[/quote]
,
Write only the answers to the questions given.
to be more exact:
in years 2002-08 time was 90' for part A and 120' for part B
red balls and 
white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...
unit. The area of the hexagon is ...
and 
.
is ...
so that 
and 
is ...
,
,
,
has an average of 
and a median of 
.
are integers greater than 
which are four consecutive quarters of an arithmetic row with 
.
and 
are squares of two consecutive natural numbers, then the smallest value of 
is ...
,
and 
.
and 
lies on the line segment 
.
and 
.
is ...
meet 
for each natural number 
.
is ....
provided that the difference of any two numbers at least 
is ...
which satisfies 
are as many as ...
and 
.
.
is a point on the side extension 
so that 
is perpendicular to 
and 
.
is ...
consists of 
with 
positive reals is ....
with 
is ....
square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of 
coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of 
be reals such that 
. Prove that
Let 
. Prove that
.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
.
.
,
are chosen such that when going around the triangle, the points occur in the order 
.![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the triplets 
and 
are equal.
and 
,
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
.![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
,![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
,
must be an integer, which is not possible.
and 
.
be reals such that 
and 
Prove that
Let 
and ![$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$](http://latex.artofproblemsolving.com/6/2/2/62211222c148c86d65526d86dde603cb6598925f.png)
![$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$](http://latex.artofproblemsolving.com/3/8/3/383d38cc87c1b1ea1e17e10e4bb95bed79329786.png)
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
.
Y by PikaPika999, Sedro
Here are two problems I made...
1) Let there be a triangle
such that 
, 
, and 
. Then, let the circumcenter of 
be 
. Furthermore, let the reflections of 
, 
, and 
across , be 
, 
, and 
respectively. Find ![$[AB'CA'BC']$](//latex.artofproblemsolving.com/c/5/4/c54d526ffe27fab853f07f87f452887ae61c4331.png)
.
2) If![\[\underbrace{\dfrac{v_2(4!)}{v_2(8!)}\cdot\dfrac{v_2(16!)}{v_2(32!)}\dots}_\text{50 terms} = \dfrac{1}{2^n}\]](//latex.artofproblemsolving.com/b/2/4/b24c2c41eec0340fd57d008ccb07c284b3dd522f.png)
find 
.
@2below Yes, that’s the intended product.
1) Let there be a triangle
such that 
, 
, and 
. Then, let the circumcenter of 
be 
. Furthermore, let the reflections of 
, 
, and 
across , be 
, 
, and 
respectively. Find ![$[AB'CA'BC']$](http://latex.artofproblemsolving.com/c/5/4/c54d526ffe27fab853f07f87f452887ae61c4331.png)
.2) If
![\[\underbrace{\dfrac{v_2(4!)}{v_2(8!)}\cdot\dfrac{v_2(16!)}{v_2(32!)}\dots}_\text{50 terms} = \dfrac{1}{2^n}\]](http://latex.artofproblemsolving.com/b/2/4/b24c2c41eec0340fd57d008ccb07c284b3dd522f.png)
find 
.@2below Yes, that’s the intended product.
This post has been edited 2 times. Last edited by Kempu33334, Apr 9, 2025, 1:39 AM
?
.![\[P = \prod_{k=1}^{50} \frac{4^k - 1}{2\cdot 4^k - 1}.\]](http://latex.artofproblemsolving.com/7/a/e/7ae5cf86184ad336d163cf38fac5e898d0394257.png)
Are we to assume 
triangles with vertex angles 
.
is 
,![$$2\left(\frac12R^2\sin\angle A+\frac12R^2\sin\angle B+\frac12R^2\sin\angle C\right)=2[ABC].$$](http://latex.artofproblemsolving.com/6/6/b/66b7fbcb609f6f8c032cab2b857fb04d9b765e6d.png)
Applying heron's with 
yields 
.
The right bound is trivial, and the left bound is true because
This gives us 
and 
.
