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
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
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
Diophantine eq.
User335559   12
N 8 minutes ago by Burak0609
Source: European Mathematical Cup 2017
Solve in integers the equation :
$x^2y+y^2=x^3$
12 replies
User335559
Jan 3, 2018
Burak0609
8 minutes ago
floor of (an+b)/(cn+d) is surjective
Miquel-point   3
N 13 minutes ago by Rohit-2006
Source: Romanian NMO 2021 grade 10 P2
Let $a,b,c,d\in\mathbb{Z}_{\ge 0}$, $d\ne 0$ and the function $f:\mathbb{Z}_{\ge 0}\to\mathbb Z_{\ge 0}$ defined by
\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]Prove that the following are equivalent:
[list=1]
[*] $f$ is surjective;
[*] $c=0$, $b<d$ and $0<a\le d$.
[/list]

Tiberiu Trif
3 replies
Miquel-point
Apr 15, 2023
Rohit-2006
13 minutes ago
evan chen??
Captainscrubz   1
N 16 minutes ago by Captainscrubz
Let point $D$ and $E$ be on sides $AB$ and $AC$ respectively in $\triangle ABC$ such that $BD=BC=CE$. Let $O_1$ be the circumcenter of $\triangle ADE$ and let $S=DC\cap EB$. Prove that $O_1S \perp BC$
1 reply
Captainscrubz
Today at 3:50 AM
Captainscrubz
16 minutes ago
SMO 2015 open q3
dominicleejun   13
N 18 minutes ago by jasperE3
Source: SMO 2015 open
Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$, where $\mathbb{R}$ is the set of real numbers, such that
$f(x)f(yf(x) - 1) = x^2 f(y) - f(x) \quad\forall x,y \in \mathbb{R}$
13 replies
dominicleejun
Mar 31, 2018
jasperE3
18 minutes ago
No more topics!
Polish MO Final 2010, 3rd problem (parallelogram and circle)
TomciO   4
N Feb 21, 2015 by JuanOrtiz
$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.
4 replies
TomciO
Nov 7, 2010
JuanOrtiz
Feb 21, 2015
Polish MO Final 2010, 3rd problem (parallelogram and circle)
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TomciO
552 posts
#1 • 2 Y
Y by Adventure10, Mango247
$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.
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jayme
9775 posts
#2 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
a shape of my synthetic proof.
1. (1) the circle passing through A, B, P, D
(2) the circle passing through CPQ
X the second point of intersection of (1) and (2)
Y the second point of intersection of CD and (1)
(3) the circle passing through C, X, Y,
T the antipole of A wrt (1).
2. According to Reim's theorem applied to (3) and (1), AX and AY are tangent to (3) at X, Y resp..
3. Consequently * T is the center of (3)
* TX = TC
* DT is the D-altitude of the triangle CDX
* DT is the A-mediator of CX
4. CX being the common chord of (2) and (3), DT goes through the center of (2)
... and we are done.
Sincerely
Jean-Louis
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hatchguy
555 posts
#3 • 3 Y
Y by Lazizbek1435, Adventure10, Mango247
Usually polish number 3 are harder...

Let $\angle DAP = \alpha$ and $\angle PAB = \beta$.

It is enough to show $DO^2 + AD^2 = AO^2$.

By power of a point $DO^2 = DQ \cdot DC - R^2$ and $AO^2 = AQ \cdot AP - R^2$.

So we must prove $AD^2 + DQ \cdot CD = AQ \cdot AP \Leftrightarrow AD^2 + DQ \cdot AB = AQ \cdot AP$. Dividing by $DQ$ and using law of sines we must prove

$\frac{AD \sin \beta}{\sin \alpha} +AB = \frac{AP \sin \alpha+\beta}{\sin \alpha}$

$\Leftrightarrow \frac{AD \cdot BP}{DP} + AB = \frac{AP \cdot BD}{DP}$

which is ptolomy's theorem
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sunken rock
4381 posts
#4 • 4 Y
Y by Lazizbek1435, Adventure10, Mango247, and 1 other user
Take $\{\ R\ \}\in BC\cap\odot (ABD)$. $A, R, B, D$ determine a a cyclic trapezoid, hence $DR=AB(=CD)$, i.e. $\triangle CRD$ is isosceles, so, obviously, we need to prove $P, R, C, Q$ are con-cyclic, which is obvious: $\angle QPR=\angle APR=\angle ABR=180^\circ-\angle BCD=180^\circ-\angle BCQ$, done.

Best regards,
sunken rock
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JuanOrtiz
366 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let X be such that ABXD is isosceles trapezoid. Notice <BXP=<PAB=<PQC so PXCQ is cyclic. So O lies on the bisector of XC, which is the perpendicular to AD through D.
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