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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Cyclic Points
IstekOlympiadTeam   38
N 6 minutes ago by eg4334
Source: EGMO 2017 Day1 P1
Let $ABCD$ be a convex quadrilateral with $\angle DAB=\angle BCD=90^{\circ}$ and $\angle ABC> \angle CDA$. Let $Q$ and $R$ be points on segments $BC$ and $CD$, respectively, such that line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$, respectively. It is given that $PQ=RS$.Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$.Prove that the points $M,N,A$ and $C$ lie on a circle.
38 replies
IstekOlympiadTeam
Apr 8, 2017
eg4334
6 minutes ago
2025 Caucasus MO Seniors P3
BR1F1SZ   1
N 19 minutes ago by iliya8788
Source: Caucasus MO
A circle is drawn on the board, and $2n$ points are marked on it, dividing it into $2n$ equal arcs. Petya and Vasya are playing the following game. Petya chooses a positive integer $d \leqslant n$ and announces this number to Vasya. To win the game, Vasya needs to color all marked points using $n$ colors, such that each color is assigned to exactly two points, and for each pair of same-colored points, one of the arcs between them contains exactly $(d - 1)$ marked points. Find all $n$ for which Petya will be able to prevent Vasya from winning.
1 reply
BR1F1SZ
Mar 26, 2025
iliya8788
19 minutes ago
RGB chessboard
BR1F1SZ   0
33 minutes ago
Source: 2025 Argentina TST P3
A $100 \times 100$ board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
[list=i]
[*]Each coloured cell has exactly three friends.
[*]All three friends of any given coloured cell lie in the same row or in the same column.
[/list]
Determine the maximum number of cells that can be coloured on the board.
0 replies
1 viewing
BR1F1SZ
33 minutes ago
0 replies
[ELMO2] The Multiplication Table
v_Enhance   26
N an hour ago by de-Kirschbaum
Source: ELMO 2015, Problem 2 (Shortlist N1)
Let $m$, $n$, and $x$ be positive integers. Prove that \[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]
Proposed by Yang Liu
26 replies
v_Enhance
Jun 27, 2015
de-Kirschbaum
an hour ago
No more topics!
Itamo 2015, problem 3
Popescu   5
N Dec 23, 2022 by parmenides51
Source: Itamo 2015
Let ABC a triangle, let K be the foot of the bisector relative to BC and J be the foot of the trisectrix relative to BC closer to the side AC (3* m(JAC)=m(CAB) ). Let C' and B' be two point on the line AJ on the side of J with respect to A, such that AC'=AC and AB=AB'. Prove that ABB'C is cyclic if and only if lines C'K and BB' are parallel.
5 replies
Popescu
May 30, 2015
parmenides51
Dec 23, 2022
Itamo 2015, problem 3
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Source: Itamo 2015
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Popescu
109 posts
#1 • 3 Y
Y by anantmudgal09, Adventure10, Mango247
Let ABC a triangle, let K be the foot of the bisector relative to BC and J be the foot of the trisectrix relative to BC closer to the side AC (3* m(JAC)=m(CAB) ). Let C' and B' be two point on the line AJ on the side of J with respect to A, such that AC'=AC and AB=AB'. Prove that ABB'C is cyclic if and only if lines C'K and BB' are parallel.
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Popescu
109 posts
#2 • 1 Y
Y by Adventure10
Can anyone solve this?
This post has been edited 1 time. Last edited by Popescu, Jun 16, 2015, 8:25 PM
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LukeMac
28 posts
#3 • 2 Y
Y by Adventure10, Mango247
That's my solution during the contest. It is different from the official one.
In order to have $C'K \parallel BB'$ o $ABB'C$ cyclic we need that $B'$ and $C'$ are both beyond $BC$ (with respect to $ABC$).
Now $C'K \parallel BB' \iff JK \cdot JB' = JB \cdot JC'$ and $ABB'C$ is cyclic $\iff AJ \cdot JB' = BJ \cdot CJ$.
So we need to prove that $ JK \cdot JB' = JB \cdot JC' \iff AJ \cdot JB' = BJ \cdot CJ $.
Now $JK=CK-CJ= \frac{AC \cdot BC}{AC+AB} - BC+JB = JB- \frac{AB \cdot BC}{AC+AB}$ and $ JB' = AB'-AJ = AB-AJ$ and $JC'=AC-AJ$.
So we need $$JB(AB^2-AC^2)=AB \cdot BC \cdot (AB-AJ) \iff AB \cdot AJ = AJ^2 + BJ \cdot CJ $$
or $$ BJ \cdot AC^2 + CJ \cdot AB^2 = AB \cdot BC \cdot AJ \iff AB \cdot BC \cdot AJ = AJ^2 \cdot BC + BJ \cdot CJ \cdot BC$$
But according to Stewart theorem we have $AB^2 \cdot CJ + AC^2 \cdot BJ = AJ^2 \cdot BC + BJ \cdot CJ \cdot BC$.
And so we finally have that $C'K \parallel BB' \iff ABB'C$ is cyclic.
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PROF65
2016 posts
#4 • 1 Y
Y by Adventure10
Let $C'' \in AB $ such that $AC''=AC;L =(ABC)\cap AK$ .we know that $AK.AL=AB.AC \implies KLB'C'$ is cyclic and $KLBC''$ is cyclic . If $B'\in ( ABC)$ then applying Reim to $ (KLB'C'),(ALB') [=ABC]$ yields $ KC'\parallel t $ which is the tangent to $(ABC)$ at $A$. let $A' $ point of $t$ in the same semi-plane bounded by $AK$ as $C$ .$\widehat{AB'B}=\widehat{ABB'}=\widehat{A'AB'}=\widehat{KC'B'} $ thus $KC'\parallel BB'$
conversly if $ KC'\parallel BB'$ then $KC'\parallel C'C'' \implies K,C',C'' $ are colinear . $KLBC''$ cyclic $ \implies \widehat{ABL}=\widehat{AKC''} =\widehat{AKC'}$ from $KLB'C' $ cyclic we deduce $ \widehat{ABL}= \widehat{C'B'L}= \widehat{AB'L}$ thus$B'\in (ABL)=(ABC)$
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PROF65
2016 posts
#5 • 3 Y
Y by v4913, Adventure10, Mango247
remark trisectrix is superfluous just cevian suffices
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parmenides51
30630 posts
#6
Y by
Let $ABC$ a triangle, let $K$ be the foot of the bisector relative to $BC$ and $J$ be the foot of the trisectrix relative to $BC$ closer to the side $AC$ ($3\angle JAC)=\angle CAB$ ). Let $C'$ and $B'$ be two point on the line $AJ$ on the side of $J$ with respect to $A$, such that $AC'=AC$ and $AB=AB'$. Prove that $ABB'C$ is cyclic if and only if lines $C'K$ and $BB'$ are parallel.
This post has been edited 2 times. Last edited by parmenides51, Jan 7, 2023, 11:59 PM
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