be a positive integer. Claire writes 
distinct positive real numbers 
in a row on a blackboard. In a 
William can erase a number 
or 
at the same location. His goal is to perform a sequence of moves such that after he is done, the number are strictly increasing from left to right.
independent of 
such that William can always reach his goal in at most 
moves.
independent of 
moves.
a friendly number if there exists a positive rational number 
such that 
.
such that 
as 
intersects 
at 
such that 
is equilateral.
Y by donotoven
These are my collection problem from my country.
Day 1 (270 mins)
1. Suppose that
is an acute triangle, 
and 
lie on 
such that 
and 
. Point 
and 
lie outside which and bisect 
and 
respectively, Lines 
intersect 
at 
. Show that
2. Find all polynomial
such that
for all real number 
.
3. Let
be set of positive integers. Find all function 
such that
for all 
4. Find all positive integers
such that 
5. Let
be real numbers. If 
then find the minimum value of
![$$K = \frac{a}{\sqrt[3]{8+b-d}} + \frac{b}{\sqrt[3]{8+c-a}} + \frac{c}{\sqrt[3]{8+d-b}} + \frac{d}{\sqrt[3]{8+a-c}}$$](//latex.artofproblemsolving.com/8/7/e/87e0736118e7d629fe838daa9ff053490a3e2878.png)
6. In a hospital, there are
doctors and 
patients. It is known that,
- 1 doctor cures exactly 100 patients, and
- for any two patients, exactly 11 doctors cure both of them
Find the value of.
Day 2 (270 minutes)
1. Let
be the circumcircle of quadrilateral 
, 
intersects 
at . 
is a midpoint of an arc 
(another side from 
and 
) such that 
and 
intersect at 
and 
respectively. Show that
2. Let
and 
be polynomial satisfy the following equation
Find the value of 
3. Find all function
, 
and satisfy the equation
for all reals 
4. Let
. Prove that
for all
and 
, There exists 
which 
and 
5. Let
and 
. Prove that
6. There are 2564 numbers (It is year 2021 but in Thailand we use 2564). Each number has possible prime divisor(s) only 2,3,5,7,11,13,17,19,23,29 and 31. Prove that there exists 2 numbers selected from those 2564 numbers such that its product is perfect square.
Day 1 (270 mins)
1. Suppose that
is an acute triangle, 
and 
lie on 
such that 
and 
. Point 
and 
lie outside which and bisect 
and 
respectively, Lines 
intersect 
at 
. Show that
2. Find all polynomial
such that
for all real number 
.3. Let
be set of positive integers. Find all function 
such that
for all 
4. Find all positive integers
such that 
5. Let
be real numbers. If 
then find the minimum value of![$$K = \frac{a}{\sqrt[3]{8+b-d}} + \frac{b}{\sqrt[3]{8+c-a}} + \frac{c}{\sqrt[3]{8+d-b}} + \frac{d}{\sqrt[3]{8+a-c}}$$](http://latex.artofproblemsolving.com/8/7/e/87e0736118e7d629fe838daa9ff053490a3e2878.png)
6. In a hospital, there are
doctors and 
patients. It is known that,- 1 doctor cures exactly 100 patients, and
- for any two patients, exactly 11 doctors cure both of them
Find the value of
.Day 2 (270 minutes)
1. Let
be the circumcircle of quadrilateral 
, 
intersects 
at . 
is a midpoint of an arc 
(another side from 
and 
) such that 
and 
intersect at 
and 
respectively. Show that
2. Let
and 
be polynomial satisfy the following equation
Find the value of 
3. Find all function
, 
and satisfy the equation
for all reals 
4. Let
. Prove thatfor all
and 
, There exists 
which 
and 
5. Let
and 
. Prove that
6. There are 2564 numbers (It is year 2021 but in Thailand we use 2564). Each number has possible prime divisor(s) only 2,3,5,7,11,13,17,19,23,29 and 31. Prove that there exists 2 numbers selected from those 2564 numbers such that its product is perfect square.
This post has been edited 3 times. Last edited by AoPSTheP, Jul 19, 2021, 3:46 PM
Reason: Add P6D2
Reason: Add P6D2
Suppose 
also 
thus
we get 
w
.
.
.
has infinity solution so 
.
so the number of doctors is 
then we get that 
doctor cures exactly 
patients, and
doctors cure both of them
where patients 
and 
are cured by doctor 
patients and each 
(as 
.
(as 
.
be the assertion of the functional equation.
gives 
.
gives 
.
for every positive integers 
.
gives us:
.
.
gives us:
.
with 
.
.
,
.
.
for every positive integer is the only solution.
is larger than or equal to...
is larger than or equal to 
where, 
are constants and you can easily find it by accurate calculations. After this, It's easy because 
and with the inequality
is larger than or equal to 
,
.
.
.
are cyclic. Finally from this and PoP we obtain 
and angle bisector theorem 
