AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be positive integers with 
. There are cupcakes of different flavors arranged around a circle and people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person 
, it is possible to partition the circle of cupcakes into groups of consecutive cupcakes so that the sum of 's scores of the cupcakes in each group is at least 
. Prove that it is possible to distribute the cupcakes to the people so that each person receives cupcakes of total score at least with respect to .
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by ?
be a sequence of real numbers with 
. Define the score 
of a permutation 
of 
to be the minima of the sum 
over all real numbers 
. attains the maxima over all permutations 
, if and only if for all 
, 
be a nonempty set of positive integers. We say that a positive integer is clean if it has a unique representation as a sum of an odd number of distinct elements from . Prove that there exist infinitely many positive integers that are not clean.
such that for any positive integer 
, satisfies 
.
and 
lie respectively on sides and 
of a triangle and 
. Segments 
and 
cross at 
. Circumscribed circles of triangles 
and 
cross again at point different from . Prove that point lies on the bisector of angle 
.
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
such that![\[
ad = bc, \quad a + d = 2^k, \quad b + c = 2^m
\]](http://latex.artofproblemsolving.com/1/1/d/11d16732e3a77af46ad17adecf104a442a5d7f67.png)
for some positive integers 
and 
.
be a triangle with incenter 
and intouch triangle 
. Let be the foot of the perpendicular from 
onto 
. Assume that 
, intersect the sides , in 
respectively. Finally, let the rays 
, 
meet the circumcircle of 
in , respectively. Prove that the tangent from to the incircle and the line 
meet on the circumcircle of .
, which is true. The base case 
gives 
. Let 
and 
. Therefore, ![$b=\frac{1}{4}*[(\sum_{i=0}^{n+1}{a_i})^2-\sum_{i=0}^{n+1}{a_i}-(\sum_{i=0}^{n}{a_i})^2+\sum_{i=0}^{n}{a_i}]=\frac{1}{4}[a_{n+1}*(2\sum_{i=0}^{n}{a_i}+a_{n+1})-a_{n+1}]=\frac{a_{n+1}}{2}[\frac{2\sum_{i=0}^{n}{a_i}+a_{n+1}-1}{2}]=\frac{a_{n+1}}{2}[\sum_{i=0}^{n}{a_i}+(\frac{a_{n+1}-1}{2})]$](http://latex.artofproblemsolving.com/7/c/e/7ce63357487ed02032f83131a48b6f2643a7ec93.png)
. Note that ![$a=\frac{a_{n+1}}{2}[(n+1)(a_{n+1}-1)]$](http://latex.artofproblemsolving.com/e/a/f/eaf8b279580b810f23ba6c7df45df7b434d4a980.png)
. Since 
. Then 
for 
with the equality case when 
for all 
.
as a base case proves the desired statement for all nonnegative integers which is stronger than for all positive integers 
At least pointing out that you are proving a stronger statement should not get docks, but should be no docks regardless.
when its actually 
Let 
and introduce a new term 
We have that 
and since 
is nonnegative we have
and since is nonnegative we have
Then, we have
Then, divide by 
and add the inequality from the inductive hypothesis, and we have the desired result.
is there an 
case when the induction looks like this or nah
rooms, with the th room having 
people. Number the people in each room 
Then 
calculates the number of ways to choose two people from the same room 
and then picking one of them to move either 
rooms down. Then, if the number of the person who went down is 
we find the person in the new room he went to with the same number 
So the end result is this matched person and the person from the intial room who did not move. It is clear that each end result from the process is unique.
counts the number of ways to choose 
people from the entire group of people directly. Since the above is an injective map the desired result follows. QED
(which we permit) there is nothing to prove. Hence to prove by induction on , it would be sufficient to verify ![\[ 2n \binom{a_n}{2} \le \binom{a_0 + a_1 + \dots + a_n}{2} - \binom{a_0 + a_1 + \dots + a_{n-1}}{2}. \]](http://latex.artofproblemsolving.com/f/0/6/f06a1f32893b335ec9495fc70e161a02716d9f1e.png)
Rearranging the terms around, that's equivalent to proving 
However, the last line is obvious because 
, and 
.
and 
for 
.
![\[ \sum_{i = 0}^n i \binom{a_i}{2} \le \dfrac{\sum_{i = 0}^n i}{(n + 1)^2} \sum_{i = 0}^n \binom{a_i}{2} = \dfrac{n}{2(n + 1)} \sum_{i = 0} \binom{a_i}{2} \le \dfrac{1}{2} \sum_{i = 0}^n \binom{a_i}{2} \le \dfrac{1}{2} \binom{ \sum_{i = 0}^n a_i }{2}\]](http://latex.artofproblemsolving.com/a/5/d/a5d9a1e9ee714c88e4fefc01b02666bac24bb87b.png)
YUH.
so the inequality becomes ![\[ \sum_{i=0}^n 2ia_i(a_i - 1) \le \left( \sum a_i \right)^2 - \sum a_i. \]](http://latex.artofproblemsolving.com/5/2/f/52f3ca3706e6b9bfd83811f052d7815ac6a9e3af.png)
Now notice that ![\[ a_ia_j \ge a_j^2 \]](http://latex.artofproblemsolving.com/e/8/a/e8abf3ee9377c2cebf58a42a045deec359bb0eaf.png)
for 
. In particular, on the left hand side the coefficient of 
is 
, and on the right hand side, the sum of the coefficients of the 
terms where 
is precisely , so we can remove all the squared terms with the cross terms on the right, requiring us to prove ![\[ \sum_{i = 0}^n -ia_i \le \sum a_i^2 - a_i. \]](http://latex.artofproblemsolving.com/a/5/6/a5668c227a06ec4600cdab63c2895bfe5c762325.png)
![\[ \ln \left( \frac{27abc}{(a+b+c)^3} \right) \le \frac{(a-b)^2+(b-c)^2+(c-a)^2}{3}. \]](http://latex.artofproblemsolving.com/d/e/2/de244e2a3c44720d4f7b27b216af6b3971aad8b2.png)
One wonders: How does one possibly solve this inequality??!?!?!?!? It is simple, dear friend. Notice that ![\[ \ln \left( \frac{27abc}{(a+b+c)^3} \right) \le 0 \le \frac{(a-b)^2+(b-c)^2+(c-a)^2}{3}. \]](http://latex.artofproblemsolving.com/2/7/e/27e4982e412a393bfdc77ad2d21385fddc99073e.png)
![\[ \sum_{i = 0}^n -ia_i \le 0 \le \sum a_i^2 - a_i. \]](http://latex.artofproblemsolving.com/b/c/7/bc77864bb989345a5acdb16656fea1e465d354f3.png)
Through the miracle work of CMJ 1226 we are done.
such that![\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/7/a/5/7a51a7be2306d18f4e80db1f5aac6e9bbf25c25c.png)
be nonnegative integers such that 
Prove that![\[
\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.
\]](http://latex.artofproblemsolving.com/c/4/9/c49d0ccfef69402d05dc44d721172498c605cf7e.png)
Note: 
for all nonnegative integers 
or 
.
balls split into 
Then, after multiplying both sides by 
-
different pairs of balls, so the inequality holds. I barely finished this writeup before time ended but I hope it gets full points.
works
right?
Equality holds when 
and 
.
for all nonnegative integers 
of nonnegative integers satisfying 
for some 
where 
Then, it follows that if we have some 
of non-negative integers satisfying 
then we will have 
It suffices to show that the desired inequality will then hold for the 
-
for any nonnegative integer 
satisfying 
It is easy to show that 
is nonnegative for all nonnegative integers 
and we also know that 
because all of 
are nonnegative. Therefore, it suffices to prove that 
This easily follows from the fact that 
for all 
concluding the inductive step.
and also 
,
condition properly.
which is already more than what we needed thus we are done, equality holds when 
.