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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Bounding is hard
whatshisbucket   20
N 22 minutes ago by torch
Source: ELMO 2018 #5, 2018 ELMO SL A2
Let $a_1,a_2,\dots,a_m$ be a finite sequence of positive integers. Prove that there exist nonnegative integers $b,c,$ and $N$ such that $$\left\lfloor \sum_{i=1}^m \sqrt{n+a_i} \right\rfloor =\left\lfloor \sqrt{bn+c} \right\rfloor$$holds for all integers $n>N.$

Proposed by Carl Schildkraut
20 replies
1 viewing
whatshisbucket
Jun 28, 2018
torch
22 minutes ago
Yet another domino problem
juckter   14
N 27 minutes ago by math-olympiad-clown
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
14 replies
juckter
Apr 9, 2019
math-olympiad-clown
27 minutes ago
BMO 2024 SL A3
MuradSafarli   6
N 29 minutes ago by quacksaysduck

A3.
Find all triples \((a, b, c)\) of positive real numbers that satisfy the system:
\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]
6 replies
MuradSafarli
Apr 27, 2025
quacksaysduck
29 minutes ago
BMO 2024 SL A1
MuradSafarli   8
N an hour ago by ja.
A1.

Let \( u, v, w \) be positive reals. Prove that there is a cyclic permutation \( (x, y, z) \) of \( (u, v, w) \) such that the inequality:

\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]
holds for all positive real numbers \( a, b \) and \( c \).
8 replies
MuradSafarli
Apr 27, 2025
ja.
an hour ago
No more topics!
Parallel
Fang-jh   5
N Oct 18, 2014 by TelvCohl
Source: Chinese TST
Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.
5 replies
Fang-jh
Apr 4, 2008
TelvCohl
Oct 18, 2014
Source: Chinese TST
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Fang-jh
237 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.
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yetti
2643 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $ K$ be the circumcenter of the $ \triangle ABC.$ Let $ \triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the pedal triangles of $ P, Q$; the midpoint $ M$ of $ (PQ)$ is their common circumcenter. The midpoint $ N_a$ of $ AQ$ is the circumcenter of the $ \triangle AQ_bQ_c,$ on account of the rigth angles $ \angle QQ_bA, \angle QQ_cA.$ The isogonal $ AP$ of $ AQ \equiv AN_a$ with respect to the angle $ \angle CAB \equiv \angle Q_cAQ_b$ is its A-altitude $ \Longrightarrow$ $ Q_bQ_c \perp AP.$ But $ O_2O_3 \perp AP$ as well $ \Longrightarrow$ $ Q_bQ_c \parallel O_2O_3$ and similarly, $ Q_cQ_a \parallel O_3O_1, Q_aQ_b \parallel O_1O_2.$ The $ \triangle O_aQ_bO_c \sim \triangle O_1O_2O_3$ are centrally similar, having the correspnding sides parallel. $ M, O$ are their corresponding circumcenters and $ Q, K$ are their corresponding points, the concurrency points of their corresponding cevians $ (Q_aQ \parallel O_1K) \perp BC,$ $ (Q_bQ \parallel O_2K) \perp CA,$ $ (Q_cQ \parallel O_3K) \perp AB.$ Their corresponding lines $ QM \parallel KO$ are parallel. In ezactly the same way, $ PM \parallel KO'$ $ \Longrightarrow$ $ OO' \parallel PQ.$
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zhuyuanzhang
1 post
#3 • 1 Y
Y by Adventure10
yetti wrote:
Let $ K$ be the circumcenter of the $ \triangle ABC.$ Let $ \triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the pedal triangles of $ P, Q$; the midpoint $ M$ of $ (PQ)$ is their common circumcenter. The midpoint $ N_a$ of $ AQ$ is the circumcenter of the $ \triangle AQ_bQ_c,$ on account of the rigth angles $ \angle QQ_bA, \angle QQ_cA.$ The isogonal $ AP$ of $ AQ \equiv AN_a$ with respect to the angle $ \angle CAB \equiv \angle Q_cAQ_b$ is its A-altitude $ \Longrightarrow$ $ Q_bQ_c \perp AP.$ But $ O_2O_3 \perp AP$ as well $ \Longrightarrow$ $ Q_bQ_c \parallel O_2O_3$ and similarly, $ Q_cQ_a \parallel O_3O_1, Q_aQ_b \parallel O_1O_2.$ The $ \triangle O_aQ_bO_c \sim \triangle O_1O_2O_3$ are centrally similar, having the correspnding sides parallel. $ M, O$ are their corresponding circumcenters and $ Q, K$ are their corresponding points, the concurrency points of their corresponding cevians $ (Q_aQ \parallel O_1K) \perp BC,$ $ (Q_bQ \parallel O_2K) \perp CA,$ $ (Q_cQ \parallel O_3K) \perp AB.$ Their corresponding lines $ QM \parallel KO$ are parallel. In ezactly the same way, $ PM \parallel KO'$ $ \Longrightarrow$ $ OO' \parallel PQ.$

이 자의 풀이를 보지 말거라. 짐이 더 좋은 풀이를 올리리라.
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orl
3647 posts
#4 • 2 Y
Y by Adventure10, Mango247
Could you please use English instead of Korean here. Thanks.
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mihai miculita
666 posts
#5 • 2 Y
Y by Adventure10, Mango247
$ \mbox{Prove that circumcircle of the } \triangle{ABC}, \triangle{O_1O_2O_3} \mbox{ and } \triangle{O'_1O'_2O'_3} \mbox{ coaxial circles.}$
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TelvCohl
2312 posts
#6 • 5 Y
Y by buratinogigle, baopbc, Adventure10, Mango247, and 1 other user
My solution :

Let $ T $ be the circumcenter of $ \triangle ABC $ .
Let $ D \equiv O_2O_3 \cap O_2'O_3', E \equiv O_3O_1 \cap O_3'O_1', F \equiv O_1O_2 \cap O_1'O_2' $ .

Easy to see $ T=O_1O_1' \cap O_2O_2' \cap O_3O_3' $.
Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively ,
so $ \angle O_2O_3O_3' =180^{\circ} - \angle BAP =180^{\circ} - \angle QAC =\angle O_2O_2'O_3' $ .
i.e. $ O_2, O_3, O_2', O_3' $ are concyclic
Similarly, we can get $ O_3, O_1, O_3', O_1' $ are concyclic and $ O_1, O_2, O_1', O_2' $ are concyclic.

Since $ O_2D \cdot DO_3=O_2'D \cdot  DO_3', O_3E \cdot EO_1=O_3'E \cdot EO_1', O_1F \cdot FO_2=O_1'F \cdot FO_2' $ ,
so $ D, E, F $ are collinear at the radical axis of $ (O_1O_2O_3) $ and $ (O_1'O_2'O_3') $ . i.e. $ \overline{DEF} $ is perpendicular to $ OO' $ ... $ (1) $

Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively ,
so $ D $ is the circumcenter of $ \triangle APQ $ . i.e. $ D $ lie on the perpendicular bisector of $ PQ $
Similarly, we can get $ E, F $ lie on the perpendicular bisector of $ PQ $ . i.e. $ \overline{DEF} $ is the perpendicular bisector of $ PQ $ ... $ (2) $

From $ (1) $ and $ (2) $ we get $ OO' \parallel PQ $ .

Q.E.D

Remark :

$ (1) $
Since $ TO_1 \cdot TO_1'=TO_2 \cdot TO_2'=TO_3 \cdot TO_3' $ ,
so $ (O_1'O_2'O_3') $ is the image of $ (O_1O_2O_3) $ under the inversion WRT $ (O) $ .
i.e. $ (O), (O_1O_2O_3), (O_1'O_2'O_3') $ are coaxial

$ (2) $
This problem can be generalized to the quadrilateral :

Let $ P, Q $ be the isogonal conjugate of $ ABCD $ .
Let $ O_1, O_2, O_3, O_4 $ be the circumcenter of $ \triangle ABP, \triangle BCP, \triangle CDP, \triangle DAP $ .
Let $ O_1', O_2', O_3', O_4' $ be the circumcenter of $ \triangle ABQ, \triangle BCQ, \triangle CDQ, \triangle DAQ $ .
Let $ O, O' $ be the circumcenter of $ O_1O_2O_3O_4, O_1'O_2'O_3'O_4' $ .

Then $ OO' \parallel PQ $
This post has been edited 1 time. Last edited by TelvCohl, Mar 20, 2016, 12:31 AM
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