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Easy but Nice 12
TelvCohl   2
N an hour ago by AuroralMoss
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
2 replies
TelvCohl
Mar 8, 2025
AuroralMoss
an hour ago
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Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   8
N an hour ago by atdaotlohbh
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
8 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
an hour ago
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Balkan Mathematical Olympiad
ABCD1728   1
N an hour ago by ABCD1728
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!
1 reply
ABCD1728
May 24, 2025
ABCD1728
an hour ago
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Unexpecredly Quick-Solve Inequality
Primeniyazidayi   2
N an hour ago by sqing
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
2 replies
Primeniyazidayi
4 hours ago
sqing
an hour ago
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cute geo
Royal_mhyasd   0
an hour ago
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
0 replies
1 viewing
Royal_mhyasd
an hour ago
0 replies
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Nice inequality
TUAN2k8   2
N an hour ago by TUAN2k8
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
2 replies
TUAN2k8
Today at 2:03 AM
TUAN2k8
an hour ago
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An NT for a break
reni_wee   1
N 2 hours ago by reni_wee
Source: ONTCP 2.4.1
Prove that there are no positive integers $x,k$ and $n \geq 2$ such that $x^2+1 = k(2^n -1)$.
1 reply
reni_wee
2 hours ago
reni_wee
2 hours ago
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p divides x^x-c
mistakesinsolutions   6
N 2 hours ago by reni_wee
Show that for integer c and a prime p, $ p |x^x-c $ has a solution
6 replies
mistakesinsolutions
Jun 13, 2023
reni_wee
2 hours ago
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exponential diophantine in integers
skellyrah   1
N 2 hours ago by skellyrah
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
1 reply
skellyrah
Yesterday at 7:04 PM
skellyrah
2 hours ago
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IMO 2017 Problem 4
Amir Hossein   117
N 2 hours ago by ezpotd
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
117 replies
Amir Hossein
Jul 19, 2017
ezpotd
2 hours ago
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Angle EBA is equal to Angle DCB
WakeUp   6
N Apr 10, 2025 by Nari_Tom
Source: Baltic Way 2011
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
6 replies
WakeUp
Nov 6, 2011
Nari_Tom
Apr 10, 2025
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Angle EBA is equal to Angle DCB
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Source: Baltic Way 2011
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WakeUp
1347 posts
WakeUp
#1 h  Nov 6, 2011, 1:25 PM • 3 Y
Y by anantmudgal09, Adventure10, Mango247
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
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newsun
1418 posts
newsun
#2 h  Nov 6, 2011, 1:58 PM • 3 Y
Y by IstekOlympiadTeam, Adventure10, Mango247
WakeUp wrote:
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
Let $F$ on the side $CD$ such that $DE=DF$ then we have $ \triangle{DEB} = \triangle{DFB} $
Therefore, $BE=BF$ and $\angle{BEA} = \angle{BFC} $
It is sufficient to show that $ \triangle{AEB} \sim \triangle{BFC} $
indeed,
$ AE\cdot ED + BE^2=CD\cdot AE \Leftrightarrow AE(CD-DE) = BE.BF \Leftrightarrow AE(CD-DF) = BE.BF \Leftrightarrow AE.FC = BE.BF \Leftrightarrow \frac{AE}{EB} = \frac{BF}{FC} $ , as required.
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Kimchiks926
256 posts
Kimchiks926
#3 h  Oct 4, 2020, 6:50 AM
Y by
Solved with @blastoor
Let $BE$ intersect $\odot(ABD)$ at point $X \neq B$. From PoP we have that $AE \cdot ED = BE \cdot EX$. Thus original equality becomes:
\begin{align*} AE \cdot ED + BE^2 = CD \cdot AE \\ BE \cdot EX + BE^2 = CD \cdot AE \\ BE \cdot BX = CD \cdot AE \qquad (2) \\ \end{align*}
Claim: $\triangle BXA \sim \triangle CDB$
Proof: It is easy to see that $\triangle AEX \sim \triangle BED $ and that $\angle AXB =\angle ABD = \angle BDC$. Consequently we have:
$$ \frac{AE}{AX} = \frac{BE}{BD} \implies BE \cdot AX = AE \cdot BD \qquad (20 $$Dividing relations $(1)$ and $(2)$ we get that:
$$ \frac{BX}{AX} = \frac{CD}{BD} $$This implies that $\triangle BXA \sim \triangle CDB$ and thus the claim is proved.

From claim follows $\angle XBA =\angle EBA =\angle DCB$ as desired.
This post has been edited 1 time. Last edited by Kimchiks926, Oct 13, 2020, 1:15 PM
Reason: typo
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rafaello
1079 posts
rafaello
#4 h  Aug 13, 2021, 9:14 PM
Y by
Construct $L$ on $AD$ such that $DL=DC$. Thus,
$$BE^2=CD\cdot AE-AE\cdot AD=EA\cdot EL,$$thus $BE$ is tangent to $(ABL)$.
As $BD$ is the angle bisector of $\triangle CDL$, we also have $BD$ the perpendicular bisector of $LC$.
Hence, $$\measuredangle EBA=\measuredangle BLD=\measuredangle DCB,$$we are done.
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HamstPan38825
8868 posts
HamstPan38825
#5 h  Sep 13, 2022, 5:18 PM
Y by
We prove the converse, as such a point $E$ is clearly unique. Let $E'$ be the reflection of $E$ over $\overline{BD}$ and define $A'$ similarly. Then $$AE \cdot ED + BE^2 = DE' \cdot E'A' + BE'^2.$$Because $\overline{BE'}$ is tangent to $(A'BC)$ by the angle condition, $$DE' \cdot E'A' + BE'^2 = DE' \cdot E'A' + E'A' \cdot E'C = EA \cdot CD,$$as needed.
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AshAuktober
1009 posts
AshAuktober
#6 h  Dec 18, 2024, 5:51 PM
Y by
Reflect $C$ about $BD$ to get $C' \in \overrightarrow{DA}$. Then the condition becomes $EA \cdot EC' = EB^2$, and what is to be proven is the Alternate Segment Theorem, so we're done.
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Nari_Tom
117 posts
Nari_Tom
#7 h  Apr 10, 2025, 9:38 AM
Y by
I had following simple solution. If we lock the points $B, E, D,C$, then there is will be unique $A$ which requires the given condition. So let's assume that $\angle EBA=\angle DCB$, and prove the given condition. Let $B'$ be the reflection of $B$ over $E$. Let $F=(ABB') \cap AE$. $G=BD \cap B'F$.
Let $H$ be the midpoint of $BG$. So it's clear that $2*EH=BD+DG$. On the other hand we have: $\triangle FDG \sim \triangle FEH \sim \triangle CDB$. Which means that $2*EH=BD+DG$ $\implies$ $2*EF=DF+DC$ $\implies$ $CD-DE=ED+DF=EF$. On the other hand by Thales theorem: $BE*BE'=AE*EF$ $\implies$ $BE^2=AE(CD-DE)$, and conclusion follows.
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