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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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Abelkonkurransen 2025 2a
Lil_flip38   2
N a minute ago by MathLuis
Source: Abelkonkurransen
A teacher asks each of eleven pupils to write a positive integer with at most four digits, each on a separate yellow sticky note. Show that if all the numbers are different, the teacher can always submit two or more of the eleven stickers so that the average of the numbers on the selected notes are not an integer.
2 replies
Lil_flip38
Mar 20, 2025
MathLuis
a minute ago
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Number Theory Chain!
JetFire008   34
N 8 minutes ago by cubres
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
34 replies
JetFire008
Apr 7, 2025
cubres
8 minutes ago
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Easy function in turkey TST
egxa   9
N 21 minutes ago by Levieee
Source: 2024 Turkey TST P2
Find all $f:\mathbb{R}\to\mathbb{R}$ functions such that
$$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$$for all real numbers $x,y$
9 replies
egxa
Mar 18, 2024
Levieee
21 minutes ago
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2023 Iran MO 2nd round P6
Amiralizakeri2007   3
N 30 minutes ago by iliya8788
Source: 2023 Iran MO
6. Circles $W_{1}$ and $W_{2}$ with equal radii are given. Let $P$,$Q$ be the intersection of the circles.
points $B$ and $C$ are on $W_{1}$ and $W_{2}$ such that they are inside $W_{2}$ and $W_{1}$ respectively.
Points $X$,$Y$ $\neq$ $P$ are on $W_{1}$ and $W_{2}$ respectively, such that $\angle{BPQ}=\angle{BYQ}$ and $\angle{CPQ}=\angle{CXQ}$.Denote by $S$ as the other intersection of $(YPB)$ and $(XPC)$. Prove that $QS,BC,XY$ are concurrent.
3 replies
Amiralizakeri2007
May 17, 2023
iliya8788
30 minutes ago
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No more topics!
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Interior points of the sides minimizing the perimeter iff cyclic
Tintarn   4
N Mar 30, 2022 by Tintarn
Source: Germany 2021, Problem 2
Let $P$ on $AB$, $Q$ on $BC$, $R$ on $CD$ and $S$ on $AD$ be points on the sides of a convex quadrilateral $ABCD$. Show that the following are equivalent:

(1) There is a choice of $P,Q,R,S$, for which all of them are interior points of their side, such that $PQRS$ has minimal perimeter.

(2) $ABCD$ is a cyclic quadrilateral with circumcenter in its interior.
4 replies
Tintarn
Jun 16, 2021
Tintarn
Mar 30, 2022
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Interior points of the sides minimizing the perimeter iff cyclic
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Reveal topic tags
geometry
perimeter
geometry proposed
cyclic quadrilateral
interior
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Source: Germany 2021, Problem 2
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Tintarn
9034 posts
Tintarn
#1 h  Jun 16, 2021, 5:08 AM • 2 Y
Y by samrocksnature, HWenslawski
Let $P$ on $AB$, $Q$ on $BC$, $R$ on $CD$ and $S$ on $AD$ be points on the sides of a convex quadrilateral $ABCD$. Show that the following are equivalent:

(1) There is a choice of $P,Q,R,S$, for which all of them are interior points of their side, such that $PQRS$ has minimal perimeter.

(2) $ABCD$ is a cyclic quadrilateral with circumcenter in its interior.
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gnoka
245 posts
gnoka
#2 h  Jun 21, 2021, 12:43 AM • 1 Y
Y by samrocksnature
I'm sorry I can't understand the firse statement.
I guess you mean P,Q,R,S is on the line of the four sides oringinly, the first statement means the minimal perimeter holds when they are all on the segments of the sides.
Is that right?
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Tintarn
9034 posts
Tintarn
#3 h  Jun 21, 2021, 6:42 PM • 2 Y
Y by gnoka, Mango247
No no. They are always on the segments (or as the problem says "on the sides") but the condition (1) is about whether for the choice with the minimal parameters among all such $PQRS$, these four points are interior points (as opposed to being one of the endpoints of the sides/segments).
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minusonetwelth
225 posts
minusonetwelth
#4 h  Mar 30, 2022, 3:03 PM
Y by
Okay if I get the problem statement right, then we actually only consider the sides an not the lines on which the sides lie, so with "interior point", we mean that it lies on one of the sides and it not equal to $A$, $B$, $C$ or $D$. However, then $(1)\Longrightarrow (2)$ does not make sense.

$(2) \Longrightarrow (1)$: Wlog, (or OBdA, as we germans say), let $P\in \overline{AB}$, $Q\in \overline{BC}$, $R\in \overline{CD}$ and $S\in \overline{DA}$, so that non of the points lie on and assume that $A$, $B$, $C$ or $D$. Let $PQRS$ now be any such quadrilateral. By the triangle inequality, we get $PB+BQ>PQ$. We get similar results for the other points. Adding these inequalities up gives
$$PB+BQ+CQ+CR+DR+DS+AS+AP=AB+BC+CD+DA>PQ+QR+RS+SP$$Therefore, the perimeter of $ABCD$ is larger than any quadrilateral inscribed on the sides, so it can be the minimal one.
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Tintarn
9034 posts
Tintarn
#5 h  Mar 30, 2022, 3:13 PM
Y by
Joel.Gerlach wrote:
However, then $(1)\Longrightarrow (2)$ does not make sense.
I don't understand. Why does it not make sense?
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