is the incenter. Line 
intersects circumcircle of 
,
.
are points chosen on 
such that 
.
is called stylish if the set 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
.
is stylish because the set 
can be partitioned as 
,
,
.
Doing this, we get that 
Now, we can solve this by factoring. By using RRT, we get that 
is a factor. Upon division, we get 
is the factored form. Now, we know that our only positive solutions are 
so our values of 
are 
Summing these gives our answer of 
to 
in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
be positive real numbers such that 
and 
Determine the value of 
Furthermore, note that 
Thus, the answer is just 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
given that 
We know from 
that 
so substituting this gives us 
Now, plugging this in to 
and 
gives us 
Solving these equations gives us 
Thus, by Vieta's, the sum of the roots is 
thus by median properties (the 2:1 ratio split thingy), we know that 
(For reference, I missed this part in-contest) Now, we can calculate 
by the Pythagorean Theorem to be 
and thus 
Now, by Pythagorean theorem, we know that the altitude from 
is 
so our area is just 
Thus, simplifying and substituting 
we get that 
Simple guess and check gives us 
as our solutions. My dumb self forgot the former solution. Thus, 
giving us 
as our sum.![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
.
we get that 
Now if we try finding 
we will realize that 
this 
pattern continues until we once again reach 
thus 
Now return to the problem statement; 
but each inner function will just become 
hence the answer is 
be real numbers such that 
Find 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
are on circle 
such that 
and 
Determine the path length from 
formed by segment 
is the diameter, thus we can draw 
and so by Special Right Triangles we do NOT in fact get 
but rather 
is also an integer.
.
to be an integer. Plugging in factors of 
we see that there are 
integers that work.
is a multiple of 
Plugging in numbers gives us 
as our smallest value.
are real numbers such that 
and 
Calculate 
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
will follow the rules such that![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](http://latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?
Furthermore, we know that 
so our answer is just 
in the summation. Essentially, we have 
.
In other words, we want to find 
.
.
in such a polynomial is 
.
,
.
.
,
.
and 
,
and divide by 
.
and 
in our sum, subtract by 
.
term look kind of weird? Well, we actually notice 
,
.
,
,
is our answer!
,
terms. Drawing it out on the unit circle (which I heavily advise), it cycles between 
and 
.
,
,
.
,
for when 
terms! Anyways, the answer is 
.
so our coefficient is just 
Find 
Thus, our answer is just 
on the positive integers using the rule that for 
For all prime 
and for all other 
Find the smallest possible value of 
can be written as the sum of two distinct, non-negative integer powers of 
etc.
we get that it is 
Thus, we note that for an integer 
Since 
we know that all values of 
satisfy our conditions. Thus, our answer is just 
be the set of positive integers of 
for some other positive integer 
Find the only three-digit value of 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
Find the value of 
and let 
be our three-digit number that we remove. Thus, we can write 
By our condition, we know that 
or that 
Furthermore, we know that 
does not work, so we try 
We know that 
by the fact that it must be divisible by a 
Noting that 
doesn't work, we try 
or that 
This, miraculously, works, and thus our value of 
be odd positive integers such that 
and both 
and 
are powers of 2. Prove that 
Y by ehuseyinyigit
Let 
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
Attachments:
This post has been edited 2 times. Last edited by m4thbl3nd3r, Mar 28, 2025, 5:00 PM
antipode of 
can be easly MMP'd as 
boh have degree one by simply checking where your cases are 
and 
also has degree one snd so does 
colinear is degree 4 and the cases are 
and 
so you are done.
which is the Miquel point of 
.
so 
collinear and passes through the 
.
Note that 
and
.
Since 
we are done.
)
be diameter of 
,
.
is collinear and 
(Because 
)
.
and 
.
and we're done
