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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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0 replies
jlacosta
Apr 2, 2025
0 replies
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IMO ShortList 1998, combinatorics theory problem 1
orl   44
N 6 minutes ago by YaoAOPS
Source: IMO ShortList 1998, combinatorics theory problem 1
A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)
44 replies
orl
Oct 22, 2004
YaoAOPS
6 minutes ago
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A game optimization on a graph
Assassino9931   2
N 7 minutes ago by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[ \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r; \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bobby has a winning strategy.
2 replies
Assassino9931
Apr 8, 2025
dgrozev
7 minutes ago
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Composite sum
rohitsingh0812   39
N 16 minutes ago by YaoAOPS
Source: INDIA IMOTC-2006 TST1 PROBLEM-2; IMO Shortlist 2005 problem N3
Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers and let $ S = a+b+c+d+e+f$.
Suppose that the number $ S$ divides $ abc+def$ and $ ab+bc+ca-de-ef-df$. Prove that $ S$ is composite.
39 replies
rohitsingh0812
Jun 3, 2006
YaoAOPS
16 minutes ago
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Problem 1
SpectralS   145
N 18 minutes ago by IndexLibrorumProhibitorum
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
145 replies
SpectralS
Jul 10, 2012
IndexLibrorumProhibitorum
18 minutes ago
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Binomial Sum
P162008   0
2 hours ago
Compute $\sum_{r=0}^{n} \sum_{k=0}^{r} (-1)^k (k + 1)(k + 2) \binom {n + 5}{r - k}$
0 replies
P162008
2 hours ago
0 replies
J
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Triple Sum
P162008   0
3 hours ago
Find the value of

$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{k.2^n + 2m + 1}$
0 replies
P162008
3 hours ago
0 replies
J
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Binomial Sum
P162008   0
3 hours ago
The numbers $p$ and $q$ are defined in the following manner:

$p = 99^{98} - \frac{99}{1} 98^{98} + \frac{99.98}{1.2} 97^{98} - \frac{99.98.97}{1.2.3} 96^{98} + .... + 99$

$q = 99^{100} - \frac{99}{1} 98^{100} + \frac{99.98}{1.2} 97^{100} - \frac{99.98.97}{1.2.3} 96^{100} + .... + 99$

If $p + q = k(99!)$ then find the value of $\frac{k}{10}.$
0 replies
P162008
3 hours ago
0 replies
J
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Polynomial Limit
P162008   0
3 hours ago
If $P_{n}(x) = \prod_{k=0}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$
0 replies
P162008
3 hours ago
0 replies
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Telescopic Sum
P162008   0
3 hours ago
Compute the value of $\Omega = \sum_{r=1}^{\infty} \frac{14 - 9r - 90r^2 - 36r^3}{7^r r(r + 1)(r + 2)(4r^2 - 1)}$
0 replies
P162008
3 hours ago
0 replies
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Theory of Equations
P162008   0
4 hours ago
Let $a,b,c,d$ and $e\in [-2,2]$ such that $\sum_{cyc} a = 0, \sum_{cyc} a^3 = 0, \sum_{cyc} a^5 = 10.$ Find the value of $\sum_{cyc} a^2.$
0 replies
P162008
4 hours ago
0 replies
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CHINA TST 2017 P6 DAY1
lingaguliguli   0
6 hours ago
When i search the china TST 2017 problem 6 day I i crossed out this lemme, but don't know to prove it, anyone have suggestion? tks
Given a fixed number n, and a prime p. Let f(x)=(x+a_1)(x+a_2)...(x+a_n) in which a_1,a_2,...a_n are positive intergers. Show that there exist an interger M so that 0<v_p((f(M))< n + v_p(n!)
0 replies
lingaguliguli
6 hours ago
0 replies
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Math and physics camp
Snezana242   0
Today at 8:53 AM
Discover IMPSC 2025: International Math & Physics Summer Camp!

Are you a high school student (grades 9–12) with a passion for Physics and Math?
Join the IMPSC 2025, an online summer camp led by top IIT professors, offering a college-level education in Physics and Math.

What Can You Expect?

Learn advanced topics from renowned IIT professors

Connect with students worldwide

Strengthen your college applications

Get a recommendation letter for top universities!

How to Apply & More Info
For all the details you need about the camp, dates, application process, and more, visit our official website:
https://www.imc-impea.org/IMC/index.php

Don't miss out on this opportunity to elevate your academic journey!
Apply now and take your education to the next level.
0 replies
Snezana242
Today at 8:53 AM
0 replies
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Combinatoric
spiderman0   1
N Today at 6:44 AM by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
1 reply
spiderman0
Yesterday at 7:46 AM
MathBot101101
Today at 6:44 AM
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Combinatorial proof
MathBot101101   10
N Today at 6:20 AM by MathBot101101
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
10 replies
MathBot101101
Apr 20, 2025
MathBot101101
Today at 6:20 AM
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2 var inquality
sqing   2
N Apr 4, 2025 by sqing
Source: Own
Let $ a,b \ge 0 $ and $ a+b=2. $ Prove that
$$\sqrt{ a^2+b+6}+\sqrt{ b^2+a+6}\leq 8\sqrt{\frac{2- ab}{ab+1}} $$$$\sqrt{2a^2+b+1}+\sqrt{2b^2+a+1}\leq 4\sqrt{\frac{5-2ab}{ab+2}} $$$$\sqrt{2a^2+b}+\sqrt{2b^2+a}\leq 2\sqrt{\frac{3(5-2ab)}{ab+2}} $$
2 replies
sqing
Apr 2, 2025
sqing
Apr 4, 2025
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2 var inquality
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Reveal topic tags
inequalities proposed
inequalities
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Source: Own
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sqing
41795 posts
sqing
#1 h  Apr 2, 2025, 1:11 PM
Y by
Let $ a,b \ge 0 $ and $ a+b=2. $ Prove that
$$\sqrt{ a^2+b+6}+\sqrt{ b^2+a+6}\leq 8\sqrt{\frac{2- ab}{ab+1}} $$$$\sqrt{2a^2+b+1}+\sqrt{2b^2+a+1}\leq 4\sqrt{\frac{5-2ab}{ab+2}} $$$$\sqrt{2a^2+b}+\sqrt{2b^2+a}\leq 2\sqrt{\frac{3(5-2ab)}{ab+2}} $$
This post has been edited 2 times. Last edited by sqing, Apr 2, 2025, 1:25 PM
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Natrium
49 posts
Natrium
#2 h  Apr 4, 2025, 12:19 PM
Y by
To prove
$$\sqrt{ a^2+b+6}+\sqrt{ b^2+a+6}\leq 8\sqrt{\frac{2- ab}{ab+1}}\text{,} $$we see that the given inequality is equivalent with
$$\sqrt{5a^2+4b^2+7ab}+\sqrt{4a^2+5b^2+7ab}\le 8(a+b)\sqrt{\frac{a^2+b^2}{a^2+b^2+6ab}}\text{.}$$Applying AM-QM to the left-hand-side:
$\sqrt{5a^2+4b^2+7ab}+\sqrt{4a^2+5b^2+7ab}\le 2\sqrt{\frac{9a^2+9b^2+14ab}{2}}\text{,}$ so it's enough to prove
$$\sqrt{\frac{9a^2+9b^2+14ab}{2}}\le 4(a+b)\sqrt{\frac{a^2+b^2}{a^2+b^2+6ab}}\text{.}$$
This is equivalent with $23a^4+23b^4\ge 4a^3b+4ab^3+38a^2b^2$, which follows by straightforward AM-GM. So we are done!
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sqing
41795 posts
sqing
#3 h  Apr 4, 2025, 12:26 PM
Y by
Very nice.Thanks.
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