[/quote]
,
be positive real numbers such that 
.
be a convex quadrilateral with ![\[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]](http://latex.artofproblemsolving.com/b/4/1/b416d5c0b29c3a69a3e87cd22b6000aa8e70d456.png)
The line through 
,
,
at point 
.
,
,
,
,
can be divided into two groups, so that the angles in each group have a sum of 
.
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples 
are integers.
that satisfies 
for all real 
be an acute triangle with 
,
be the center of its circumcircle. Let 
be the reflection of 
with respect to 
.
at 
,
intersects the line through 
.
be a point on the ray 
,
.
.
and let 
be positive integers such that 
.
and determine when equality holds.
grid contains a number from 
to 
;
that satisfies the following condition:
be two triangles inscribed in the same circle, centered at 
.
with respect to triangle 
.
.
lie in this order on the circle and form a convex, non-degenerate hexagon.
be a sequence of strictly positive real numbers. For each nonzero positive integer 
,![\[
s_n = a_1 + a_2 + \cdots + a_n \quad \text{and} \quad
\sigma_n = \frac{a_1}{1 + a_1} + \frac{a_2}{1 + a_2} + \cdots + \frac{a_n}{1 + a_n}.
\]](http://latex.artofproblemsolving.com/1/9/1/191a2ebd41e3d236c90e9022ed65323461083815.png)
Show that if the sequence 
is unbounded, then the sequence 
is also unbounded.
and 
Show that
and 
Show that
Let 
Show that
be nonzero vectors in 
-
are real numbers. Show that there exists 
such that![\[\left|\left\langle\sum_{i=1}^t\varepsilon_iv_i,\frac{v_k}{|v_k|}\right\rangle-m_k\right|\ge |v_k|\]](http://latex.artofproblemsolving.com/6/c/6/6c6f82687e0fbf781f18caed5ba6fdbe9bc1125b.png)
holds for all 
,
of 
.
and Anna plays first. The game is won by whom who shall be first able to write a number bigger or equal to 
.
Y by RainbowJessa, radian_51, farhad.fritl, dangerousliri
In each cell of a 
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
, and the sum of the numbers in each column is equal to . Define 
to be the largest value in row 
, and let 
. Similarly, define 
to be the largest value in column , and let 
.
What is the largest possible value of
?
Proposed by Paulius Aleknavičius, Lithuania, and Anghel David Andrei, Romania
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
, and the sum of the numbers in each column is equal to . Define 
to be the largest value in row 
, and let 
. Similarly, define 
to be the largest value in column , and let 
.What is the largest possible value of
?Proposed by Paulius Aleknavičius, Lithuania, and Anghel David Andrei, Romania
This post has been edited 2 times. Last edited by Lukaluce, Apr 29, 2025, 10:37 AM
is 
.
columns into 
squares, and put 
in each of the main diagonals of the squares and 
s
.
and 
,
.
.
be the number of circles in column 
be the sum of the circles in column 
.
,
as 
is equivalent to 
since each circled number must be at least 
.
rectangle with 
rectangle. Fill the 
,
occurences of 
with ![\[
\frac{R}{C} = \frac{2025 \cdot \frac{1}{45}}{45 \cdot \frac{1}{45} + \frac{2025 - 45}{45}} = \frac{2025}{89}
\]](http://latex.artofproblemsolving.com/c/b/3/cb30fdd6325affe7ccf8b8b9fe3364965db08e44.png)
remains the same and 
remains the same or decreases. 
.
.
.![\[
a_1r_1 + a_2r_2 + \dots + a_ir_i \le N \cdot \left(r_1 + r_2 + r_3 + \dots + \frac{2025 - i}{2025}\right)
\]](http://latex.artofproblemsolving.com/1/d/5/1d586dac5b2dd959c42ceecd6f13cccf31a0d9d2.png)
This is tightest when the 
have 
and the 
have 
.![\[
j + \frac{t}{2025} \le N \cdot \left(\frac{1}{a_1} + \dots + \frac{1}{a_j} + \frac{i-j}{2025} + \frac{2025 - i}{2025}\right)
\]](http://latex.artofproblemsolving.com/1/3/4/134b1b2312664e85d716e7095387c4c850265866.png)
where 
.
,![\[
\frac{j}{N} \le \frac{1}{a_1} + \dots + \frac{1}{a_j} + \frac{2025 - i}{2025}
\]](http://latex.artofproblemsolving.com/c/c/b/ccb98feb86ff336cda077524517d61d7338a5960.png)
which with AM-HM and applying 
becomes![\[
\frac{j}{N} \le \frac{j^2}{a+j} + \frac{a}{2025}
\]](http://latex.artofproblemsolving.com/e/a/1/ea1ea3b356d56851573e1e7569ee2debe1e97438.png)
with 
.
as the only root. If 
,![\[
\frac{j}{N} \le \frac{j^2}{45j} + \frac{44j}{2025} \iff
\frac{89j}{2025} \le \frac{45j}{2025j} + \frac{44j}{2025}
\]](http://latex.artofproblemsolving.com/9/e/b/9eb5abf9121eeae979999aa3df959f986e75b35d.png)
which holds. If 
,![\[
\frac{j}{N} \le \frac{j^2}{2025} + \frac{2025-j}{2025} \iff
89j \le j^2 + 2025 - j \iff (j-45)^2 \ge 0
\]](http://latex.artofproblemsolving.com/7/0/9/709b85997c99d6f948062143604dd238d5b5bc8c.png)
which holds.
.
,
(everything should be divided by 
)
columns that contains scorers. Note that the non-scorers in a column with more than 
scorers should be distributed to the scorers to maximize the ratio, the converse is true (distribute among non-scorers if there aren't enough scorers).![\[\frac{n^2a+n^2-s}{n^2-a+\frac{n^2a^2}{s}}\le\frac{n^2}{2n-1}.\]](http://latex.artofproblemsolving.com/c/6/e/c6ecd4d4270253776f6af21ed254ba252c09f0d1.png)
After more bashing, we realize that this is optimized when 
,
the number of yellow cells it has.
and all we want is for this to be in terms of 
is trivially true by double counting so now we will split sum of 
and therefore 
And now here we need 
and thus 
but also trivially 
which should lead to picking 
for some 
,
and therefore 
so our pick here will be 
to get rid of each 
in which case we get that 
this so that we don't have the need to summon each 
so tracing back the equality this gives that in each of the columns where 
we have that the only nonzero cells are the ones in yellow, also from the AM-GM we have that 
if 
'
.
.
,
in each such column, we guarantee that each heavy column carries the minimal possible maximum to serve 45 rows. (A short optimization shows that if one designates 
columns to serve as the targets for high entries and distributes the mass evenly among them so that every heavy row gets exactly 
in the column it uses, then one may obtain a ratio of![\[ \frac{n}{\,k+r-1}\,, \]](http://latex.artofproblemsolving.com/b/5/c/b5c9448f65ff9408fbdbb06068b0f8da03f56b39.png)
with the constraint 
.
;
,![\[ \frac{2025}{45+45-1}=\frac{2025}{89}\,. \]](http://latex.artofproblemsolving.com/d/d/8/dd8e174440dcffbb58f3bc3039935de45724c965.png)
)
block with the “flat” matrix on the remaining cells.)
rows into 
.
.
columns that have 
numbers in those columns and the sum in each row is 
.
because they are contained in 
.
'
.
,
.
which will get us to 
which yields the needed answer.
