AoPS Academy
be positive real numbers. Prove that:![$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$](http://latex.artofproblemsolving.com/2/6/e/26ebcdae35153bee73abbcfaf4a767d3a32b8912.png)
, 
is the incenter. Line 
intersects circumcircle of at 
, and it is given that 
. 
are points chosen on 
such that 
. Prove that 
be a positive integer. In an 
table, an upright path is a sequence of adjacent cells starting from the southwest corner to the northeast corner such that the next cell is either on the top or on the right of the previous cell. Find the smallest number of grids one needs to color in an table such that there exists only one possible upright path not containing any colored cells.
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 
. For example, 
is stylish because the set 
can be partitioned as 
, with sums , 
, and respectively. Determine the number of stylish numbers less than 
.
Hint
to 
be positive real numbers such that 
and 
Determine the value of 
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
given that 
Hint
![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
.
be real numbers such that 
Find 
are on circle 
such that 
and 
Determine the path length from 
to 
formed by segment and arc 
such that the expression 
is also an integer. such that 
is a multiple of 
and 
are real numbers such that 
and 
Calculate 
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?
The probability that at least two people choose the same number can be written as 
Find 
on the positive integers using the rule that for 
For all prime , 
and for all other 
Find the smallest possible value of such that 
can be written as the sum of two distinct, non-negative integer powers of 
be the set of positive integers of such that 
for some other positive integer 
Find the only three-digit value of in 
be a positive integer and let be the integer that is formed by removing the first three digits from 
Find the value of with least value such that 
be odd positive integers such that 
and both 
and 
are powers of 2. Prove that 
points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least 
of the other points in its interior.
points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least of the other points in its interior.
on the convex hull. between all other points 
, there exists a unique one such that 
is minimal (because no four are collinear). it is clear now that 
is nice.
are points such that 
such that is not an edge of the convex hull and there exists a such that 
is nice, then there is another point on the other side of such that 
is also nice., divide the polygon into two parts 
such that 
and call the circumcircle of 
, 
. it is clear that except the rest of 
lies strictly inside . now continuously change such that the side not containing shrinks and the side containing grows. we know it still contains 
. do this until touches at a third point. if we choose this point as it is clear that 
is nice.(and of course is vertex and not in the middle of an edge) so the lemma is proven. as long as the polygon isn't fully triangulated, we can add another triangle to it. note that because we are adding triangles in the parts of the polygon that were previously untouched, this actually does create a triangulation.
vertices, there exists an edge in the triangulation such that the endpoints are at least 
points apart on the perimeter both clockwise and counter-clockwise.-gon on the vertices of a regular -gon (such that the order of the points on the convex hull remains the same) so its enough to prove it for a regular polygon. which can be done by taking the longest edge in the triangulation. if it's ends are less than 
apart, then one of the two triangles on each side of it has a longer edge, a contradiction (look on the perimeter of its circumcircle) .
points! so we are done.
？ 
？ 
and 
such that
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 
Summing these gives our answer of 
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.
Furthermore, note that 
Thus, the answer is just 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
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 
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.
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 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
is the diameter, thus we can draw 
and so by Special Right Triangles we do NOT in fact get 
but rather 
.
to be an integer. Plugging in factors of 
we see that there are 
integers that work.
Plugging in numbers gives us 
as our smallest value.
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
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 
Thus, our answer is just 
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 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
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 
contains all the other points of ![[asy]
unitsize(2cm);
pair A = (Cos(100),Sin(100));
pair B = (Cos(220),Sin(220));
pair C = (Cos(320), Sin(320));
pair X1 = (-0.44728,-0.74601);
pair X2 = (-0.16890,-0.77087);
pair X3 = (0.20890,-0.75098);
pair X4 = (0.50717,-0.71121);
pair Y1 = (0.79549,-0.08983);
pair Y2 = (0.53699,0.69064);
pair Z1 = (-0.58647,0.75029);
dot(A);dot(B);dot(C);dot(X1);dot(X2);dot(X3);dot(X4);dot(Y1);dot(Y2);dot(Z1);
label("$A$",1.15*A);label("$B$",1.15*B);label("$C$",1.15*C);
draw(A--Z1--B--X1--X2--X3--X4--C--Y1--Y2--cycle);
draw(circle((0,0),1));
[/asy]](http://latex.artofproblemsolving.com/4/5/5/45547ca1996d95c7562d293d7205cb53fc337fd7.png)
and significance 
of smallest area which contains all of 
is a large triple), or there exist two distinct elements 
of 
which minimizes the undirected angle 
.
is a large triple).![[asy]
unitsize(2cm);
real x = 1.0/15;
fill(circle((20*x,0),25*x),gray);
clip(currentpicture,circle((-20*x,0),25*x),invisible);
draw(circle((0,0),1));
draw(circle((20*x,0),25*x),dashed);
pair A = (0,1); dot(A); label("$A$",1.15*A);
pair B = (0,-1); dot(B); label("$B$",1.15*B);
pair D = (-4*x,-7*x); dot(D); label("$D$",D,(-0.45,0));
label("$\mathcal{C}$",(-1.15,0));
dot((-0.1,0.6)); dot((-0.14,0.49)); dot((0.18,0.45)); dot((-0.21,0.37)); dot((0.04,0.35)); dot((0.23,0.20)); dot((-0.04,0.18)); dot((-0.17,-0.03));
label("$S$",(0,-0.3));
[/asy]](http://latex.artofproblemsolving.com/e/b/9/eb9e25b28074629cc0f2d2eaeccbb58b16a2e4d6.png)
,
first).
.
not containing 
not containing 
.
and 
,
is minimized. Then 
is also a large triple, and the old arc has been split into two new arcs.![[asy]
unitsize(2cm);
pair A = (Cos(100),Sin(100));
pair B = (Cos(220),Sin(220));
pair C = (Cos(320), Sin(320));
pair X1 = (-0.44728,-0.74601);
pair X2 = (-0.16890,-0.77087);
pair X3 = (0.20890,-0.75098);
pair X4 = (0.50717,-0.71121);
pair Y1 = (0.79549,-0.08983);
pair Y2 = (0.53699,0.69064);
pair Z1 = (-0.58647,0.75029);
dot(A);dot(B);dot(C);dot(X1);dot(X2);dot(X3);dot(X4);dot(Y1);dot(Y2);dot(Z1);
label("$A$",1.15*A);label("$B$",1.15*B);label("$C$",1.15*C);
draw(A--Z1--B--X1--X2--X3--X4--C--Y1--Y2--cycle);
draw(circle((0,0),1));
draw(circle((0,1.17905),1.976),dashed);
label("$D$",0.85*X1);
[/asy]](http://latex.artofproblemsolving.com/6/5/a/65adfd4cde038673a5b6524c62356035f62ef6c4.png)
,
.
,
.
.
be a large triple with significance at most 
.
not containing 
,
.
as the two desired points in the problem statement. Indeed, any circle through 
points of 
,
which are all good triangles
,
is the smallest
is a good triangle
,
be a diagonal of 
on the other side of 
wrt 
is the smallest
,
,
points.