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A7456321   11
N 23 minutes ago by Bocabulary142857
What is your favorite math topic/subject?

If you don't know why you are here, go binge watch something!

If you forgot why you are here, go to a hospital! :)

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11 replies
A7456321
Friday at 11:53 PM
Bocabulary142857
23 minutes ago
PHP on subsets
SYBARUPEMULA   0
an hour ago
Source: inspired by Romania 2009
Let $X$ be a set of $102$ elements. Let $A_1, A_2, A_3, ..., A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $100$ elements. Show that there's a member of $X$ that occurs in at least $7$ different subsets $A_j$.
0 replies
SYBARUPEMULA
an hour ago
0 replies
Moscow Geometry Problems
nataliaonline75   1
N an hour ago by MathLuis
Source: MMO 2003 10.4
Let M be the intersection point of the medians of ABC. On the perpendiculars dropped from M to sides BC, AC, AB, points A1, B1, C1 are taken, respectively, with A1B1 perpendicular to MC and A1C1 perpendicular to MB. prove that M is the intersections pf the medians in A1B1C1.
Any solutions without vectors? :)
1 reply
1 viewing
nataliaonline75
Jul 9, 2024
MathLuis
an hour ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   1
N an hour ago by nataliaonline75
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
1 reply
1 viewing
OgnjenTesic
May 22, 2025
nataliaonline75
an hour ago
Another OI geometry problem
chengbilly   7
N an hour ago by MathLuis
Source: own
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$. $H$ the orthocenter of triangle $BIC$ . The incircle of $ABC$ touches side $AC,AB$ at $E,F$ respectively. Suppose that $\odot (AEF)$ and $\odot (AIO)$ intersects $\odot (ABC)$ at $S$ and $T$ respectively (differ from $A$). Prove that $T,H,I,S$ lies on a circle.

($\odot (ABC)$ denote the circumcircle of $ABC$)
7 replies
chengbilly
Apr 14, 2021
MathLuis
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   2
N an hour ago by nataliaonline75
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
nataliaonline75
an hour ago
Infinite Pairs of Integers
steven_zhang123   0
2 hours ago
Source: 2025 Spring NSMO P6
Given a positive integer \( k \), prove that there exist infinitely many pairs of positive integers \((m, n)\) (\(m < n\)) such that
\[
\tau(m^k) \tau(m^k + 1) \cdots \tau(n^k - 1) \tau(n^k)
\]is a perfect square, where \(\tau(n)\) denotes the number of positive divisors of \(n\).
Proposed by Dong Zhenyu
0 replies
steven_zhang123
2 hours ago
0 replies
power sum system of equations in 3 variables
Stear14   0
2 hours ago
Given that
$x^2+y^2+z^2=8\ ,$
$x^3+y^3+z^3=15\ ,$
$x^5+y^5+z^5=100\ .$

Find the value of $\ x+y+z\ .$
0 replies
Stear14
2 hours ago
0 replies
Game on 6 by 6 grid
billzhao   26
N 2 hours ago by Sleepy_Head
Source: USAMO 2004, problem 4
Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.
26 replies
billzhao
Apr 29, 2004
Sleepy_Head
2 hours ago
USA GEO 2003
dreammath   21
N 3 hours ago by lpieleanu
Source: TST USA 2003
Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]  \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)
21 replies
dreammath
Feb 16, 2004
lpieleanu
3 hours ago
Balkan Mathematical Olympiad
ABCD1728   0
3 hours ago
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
0 replies
ABCD1728
3 hours ago
0 replies
geometry problem
kjhgyuio   8
N Apr 23, 2025 by EthanNg6
........
8 replies
kjhgyuio
Apr 20, 2025
EthanNg6
Apr 23, 2025
geometry problem
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kjhgyuio
69 posts
#1
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jb2015007
1969 posts
#2
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$          $
This post has been edited 2 times. Last edited by jb2015007, Apr 20, 2025, 10:30 PM
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kjhgyuio
69 posts
#3
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should i release answer?
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Shan3t
416 posts
#4 • 1 Y
Y by HacheB2031
25 right????

bc $\angle ACE = 90^\circ$
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K1mchi_
142 posts
#5
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Shan3t wrote:
25 right????

bc $\angle ACE = 90^\circ$

they’re similar triangles
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Pengu14
636 posts
#6
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K1mchi_ wrote:
Shan3t wrote:
25 right????

bc $\angle ACE = 90^\circ$

they’re similar triangles

And?
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MathWinner121
533 posts
#7
Y by
Pengu14 wrote:
K1mchi_ wrote:
Shan3t wrote:
25 right????

bc $\angle ACE = 90^\circ$

they’re similar triangles

And?

Yeah. We don't have a second side length. Just the ratio.
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Shan3t
416 posts
#8
Y by
K1mchi_ wrote:
Shan3t wrote:
25 right????

bc $\angle ACE = 90^\circ$

they’re similar triangles

yea but $\angle ACE = 90^\circ$, so Pythagoras
This post has been edited 1 time. Last edited by Shan3t, Apr 21, 2025, 12:56 AM
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EthanNg6
35 posts
#9
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????????????
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