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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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[*]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
Problem 4 (second day)
darij grinberg   92
N 4 minutes ago by cubres
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
92 replies
+1 w
darij grinberg
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4 minutes ago
functional equation interesting
skellyrah   2
N 10 minutes ago by jasperE3
find all functions IR->IR such that $$xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2  + (x+1)f(x)$$
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tastymath75025   31
N an hour ago by cj13609517288
Source: 2019 ELMO Shortlist G3
Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$. The circumcircles of $ABC$ and $A'EF$ meet at $G$, and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$, where $M$ is the midpoint of $EF$. Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.

Proposed by Ankit Bisain
31 replies
tastymath75025
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an hour ago
\frac{1}{5-2a}
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Yesterday at 9:56 AM
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Vivacious_Owl   2
N 5 hours ago by Vivacious_Owl
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Could you help me with some 3D geometry books? Or any book with 3D geometry information, specially if it's focuses on math olympiads (like Putnam).
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Saucepan_man02   6
N Today at 9:10 AM by kiyoras_2001
Hello

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Valentin Vornicu   13
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Find the set of $x\in\mathbb{F}_{5^5}$ such that the equation in the unknown $y\in \mathbb{F}_{5^5}$:
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N Yesterday at 5:47 PM by loup blanc
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Yesterday at 5:47 PM
Cute inequality in equilateral triangle
Miquel-point   1
N Apr 7, 2025 by Quantum-Phantom
Source: Romanian IMO TST 1981, Day 3 P5
Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$.

Laurențiu Panaitopol
1 reply
Miquel-point
Apr 6, 2025
Quantum-Phantom
Apr 7, 2025
Cute inequality in equilateral triangle
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Source: Romanian IMO TST 1981, Day 3 P5
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Miquel-point
472 posts
#1 • 2 Y
Y by PikaPika999, kiyoras_2001
Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$.

Laurențiu Panaitopol
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Quantum-Phantom
267 posts
#2
Y by
I think we should not allow $M$ to be on the three medians of the triangle.

It suffices to prove that $A'A''+B'B''>C'C''$. By Ceva's theorem, if $\tfrac{AB'}{B'C}=a$ and $\tfrac{BA'}{A'C}=b$, then $\tfrac{AC'}{C'B}=\tfrac ab$ ($a$, $b\ne1$, $a\ne b$). Let the side length be $1$, then
\[A'A''=\left|BA'-BA''\right|=\left|\frac b{b+1}-\frac12\right|=\frac{|b-1|}{2(b+1)}.\]Similarly we obtain $B'B''=\tfrac{|a-1|}{2(a+1)}$ and $C'C''=\tfrac{|a-b|}{2(a+b)}$. Multiplying by $2$, we need to show that
\[\frac{|a-1|}{a+1}+\frac{|b-1|}{b+1}=\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}>\frac{|a-b|}{a+b}.\tag{*}\]On the one hand,
\begin{align*}
\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}&\ge\frac{|(a+1)(b-1)+(b+1)(a-1)|}{(a+1)(b+1)}\\&=\frac{2|ab-1|}{(a+1)(b+1)}\overset{(1)}>\frac{|a-b|}{a+b},
\end{align*}where
\begin{align*}
(1)&\Leftrightarrow4(ab-1)^2(a+b)^2>(a-b)^2(a+1)^2(b+1)^2\\
&\Leftrightarrow(a-1)(b-1)\left(b^2+3ab+a+3b\right)\left(a^2+3ab+b+3a\right)>0.
\end{align*}On the other hand,
\begin{align*}
\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}&\ge\frac{|(a+1)(b-1)-(b+1)(a-1)|}{(a+1)(b+1)}\\&=\frac{2|a-b|}{(a+1)(b+1)}\overset{(2)}>\frac{|a-b|}{a+b},
\end{align*}where
\[(2)\Leftrightarrow2(a+b)>(a+1)(b+1)\Leftrightarrow(a-1)(b-1)<0.\]At least one of $(1)$ and $(2)$ holds, so we are done.

Are there any other nice proofs of $(^*)$?
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