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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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0 replies
jlacosta
May 1, 2025
0 replies
A tangent problem
hn111009   0
7 minutes ago
Source: Own
Let quadrilateral $ABCD$ with $P$ be the intersection of $AC$ and $BD.$ Let $\odot(APD)$ meet again $\odot(BPC)$ at $Q.$ Called $M$ be the midpoint of $BD.$ Assume that $\angle{DPQ}=\angle{CPM}.$ Prove that $AB$ is the tangent of $\odot(APD)$ and $BC$ is the tangent of $\odot(AQB).$
0 replies
hn111009
7 minutes ago
0 replies
Inspired by Bet667
sqing   4
N 10 minutes ago by ytChen
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
4 replies
sqing
Thursday at 1:03 PM
ytChen
10 minutes ago
3-var inequality
sqing   4
N 10 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
4 replies
sqing
May 7, 2025
sqing
10 minutes ago
Inspired by Kosovo 2010
sqing   2
N 12 minutes ago by sqing
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
2 replies
sqing
Yesterday at 3:56 AM
sqing
12 minutes ago
Triangle on a tetrahedron
vanstraelen   2
N Yesterday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Yesterday at 2:43 PM
ReticulatedPython
Yesterday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Yesterday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Yesterday at 3:08 PM
vanstraelen
Yesterday at 6:33 PM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   3
N Yesterday at 5:13 PM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
3 replies
parmenides51
Dec 4, 2023
jasperE3
Yesterday at 5:13 PM
Geometry
AlexCenteno2007   3
N Yesterday at 4:18 PM by AlexCenteno2007
Let ABC be an acute triangle and let D, E and F be the feet of the altitudes from A, B and C respectively. The straight line EF and the circumcircle of ABC intersect at P such that F is between E and P, the straight lines BP and DF intersect at Q. Show that if ED = EP then CQ and DP are parallel.
3 replies
AlexCenteno2007
Apr 28, 2025
AlexCenteno2007
Yesterday at 4:18 PM
Cube Sphere
vanstraelen   4
N Yesterday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Yesterday at 1:10 PM
pieMax2713
Yesterday at 2:37 PM
Square number
linkxink0603   3
N Yesterday at 2:36 PM by Zok_G8D
Find m is positive interger such that m^4+3^m is square number
3 replies
linkxink0603
Yesterday at 11:20 AM
Zok_G8D
Yesterday at 2:36 PM
Combinatorics
AlexCenteno2007   0
Yesterday at 2:05 PM
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
0 replies
AlexCenteno2007
Yesterday at 2:05 PM
0 replies
How many pairs
Ecrin_eren   6
N Yesterday at 12:57 PM by Ecrin_eren


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



6 replies
Ecrin_eren
May 2, 2025
Ecrin_eren
Yesterday at 12:57 PM
parallelogram in a tetrahedron
vanstraelen   1
N Yesterday at 12:19 PM by vanstraelen
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
1 reply
vanstraelen
May 5, 2025
vanstraelen
Yesterday at 12:19 PM
Find max
tranlenhanhbnd   0
Yesterday at 11:50 AM
Let $x,y,z>0$ and $x^2+y^2+z^2=1$. Find max
$D=\dfrac{x}{\sqrt{2 y z+1}}+\dfrac{y}{\sqrt{2 z x+1}}+\dfrac{z}{\sqrt{2 x y+1}}$.
0 replies
tranlenhanhbnd
Yesterday at 11:50 AM
0 replies
Hard FE R^+
DNCT1   5
N Apr 21, 2025 by jasperE3
Find all functions $f:\mathbb{R^+}\to\mathbb{R^+}$ such that
$$f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$$
5 replies
DNCT1
Dec 30, 2020
jasperE3
Apr 21, 2025
Hard FE R^+
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DNCT1
235 posts
#1 • 2 Y
Y by toanhocmuonmau123, Mango247
Find all functions $f:\mathbb{R^+}\to\mathbb{R^+}$ such that
$$f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$$
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DNCT1
235 posts
#2 • 2 Y
Y by Mango247, Mango247
I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?
This post has been edited 1 time. Last edited by DNCT1, Dec 31, 2020, 1:29 AM
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TuZo
19351 posts
#3
Y by
DNCT1 wrote:
I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?

How you proved this?
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Mgh
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#4
Y by
TuZo wrote:
DNCT1 wrote:
I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?

How you proved this?

Consider there is a positive integer x fod which $ x-f (x) $ is greter than zero, now $ p (x, x-f (x) $gives contradiction. I thinks it's good to write it:$ f (x)=x+g (x) $
This post has been edited 1 time. Last edited by Mgh, Dec 31, 2020, 4:23 PM
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DNCT1
235 posts
#5
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Well, here's a solution which I have
Let $P(x,y)$ be the assertion of $f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$

Assume there exist $x_0\in\mathbb{R^+}$ such that $f(x_0)<x_0$ so
$$P(x_0,x_0-f(x_0))\implies f(x_0-f(x_0))=0\quad(\text{impossible})$$Hence we have $f(x)\geq x\quad\forall x\in\mathbb{R^+}$.

By $P(x,y)$ there exxists $a,b\in\mathbb{R^+}$ such that
$$f(x+a)=f(x)+b\quad\forall x\in\mathbb{R^+}.$$So we also have $f(x+na)=f(x)+nb\quad\forall x\in\mathbb{R^+},n\in\mathbb{N^+}$
$$P(x+a,y)\implies f(3x+3a+f(x)+b+y)=f(4x)+f(y)+4b\quad\forall x,y\in\mathbb{R^+}$$$$P(x,y+4a)\implies f(3x+f(x)+y+4a)=f(4x)+f(y)+4b\quad\forall x,y\in\mathbb{R^+}$$And by two equations we have
$$f(3x+3a+f(x)+b+y)=f(3x+f(x)+y+4a)\quad\forall x,y\in\mathbb{R^+}$$If $a\neq b$ we assume that $a<b$ and $y$ such that $y>4a+3+f(1)$
$$P(1,y-4a-3-f(1))\implies f(y)=f(y+b-a)\quad\forall y\in\mathbb{R^+},y>4a+3+f(1)$$Hence $f$ must be period $a-b$ with $x>4a+3+f(1)$ but we have $f(x)\geq x\quad\forall x\in\mathbb{R^+}$ that implies $f$ must be period $0$ (absurd).$\quad$ Hence $a=b$.
Otherwise we have $f(h)=f(h+n(b-a))\quad\forall n\in\mathbb{N^+}$ for some $h>4a+3+f(1)$ so $d=f(h)\geq h+n(b-a)$ so choose $n\in\mathbb{N^+}$ such that $n>\frac{d}{b-a}$ we get the contracdition.
And so we have
$$f(4x)=f(x)+3x\quad\forall x\in\mathbb{R^+}$$$P(x,y)$ becomes
$$f(f(x)+y)=f(x)+f(y)\quad\forall x,y\in\mathbb{R^+}$$$$P(x,y-3x)\implies f(f(x)+y)=f(4x)+f(y-3x)=f(x)+3x+f(y-3x)\quad\forall x,y\in\mathbb{R^+}, y>3x$$And so $$3x+f(y-3x)=f(y)\quad\forall x,y\in\mathbb{R^+}, y>3x$$$$\implies f(y)-f(y-x)=x\quad\forall x,y\in\mathbb{R^+}, y>x (1)$$Let $y\to x+1$ in $(1)$ we have $f$ is linear $\forall x>1$ or $f(x)=x+f(1)-1\quad\forall x>1$
By $(1)$ we let $(x,y)\to (1-x,1)$ we have $f(x)=x+f(1)-1\quad\forall 0<x<1$ and it's also true with $x=1$.
And so
$$\boxed{\text{Solution:}\quad f(x)=x+f(1)-1\quad\forall x\in\mathbb{R^+}}$$
This post has been edited 2 times. Last edited by DNCT1, Jan 1, 2021, 5:02 PM
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jasperE3
11317 posts
#6 • 1 Y
Y by truongphatt2668
Let $P(x,y)$ be the assertion $f(3x+f(x)+y)=f(4x)+f(y)$.

Claim 1: $f(x)\ge x$ for all $x$
Suppose $f(u)<u$ for some $u$, then:
$P(u,u-f(u))\Rightarrow f(u-f(u))=0$ which is impossible.

Claim 2: if $f(x+a)=f(x)+b$ for all $x>0$, given some $a,b>0$, then $f(x+b)=f(x)+b$
Note that $f(x+na)=f(x)+nb$ for $n\in\mathbb N$, by simple induction. Then:
$P(x+a,y)\Rightarrow f(3x+f(x)+y+b)=f(4x)+f(y)+b$
$P(x,y+b)\Rightarrow f(3x+f(x)+y+b)=f(4x)+f(y+b)$
and the claim is proven.

Claim 3: $f$ is injective.
Suppose $f(a)=f(b)$ for some $a>b>0$.
$P(x,a)\Rightarrow f(3x+f(x)+a)=f(4x)+f(a)$
$P(x,b)\Rightarrow f(3x+f(x)+b)=f(4x)+f(a)$
Fix $x$ and apply claim $2$ to $P(x,y)$, we get $f(f(4x)+y)=f(4x)+f(y)$, which implies $f(f(x)+y)=f(x)+f(y)$. Then:
$$f(x)+f(3x+a)=f(3x+f(x)+a)=f(3x+f(x)+b)=f(x)+f(3x+b)$$so $f(3x+a)=f(3x+b)$. This transforms to $f(x+a-b)=f(x)$ for all $x>b$. Let $x>b$, then:
$P(x,y+a-b)\Rightarrow f(3x+f(x)+y)=f(3x+f(x)+y+a-b)=f(4x)+f(y+a-b)\Rightarrow f(y)=f(y+a-b)$ so $f$ is periodic in general
By induction, $f(x)=f(x+n(a-b))$ for $n\in\mathbb N$, but by Claim 1 we have:
$$f(x)\ge x+n(a-b),$$taking $n\to\infty$ gives a contradiction.

Finish:
Recall that $f(f(x)+y)=f(x)+f(y)$, so swapping $x,y$ and using injectivity gives $\boxed{f(x)=x+c}$ which works for any constant $c>0$.
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