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
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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
BMO 2024 SL A1
MuradSafarli   2
N a few seconds ago by Sadigly

---

**A1.**

Let \( u, v, w \) be positive reals. Prove that there is a cyclic permutation \( (x, y, z) \) of \( (u, v, w) \) such that the inequality:

\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]
holds for all positive real numbers \( a, b \) and \( c \).

---

2 replies
MuradSafarli
a minute ago
Sadigly
a few seconds ago
all functions satisfying f(x+yf(x))+y = xy + f(x+y)
falantrng   5
N 4 minutes ago by MuradSafarli
Source: Balkan MO 2025 P3
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[f(x+yf(x))+y = xy + f(x+y).\]
Proposed by Giannis Galamatis, Greece
5 replies
falantrng
33 minutes ago
MuradSafarli
4 minutes ago
Interesting inequality
sqing   2
N 4 minutes ago by sqing
Let $ a,b> 0 $ and $ a+b+ab=1. $ Prove that
$$\frac{1}{1+a^2} + \frac{1}{1+b^2}  +a+b\leq \frac{5}{\sqrt{2}}-1 $$
2 replies
sqing
Today at 3:35 AM
sqing
4 minutes ago
Arbitrary point on BC and its relation with orthocenter
falantrng   5
N 12 minutes ago by ErTeeEs06
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.
5 replies
falantrng
38 minutes ago
ErTeeEs06
12 minutes ago
No more topics!
2014 China Second Round Olympiad Second Part Problem 2
abccsss   3
N Jan 3, 2025 by Captainscrubz
Source: 2014 China Second Round Olympiad
Let $ABC$ be an acute triangle such that $\angle BAC \neq 60^\circ$. Let $D,E$ be points such that $BD,CE$ are tangent to the circumcircle of $ABC$ and $BD=CE=BC$ ($A$ is on one side of line $BC$ and $D,E$ are on the other side). Let $F,G$ be intersections of line $DE$ and lines $AB,AC$. Let $M$ be intersection of $CF$ and $BD$, and $N$ be intersection of $CE$ and $BG$. Prove that $AM=AN$.
3 replies
abccsss
Aug 4, 2015
Captainscrubz
Jan 3, 2025
2014 China Second Round Olympiad Second Part Problem 2
G H J
G H BBookmark kLocked kLocked NReply
Source: 2014 China Second Round Olympiad
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abccsss
15 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an acute triangle such that $\angle BAC \neq 60^\circ$. Let $D,E$ be points such that $BD,CE$ are tangent to the circumcircle of $ABC$ and $BD=CE=BC$ ($A$ is on one side of line $BC$ and $D,E$ are on the other side). Let $F,G$ be intersections of line $DE$ and lines $AB,AC$. Let $M$ be intersection of $CF$ and $BD$, and $N$ be intersection of $CE$ and $BG$. Prove that $AM=AN$.
Z K Y
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Luis González
4148 posts
#2 • 2 Y
Y by Adventure10, Mango247
It's a particular case of http://www.artofproblemsolving.com/community/c6h610465, when the points M and N in that configuration coincide with B and C.
Z K Y
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Pluto04
797 posts
#3 • 2 Y
Y by Adventure10, Mango247
AM=AN only if AB= AC.

It is trivial to observe that B,D, E, C are con cyclic. Also observe that BC∥FG because BC∥DE due to the fact that BC=BD=CE.

Now since AB =AC and B,D, E, C are con cyclic, by chasing some angles we get,
∠ABM =∠AN and hence ∆ABM≅∆ACNsave.

So AM =AN and we are done.

Please refer attachment for diagram
Attachments:
Image_5058.pdf (186kb)
Z K Y
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Captainscrubz
58 posts
#6 • 1 Y
Y by MrdiuryPeter
Pluto04 wrote:
AM=AN only if AB= AC.

It is trivial to observe that B,D, E, C are con cyclic. Also observe that BC∥FG because BC∥DE due to the fact that BC=BD=CE.

Now since AB =AC and B,D, E, C are con cyclic, by chasing some angles we get,
∠ABM =∠AN and hence ∆ABM≅∆ACNsave.

So AM =AN and we are done.

Please refer attachment for diagram

It is also possible when $AB$ is not equal to $AC$
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