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integral points
jhz   1
N 20 minutes ago by gaussiemann144
Source: 2025 CTST P17
Prove: there exist integer $x_1,x_2,\cdots x_{10},y_1,y_2,\cdots y_{10}$ satisfying the following conditions:
$(1)$ $|x_i|,|y_i|\le 10^{10} $ for all $1\le i \le 10$
$(2)$ Define the set \[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]then \(|S| = 1024\)and any rectangular strip of width 1 covers at most two points of S.
1 reply
jhz
5 hours ago
gaussiemann144
20 minutes ago
Probability-hard
Noname23   2
N 21 minutes ago by Noname23
problem
2 replies
Noname23
an hour ago
Noname23
21 minutes ago
7 triangles in a square
gghx   2
N an hour ago by lightsynth123
Source: SMO junior 2024 Q3
Seven triangles of area $7$ lie in a square of area $27$. Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$.
2 replies
gghx
Oct 12, 2024
lightsynth123
an hour ago
n-variable inequality
ABCDE   65
N 2 hours ago by LMat
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
65 replies
ABCDE
Jul 7, 2016
LMat
2 hours ago
2 degree polynomial
PrimeSol   3
N 2 hours ago by PrimeSol
Let $P_{1}(x)= x^2 +b_{1}x +c_{1}, ... , P_{n}(x)=x^2+ b_{n}x+c_{n}$, $P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$ $\forall i,j ,1 \leq i<j \leq n : P_{i}(x) \ne P_{j}(x)$.
$\forall i,j, 1\leq i<j \leq n : Q_{i,j}(x)= P_{i}(x) + P_{j}(x)$ polynomial with only one root.
$max(n)=?$
3 replies
PrimeSol
Mar 24, 2025
PrimeSol
2 hours ago
Additive Combinatorics!
EthanWYX2009   3
N 2 hours ago by flower417477
Source: 2025 TST 15
Let \( X \) be a finite set of real numbers, \( d \) be a real number, and \(\lambda_1, \lambda_2, \cdots, \lambda_{2025}\) be 2025 non-zero real numbers. Define
\[A = 
\left\{ 
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d 
\right\},\]\[B = 
\left\{ 
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0 
\right\},\]\[C = 
\left\{ 
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0 
\right\}.\]Show that \( |A|^2 \leq |B| \cdot |C| \).
3 replies
EthanWYX2009
Yesterday at 12:49 AM
flower417477
2 hours ago
Inspired by IMO 1984
sqing   0
3 hours ago
Source: Own
Let $ a,b,c\geq 0 $ and $a^2+b^2+ ab +24abc\geq\frac{81}{64}$. Prove that
$$a+b+\frac{9}{5}c\geq\frac{9}{8}$$$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$$$a+b+\frac{8}{5}c\geq  \frac{9\sqrt [3]{25}-4}{20}$$Let $ a,b,c\geq 0 $ and $ a^2+b^2+ ab +18abc\geq\frac{343}{324} $. Prove that
$$a+b+\frac{6}{5}c\geq\frac{7\sqrt 7}{18}$$$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$
0 replies
1 viewing
sqing
3 hours ago
0 replies
equal angles
jhz   2
N 3 hours ago by YaoAOPS
Source: 2025 CTST P16
In convex quadrilateral $ABCD, AB \perp AD, AD = DC$. Let $ E$ be a point on side $BC$, and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$. Let $O$ be the circumcenter of $\triangle CDE$, and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$. Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$
2 replies
jhz
6 hours ago
YaoAOPS
3 hours ago
Flee Jumping on Number Line
utkarshgupta   23
N 3 hours ago by Ilikeminecraft
Source: All Russian Olympiad 2015 11.5
An immortal flea jumps on whole points of the number line, beginning with $0$. The length of the first jump is $3$, the second $5$, the third $9$, and so on. The length of $k^{\text{th}}$ jump is equal to $2^k + 1$. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
23 replies
utkarshgupta
Dec 11, 2015
Ilikeminecraft
3 hours ago
Smallest value of |253^m - 40^n|
MS_Kekas   3
N 3 hours ago by imagien_bad
Source: Kyiv City MO 2024 Round 1, Problem 9.5
Find the smallest value of the expression $|253^m - 40^n|$ over all pairs of positive integers $(m, n)$.

Proposed by Oleksii Masalitin
3 replies
MS_Kekas
Jan 28, 2024
imagien_bad
3 hours ago
Double factorial inequality
Snoop76   3
N Yesterday at 8:02 AM by Snoop76
Source: own
Show that: $$2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}>\sum_{k=0}^n (2k+1)!!{n\choose k}$$Note: consider $(-1)!!=1$ and $n>1$
3 replies
Snoop76
Feb 7, 2025
Snoop76
Yesterday at 8:02 AM
Double factorial inequality
G H J
Source: own
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Snoop76
312 posts
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Show that: $$2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}>\sum_{k=0}^n (2k+1)!!{n\choose k}$$Note: consider $(-1)!!=1$ and $n>1$
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Snoop76
312 posts
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bumppppp
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Snoop76
312 posts
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bumppppp
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312 posts
#10 • 1 Y
Y by MihaiT
We note: $a_n=\sum_{k=0}^n (2k-1)!!{n\choose k}$, $b_n=\sum_{k=0}^n (2k+1)!!{n\choose k}$ and $c_n=2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}-\sum_{k=0}^n (2k+1)!!{n\choose k}$

Using Pascal identity $a_{n+1}=\sum_{k=0}^{n+1} (2k-1)!!{n+1\choose k}=\sum_{k=0}^{n+1} (2k-1)!!{n\choose k}+\sum_{k=0}^{n+1} (2k-1)!!{n\choose k-1}$ or

$a_{n+1}=a_n+b_n$

$b_{n+1}-a_{n+1}=\sum_{k=0}^{n+1} (2k+1)!!{n+1\choose k}-\sum_{k=0}^{n+1} (2k-1)!!{n+1\choose k}=\sum_{k=0}^{n+1}2k(2k-1)!!{n+1\choose k}$ or

$b_{n+1}-a_{n+1}=2(n+1)\sum_{k=0}^{n+1} (2k-1)!!{n\choose k-1}=2(n+1)b_n$

We have: $c_{n+1}-c_n=2(n+1)a_{n+1}-b_{n+1}-2na_n+b_n=2n(a_n+b_n)-2(n+1)b_n+a_n+b_n-2na_n+b_n$ or

$c_{n+1}-c_n=a_n>0$, $a_0=1$ results with simple induction that $c_n>0$ for $n>1$.
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