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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
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BMO Shortlist 2021 A5
Lukaluce   17
N 7 minutes ago by jasperE3
Source: BMO Shortlist 2021
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$f(xf(x + y)) = yf(x) + 1$$holds for all $x, y \in \mathbb{R}^{+}$.

Proposed by Nikola Velov, North Macedonia
17 replies
Lukaluce
May 8, 2022
jasperE3
7 minutes ago
J
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Function equation
luci1337   3
N 24 minutes ago by jasperE3
find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number
3 replies
luci1337
Yesterday at 3:01 PM
jasperE3
24 minutes ago
J
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Circumcenter of reflection of collinear points over sides
a1267ab   27
N 24 minutes ago by Giant_PT
Source: USA TST 2025
Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY=AZ$, $BZ=BX$, and $CX=CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$.

Michael Ren
27 replies
a1267ab
Jan 11, 2025
Giant_PT
24 minutes ago
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a+b+c=abc
KhuongTrang   1
N 33 minutes ago by KhuongTrang
Source: own
Problem. Let $a,b,c$ be three positive real numbers satisfying $a+b+c=abc.$ Prove that$$\sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}\ge4\cdot \frac{a^2b^2c^2-3}{ab+bc+ca-3}-7.$$There is a very elegant proof :-D Could anyone think of it?
1 reply
KhuongTrang
Wednesday at 11:51 AM
KhuongTrang
33 minutes ago
J
No more topics!
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geometry
TSandino   1
N Nov 10, 2016 by george_54
Source: Rioplatense Mathematical Olympiad, Level 1 2008
Let $ABC$ be an obtuse triangle at $C$ so that, $2 \angle BAC=\angle ABC$. $P$ is a point at side $AB$ so that $BP=2BC$. $M$ is the midpoint of $AB$ ($M$ is between $P$ and $B$). Prove that the perpendicular to the side $AC$ from $M$ cuts $PC$ at his midpoint.
1 reply
TSandino
Nov 9, 2016
george_54
Nov 10, 2016
J
geometry
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Obtuse triangle
midpoints
geometry
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G H BBookmark kLocked kLocked NReply
Source: Rioplatense Mathematical Olympiad, Level 1 2008
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TSandino
19 posts
TSandino
#1 h  Nov 9, 2016, 11:47 PM • 1 Y
Y by Adventure10
Let $ABC$ be an obtuse triangle at $C$ so that, $2 \angle BAC=\angle ABC$. $P$ is a point at side $AB$ so that $BP=2BC$. $M$ is the midpoint of $AB$ ($M$ is between $P$ and $B$). Prove that the perpendicular to the side $AC$ from $M$ cuts $PC$ at his midpoint.
This post has been edited 2 times. Last edited by TSandino, Nov 11, 2016, 3:14 PM
Reason: wrong data
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george_54
1585 posts
george_54
#2 h  Nov 10, 2016, 5:52 PM • 2 Y
Y by TSandino, Adventure10
TSandino wrote:
... is a point at side $BC$ so that ...

It is at side $AB$, not $BC$!
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