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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Thailand geometry
EeEeRUT   0
a minute ago
Source: Thailand MO 2025 P7
Let $ABC$ be a triangle with $AB < AC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects $BC$ at $D$. The angle bisector of $\angle BAC$ intersect $BC$ at $E$. Suppose that the perpendicular bisector of $AE$ intersect $AB, AC$ at $P,Q$, respectively. Show that $$\sqrt{\frac{BP}{CQ}} = \frac{AC \cdot BD}{AB \cdot CD}$$
0 replies
EeEeRUT
a minute ago
0 replies
ISI UGB 2025 P2
SomeonecoolLovesMaths   9
N 5 minutes ago by SatisfiedMagma
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
9 replies
SomeonecoolLovesMaths
May 11, 2025
SatisfiedMagma
5 minutes ago
inequality
xytunghoanh   6
N 21 minutes ago by sqing
For $a,b,c\ge 0$. Let $a+b+c=3$.
Prove or disprove
\[\sum ab +\sum ab^2 \le 6\]
6 replies
1 viewing
xytunghoanh
4 hours ago
sqing
21 minutes ago
Inspired by old results
sqing   2
N 23 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $ . Prove that
$$\frac{a+kb}{b+c}+\frac{b+kc}{c+a}+\frac{c+ka}{a+b}\geq \frac{3(k+1)}{2}$$W here $-1 \leq k \leq  \frac{537}{90}.$
2 replies
sqing
5 hours ago
sqing
23 minutes ago
No more topics!
Navid FE on R+
Assassino9931   2
N Apr 12, 2025 by internationalnick123456
Source: Bulgaria Balkan MO TST 2025
Determine all functions $f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that
\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]for any positive real numbers $x,y$.
2 replies
Assassino9931
Apr 9, 2025
internationalnick123456
Apr 12, 2025
Navid FE on R+
G H J
Source: Bulgaria Balkan MO TST 2025
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Assassino9931
1353 posts
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Determine all functions $f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that
\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]for any positive real numbers $x,y$.
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luutrongphuc
53 posts
#4
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very hard
This post has been edited 1 time. Last edited by luutrongphuc, Apr 12, 2025, 7:06 AM
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internationalnick123456
135 posts
#5 • 2 Y
Y by luutrongphuc, MuradSafarli
Claim 1. $4f(y)\geq y,\forall y>0$.
Proof. By hypothesis, $f(x + 4f(y)) > f(y),\forall x,y>0$.
Hence, if there exists $y>0$ s.t. $4f(y)<y$, setting $x=y-4f(y)$ in the above inequality we get $f(y)>f(y)$, absurd.
Claim 2. $4f(x)\geq f(y),\forall x>y>0$.
Proof. Applying Claim 1 to the given functional equation, we obtain $$xf(x+3y)+f(x)f(y)\geq \dfrac{f(x)f(x + 4f(y))}{4} ,\forall x,y>0$$$\Rightarrow 4f(x+3y)\geq f(x),\forall x,y>0\Rightarrow 4f(x)\geq f(y),\forall x>y>0$.
Claim 3. $\inf f = 0$.
Proof. Assume there exists a constant $c>0$ such that $f(x)\geq c,\forall x>0$.
We have $$f(x + 4f(y)) = \dfrac{x}{f(x)}f(x + 3y) + f(y),\forall x,y>0$$Since $f(x)\geq c,\forall x>0$, letting $x\to 0^+$ we have $\lim_{x\to 4f(y)^{+}}f(x)=f(y) ,\forall y>0$.
Moreover, $f(x)f(x + 4f(y))\geq xf(x + 3y) + cf(x),\forall x,y>0$
$\Rightarrow f(4f(x))f(4f(x) + 4f(y)) \geq 4f(x)f(4f(x) + 3y) + cf(4f(x)),\forall x,y>0$.
Therefore, letting $y\to 0^+$ we have $f(4f(x))f(x) \geq 4f^2(x) + cf(4f(x)),\forall x>0\Rightarrow f(4f(x))>4f(x)$.
This contradicts Claim 2 since $f(4f(x))\leq 4\lim_{y\to 4f(x)^{+}}f(y) = 4f(x)$.
Claim 4. $\lim_{x\to 0^{+}}f(x) = 0$.
Proof. For all $\varepsilon>0$, by Claim 3, there exists some $\delta >0$ s.t. $f(\delta) < \varepsilon/4$.
By Claim 2, $f(x)\leq 4f(\delta) <\varepsilon,\forall x<\delta$. Hence, $\lim_{x\to 0^+}f(x) = 0$.
Claim 5. $f(x) = x,\forall x>0$.
Proof. From the given functional equation, we obtain $$\dfrac{f(x + 4f(y))}{f(x + 3y)} = \dfrac{x}{f(x)} + \dfrac{f(y)}{f(x + 3y)},\forall x,y>0$$Hence, $\lim_{y\to 0^+}\dfrac{f(x + 4f(y))}{f(x + 3y)} = \dfrac{x}{f(x)}$. Define $$T_0(y) = y, \quad T_{n}(y) = 4f\left(\dfrac{T_{n-1}(y)}{3}\right),\forall n\in\mathbb N$$Then, $$\lim_{y\to 0^+}\dfrac{f(x + T_n(y))}{f(x+T_{n-1}(y))} =\dfrac{x}{f(x)},\forall y>0,n\in\mathbb N$$$\Rightarrow \lim_{y\to 0^{+}}\dfrac{f(x +T_n(y))}{f(x + y)} = \left(\dfrac{x}{f(x)}\right)^n,\forall y>0,n\in\mathbb N$.
Notice that as $y\to 0^+$, $f(x + y)$ is bounded below by $\dfrac{f(x)}{4}$ and above by $4f(x+1)$. Consequently, $\left(\dfrac{x}{f(x)}\right)^n$ remains bounded by a constant depending $x$. This leads to the conclusion that $f(x)=x,\forall x>0$. (qed)

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This post has been edited 4 times. Last edited by internationalnick123456, Apr 12, 2025, 12:25 PM
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