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AM=CN in Russia
mathuz   25
N 8 minutes ago by Ilikeminecraft
Source: AllRussian-2014, Grade 11, day1, P4
Given a triangle $ABC$ with $AB>BC$, $ \Omega $ is circumcircle. Let $M$, $N$ are lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. $K(.)=MN\cap AC$ and $P$ is incenter of the triangle $AMK$, $Q$ is K-excenter of the triangle $CNK$ (opposite to $K$ and tangents to $CN$). If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.

M. Kungodjin
25 replies
mathuz
Apr 29, 2014
Ilikeminecraft
8 minutes ago
Simson lines on OH circle
DVDTSB   2
N 17 minutes ago by SomeonesPenguin
Source: Romania TST 2025 Day 2 P4
Let \( ABC \) and \( DEF \) be two triangles inscribed in the same circle, centered at \( O \), and sharing the same orthocenter \( H \ne O \). The Simson lines of the points \( D, E, F \) with respect to triangle \( ABC \) form a non-degenerate triangle \( \Delta \).
Prove that the orthocenter of \( \Delta \) lies on the circle with diameter \( OH \).

Note. Assume that the points \( A, F, B, D, C, E \) lie in this order on the circle and form a convex, non-degenerate hexagon.

Proposed by Andrei Chiriță
2 replies
+2 w
DVDTSB
3 hours ago
SomeonesPenguin
17 minutes ago
Tangents inducing isogonals
nikolapavlovic   55
N 24 minutes ago by SimplisticFormulas
Source: Serbian MO 2017 6
Let $k$ be the circumcircle of $\triangle ABC$ and let $k_a$ be A-excircle .Let the two common tangents of $k,k_a$ cut $BC$ in $P,Q$.Prove that $\measuredangle PAB=\measuredangle CAQ$.
55 replies
nikolapavlovic
Apr 2, 2017
SimplisticFormulas
24 minutes ago
Find all integers satisfying this equation
Sadigly   2
N 31 minutes ago by ilovenumbertheories
Source: Azerbaijan NMO 2019
Find all $x;y\in\mathbb{Z}$ satisfying the following condition: $$x^3=y^4+9x^2$$
2 replies
Sadigly
Sunday at 8:30 PM
ilovenumbertheories
31 minutes ago
Locus of the circumcenter of triangle PST
v_Enhance   14
N 41 minutes ago by Ilikeminecraft
Source: USA TSTST 2013, Problem 4
Circle $\omega$, centered at $X$, is internally tangent to circle $\Omega$, centered at $Y$, at $T$. Let $P$ and $S$ be variable points on $\Omega$ and $\omega$, respectively, such that line $PS$ is tangent to $\omega$ (at $S$). Determine the locus of $O$ -- the circumcenter of triangle $PST$.
14 replies
v_Enhance
Aug 13, 2013
Ilikeminecraft
41 minutes ago
Problem 6 (Second Day)
darij grinberg   43
N an hour ago by cj13609517288
Source: IMO 2004 Athens
We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity.

Find all positive integers $n$ such that $n$ has a multiple which is alternating.
43 replies
darij grinberg
Jul 13, 2004
cj13609517288
an hour ago
Dou Fang Geometry in Taiwan TST
Li4   8
N an hour ago by X.Allaberdiyev
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
8 replies
Li4
Apr 26, 2025
X.Allaberdiyev
an hour ago
geometry
EeEeRUT   4
N an hour ago by Tkn
Source: Thailand MO 2025 P4
Let $D,E$ and $F$ be touch points of the incenter of $\triangle ABC$ at $BC, CA$ and $AB$, respectively. Let $P,Q$ and $R$ be the circumcenter of triangles $AFE, BDF$ and $CED$, respectively. Show that $DP, EQ$ and $FR$ concurrent.
4 replies
EeEeRUT
Today at 6:44 AM
Tkn
an hour ago
min A=x+1/x+y+1/y if 2(x+y)=1+xy for x,y>0 , 2020 ISL A3 for juniors
parmenides51   14
N an hour ago by GayypowwAyly
Source: 2021 Greece JMO p1 (serves also as JBMO TST) / based on 2020 IMO ISL A3
If positive reals $x,y$ are such that $2(x+y)=1+xy$, find the minimum value of expression $$A=x+\frac{1}{x}+y+\frac{1}{y}$$
14 replies
parmenides51
Jul 21, 2021
GayypowwAyly
an hour ago
Unbounded Sequences
DVDTSB   3
N an hour ago by Ciobi_
Source: Romania TST 2025 Day 2 P2
Let \( a_1, a_2, \ldots, a_n, \ldots \) be a sequence of strictly positive real numbers. For each nonzero positive integer \( n \), define
\[
s_n = a_1 + a_2 + \cdots + a_n \quad \text{and} \quad
\sigma_n = \frac{a_1}{1 + a_1} + \frac{a_2}{1 + a_2} + \cdots + \frac{a_n}{1 + a_n}.
\]Show that if the sequence \( s_1, s_2, \ldots, s_n, \ldots \) is unbounded, then the sequence \( \sigma_1, \sigma_2, \ldots, \sigma_n, \ldots \) is also unbounded.

Proposed by The Problem Selection Committee
3 replies
DVDTSB
3 hours ago
Ciobi_
an hour ago
Long and wacky inequality
Royal_mhyasd   1
N an hour ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
1 reply
Royal_mhyasd
Yesterday at 7:01 PM
Royal_mhyasd
an hour ago
Number Theory
adorefunctionalequation   3
N an hour ago by MITDragon
Find all integers k such that k(k+15) is perfect square
3 replies
adorefunctionalequation
Jan 9, 2023
MITDragon
an hour ago
PQ bisects AC if <BCD=90^o, A, B,C,D concyclic
parmenides51   2
N Apr 22, 2025 by venhancefan777
Source: Mathematics Regional Olympiad of Mexico Northeast 2020 P2
Let $A$, $B$, $C$ and $D$ be points on the same circumference with $\angle BCD=90^\circ$. Let $P$ and $Q$ be the projections of $A$ onto $BD$ and $CD$, respectively. Prove that $PQ$ cuts the segment $AC$ into equal parts.
2 replies
parmenides51
Sep 7, 2022
venhancefan777
Apr 22, 2025
PQ bisects AC if <BCD=90^o, A, B,C,D concyclic
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G H BBookmark kLocked kLocked NReply
Source: Mathematics Regional Olympiad of Mexico Northeast 2020 P2
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parmenides51
30652 posts
#1 • 1 Y
Y by Rounak_iitr
Let $A$, $B$, $C$ and $D$ be points on the same circumference with $\angle BCD=90^\circ$. Let $P$ and $Q$ be the projections of $A$ onto $BD$ and $CD$, respectively. Prove that $PQ$ cuts the segment $AC$ into equal parts.
This post has been edited 1 time. Last edited by parmenides51, Sep 7, 2022, 12:40 AM
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Tafi_ak
309 posts
#2
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Since $C$ is the orthocenter of $\triangle BCD$ therefore $PQ$ bisects $AC$ by the Simson line theorem.
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venhancefan777
2 posts
#4
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APQD is clearly cyclic, then PQD=PAD, also since BD is a diameter, PAD=ABD. Now ACQ=ACD and ACD=ABD since they open the same arc. Now we have that PQD=GQC and ACQ=GCQ but we knew PQD=ABD and ACQ=ABD so GQC=GCQ, CGQ is isosceles and then clearly AG=GC QED.
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