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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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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
IMO 2014 Problem 2
v_Enhance   60
N a minute ago by math-olympiad-clown
Source: 0
Let $n \ge 2$ be an integer. Consider an $n \times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there is a $k \times k$ square which does not contain a rook on any of its $k^2$ unit squares.
60 replies
v_Enhance
Jul 8, 2014
math-olympiad-clown
a minute ago
Counting friends in two ways
joybangla   18
N 5 minutes ago by Mathworld314
Source: ISI Entrance 2014, P1
Suppose a class contains $100$ students. Let, for $1\le i\le 100$, the $i^{\text{th}}$ student have $a_i$ many friends. For $0\le j\le 99$ let us define $c_j$ to be the number of students who have strictly more than $j$ friends. Show that \begin{align*} & \sum_{i=1}^{100}a_i=\sum_{j=0}^{99}c_j \end{align*}
18 replies
joybangla
May 11, 2014
Mathworld314
5 minutes ago
Function equation
luci1337   0
23 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
23 minutes ago
0 replies
inequalities
pennypc123456789   2
N an hour 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
2 viewing
pennypc123456789
2 hours ago
arqady
an hour ago
No more topics!
right angle wanted, 3 circles related
parmenides51   2
N Oct 24, 2021 by ike.chen
Source: 2016 Saudi Arabia BMO TST , level 4, II p2
A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\angle OMB = 90^o$.
2 replies
parmenides51
Jul 26, 2020
ike.chen
Oct 24, 2021
right angle wanted, 3 circles related
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G H BBookmark kLocked kLocked NReply
Source: 2016 Saudi Arabia BMO TST , level 4, II p2
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parmenides51
30629 posts
#1 • 1 Y
Y by Mango247
A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\angle OMB = 90^o$.
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JustinLee2017
1703 posts
#3
Y by
Let $X'$ denote the image of $X$. An inversion $\Psi$ around $B$ causes $A',B',C'$ to be collinear, and $K'M'N'$ to be collinear. Then, $\Psi$ also sends the circumcircle of $ACNK$ to the circumcircle $\gamma$ of $A'C'N'K'$. As such, $O'$ lies on the midpoint of the tangents from $\gamma$ to $B$.
$M'$ lies on the polar of $B$, hence $\angle M'O'B = 90^{\circ}$. Inverting back implies the result. $\quad \blacksquare$

[asy]
import graph; size(10cm); 
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 = -9.627272727272732, xmax = 11.245454545454548, ymin = -6.2381818181818245, ymax = 7.652727272727277;  /* image dimensions */
pen ffxfqq = rgb(1,0.4980392156862745,0); pen yqqqqq = rgb(0.5019607843137255,0,0); 
 /* draw figures */
draw(circle((-1.1727272727272753,0.16181818181818255), 3.4075754207256947), linewidth(1) + ffxfqq); 
draw((-3.2454545454545487,5.052727272727277)--(-4.136876511792053,-1.51907132570026), linewidth(1) + red); 
draw((-4.136876511792053,-1.51907132570026)--(1.1909090909090891,-2.292727272727274), linewidth(1) + red); 
draw((1.1909090909090891,-2.292727272727274)--(-3.2454545454545487,5.052727272727277), linewidth(1) + red); 
draw(circle((-0.9928108389069386,1.4008136157456776), 4.290789790551068), linewidth(1) + ffxfqq); 
draw(circle((-3.4253709792748843,3.8137318387997823), 1.2519902589284826), linewidth(1) + red); 
draw((-4.662867079350029,3.6237765526121537)--(-1.1727272727272753,0.16181818181818255), linewidth(1) + yqqqqq); 
draw((-4.662867079350029,3.6237765526121537)--(-3.2454545454545487,5.052727272727277), linewidth(1) + yqqqqq); 
 /* dots and labels */
dot((-1.1727272727272753,0.16181818181818255),dotstyle); 
label("$O$", (-1.1,0.3436363636363624), NE * labelscalefactor); 
dot((1.1909090909090891,-2.292727272727274),dotstyle); 
label("$A$", (1.3,-2.7290909090909126), NE * labelscalefactor); 
dot((-4.136876511792053,-1.51907132570026),dotstyle); 
label("$C$", (-4.063636363636367,-1.3290909090909115), NE * labelscalefactor); 
dot((-3.2454545454545487,5.052727272727277),dotstyle); 
label("$B$", (-3.172727272727276,5.234545454545457), NE * labelscalefactor); 
dot((-2.2450261848480517,3.3962803150017655),linewidth(4pt) + dotstyle); 
label("$K$", (-2.1727272727272755,3.543636363636365), NE * labelscalefactor); 
dot((-3.5820063289067603,2.57157847640949),linewidth(4pt) + dotstyle); 
label("$N$", (-3.9909090909090943,2.8709090909090915), NE * labelscalefactor); 
dot((-4.662867079350029,3.6237765526121537),linewidth(4pt) + dotstyle); 
label("$M$", (-5.318181818181822,3.6890909090909103), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
This post has been edited 3 times. Last edited by JustinLee2017, Oct 23, 2021, 4:16 AM
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ike.chen
1162 posts
#4
Y by
https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems See problem 5 from here.

The problem is also trivial by Master Miquel.
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