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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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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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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
J
No more topics!
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J
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8 points concyclic
parmenides51   1
N Oct 6, 2021 by JuanDelPan
Source: Mexican Mathematical Olympiad 1991 OMM P4
The diagonals $AC$ and $BD$ of a convex quarilateral $ABCD$ are orthogonal. Let $M,N,R,S$ be the midpoints of the sides $AB,BC,CD$ and $DA$ respectively, and let $W,X,Y,Z$ be the projections of the points $M,N,R$ and $S$ on the lines $CD,DA,AB$ and $BC$, respectively. Prove that the points $M,N,R,S,W,X,Y$ and $Z$ lie on a circle.
1 reply
parmenides51
Jul 29, 2018
JuanDelPan
Oct 6, 2021
J
8 points concyclic
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Reveal topic tags
geometry
orthogonal
convex quadrilateral
Concyclic
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Source: Mexican Mathematical Olympiad 1991 OMM P4
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parmenides51
30629 posts
parmenides51
#1 h  Jul 29, 2018, 7:36 AM • 3 Y
Y by HWenslawski, Adventure10, Mango247
The diagonals $AC$ and $BD$ of a convex quarilateral $ABCD$ are orthogonal. Let $M,N,R,S$ be the midpoints of the sides $AB,BC,CD$ and $DA$ respectively, and let $W,X,Y,Z$ be the projections of the points $M,N,R$ and $S$ on the lines $CD,DA,AB$ and $BC$, respectively. Prove that the points $M,N,R,S,W,X,Y$ and $Z$ lie on a circle.
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JuanDelPan
122 posts
JuanDelPan
#2 h  Oct 6, 2021, 9:58 PM • 1 Y
Y by HWenslawski
Since $M,N,R,S$ are midpoints, $AC||MN||SR$ and $BD||MS||NR||$. This means that $MNRS$ is a rectangle and $\angle NMS=90^{\circ}$. Also, since $W$ is the projection of $N$ to $AD$, $\angle NWS=90^{\circ}$, so $W$ lies on the circumcircle of $MNRS$. $X,Y,Z$ are defined similarly, so we can repeat this process (it is analogous) to see that $X,Y,Z$ also lie on the circumcircle of $MNRS$.
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