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Nine point circle + Perpendicularities
YaoAOPS   18
N 36 minutes ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
36 minutes ago
Inequality conjecture
RainbowNeos   0
an hour ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
an hour ago
0 replies
inequality 2905
pennypc123456789   0
an hour ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
1 viewing
pennypc123456789
an hour ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N an hour ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
1 viewing
sqing
Today at 3:43 AM
sqing
an hour ago
Inspired by qrxz17
sqing   7
N an hour ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
1 viewing
sqing
5 hours ago
sqing
an hour ago
Geometry problem
Whatisthepurposeoflife   2
N an hour ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
an hour ago
A Sequence of +1's and -1's
ike.chen   36
N an hour ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
an hour ago
Basic ideas in junior diophantine equations
Maths_VC   1
N 2 hours ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
2 hours ago
Inspired by qrxz17
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
sqing
5 hours ago
sqing
2 hours ago
Hardest in ARO 2008
discredit   30
N 2 hours ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
2 hours ago
Find the value
sqing   12
N 2 hours ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
2 hours ago
Trillium geometry
Assassino9931   4
N Apr 28, 2025 by Rayvhs
Source: Bulgaria EGMO TST 2018 Day 2 Problem 1
The angle bisectors at $A$ and $C$ in a non-isosceles triangle $ABC$ with incenter $I$ intersect its circumcircle $k$ at $A_0$ and $C_0$, respectively. The line through $I$, parallel to $AC$, intersects $A_0C_0$ at $P$. Prove that $PB$ is tangent to $k$.
4 replies
Assassino9931
Feb 3, 2023
Rayvhs
Apr 28, 2025
Trillium geometry
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G H BBookmark kLocked kLocked NReply
Source: Bulgaria EGMO TST 2018 Day 2 Problem 1
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Assassino9931
1375 posts
#1
Y by
The angle bisectors at $A$ and $C$ in a non-isosceles triangle $ABC$ with incenter $I$ intersect its circumcircle $k$ at $A_0$ and $C_0$, respectively. The line through $I$, parallel to $AC$, intersects $A_0C_0$ at $P$. Prove that $PB$ is tangent to $k$.
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PROF65
2016 posts
#2 • 1 Y
Y by GuvercinciHoca
Lemma : $ABC$ is a triangle with incenter $I$ ; $A'$ the midpoint of arc $BC$ not containing $A$.Let the bissectors of $\angle AA'B,\angle
AA'C$ cut respectively $BA,CA$ at $U,V$ then $I,U$ and $V$ are collinear and $UV\parallel BC$
Proof :
Let the parallel to $BC$ through $I$ cut $AB,AC$ at $ U',V'$ ; it ' s clear that $IU'B,IV'C$ are isoceles besides $A'I=A'B=A'C$ we get $A'IU=A'BU$ thus $A'U' $ is bisector of $ \angle IA'B$ hence $U'=U$ idem $V'=V$
Back to the problem :
Let $B'$ the midpoint of arc $BC$ not containing $B$; $Q=BB\cap A_0C_0$ we need to show that $Q=P$
Lemma ascertains that if $B'A_0\cap BC=V,B'C_0\cap BA=U$ then $I\in UV\parallel AC$.
applying pascal to $BBCC_0A_0B'$we get $Q-V-I$ are collinear which closes the proof .
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Assassino9931
1375 posts
#3
Y by
No need to use harsh theory.

Angle chasing gives $C_0I = C_0B$ and $A_0I = A_0B$ - so $PC_0A_0$ is the perpendicular bisector of $BI$, whence $\triangle PC_0B \cong \triangle PC_0I$. Hence $\angle PBC_0 = \angle PIC_0 = \angle ACI$ (the latter is from the line parallel to $AC$) and $\angle ABP = \angle PBC_0 + \angle ABC_0 = 2\angle ACI = \angle ACB$, which solves the problem.
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Davud29_09
20 posts
#4 • 1 Y
Y by MuradSafarli
AC₀BA₀C is cyclic.Firstly we note that <ACC₀=<BCC₀=α and <BAA₀=<CAA₀=β.From cyclic <C₀BA=<C₀AB=<C₀A₀A=<C₀A₀B=α similarly <A₀BC=<A₀C₀C=<BC₀A₀=β.PI parallel to AC and we get from this C₀IP=α.We get <A₀PI=β-α and <CBI=90-α-β from angle chasing.Then we get A₀C₀ perpendicular to BI and bisects BI.Then we get PB=PI.So we get <BPA₀=β-α and <PBC₀=α. From <PBC₀=<BA₀C₀ we get PB touch k.We are done.
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Rayvhs
27 posts
#5
Y by
A very simple way to solve it.
Let $\angle ACB =\gamma$.
We have that $\angle C_0BA = \angle C_0AB = \angle C_0A_0A = \frac{\gamma}{2}$.
Since $PI \parallel AC$, $\angle C_0IP = \frac{\gamma}{2}$.
That means $PI$ is tangent to $\odot(C_0A_0I)$, hence
\[PI^2 = PC_0 \times PA_0.\]But we notice that $I$ is the symmetric point of $B$ wrt $C_0A_0$.
Thus, $PC_0A_0$ is the perpendicular bisector of $BI$, so $PB = PI$.
Then
\[PB^2 = PC_0 \times PA_0,\]which implies that $\angle PBC_0 = \angle BA_0C_0 = \frac{\gamma}{2}$.
Therefore, $\angle PBA = \gamma$.
This post has been edited 1 time. Last edited by Rayvhs, Apr 28, 2025, 8:45 PM
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