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Help my diagram has too many points
MarkBcc168   27
N 19 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
19 minutes ago
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Geometry, SMO 2016, not easy
Zoom   18
N an hour 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
an hour ago
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A touching question on perpendicular lines
Tintarn   2
N an hour 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
an hour ago
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Woaah a lot of external tangents
egxa   2
N 2 hours ago by soryn
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.
2 replies
egxa
Apr 18, 2025
soryn
2 hours ago
J
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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
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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
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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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Apple sharing in Iran
mojyla222   3
N 3 hours ago by math-helli
Source: Iran 2025 second round p6
Ali is hosting a large party. Together with his $n-1$ friends, $n$ people are seated around a circular table in a fixed order. Ali places $n$ apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction).

Find all values of $n$ such that after some number of steps, the situation reaches a point where each person has exactly one apple.
3 replies
mojyla222
Apr 20, 2025
math-helli
3 hours ago
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A metrical relation in a cyclical quadrilateral.
Virgil Nicula   5
N Mar 10, 2016 by PROF65
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+MA\cdot MB\right)\left(PN^2+NC\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
5 replies
Virgil Nicula
Mar 9, 2016
PROF65
Mar 10, 2016
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A metrical relation in a cyclical quadrilateral.
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Virgil Nicula
7054 posts
Virgil Nicula
#1 h  Mar 9, 2016, 1:53 PM • 2 Y
Y by Adventure10, Mango247
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+MA\cdot MB\right)\left(PN^2+NC\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
This post has been edited 2 times. Last edited by Virgil Nicula, Mar 9, 2016, 1:58 PM
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uvwmethod
43 posts
uvwmethod
#2 h  Mar 9, 2016, 3:27 PM • 1 Y
Y by Adventure10
Hint:Easy by applying steward to $\triangle PAB$ and $\triangle PCD$ and using the fact $PA*PB=PC*PD$
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Virgil Nicula
7054 posts
Virgil Nicula
#3 h  Mar 10, 2016, 1:38 AM • 2 Y
Y by Adventure10, Mango247
Why is easily and $PA\cdot PB=PC\cdot PD$ ?!
This post has been edited 2 times. Last edited by Virgil Nicula, Mar 10, 2016, 1:39 AM
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suli
1498 posts
suli
#4 h  Mar 10, 2016, 3:52 AM • 2 Y
Y by Adventure10, Mango247
$PA * PB = PC * PD$ is false. $PA * PC = PB * PD$ is true by Power of a Point.

But instead of listening to the Steward, let's try a synthetic solution.

Let the circumcircles of $DPC$ and $APB$ intersect line $MN$ at $R$ and $S$ respectively. Then
$$\angle DRP = \angle DCP = \angle DBM,$$so $DRBM$ is cyclic. Thus $\angle DP \cdot PB = RP \cdot PM$. Similarly, $ASCN$ is cyclic, so $AP \cdot PC = SP \cdot PN$.

Thus
$$AP \cdot BP \cdot CP \cdot DP = NP \cdot RP \cdot MP \cdot SP = (PN^2 + PN \cdot NR)(PM^2 + PM \cdot MS) = (PN^2 + ND \cdot NC)(PM^2 + MA \cdot MB)$$by use of Power of a Point.
This post has been edited 1 time. Last edited by suli, Mar 10, 2016, 4:29 AM
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Virgil Nicula
7054 posts
Virgil Nicula
#5 h  Mar 10, 2016, 2:02 PM • 2 Y
Y by Adventure10, Mango247
@ Suli. Thank you very much for your very nice proof! See PP11 and its easy extension from here
This post has been edited 1 time. Last edited by Virgil Nicula, Mar 10, 2016, 2:03 PM
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PROF65
2016 posts
PROF65
#6 h  Mar 10, 2016, 6:16 PM • 2 Y
Y by Adventure10, Mango247
Virgil Nicula wrote:
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+\underline{AM}\cdot MB\right)\left(PN^2+\underline{CN}\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
the distances are signed .
@ Suli. your approch is good apart from ,i think just typo , the last equality which should be $(PN^2+DN\cdot NC)(PM^2+AM\cdot MB)$
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