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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 involving Power of centroid
Mahdi_Mashayekhi   0
3 minutes ago
Given is an triangle $ABC$ with centroid $G$. Let $p$ be the power of $G$ w.r.t circumcircle of $ABC$ and $q$ be the power of $G$ w.r.t incircle of $ABC$. prove that $\frac{a^2+b^2+c^2}{12} \le q-p < \frac{a^2+b^2+c^2}{3}$.
0 replies
Mahdi_Mashayekhi
3 minutes ago
0 replies
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EXTENSION OF BABBAGES THEOREM
Mathgloggers   0
4 minutes ago
A few days ago I came across this question while solving the usage 2025 p-5.


$\boxed{\sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{k=p+1}^{2p-1} \frac{1}{k} \equiv \sum_{k=2p+1}^{3p-1}\frac{1}{k} \equiv.....\sum_{k=p(p-1)+1}^{p^2-1}\frac{1}{k} \equiv 0(mod p^2)}$
Prove this for $p$ in prime and $k \in Z^{+}$
0 replies
Mathgloggers
4 minutes ago
0 replies
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2020 EGMO P2: Sum inequality with permutations
alifenix-   28
N 9 minutes ago by math-olympiad-clown
Source: 2020 EGMO P2
Find all lists $(x_1, x_2, \ldots, x_{2020})$ of non-negative real numbers such that the following three conditions are all satisfied:

[list]
[*] $x_1 \le x_2 \le \ldots \le x_{2020}$;
[*] $x_{2020} \le x_1 + 1$;
[*] there is a permutation $(y_1, y_2, \ldots, y_{2020})$ of $(x_1, x_2, \ldots, x_{2020})$ such that $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3.$$[/list]

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2, 1, 2)$ is a permutation of $(1, 2, 2)$, and they are both permutations of $(2, 2, 1)$. Note that any list is a permutation of itself.
28 replies
alifenix-
Apr 18, 2020
math-olympiad-clown
9 minutes ago
J
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inequality problem
pennypc123456789   0
11 minutes ago
Given $a,b,c$ be positive real numbers . Prove that
$$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \ge \frac{6abc }{(a+b)(b+c)(a+c)}$$
0 replies
pennypc123456789
11 minutes ago
0 replies
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No more topics!
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concyclic wanted, intersections of circumcircles with lines parallel
parmenides51   1
N Jul 20, 2020 by CROWmatician
Source: 2017 France JBMO TST 4.3
Let $ABC$ be a triangle. Let $D$ and $E$ be two points of $[AC]$ so that $D$ is lies between $C$ and $E$. Let $F$ be the intersection of the circle circumscribed to $ABD$ with the parallel to $BC$ through $E$ such that $E, F$ lie in different half-planes wrt $AB$. Let $G$ the intersection of the circle circumscribed to triangle $BCD$ with the line parallel to $AB$ through $E$, so that $E,G$ lie in different half-planes wrt $BC$. Prove that the points $D, E, F, G$ are concyclic.
1 reply
parmenides51
Jul 20, 2020
CROWmatician
Jul 20, 2020
J
concyclic wanted, intersections of circumcircles with lines parallel
G H J
Reveal topic tags
geometry
circumcircle
Concyclic
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G H BBookmark kLocked kLocked NReply
Source: 2017 France JBMO TST 4.3
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parmenides51
30650 posts
parmenides51
#1 h  Jul 20, 2020, 5:18 AM • 1 Y
Y by Mango247
Let $ABC$ be a triangle. Let $D$ and $E$ be two points of $[AC]$ so that $D$ is lies between $C$ and $E$. Let $F$ be the intersection of the circle circumscribed to $ABD$ with the parallel to $BC$ through $E$ such that $E, F$ lie in different half-planes wrt $AB$. Let $G$ the intersection of the circle circumscribed to triangle $BCD$ with the line parallel to $AB$ through $E$, so that $E,G$ lie in different half-planes wrt $BC$. Prove that the points $D, E, F, G$ are concyclic.
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CROWmatician
272 posts
CROWmatician
#2 h  Jul 20, 2020, 6:50 AM • 1 Y
Y by Mango247
Obviously $F,B,$ and $G$ are collinear, and by angle chasing we know that $FA||GC$, then: $$\angle BDG= \angle BCG= \angle AFE$$and $$\angle FDB=\angle FAB$$then, let $X$ be the intersection point of $\overline{FE}$ and $\overline{AB}$. Finally we have $$\angle FDG=\angle BDG+\angle FDB=\angle AFE+\angle FAB+\angle AFE=\angle FXB=\angle FEG$$which tells us that $DEFG$ is cyclic quadrillateral.
$\blacksquare$
This post has been edited 1 time. Last edited by CROWmatician, Jul 20, 2020, 6:51 AM
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