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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
Disjoint Pairs
MithsApprentice   42
N 24 minutes ago by endless_abyss
Source: USAMO 1998
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
42 replies
MithsApprentice
Oct 9, 2005
endless_abyss
24 minutes ago
FE with gcd
a_507_bc   8
N 26 minutes ago by Tkn
Source: Nordic MC 2023 P2
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $$\gcd(f(x),y)f(xy)=f(x)f(y)$$for all positive integers $x, y$.
8 replies
+1 w
a_507_bc
Apr 21, 2023
Tkn
26 minutes ago
2014 JBMO Shortlist G1
parmenides51   19
N 36 minutes ago by tilya_TASh
Source: 2014 JBMO Shortlist G1
Let ${ABC}$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$. Prove that $BD+DA=BC$.
19 replies
parmenides51
Oct 8, 2017
tilya_TASh
36 minutes ago
Stars and bars i think
RenheMiResembleRice   1
N an hour ago by NicoN9
Source: Diao Luo
Solve the following attached with steps
1 reply
RenheMiResembleRice
an hour ago
NicoN9
an hour ago
Square number
linkxink0603   4
N 5 hours ago by pooh123
Find m is positive interger such that m^4+3^m is square number
4 replies
linkxink0603
Yesterday at 11:20 AM
pooh123
5 hours ago
Inequalities
sqing   7
N Today at 2:01 AM by sqing
Let $ a,b>0, a^2+ab+b^2 \geq 6  $. Prove that
$$a^4+ab+b^4\geq 10$$Let $ a,b>0, a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$a^4+ab+b^4  \leq 10$$Let $ a,b>0,  a^2+ab+b^2 \geq \frac{15}{2}  $. Prove that
$$ a^4-ab+b^4\geq 10$$Let $ a,b>0,  a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$-\frac{1}{8}\leq  a^4-ab+b^4\leq 10$$
7 replies
sqing
Thursday at 2:42 PM
sqing
Today at 2:01 AM
Compilation of functions problems
Saucepan_man02   2
N Today at 12:45 AM by Saucepan_man02
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
2 replies
Saucepan_man02
May 7, 2025
Saucepan_man02
Today at 12:45 AM
How many triangles
Ecrin_eren   5
N Today at 12:10 AM by jasperE3


"Inside a triangle, 2025 points are placed, and each point is connected to the vertices of the smallest triangle that contains it. In the final state, how many small triangles are formed?"


5 replies
Ecrin_eren
May 2, 2025
jasperE3
Today at 12:10 AM
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
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
Can this sequence be bounded?
darij grinberg   70
N Thursday at 8:26 PM by ezpotd
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
70 replies
darij grinberg
Jan 19, 2005
ezpotd
Thursday at 8:26 PM
Can this sequence be bounded?
G H J
G H BBookmark kLocked kLocked NReply
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
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darij grinberg
6555 posts
#1 • 8 Y
Y by Davi-8191, Adventure10, jhu08, megarnie, HWenslawski, ImSh95, Mango247, and 1 other user
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
This post has been edited 1 time. Last edited by djmathman, Sep 27, 2015, 2:12 PM
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Valentin Vornicu
7301 posts
#2 • 6 Y
Y by jhu08, megarnie, Adventure10, AlphaBetaGammaOmega, ImSh95, Mango247
Yes, actually this is a Romanian proposal by Mihai Bălună. Have fun with this nice problem.
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Peter Scholze
644 posts
#3 • 5 Y
Y by Adventure10, jhu08, megarnie, ImSh95, Mango247
? yes?
there are thousands of proofs given(in particular, by darij and me), that this sequence is always unbounded.
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darij grinberg
6555 posts
#4 • 4 Y
Y by jhu08, Adventure10, ImSh95, Mango247
Well, it's not exactly that I didn't give a proof, but it wasn't quite correct either... So, the problem is not too trivial...

darij
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zhaobin
2382 posts
#5 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
I think a_i must be position.

BTW,I am intrested in this problem.can you post the solution.
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Xell
40 posts
#6 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
First, some basic facts: all the terms of the sequence are positive, two consecutive terms are always different, we have $u_{n+2}=u_{n+1}-u_{n}$ or $u_{n+2}=u_{n+1}+u_{n}$, but if $u_{n}>u_{n+1}$, then necessarily, $u_{n+2}=u_{n+1}+u_{n}$.

I will then prove the following result: if $u_{n}<u_{n+1}$ for some integer n, then for some $m>n$ we have $u_{n+1}<u_{m}$ and $u_{n}+u_{n+1} < u_{m+1}$.

Property 1: let a,b be two positive real numbers with $a<b$ and suppose that $u_{n}=b-a$,$u_{n+1}=2kb-(2k-1)a$ for some integer k>0. If $u_{n+2}=u_{n+1}-u_{n}$ then we have $u_{n+2}<u_{n+1}$ so, from a previous remark, we have necessarily, $u_{n+3}=u_{n+1}+u_{n+2}$. We check that in that case, $u_{n+2}>b$ and $u_{n+3}>a+b$.

Property 2: let a,b be two positive real numbers with $a<b$ and suppose that $u_{n}=2kb-(2k-1)a$,$u_{n+1}=(2k+1)b-2ka$ for some integer k>0. If $u_{n+2}=u_{n+1}+u_{n}$, then We check that $u_{n+1}> b$ and $u_{n+2}> a+b$.

Suppose now that $a=u_{n}<u_{n+1}=b$. If $u_{n+2}=a+b$, the result is proved. So suppose that $u_{n+2}=b-a$. Then necessarily, $u_{n+3}=u_{n+2}+u_{n+1}=2b-a$. Property 1 (with k=1) shows that the result is proved if $u_{n+4}=u_{n+3}-u_{n+2}$, so suppose that $u_{n+4}=u_{n+3}+u_{n+2}=3b-2a$. Property 2 (with k=1) proves the result if $u_{n+5}=u_{n+4}+u_{n+3}$ so suppose that $u_{n+5}=u_{n+4}-u_{n+3}=b-a$.$u_{n+4}>u_{n+5}$, so $u_{n+6}=u_{n+5}+u_{n+4}=4b-3a$. And so on (= induction) we show that the result is proved by using property 1 and 2 unless we have the following sequence:
a,b,b-a,2b-a,3b-2a,b-a,4b-3a,5b-4a,...,b-a,2kb-(2k-1)a,(2k+1)b-2ka,b-a,...
But in that last case, 2kb-(2k-1)a, can be as high as we want, for example higher than b, and (2k+1)b-2ka higher than a+b.
The result is therefore proved.

Now, there are always consecutive terms such that $u_{n}<u_{n+1}$ (if $u_{n}>u_{n+1}$ for some n, then $u_{n+2}=u_{n}+u_{n+1}>u_{n+1}$). Name $a=u_{n}<u_{n+1}=b$. Then an easy induction using the result shows that one can find terms higher than $k(a+b)$ for every integer k, which shows that such a sequence can not be bounded.
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k2c901_1
146 posts
#7 • 3 Y
Y by Adventure10, jhu08, ImSh95
This problem was used as problem 1 of the final exam of the 3rd TST of Taiwan 2005.
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indybar
398 posts
#8 • 5 Y
Y by ultralako, Adventure10, jhu08, ImSh95, Mango247
I'm not really clear with Xell's solution. Can anybody elaborate?
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bomb
365 posts
#9 • 3 Y
Y by Adventure10, jhu08, ImSh95
Wasnt this posted before???

Bomb
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Philip_Leszczynski
327 posts
#10 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
Since $a_n = |a_{n+1}-a_{n+2}|$, we see that all elements of $\{a_n\}$ are nonnegative.
We first show that all elements of $\{a_n\}$ are positive.

Suppose for the sake of contradiction that there exists an $i \ge 0$ such that $a_{i+2}=0$.
(Obviously $a_0,a_1 \ne 0$ since this is given.)
$a_i = |a_{i+1} - a_{i+2}|$, so $a_i = a_{i+1}$. Let $x=a_i=a_{i+1}$.
We will show that for any $j \ge 0$, the set $\{a_k|j \le k \le i_1 \}$ has only two distinct elements: $x$ and $0$.
For the base case we have $j=i$. We will induct backwards on $j$, from $i$ to $0$.
By the inductive hypothesis, $a_{j+1} \epsilon \{0,x\}$, $a_{j+2} \epsilon \{0,x\}$.
Since $a_j = |a_{j+1} - a_{j+2}|$, $a_j \epsilon \{|0-0|,|0-x|,|x-0|,|x-x|\} = \{0,x\}$.
(Remember that $x$ is nonnegative.)
This completes the induction. So $a_0 \epsilon \{0,x\}$ and $a_1 \epsilon \{0,x\}$.
Since we are given that $a_0$ and $a_1$ are positive, $a_0=a_1=x$. (If $x=0$, then we have our contradiction.)
But we are also given that $a_0$ and $a_1$ are different, so we have our contradiction.
Therefore, all elements of $\{a_n\}$ are positive, and as a corollary, no two consecutive elements of $\{a_n\}$ are equal.

We will now show that $a_0$ cannot be the largest element of $\{a_n\}$.
Suppose for the sake of contradiction that it is.
Let $a_0 = x/k$, and let $a_1 = 1/k$. Then $k>0$ and $x>1$.
$ka_0 = x$
$ka_1 = 1$
$ka_2 = b$
$a_0 = |a_1 - a_2|$, so $ka_0 = k|a_1 - a_2| = |ka_1 - ka_2| = |1-b|$.
Then $x=|1-b|$. So $x=1-b$ or $x=b-1$. The former is impossible since it would imply that $b=1-x$, which would make $b$ negative.
So $x=b-1$, and $b=x+1$. $b>x$. So $a_2 = b/k > x/k = a_0$. This contradicts the assumption that $a_0$ is the largest element of $\{a_n\}$.

We will now suppose, for the sake of contradiction, that $a_{i+1}$, for $i \ge 0$, is the largest element of $\{a_n\}$.
Let $a_0 = 1/k$, and let $a_1 = x/k$. Then $k>0$ and $x>1$.
$ka_i = 1$
$ka_{i+1} = x$
$ka_{i+2} = x-1$ (It can't be $1-x$, since this is negative.)
$ka_{i+3} = b$
$ka_{i+1} = k|a_{i+2} - a_{i+3}| = |ka_{i+2} - ka_{i+3}| = |x-1-b|$
$x=|x-1-b|$
The option $x=x-1-b$ is impossible, since it implies $b=-1$.
Then $x=b+1-x$, and $2x=b+1$, and $b=2x-1$.
Suppose $2x-1<x$. Then $x-1<0$, so $x<1$. Contradiction.
So $a_{i+3} = b/k > x/k = a_{i+1}$. This contradicts our assumption that $a_{i+1}$ is the largest of $\{a_n\}$.

We conclude that $\{a_n\}$ does not have a largest element, and it is therefore not bounded.

QED.
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darktreb
732 posts
#11 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
This was also the first problem of the first MOP test for the Black group (USAMO winners) in 2005.
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Valentin Vornicu
7301 posts
#12 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
darktreb wrote:
This was also the first problem of the first MOP test for the Black group (USAMO winners) in 2005.
Also in the Romanian TST in 2005 :) Spread problem :D
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nipolo
7 posts
#13 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
What a easy problem????

There is no bounded.It is unlimited or it has at least 2 limits
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ZetaX
7579 posts
#14 • 4 Y
Y by Adventure10, jhu08, ImSh95, Assassino9931
Philip_Leszczynski wrote:
[...]
We conclude that $\{a_n\}$ does not have a largest element, and it is therefore not bounded.

QED.
You can't make that assumption, e.g the sequence $a_i = 1- \frac{1}{i}$ doesn't have a largest element and is bounded.
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me@home
2349 posts
#15 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
I'm guessing nipolo's brief "proof" is incorrect? Or if it is correct, can someone elaborate?
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