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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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Problem 5
blug   1
N 39 minutes ago by atdaotlohbh
Source: Polish Math Olympiad 2025 Finals P5
Convex quadrilateral $ABCD$ is described on a circle $\omega$, and is not a trapezius inscribed in a circle. Let the tangency points of $\omega$ and $AB, BC, CD, DA$ be $K, L, M, N$ respectively. A circle with a center $I_K$, different from $\omega$ is tangent to the segement $AB$ and lines $AD, BC$. A circle with center $I_L$, different from $\omega$ is tangent to segment $BC$ and lines $AB, CD$. A circle with center $I_M$, different from $\omega$ is tangent to segment $CD$ and lines $AD, BC$. A circle with center $I_N$, different from $\omega$ is tangent to segment $AD$ and lines $AB, CD$. Prove that the lines $I_KK, I_LL, I_MM, I_NN$ are concurrent.
1 reply
blug
Yesterday at 12:08 PM
atdaotlohbh
39 minutes ago
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¿10^n-1 is a divisor of 11^n-1?
EmersonSoriano   1
N 44 minutes ago by Filipjack
Source: 2017 Peru Southern Cone TST P2
Determine if there exists a positive integer $n$ such that $10^n - 1$ is a divisor of $11^n - 1$.
1 reply
EmersonSoriano
3 hours ago
Filipjack
44 minutes ago
J
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D1018 : Can you do that ?
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
We can find $A,B,C$, such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$.

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$?
1 reply
Dattier
Mar 24, 2025
Dattier
an hour ago
J
H
A special family of subsets of {1, 2, ..., 100}.
EmersonSoriano   0
2 hours ago
Source: 2017 Peru Southern Cone TST P10
Miguel has a list consisting of several subsets of 10 elements from the set ${1,2,\dots,100}$. He says to Cecilia: "If you pick any subset of 10 elements from ${1,2,\dots,100}$, it will be disjoint with at least one subset from my list." What is the smallest possible number of subsets Miguel's list can have if what he tells Cecilia is true?
0 replies
EmersonSoriano
2 hours ago
0 replies
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No more topics!
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perpendiculars from projections are concurrent
parmenides51   2
N Jul 22, 2018 by jayme
Source: Korean Mathematical Olympiad 1996, Final Round P3 FKMO
Let $\ell$ be a line having no common points with a triangle $ABC$. Let $L,M,N$ be the projections of $A,B,C$ onto $\ell$, and let $LX,MY,NZ$ be the perpendiculars from $L,M,N$ to $BC,CA,AB$, respectively. Prove that these three perpendiculars are concurrent.
2 replies
parmenides51
Jul 21, 2018
jayme
Jul 22, 2018
J
perpendiculars from projections are concurrent
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Reveal topic tags
geometry
perpendicular lines
projection
concurrency
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Source: Korean Mathematical Olympiad 1996, Final Round P3 FKMO
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parmenides51
30629 posts
parmenides51
#1 h  Jul 21, 2018, 11:10 PM • 1 Y
Y by Adventure10
Let $\ell$ be a line having no common points with a triangle $ABC$. Let $L,M,N$ be the projections of $A,B,C$ onto $\ell$, and let $LX,MY,NZ$ be the perpendiculars from $L,M,N$ to $BC,CA,AB$, respectively. Prove that these three perpendiculars are concurrent.
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enhanced
515 posts
enhanced
#2 h  Jul 22, 2018, 12:29 AM • 2 Y
Y by Adventure10, Mango247
Its too well known they concur at the orthopole of $\ell$
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jayme
9775 posts
jayme
#3 h  Jul 22, 2018, 7:34 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
for a prove and more

http://jl.ayme.pagesperso-orange.fr/Docs/Orthopole.pdf

Sincerely
Jean-Louis
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