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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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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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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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length of segment of intersections of pairs of external angle bisectors = s
parmenides51   1
N Jul 28, 2018 by PavelMath
Source: Mexican Mathematical Olympiad 1999 OMM P5
In a quadrilateral $ABCD$ with $AB // CD$, the external bisectors of the angles at $B$ and $C$ meet at $P$, while the external bisectors of the angles at $A$ and $D$ meet at $Q$. Prove that the length of $PQ$ equals the semiperimeter of $ABCD$.
1 reply
parmenides51
Jul 28, 2018
PavelMath
Jul 28, 2018
J
length of segment of intersections of pairs of external angle bisectors = s
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geometry
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Source: Mexican Mathematical Olympiad 1999 OMM P5
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parmenides51
30629 posts
parmenides51
#1 h  Jul 28, 2018, 10:30 AM • 2 Y
Y by Adventure10, Mango247
In a quadrilateral $ABCD$ with $AB // CD$, the external bisectors of the angles at $B$ and $C$ meet at $P$, while the external bisectors of the angles at $A$ and $D$ meet at $Q$. Prove that the length of $PQ$ equals the semiperimeter of $ABCD$.
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PavelMath
185 posts
PavelMath
#2 h  Jul 28, 2018, 10:35 AM • 2 Y
Y by Adventure10, Mango247
parmenides51 wrote:
In a quadrilateral $ABCD$ with $AB // CD$, the external bisectors of the angles at $B$ and $C$ meet at $P$, while the external bisectors of the angles at $A$ and $D$ meet at $Q$. Prove that the length of $PQ$ equals the semiperimeter of $ABCD$.

Just note that $PQ$ lies on the middle line of the trapezoid $ABCD$. As $\angle AQD=\angle BPC=90^\circ$, then $QM=AD/2$ and $PN=BC/2$ (here $M$ and $N$ are the midpoints of $AD$ and $BC$). Finally, $MN=(AB+CD)/2$. Combining, we get our statement.
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