Stay ahead of learning milestones! Enroll in a class over the summer!

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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]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
Problem 5
blug   1
N 21 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
21 minutes ago
¿10^n-1 is a divisor of 11^n-1?
EmersonSoriano   1
N 26 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
26 minutes ago
D1018 : Can you do that ?
Dattier   1
N 28 minutes 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
28 minutes ago
A special family of subsets of {1, 2, ..., 100}.
EmersonSoriano   0
an hour 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
an hour ago
0 replies
No more topics!
Turkey Egmo Tst 2018 p1
KereMath   2
N Mar 28, 2020 by electrovector
Let $ABCD$ be a cyclic quadrilateral and $w$ be its circumcircle. For a given point $E$ inside $w$, $DE$ intersects $AB$ at $F$ inside $w$. Let $l$ be a line passes through $E$ and tangent to circle $AEF$. Let $G$ be any point on $l$ and inside the quadrilateral $ABCD$. Show that if $\angle GAD =\angle BAE$ and $\angle GCB + \angle GAB = \angle EAD + \angle AGD +  \angle ABE$ then $BC$, $AD$ and $EG$ are concurrent.
2 replies
KereMath
Feb 9, 2018
electrovector
Mar 28, 2020
Turkey Egmo Tst 2018 p1
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KereMath
160 posts
#1 • 1 Y
Y by Adventure10
Let $ABCD$ be a cyclic quadrilateral and $w$ be its circumcircle. For a given point $E$ inside $w$, $DE$ intersects $AB$ at $F$ inside $w$. Let $l$ be a line passes through $E$ and tangent to circle $AEF$. Let $G$ be any point on $l$ and inside the quadrilateral $ABCD$. Show that if $\angle GAD =\angle BAE$ and $\angle GCB + \angle GAB = \angle EAD + \angle AGD +  \angle ABE$ then $BC$, $AD$ and $EG$ are concurrent.
This post has been edited 2 times. Last edited by KereMath, Feb 9, 2018, 6:59 PM
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falantrng
249 posts
#2 • 1 Y
Y by Adventure10
We can find easily the circles $(GEAD),(BEGC)$ are cyclic.Then from radical axis theorem on $(GEAD),(ABCD),(BEGC),$ the lines $BC,AD,EG$ are concurrent.
This post has been edited 1 time. Last edited by falantrng, Feb 10, 2018, 3:58 PM
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electrovector
479 posts
#3
Y by
KereMath wrote:
Let $ABCD$ be a cyclic quadriteral whose sides $BC$ and $AD$ are not parallel. Let $E$ be a point inside the circumcircle of $ABCD$ which is on the opposite side of the line $AD$ with respect to the point $C$. The lines $DE$ and $AB$ meet at $F$. Let $G$ be a point inside $ABCD$ and also on the line which is tangent to the circumcircle of triangle $AEF$ at $E$. If
$$ \angle GAD = \angle BAE $$and $$ \angle GCB + \angle GBA = \angle EAD + \angle AGD + \angle ABE$$then show that the lines $BC$, $AD$ and $EG$ are concurrent.
This post has been edited 2 times. Last edited by electrovector, Mar 28, 2020, 4:15 PM
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