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Trigonometric Product
Henryfamz   0
38 minutes ago
Compute $$\prod_{n=1}^{45}\sin(2n-1)$$
0 replies
Henryfamz
38 minutes ago
0 replies
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3 variable FE with divisibility condition
pithon_with_an_i   1
N 43 minutes ago by Primeniyazidayi
Source: Revenge JOM 2025 Problem 1, Revenge JOMSL 2025 N2, Own
Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $$f(a)+f(b)+f(c) \mid a^2 + af(b) + cf(a)$$for all $a,b,c \in \mathbb{N}$.
1 reply
pithon_with_an_i
an hour ago
Primeniyazidayi
43 minutes ago
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Gives typical russian combinatorics vibes
Sadigly   1
N an hour ago by Sadigly
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
1 reply
Sadigly
May 8, 2025
Sadigly
an hour ago
J
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Roots of unity
Henryfamz   0
an hour ago
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
0 replies
Henryfamz
an hour ago
0 replies
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Thailand MO 2025 P3
Kaimiaku   4
N an hour ago by mihaig
Let $a,b,c,x,y,z$ be positive real numbers such that $ay+bz+cx \le az+bx+cy$. Prove that $$ \frac{xy}{ax+bx+cy}+\frac{yz}{by+cy+az}+\frac{zx}{cz+az+bx} \le \frac{x+y+z}{a+b+c}$$
4 replies
Kaimiaku
Today at 6:48 AM
mihaig
an hour ago
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Tangents involving a centroid with an isosceles triangle result
pithon_with_an_i   1
N an hour ago by wassupevery1
Source: Revenge JOM 2025 Problem 5, Revenge JOMSL 2025 G5, Own
A triangle $ABC$ has centroid $G$. A line parallel to $BC$ passing through $G$ intersects the circumcircle of $ABC$ at a point $D$. Let lines $AD$ and $BC$ intersect at $E$. Suppose a point $P$ is chosen on $BC$ such that the tangent of the circumcircle of $DEP$ at $D$, the tangent of the circumcircle of $ABC$ at $A$ and $BC$ concur. Prove that $GP = PD$.

Remark 1
Remark 2
1 reply
pithon_with_an_i
an hour ago
wassupevery1
an hour ago
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Beautiful numbers in base b
v_Enhance   20
N an hour ago by cursed_tangent1434
Source: USEMO 2023, problem 1
A positive integer $n$ is called beautiful if, for every integer $4 \le b \le 10000$, the base-$b$ representation of $n$ contains the consecutive digits $2$, $0$, $2$, $3$ (in this order, from left to right). Determine whether the set of all beautiful integers is finite.

Oleg Kryzhanovsky
20 replies
v_Enhance
Oct 21, 2023
cursed_tangent1434
an hour ago
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Gcd(m,n) and Lcm(m,n)&F.E.
Jackson0423   0
an hour ago
Source: 2012 KMO Second Round

Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all positive integers \( m, n \),
\[ f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)), \]where \( \mathrm{lcm}(m, n) \) and \( \gcd(m, n) \) denote the least common multiple and the greatest common divisor of \( m \) and \( n \), respectively.
0 replies
+1 w
Jackson0423
an hour ago
0 replies
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f(f(n))=2n+2
Jackson0423   0
an hour ago
Source: 2013 KMO Second Round

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function satisfying the following conditions for all \( n \in \mathbb{N} \):
\[ \begin{cases} f(n+1) > f(n) \\ f(f(n)) = 2n + 2 \end{cases} \]Find the value of \( f(2013) \).
0 replies
Jackson0423
an hour ago
0 replies
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Thailand MO 2025 P2
Kaimiaku   2
N an hour ago by carefully
A school sent students to compete in an academic olympiad in $11$ differents subjects, each consist of $5$ students. Given that for any $2$ different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least $4$ different subjects.
2 replies
Kaimiaku
Today at 7:38 AM
carefully
an hour ago
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a geometry problem
lumos123   5
N Dec 5, 2023 by DomX
Given triangle $ABC$ inscribed in $(O)$. Let $P$ be the intersection point of the two tangents at $B$ and $C$ of $(O).$ On $AP$ take $E, F$ so that $BE // AC$, $CF // AB$. $BE$ intersects $CF$ at $D$. Call$ I $the center $(DEF).$
a. Prove: $AI$ perpendicular to $BC$
b. $CD, BD$ cut $(ADE), (ADF)$ at $M, N$. Prove that $AP$ perpendicular to $MN$
5 replies
lumos123
Dec 4, 2023
DomX
Dec 5, 2023
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a geometry problem
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lumos123
40 posts
lumos123
#1 h  Dec 4, 2023, 4:13 AM
Y by
Given triangle $ABC$ inscribed in $(O)$. Let $P$ be the intersection point of the two tangents at $B$ and $C$ of $(O).$ On $AP$ take $E, F$ so that $BE // AC$, $CF // AB$. $BE$ intersects $CF$ at $D$. Call$ I $the center $(DEF).$
a. Prove: $AI$ perpendicular to $BC$
b. $CD, BD$ cut $(ADE), (ADF)$ at $M, N$. Prove that $AP$ perpendicular to $MN$
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club52
724 posts
club52
#2 h  Dec 4, 2023, 4:34 AM
Y by
The wording of the question is not very clear to me. I will try to make some clarifications.

Does "Given triangle $ABC$ inscribed in $(O)$." mean "Given triangle $ABC$ inscribed in $\odot O$, where $O$ is the circumcenter of $\triangle ABC$..."?

What does $BE//AC, CF//AB$ mean?

Does "Call $I$ the center $(DEF).$" mean "Call $I$ the incenter of $(DEF)$."?

What does part b. of this problem mean?

Edit: Also, here is a fun fact about this problem:

Click to reveal hidden text
This post has been edited 1 time. Last edited by club52, Dec 4, 2023, 4:39 AM
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Orthogonal.
593 posts
Orthogonal.
#3 h  Dec 4, 2023, 4:42 AM
Y by
club52 wrote:

Does "Given triangle $ABC$ inscribed in $(O)$." mean "Given triangle $ABC$ inscribed in $\odot O$, where $O$ is the circumcenter of $\triangle ABC$..."?
Yes.
Quote:
What does $BE//AC, CF//AB$ mean?
BE is parallel to AC, CF is parallel to AB
Quote:
Does "Call $I$ the center $(DEF).$" mean "Call $I$ the incenter of $(DEF)$."?

It quite literally means that I is the center of circle (DEF)
Quote:
What does part b. of this problem mean?
If CD intersects (ADE) at M and BD intersects (ADF) at N then prove that AP is perpendicular to MN
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lumos123
40 posts
lumos123
#4 h  Dec 4, 2023, 5:51 AM
Y by
Orthogonal. wrote:
club52 wrote:

Does "Given triangle $ABC$ inscribed in $(O)$." mean "Given triangle $ABC$ inscribed in $\odot O$, where $O$ is the circumcenter of $\triangle ABC$..."?
Yes.
Quote:
What does $BE//AC, CF//AB$ mean?
BE is parallel to AC, CF is parallel to AB
Quote:
Does "Call $I$ the center $(DEF).$" mean "Call $I$ the incenter of $(DEF)$."?

It quite literally means that I is the center of circle (DEF)
Quote:
What does part b. of this problem mean?
If CD intersects (ADE) at M and BD intersects (ADF) at N then prove that AP is perpendicular to MN

Yes it's right, thank you for your edition
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ancamagelqueme
104 posts
ancamagelqueme
#5 h  Dec 5, 2023, 11:04 AM
Y by
lumos123 wrote:
Given triangle $ABC$ inscribed in $(O)$. Let $P$ be the intersection point of the two tangents at $B$ and $C$ of $(O).$ On $AP$ take $E, F$ so that $BE // AC$, $CF // AB$. $BE$ intersects $CF$ at $D$. Call$ I $the center $(DEF).$
a. Prove: $AI$ perpendicular to $BC$
b. $CD, BD$ cut $(ADE), (ADF)$ at $M, N$. Prove that $AP$ perpendicular to $MN$

As a consequence of this topic, a triangle center results that does not currently appear in ETC.

We call $\ell_a$ the line $MN$ and define $\ell_b$ and $\ell_c$ cyclically. Let $A'B'C'$ be the triangle formed by the lines $\ell_a$, $\ell_b$, $\ell_c$. Since $AP$ (A-simmedian) is perpendicular to $\ell_a$, the triangles $ABC$ and $A'B'C'$ are orthologic. The orthologic center of $A'B'C'$ with respect to $ABC$ is the reflection of the anticomplement of the retrocenter at the De Longchamps point, $W=2X_{20}-X_{193}$.

$W = 3 a^6+9 a^4 (b^2+c^2)-a^2 (11 b^4+2 b^2 c^2+11 c^4)-(b^2-c^2)^2 (b^2+c^2) : :$ (barycentrics)
Figure
This post has been edited 1 time. Last edited by ancamagelqueme, Dec 5, 2023, 11:08 AM
Reason: Modified link
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DomX
8 posts
DomX
#6 h  Dec 5, 2023, 3:22 PM • 1 Y
Y by Vahe_Arsenyan
We first notice that $AD$ is tangent to circle $(DEF)$, as $\angle DFE=\angle CFA= 180-\angle ACF-\angle FAC= 180 -\angle DBA-\angle BAD= \angle ADE$.

Now, let $D'$ be the the other point in $(DEF)$ such that $AD'$ is tangent to the circle. Notice that $DD'$ is perpendicular to $AI$. So we must prove that $DD'$ is paral·lel to $BC$. Let $M_A$ be the midpoint of $BC$, and $AM_AD$ are collinear. By projections $-1=(D,D',E,F)=(M_A, P_\infty;B,C)$, proving the first part of the problem.

Now, for the next part, the sides $AB$ and $AC$ of the triangle are tangent to the circles $(AED)$ and $(AFD)$ respectively, as $\angle ADE= \angle DAC =\angle BAE$ (a simmilar logic proves the other tangency). And, as $DM||BA$ and $BA$ is tangent, $AD=AM$. Once again, a simmilar logic proves $AN=AD=AM$.

We will now prove that $EF$ is the perpendicular bisector of $MN$. Notice that $\angle ANE=\angle AND=\angle NDA =\angle EDA=\angle EMA$ so $ME=NE$. We also prove the same for point F, and so line $AEFP$ is the perpendicular bisector of $MN$, solving the problem.
This post has been edited 1 time. Last edited by DomX, Dec 5, 2023, 3:55 PM
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