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.
, find 
be an acute triangle with circumcircle 
. Let 
be a variable point on the arc 
of not containing 
. Squares 
and 
are constructed such that , 
, 
lie on the same side of line 
and , 
, 
lie on the same side of line 
. Let 
be the intersection of lines 
and 
. Show that as varies, lies on a fixed circle.
has side lengths 
, 
, and 
. Select a point 
inside , and construct the incenters of 
, 
, and 
and denote them as 
, 
, 
. What is the maximum area of the triangle 
?
be a convex pentagon such that 
. Assume that there is a point 
inside with 
and 
. Let line 
intersect lines 
and 
at points and 
, respectively. Assume that the points 
occur on their line in that order. Let line 
intersect and 
at points 
and 
, respectively. Assume that the points 
occur on their line in that order. Prove that the points 
lie on a circle.
. Nick intends to find a seven-digit number divisible by 
such that its decimal representation contains no figures besides these two. Is this possible for each teacher’s choice? (4 marks)
and for all 
, 
Determine the smallest 
such that 
.
coins numbered from 
to 
. These coins are placed around a circle, not necesarily in order.
, we will jump to the one places from it, always in a clockwise order, beginning with coin number 1. For an example, see the figure below. for which there exists an arrangement of the coins in which every coin will be visited.
. be a polynomial of degree with only real zeros and real coefficients.
we have 
. When does equality occur?
be an 
square matrix with eigenvalues 
(each 
having algebraic multiplicity 
, so that 
). Let 
be a polynomial. It is known that if 
is an eigenvalue of then 
is an eigenvalue of 
. of the form 
(counting multiplicities)?
vs. equation 
)
sort of means 
times of coin flips, and the 
-th power kind of tells you that there is 2018 independent events. heads, and see if the time that the -th heads occurred minus the time that the 
-th heads occured is in the set 
. Then there is some work to relate this to the original problem, but it is pretty straight forward.
is not reduced, so I seriously thought that 
, and I somehow got that 
. I am really expecting a 1 on this now oops :p
sort of means times of coin flips, and the -th power kind of tells you that there is 2018 independent events. heads, and see if the time that the -th heads occurred minus the time that the -th heads occured is in the set . Then there is some work to relate this to the original problem, but it is pretty straight forward. is not reduced, so I seriously thought that 
, and I somehow got that . I am really expecting a 1 on this now oops :p
is the coefficient of 
in the expansion of 
. To extract this coefficient we can use the residue theorem from complex analysis:
where 
is any contour enclosing the origin. Choosing to be the circle of radius 
centered at the origin, we can use the triangle inequality to bound this as:
as desired.
be the outcomes. Define 
. Then, we wish to prove that
However, we can use Markov's inequality in way that a Chernoff bound is derived:![$$P(X \le 3860) = P(2^{-X} \ge 2^{-3860}) \le \mathbb{E}[2^{-X}] \cdot 2^{3860}$$](http://latex.artofproblemsolving.com/f/1/0/f10a3aea5620c8a07c18b181dbd0936256fbf1c4.png)
Note that 
, and each of the factors are i.i.d. Therefore, we can break the expectation across the product.![$$\mathbb{E}[2^{-X_1}] = \frac{1}{7} \left( 2^{-1} + 2^{-2} + \cdots + 2^{-6} + 2^{-10}\right) = \frac{1}{7} \cdot \frac{2018}{2048}$$](http://latex.artofproblemsolving.com/e/8/b/e8be8ce25dba0bc0882ff79b9e397d9355b040f3.png)
Raise this result to the 2018th power, and we get the desired result.
is the coefficient of in the expansion of . To extract this coefficient we can use the residue theorem from complex analysis:where is any contour enclosing the origin. Choosing to be the circle of radius centered at the origin, we can use the triangle inequality to bound this as:as desired.
?
(which is otherwise irrelevant to the problem) was chosen for the following reason: as a function of , the upper bound 
is minimized when![\[\frac{x(1+2x+\dotsb+6x^5+10x^9)}{x+x^2+\dotsb+x^6+x^{10}}=\frac{n}{k}.\]](http://latex.artofproblemsolving.com/c/7/c/c7cb9a6c89896713c38293a97adf1c78ffcb2246.png)
In order for this to hold for 
, 
, one must take .
term in![\[(x^1+x^2+x^3+x^4+x^5+x^6+x^{10})^{2018}.\]](http://latex.artofproblemsolving.com/4/e/0/4e03315c499c5fa90d7d7c7ec3b3e47f2bb5ae24.png)
Let the desired number be 
. Note that for positive , we have![\[Nx^{3860} < (x^1+x^2+x^3+x^4+x^5+x^6+x^{10})^{2018}.\]](http://latex.artofproblemsolving.com/d/2/3/d231cc04898a9e19ff96b5cdfd446d07bffc6f07.png)
Taking 
and rearranging gives the desired result.
term inLet the desired number be . Note that for positive , we haveTaking and rearranging gives the desired result.
. Then, 
is equal to the given expression, and is at least 
, where 
is the th degree coefficient and counts the number of satisfactory tuples.
be the coefficient of in 
Note that 
for all positive . If 
, the conclusion follows.
analytic at 
with non-negative coefficients, we can estimate the coefficient on 
for some given by
less than the radius of convergence and ignoring all other terms (e.g. how #13 frames it) through a saddle point of 
on 
and using the Estimation Lemma (e.g. how #7 frames it)
,
![\[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]](http://latex.artofproblemsolving.com/b/7/2/b72b5e302cecdf2649c43a078aea3aa9a88ef2cc.png)
find 
and 
(with multiplicity 3),
(with multiplicity 2),
(with multiplicity 1),
(with multiplicity 3), 
(with multiplicity 2), and 
(with multiplicity 1)?
and sum to 3860. Prove that the cardinality of ![\[2^{3860} \cdot \left(\frac{2018}{2048}\right)^{2018}.\]](http://latex.artofproblemsolving.com/d/e/8/de857941803349e275e9c9aa968a201f2988cb8a.png)
![\[2^{-1} + 2^{-2} + 2^{-3} + 2^{-4} + 2^{-5} + 2^{-6} + 2^{-10} = \frac{2018}{2048}.\]](http://latex.artofproblemsolving.com/d/9/d/d9d6ee4cbf0813e360b26d5e58a6585573a44c36.png)
With this observation, the problem is straightforward. Let 
for convenience and note that
which solves the problem.
holds for all 
.
in ![\[ (2^{-1} + 2^{-2} + 2^{-3} + 2^{-4} + 2^{-5} + 2^{-6} + 2^{-10})^{2018} = \left(\frac{2018}{2048}\right)^{2018} \]](http://latex.artofproblemsolving.com/d/0/2/d0268aff12fae20d67e65d28129d85fa8a94bbcd.png)
As such, it follows that the coefficient can be at most the desired result, or the bound is violated.
