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
Function equation
luci1337   0
11 minutes ago
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
0 replies
luci1337
11 minutes ago
0 replies
inequalities
pennypc123456789   2
N 35 minutes ago by arqady
If $a,b,c$ are positive real numbers, then
$$
\frac{a + b}{a + 7b + c} + \dfrac{b + c}{b + 7c + a}+\dfrac{c + a}{c + 7a + b} \geq \dfrac{2}{3}$$
we can generalize this problem
2 replies
+1 w
pennypc123456789
an hour ago
arqady
35 minutes ago
Coaxal Circles
fattypiggy123   29
N 41 minutes ago by sttsmet
Source: China TSTST Test 2 Day 1 Q3
Let $ABCD$ be a quadrilateral and let $l$ be a line. Let $l$ intersect the lines $AB,CD,BC,DA,AC,BD$ at points $X,X',Y,Y',Z,Z'$ respectively. Given that these six points on $l$ are in the order $X,Y,Z,X',Y',Z'$, show that the circles with diameter $XX',YY',ZZ'$ are coaxal.
29 replies
fattypiggy123
Mar 13, 2017
sttsmet
41 minutes ago
q(x) to be the product of all primes less than p(x)
orl   16
N an hour ago by Maximilian113
Source: IMO Shortlist 1995, S3
For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.
16 replies
orl
Aug 10, 2008
Maximilian113
an hour ago
No more topics!
Locus of centers.
MexicOMM   4
N Jan 20, 2015 by utkarshgupta
Let $ABC$ be an acute triangle.
Find the locus of the centers of the rectangles which have their vertices on the sides of $ABC$.
4 replies
MexicOMM
Jul 19, 2014
utkarshgupta
Jan 20, 2015
Locus of centers.
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MexicOMM
292 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an acute triangle.
Find the locus of the centers of the rectangles which have their vertices on the sides of $ABC$.
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Luis González
4147 posts
#2 • 3 Y
Y by MexicOMM, Adventure10, Mango247
Label $PQRS$ the vertices of these rectangles and $K \equiv PR \cap QS$ their centers. There are exactly two possibilities to inscribe $PQRS$ in $\triangle ABC,$ namely two consecutive vertices $P,S$ on $BC$ or two opposite vertices $P,R$ on $BC$ and cyclically for $CA$ and $AB.$ The former case trivially gives the locus of $K$ as the union of the sidelines $BC,CA,AB.$

For the remaining case, animate $Q,R$ on $AB,AC.$ The application sending $Q$ to the midpoint $K$ of $PR$ is an affine homography, even for any fixed directions, not necessarily $QR \parallel BC$ and $QP \perp BC$ (check the general configuration at Beautiful locus 3), therefore $K$ runs on a line. Considering limiting cases $Q \equiv R \equiv A$ and $(Q \equiv B,R \equiv C),$ it follows that this line passes through the midpoint of $BC$ and the midpoint of the A-altitude.

Hence summing up, the locus of $K$ is the union of the 3 sidelines of $\triangle ABC$ and the 3 lines joining the midpoint of each side with the midpoint of its correponding altitude.
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MexicOMM
292 posts
#3 • 2 Y
Y by Adventure10, Mango247
Additional question. Is it possible that one point is the center of 3 rectangles?
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Luis González
4147 posts
#4 • 4 Y
Y by MexicOMM, mhq, Adventure10, Mango247
MexicOMM, yes it is the symmedian point L of ABC. This follows from the fact the lines joining the midpoints of the sides with the midpoints of their correponding altitudes are concurrent.

Let $\triangle A'B'C'$ be the tangential triangle of $\triangle ABC.$ $L \equiv AA' \cap BB' \cap CC'$ is the symmedian point of $\triangle ABC.$ $X$ is the foot of the A-altitude and $D,M$ are the midpoints of $BC,AX,$ resp. $DM$ is the D-median of $\triangle DAX$ and $DA' \parallel AX$ (both perpendicular to BC) $\Longrightarrow$ $D(A,X,M,A')=-1.$ Hence, if $AA'$ cuts $DM,BC$ at $L',U,$ we get $(A,U,L',A')=-1,$ but from the complete quadrilateral $BCB'C',$ we have $(A,U,L,A')=-1$ $\Longrightarrow$ $L \equiv L',$ i.e. $DM$ passes through $L.$ Similarly the lines joining the midpoints of $CA$ and $AB$ with midpoints of the B- and C- altitude pass through $L.$
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utkarshgupta
2280 posts
#5 • 2 Y
Y by Adventure10, Mango247
Canadian MO 2006
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