. WLOG assume $AB<AC$. We have,
$$\angle PQO=90^{\circ}-\angle OCB=\angle A$$. Similarly $\angle QPO=\angle A$, hence $OP=OQ$. From here we can also get $\triangle PXO \sim \triangle BOC$, which implies
$$ 2\cdot BC \cdot XP-2\cdot OP\cdot OB=0 \ \ (1)$$.Next, notice that
$$\angle EBP = 180^{\circ}-\angle OBC=90^{\circ}+\angle A$$. And,
$$\angle BPX = 180^{\circ}-\angle XPO=90^{\circ}+\angle A$$. Hence $EBPX$ is an isosceles trapezoid or $EB=XP$. Similarly we get $FC=XQ=XP$.

Now define $f(L)=pow(L,(XPB))-pow(L,(XQC))$, notice that $O$ lies inside $(XQC)$ and outside $(XPB)$.
Hence,
$$f(O)=(OP\cdot OB)-(-OP\cdot OC)=2\cdot OP\cdot OB$$. Next, by Linearity of PoP
$$f(M)=\frac{f(B)+f(C)}{2}=\frac{(0-BF\cdot BC)+(CE\cdot BC-0)}{2}=\frac{BC(CE-BF)}{2}=BC\cdot XP $$. Notice that $M$ is the midpoint of segment $OO'$.By Linearity of PoP
$$f(O')=2f(M)-f(O)=2\cdot BC\cdot XP-2\cdot OP\cdot OB$$. But we know from (1), this value is zero. Hence $O'$ lies on the radax of $(XBP)$ and $(XCQ)$ which is $XY$.
This post has been edited 2 times. Last edited by amogususususus, Jun 1, 2025, 10:27 AM
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