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.

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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 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
Pythagorean new journey
XAN4   4
N 3 minutes ago by XAN4
Source: Inspired by sarjinius
The number $4$ is written on the blackboard. Every time, Carmela can erase the number $n$ on the black board and replace it with a new number $m$, if and only if $|n^2-m^2|$ is a perfect square. Prove or disprove that all positive integers $n\geq4$ can be written exactly once on the blackboard.
4 replies
XAN4
Yesterday at 3:41 AM
XAN4
3 minutes ago
wu2481632 Mock Geometry Olympiad problems
wu2481632   14
N 4 minutes ago by bin_sherlo
To avoid clogging the fora with a horde of geometry problems, I'll post them all here.

Day I

Day II

Enjoy the problems!
14 replies
wu2481632
Mar 13, 2017
bin_sherlo
4 minutes ago
Straight line
uTOPi_a   19
N 12 minutes ago by NerdyNashville
Source: 41-st Vietnamese Mathematical Olympiad 2003
The circles $ C_{1}$ and $ C_{2}$ touch externally at $ M$ and the radius of $ C_{2}$ is larger than that of $ C_{1}$. $ A$ is any point on $ C_{2}$ which does not lie on the line joining the centers of the circles. $ B$ and $ C$ are points on $ C_{1}$ such that $ AB$ and $ AC$ are tangent to $ C_{1}$. The lines $ BM$, $ CM$ intersect $ C_{2}$ again at $ E$, $ F$ respectively. $ D$ is the intersection of the tangent at $ A$ and the line $ EF$. Show that the locus of $ D$ as $ A$ varies is a straight line.
19 replies
1 viewing
uTOPi_a
Aug 28, 2004
NerdyNashville
12 minutes ago
Number Theory Chain!
JetFire008   3
N 14 minutes ago by ExcitablePorcupine48
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
3 replies
JetFire008
5 hours ago
ExcitablePorcupine48
14 minutes ago
No more topics!
collinear wanted, starting with a cyclic quadr.
parmenides51   4
N Apr 11, 2022 by Mahdi_Mashayekhi
Source: 2021 Nordic MC p4
Let $A, B, C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \ne D$ be the point on $\omega$ such that $DX$ and $EF$ are parallel. Let $Y$ be the reflection of $D$ through $EF$ and suppose that $Y$ is inside the circle $\omega$.
Show that $A, X$, and $Y$ are collinear.
4 replies
parmenides51
May 15, 2021
Mahdi_Mashayekhi
Apr 11, 2022
collinear wanted, starting with a cyclic quadr.
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 Nordic MC p4
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parmenides51
30630 posts
#1
Y by
Let $A, B, C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \ne D$ be the point on $\omega$ such that $DX$ and $EF$ are parallel. Let $Y$ be the reflection of $D$ through $EF$ and suppose that $Y$ is inside the circle $\omega$.
Show that $A, X$, and $Y$ are collinear.
Z K Y
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EulersTurban
386 posts
#3 • 1 Y
Y by Mango247
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -12.503988125868464, xmax = 9.00006180677948, ymin = -6.927953789716436, ymax = 6.475915899243479;  /* image dimensions */
pen wwzzqq = rgb(0.4,0.6,0); 
 /* draw figures */
draw(circle((0,0), 4), linewidth(0.4) + red); 
draw((-8.154058564822963,-0.9219273465879328)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); 
draw((-1.612329064547513,3.660654994343955)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); 
draw((-1.612329064547513,3.660654994343955)--(-8.154058564822963,-0.9219273465879328), linewidth(0.4) + blue); 
draw((-2.889584353828225,2.7659179781965912)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); 
draw((-1.8968541003492212,-2.3923980697013247)--(1.6565885029890675,3.6408398113298586), linewidth(0.4) + blue); 
draw((-1.612329064547513,3.660654994343955)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); 
draw((-2.889584353828225,2.7659179781965912)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); 
draw((-8.154058564822963,-0.9219273465879328)--(-1.992057807278974,0.2639644283969095), linewidth(0.4) + blue); 
draw((-2.889584353828225,2.7659179781965912)--(1.6565885029890675,3.6408398113298586), linewidth(0.4) + blue); 
draw((-2.889584353828225,2.7659179781965912)--(-1.8968541003492212,-2.3923980697013247), linewidth(0.4) + blue); 
draw((-8.154058564822963,-0.9219273465879328)--(-1.8968541003492212,-2.3923980697013247), linewidth(0.4) + blue); 
draw((-1.8968541003492212,-2.3923980697013247)--(-1.992057807278974,0.2639644283969095), linewidth(0.4) + blue); 
draw(circle((-4.91109805798729,-1.1705409107014284), 3.252476218691984), linewidth(0.4) + linetype("2 2") + wwzzqq); 
draw((-1.8968541003492212,-2.3923980697013247)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); 
 /* dots and labels */
dot((-8.154058564822963,-0.9219273465879328),dotstyle); 
label("$E$", (-8.068175135277407,-0.6840648518735976), NE * labelscalefactor); 
dot((-2.380898965077216,-3.214237097367623),dotstyle); 
label("$A$", (-2.282332104071683,-2.9742943850591947), NE * labelscalefactor); 
dot((-0.47249042781922146,-3.9719960719541514),linewidth(4pt) + dotstyle); 
label("$B$", (-0.37782543963313225,-3.769847801849981), NE * labelscalefactor); 
dot((-1.612329064547513,3.660654994343955),dotstyle); 
label("$C$", (-1.5108863665775867,3.8963942144975965), NE * labelscalefactor); 
dot((-2.889584353828225,2.7659179781965912),linewidth(4pt) + dotstyle); 
label("$D$", (-2.788593369302184,2.9561947219266673), NE * labelscalefactor); 
dot((-1.992057807278974,0.2639644283969095),linewidth(4pt) + dotstyle); 
label("$F$", (-1.8966092353246349,0.4489960750708557), NE * labelscalefactor); 
dot((1.6565885029890675,3.6408398113298586),linewidth(4pt) + dotstyle); 
label("$X$", (1.7436503384756334,3.824071176607525), NE * labelscalefactor); 
dot((-1.8968541003492212,-2.3923980697013247),linewidth(4pt) + dotstyle); 
label("$Y$", (-1.8001785181378729,-2.202848647565099), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Classical geometry at its' finest!

Essentially we want to show that $EYAF$ is a cyclic quad.
Since if that is true we have that following:
$$\angle YAF = \angle YEF = \angle FED = \angle XDC = \angle XAC = \angle XAF$$
But showing that $EYAF$ is cyclic is pretty easy, since we have that:
$$\angle EAF = \angle EAD + \angle DAF = \angle DCB +\angle DBC = \angle EDB = \angle EDF = \angle EYF$$therefore proving that $EYAF$ is cyclic and in turn proving that $A,Y$ and $X$ are collinear.
Z K Y
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zuss77
520 posts
#4
Y by
Let $AD\cap BC=G$ and let $GX$ meet $(ABC)$ at $L$. Since $EF$ is a polar of $G$ and $DX\parallel EF$, we have $AL\parallel DX$. If $Z$ - center of this trapezoid, then $Z\in EF$ and $ZE$ bisects $\angle AZD.$
Z K Y
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edfearay123
92 posts
#5
Y by
Inversion on the circumference with center $E$ and ratio $\sqrt{EA\cdot EB}$
Z K Y
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Mahdi_Mashayekhi
689 posts
#7
Y by
$\angle BAX = \angle BDX = \angle DFE = \angle YFE$ so if $\angle BAY = \angle BAX$ then $AYFE$ is cyclic so we'll prove it in order to solve the Problem. $\angle EAF = \angle 180 - \angle BAC = \angle 180 - \angle BDC = \angle FDE = \angle FYE$ so $AYFE$ is cyclic.
Z K Y
N Quick Reply
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