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Woaah a lot of external tangents
egxa   3
N 4 minutes ago by NO_SQUARES
Source: All Russian 2025 11.7
A quadrilateral \( ABCD \) with no parallel sides is inscribed in a circle \( \Omega \). Circles \( \omega_a, \omega_b, \omega_c, \omega_d \) are inscribed in triangles \( DAB, ABC, BCD, CDA \), respectively. Common external tangents are drawn between \( \omega_a \) and \( \omega_b \), \( \omega_b \) and \( \omega_c \), \( \omega_c \) and \( \omega_d \), and \( \omega_d \) and \( \omega_a \), not containing any sides of quadrilateral \( ABCD \). A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle \( \Gamma \). Prove that the lines joining the centers of \( \omega_a \) and \( \omega_c \), \( \omega_b \) and \( \omega_d \), and the centers of \( \Omega \) and \( \Gamma \) all intersect at one point.
3 replies
egxa
Apr 18, 2025
NO_SQUARES
4 minutes ago
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2023 Hong Kong TST 3 (CHKMO) Problem 4
PikaNiko   3
N 14 minutes ago by lightsynth123
Source: 2023 Hong Kong TST 3 (CHKMO)
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
3 replies
PikaNiko
Dec 3, 2022
lightsynth123
14 minutes ago
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Help my diagram has too many points
MarkBcc168   27
N 34 minutes ago by Om245
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
27 replies
MarkBcc168
Jul 17, 2024
Om245
34 minutes ago
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Geometry, SMO 2016, not easy
Zoom   18
N 2 hours ago by SimplisticFormulas
Source: Serbia National Olympiad 2016, day 1, P3
Let $ABC$ be a triangle and $O$ its circumcentre. A line tangent to the circumcircle of the triangle $BOC$ intersects sides $AB$ at $D$ and $AC$ at $E$. Let $A'$ be the image of $A$ under $DE$. Prove that the circumcircle of the triangle $A'DE$ is tangent to the circumcircle of triangle $ABC$.
18 replies
Zoom
Apr 1, 2016
SimplisticFormulas
2 hours ago
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A touching question on perpendicular lines
Tintarn   2
N 2 hours ago by pi_quadrat_sechstel
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
2 replies
Tintarn
Mar 17, 2025
pi_quadrat_sechstel
2 hours ago
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Some nice summations
amitwa.exe   31
N 2 hours ago by soryn
Problem 1: $\Omega=\left(\sum_{0\le i\le j\le k}^{\infty} \frac{1}{3^i\cdot4^j\cdot5^k}\right)\left(\mathop{{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}}}_{i\neq j\neq k}\frac{1}{3^i\cdot3^j\cdot3^k}\right)=?$
31 replies
amitwa.exe
May 24, 2024
soryn
2 hours ago
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Interesting inequalities
sqing   0
2 hours ago
Source: Own
Let $ a,b,c\geq 0 ,b+c-ca=1 $ and $ c+a-ab=3.$ Prove that
$$a+\frac{19}{10}b-bc\leq 2-\sqrt 2$$$$a+\frac{17}{10}b+c-bc\leq 3$$$$ a^2+\frac{9}{5}b-bc\leq 6-4\sqrt 2$$$$ a^2+\frac{8}{5}b^2-bc\leq 6-4\sqrt 2$$$$a+1.974873b-bc\leq 2-\sqrt 2$$$$a+1.775917b+c-bc\leq 3$$

0 replies
sqing
2 hours ago
0 replies
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Two permutations
Nima Ahmadi Pour   12
N 3 hours ago by Zhaom
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i \]

Proposed by Ricky Liu & Zuming Feng, USA
12 replies
Nima Ahmadi Pour
Apr 24, 2006
Zhaom
3 hours ago
J
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Easy Number Theory
math_comb01   37
N 3 hours ago by John_Mgr
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
37 replies
math_comb01
Jan 21, 2024
John_Mgr
3 hours ago
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ALGEBRA INEQUALITY
Tony_stark0094   3
N 3 hours ago by sqing
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
3 replies
Tony_stark0094
Today at 12:17 AM
sqing
3 hours ago
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Inspired by hlminh
sqing   3
N 3 hours ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
3 replies
sqing
Yesterday at 4:43 AM
sqing
3 hours ago
J
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A Familiar Point
v4913   51
N 3 hours ago by xeroxia
Source: EGMO 2023/6
Let $ABC$ be a triangle with circumcircle $\Omega$. Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$). Let $I$ be the incenter of $ABC$. Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
51 replies
v4913
Apr 16, 2023
xeroxia
3 hours ago
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Find an angle
socrates   3
N Apr 6, 2025 by Nari_Tom
Source: Baltic Way 2016, Problem 18
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
3 replies
socrates
Nov 5, 2016
Nari_Tom
Apr 6, 2025
J
Find an angle
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Baltic Way
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G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2016, Problem 18
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socrates
2105 posts
socrates
#1 h  Nov 5, 2016, 10:48 PM • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
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george_54
1585 posts
george_54
#4 h  Nov 16, 2016, 9:46 PM • 3 Y
Y by StarLex1, Adventure10, Mango247
socrates wrote:
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$

Let $AB, LK$ intersect at $E$, $EB=a,AB=2a$ and $EK=KL=b, BD=2b$. I draw $BH$ perpendicular to $AB$.

It is: $EB \cdot EA = EK \cdot EL \Leftrightarrow 3{a^2} = 2{b^2} \Leftrightarrow b = \frac{{a\sqrt 6 }}{2}$

We also have, $BH = a\sqrt 3 \Leftrightarrow BD = 2b = a\sqrt 3 \sqrt 2 = BH\sqrt 2 $, thus $\omega=45^0$ and finally $\boxed{\angle ABD=75^0}$
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Reason: typo
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K.N
532 posts
K.N
#5 h  Nov 17, 2016, 4:45 AM • 1 Y
Y by Adventure10
Yes that's exactly my solution!
Easy problem
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Nari_Tom
116 posts
Nari_Tom
#6 h  Apr 6, 2025, 2:43 PM
Y by
Let $F$ be the intersection of $AL \cap BD$. Since $\triangle BAF \sim \triangle FLD$, we have $BF=2*FD$. By some angle chase $\angle BAF=\angle FDA$. So by the secant tangent theorem: $BF*BD=BA^2$ $\implies$ $\frac{2*BD^2}{3}=BA^2$, and rest is just some bash.
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