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)!

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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
Inequality from my inequality training.
Orkhan-Ashraf_2002   2
N 7 minutes ago by sqing
Let $a,b,c$ non-negative real numbers,but $ab+bc+ca\not=$0.Prove that
\[1\leq \frac{a+b}{a+4b+c}+\frac{b+c}{b+4c+a}+\frac{c+a}{c+4a+b}\leq \frac{4}{3}\]
2 replies
Orkhan-Ashraf_2002
Aug 21, 2016
sqing
7 minutes ago
inequalities
pennypc123456789   3
N 19 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
3 replies
2 viewing
pennypc123456789
2 hours ago
arqady
19 minutes ago
Using Humpty point on Trapezoid ??
FireBreathers   1
N 27 minutes ago by aidenkim119
Given a trapezoid $ABCD$ with $AD//BC$. Let point $H$ be orthocenter $ABD$ and $M$ midpoint $AD$. It is also known that $HC$ perpendicular to $BM$. Let $X$ be a point on the segment $AB$ such that $XH=BH$ and point $Y$ be the intersection of $CX$ and $BD$. Prove that $AXYD$ concyclic
1 reply
FireBreathers
4 hours ago
aidenkim119
27 minutes ago
IMO 2014 Problem 2
v_Enhance   60
N 38 minutes 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
38 minutes ago
No more topics!
Iran NMO 2008 (Second Round) - Problem6
sororak   6
N May 7, 2017 by andrei.pantea
In triangle $ABC$, $H$ is the foot of perpendicular from $A$ to $BC$. $O$ is the circumcenter of $\Delta ABC$. $T,T'$ are the feet of perpendiculars from $H$ to $AB,AC$, respectively. We know that $AC=2OT$. Prove that $AB=2OT'$.
6 replies
sororak
Sep 22, 2010
andrei.pantea
May 7, 2017
Iran NMO 2008 (Second Round) - Problem6
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sororak
337 posts
#1 • 2 Y
Y by Adventure10, Mango247
In triangle $ABC$, $H$ is the foot of perpendicular from $A$ to $BC$. $O$ is the circumcenter of $\Delta ABC$. $T,T'$ are the feet of perpendiculars from $H$ to $AB,AC$, respectively. We know that $AC=2OT$. Prove that $AB=2OT'$.
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jgnr
1343 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $M$ be the midpoint of $AC$, also let $O'$ be the point on $AO$ such that $AH=AO'$ and $T"$ be the point such that $O'T"=OT$ and $AT"\perp T"H$. Hence $\triangle AMO\cong\triangle AT"H$, hence $AM=AT"$, which implies $\triangle AT"O'\cong\triangle AMH$, therefore $\angle O'AT"=\angle HAM$. But $\angle OAT=\angle HAM$, therefore $\angle OAT"=\angle OAT$. So $T"$ and $T$ are points such that $\angle OAT"=\angle OAT$ and $\angle AT"H=\angle ATH$, thus $T"=T$. Now $O'T=OT$ and $\angle TAO=\angle TAO'$, so $O'=O$. Therefore we conclude that $AO=AH$. Let $N$ be the midpoint of $AB$. We have $\triangle ANO\cong\triangle AT'H$, thus $OT'=HN=AN=\frac12 AB$, as desired.
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Virgil Nicula
7054 posts
#3 • 1 Y
Y by Adventure10
Sorry, exist two points T'' which belong to the circle with diameter $AH$ and $O'T''=OT$ .
I didn't understand your proof. Maybe enter in some details. Thank you.
I have a proof with the power of a point w.r.t. circumcircle of $ABC$ .
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jgnr
1343 posts
#4 • 2 Y
Y by Adventure10, Mango247
From $\angle OAT"=\angle OAT$ we know that $A,T,T"$ are collinear. So $T$ and $T"$ are intersections of line $AT$ and circle with diameter $AH$, so they must coincide because the other intersection is point $A$.
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Virgil Nicula
7054 posts
#5 • 1 Y
Y by Adventure10
Quote:
Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ . Denote the midpoints $E$ , $F$

of $[AC]$ , $[AB]$ respectively, the projection $D$ of $A$ on $BC$ and the projections $X$ , $Y$ of $D$

on $AB$ , $AC$ respectively. Prove that $OX=AE\iff OY=AF\iff h_a=R$ .

Proof of PP1. Since $\begin{array}{ccccc}
DX^2=XA\cdot XB & ; & DX=h_a\cos B & ; & OE=R\cos B\\\
DY^2=YA\cdot YC & ; & DY=h_a\cos C & ; & OF=R\cos C\end{array}$

obtain that $\left\|\begin{array}{cccc}
XA\cdot XB=R^2-OX^2 & \implies & OX^2=R^2-DX^2\\\
YA\cdot YC=R^2-OY^2 & \implies & OY^2=R^2-DY^2\end{array}\right\|$ . Thus,

$OX=AE \iff  R^2-AE^2=DX^2 $ $\iff OE=DX \iff  h_a=R\ .$

$OY=AF  \iff  R^2-AF^2=DY^2 $ $\iff OF=DY \iff h_a=R\ .$

Remark. $h_a=R\iff bc=2R^2\iff \sin B\sin C=\frac 12$ .
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reveryu
218 posts
#6 • 2 Y
Y by Adventure10, Mango247
Virgil wrote:
$R^2-AE^2=DX^2 $ $\iff OE=DX$

How?
This post has been edited 1 time. Last edited by reveryu, May 7, 2017, 8:10 AM
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andrei.pantea
172 posts
#7 • 4 Y
Y by GGPiku, reveryu, Adventure10, Mango247
Pitagora
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