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.
be a group, 
a subgroup of 
and 
an endomorphism with the following property:
such that for any 
there exist 
with 
and 
, for any 
.
is abelian.
is a subgroup of , prove that 
be a quadrilateral inscribed in a circle 
. The lines 
and 
meet at 
, the lines 
and 
meet at 
, and the diagonals 
and 
meet at 
. Let 
be the midpoint of the segment 
, and let be the common point of the segment 
and the circle . Prove that the circumcircle of the triangle 
and are tangent to one another.
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
be an acute triangle with orthocenter , and let 
be a point on the side , lying strictly between 
and 
. The points and 
are the feet of the altitudes from and , respectively. Denote by 
is the circumcircle of 
, and let 
be the point on such that 
is a diameter of . Analogously, denote by 
the circumcircle of triangle 
, and let 
be the point such that 
is a diameter of . Prove that 
and are collinear.
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
. Let the Taylor series of 
be 
. Prove that the number of zeroes of (counted with multiplicities) equals 
.
be an acute triangle. Define 
as its incenter. Let 
and 
be the incircle's tangent points to 
and 
, respectively. Let 
be the midpoint of 
. Let 
be the intersection point of a perpendicular line passing through to 
. Line 
intersects the circumcircle of 
at 
. The circumcircle of 
intersects line 
at 
. Prove that quadrilateral 
is cyclic.
be an abelian additive group such that all nonzero elements have infinite order and for each prime number 
we have the inequality 
, where 
, 
(where the sum has summands) and 
is the order of the quotient group 
(the index of the subgroup 
). of finite index is isomorphic to .
, the diagonal bisects neither the angle nor the angle 
. The point lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines and meet at a point . Let be the point of the line 
, different from , such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
be an acute triangle. Let be the altitude on , and let be any interior point on . Lines 
, when extended, intersect 
at 
respectively. Prove that 
.
, 
, 
be integers for 
. Assume for each 
, at least one of , , is odd. Show that there exists integers 
such that 
is odd for at least 
values of , .
be arbitrary positive integers. For each 
let 
be an 
-tuple of integers, such that at least one of 
is odd. Prove that there exists an -tuple 
of integers such that of the numbers 
at least 
of them are odd.
, 
, as an element of 
. There are 
possible values of 
, and for each of them, we define 
to be the set of all 
's such that 
., its orthogonal complement is a 
-dimensional subspace of , so contains 
vectors. Hence, each does not annihilate exactly of the 
's. Consequently, ![\[\sum_{\mathbf{x}\in\mathbb{F}_2^n\setminus\{\mathbf{0}\}}\big|S_\mathbf{x}\big| = 2^{n-1}N\,.\]](http://latex.artofproblemsolving.com/a/e/7/ae7d0d3a097142aadcf06550142ed2fc68e89584.png)
There must then be a 
such that 
contains at least elements.
. Letting 
represent an odd number and 
represent an even number, the 
possibilities are
represent the number of ordered triples that have parity distribution . Then 
. Now, define triples 
similarly to 
and define
satisfies 
.
to be 
if . 
are odd in any solution.
there is a 
chance of the sum being odd.
are not all zero. Then there are four choices of 
such that 
.
we have that 
has dimension 
,
has dimension 
randomly (uniformly) not all zero, and let 
be Bernoulli random variables for the events 
(i.e. 
is ![$E[X_i]=\frac{4}{7}$](http://latex.artofproblemsolving.com/2/3/9/2393cdf3c54f8691c6c0f5d4d60741210238c4dc.png)
,![$E[X_1+\ldots+X_N]=E[X_1]+\ldots+E[X_N]=\frac{4N}{7}$](http://latex.artofproblemsolving.com/c/1/6/c16f21e5b79608199cb7dd3ba502080bd571cfee.png)
.
is a r.v. giving the number of possible values for 
that gives at least 
values of 
,
and 
Using 
to denote the cardinality of a set 
we partition 
into 
disjoint sets, which we count as follows:
However, one of the hypotheses of the problem was that at each 
or 
Hence, 
and 
and 
,
That gives us 
one of which is 
which would give us no odd numbers. The following table shows how many odd values of 
we get for the given choice of 
![\[\begin{array}{llccccccccccccc}(1,0,0):&\#\text{ of odd values}=&a&&+&&c&+&d&&+&&&&g\\
(0,1,0):&\#\text{ of odd values}=&a&+&b&&+&&d&&+&&f&&\\
(0,0,1):&\#\text{ of odd values}=&a&+&b&+&c&&+&&e&&&&\\
(1,1,0):&\#\text{ of odd values}=&&&b&+&c&&+&&&&f&+&g\\
(1,0,1):&\#\text{ of odd values}=&&&b&&+&&d&+&e&&+&&g\\
(0,1,1):&\#\text{ of odd values}=&&&&&c&+&d&+&e&+&f&&\\
(1,1,1):&\#\text{ of odd values}=&a&&&&&&+&&e&+&f&+&g\end{array}\]](http://latex.artofproblemsolving.com/3/0/2/30216b7efce2b26c4c9bbf751e7062a83d105f4d.png)
Hence the average of these 
which means that at least one of these numbers must be at least 
![\[(1, 1, 1)\]](http://latex.artofproblemsolving.com/f/8/3/f8320925706879c8fb6212e450664f974416aa0d.png)
![\[(1, 1, 0)\]](http://latex.artofproblemsolving.com/e/5/3/e537affbd97de76a50123b265be61b64641c7990.png)
![\[(1, 0, 1)\]](http://latex.artofproblemsolving.com/a/6/8/a68ba5cd2a133b12308752f7422b84737fce309a.png)
![\[(0, 1, 1)\]](http://latex.artofproblemsolving.com/2/8/4/28457b2c6f9e37d09df669d99058b6433366e37d.png)
![\[(1, 0, 0)\]](http://latex.artofproblemsolving.com/3/c/c/3cc44dfc830033cc711082a9031ec05d9b591909.png)
![\[(0, 1, 0)\]](http://latex.artofproblemsolving.com/5/7/b/57b068a6761bf4b2e548e617c9d062eadf5c4027.png)
![\[(0, 0, 1).\]](http://latex.artofproblemsolving.com/4/1/3/413ad5190515e89657fbc859e80ebb5996bb4fcc.png)
top-down. Let 
represent the number of ordered triples 
.
as follows:![\[r_1 = (1, 1, 1)\]](http://latex.artofproblemsolving.com/a/3/4/a345e8ef231c31fd8c67dfe86252b06836e8f871.png)
![\[r_2 = (1, 1, 0)\]](http://latex.artofproblemsolving.com/f/9/1/f914d5eb56f802bcc8bea17f35117291410cf170.png)
![\[r_3 = (1, 0, 1)\]](http://latex.artofproblemsolving.com/9/a/e/9ae73065b18b465a667dc300439cd11ab4506f3c.png)
![\[r_4 = (0, 1, 1)\]](http://latex.artofproblemsolving.com/5/0/4/50486e1325aa19c59f85d5b63633d92a0ec1f872.png)
![\[r_5 = (1, 0, 0)\]](http://latex.artofproblemsolving.com/2/e/5/2e51aa6667c78d698b84b5f0f65f7382562d8f05.png)
![\[r_6 = (0, 1, 0)\]](http://latex.artofproblemsolving.com/e/6/4/e64e51121395e099fbdc4d30cf322046b5ca603b.png)
![\[r_7 = (0, 0, 1).\]](http://latex.artofproblemsolving.com/5/e/8/5e8f890e1f03ab2683373f3a4de17cb2dd9378db.png)
![\[s_1 = a_1 + a_5 + a_6 + a_7\]](http://latex.artofproblemsolving.com/3/5/b/35be8dbaffbacfbb132443cfd63e4402381dbe56.png)
![\[s_2 = a_3 + a_4 + a_5 + a_6\]](http://latex.artofproblemsolving.com/8/4/d/84dcb73d60027934ce1ed24610527ca06067852f.png)
![\[s_3 = a_2 + a_4 + a_5 + a_7\]](http://latex.artofproblemsolving.com/8/5/c/85c1cba7a9ed7b9189a1fa2da9c144b61098c568.png)
![\[s_4 = a_2 + a_3 + a_6 + a_7\]](http://latex.artofproblemsolving.com/1/6/5/165bb84c44d457f2b9e54b1f06cb8e0e3d361c38.png)
![\[s_5 = a_1 + a_2 + a_3 + a_5\]](http://latex.artofproblemsolving.com/3/6/2/3621733745cbefe554f8c1b1b41dd50b8c12edde.png)
![\[s_6 = a_1 + a_2 + a_4 + a_6\]](http://latex.artofproblemsolving.com/9/3/f/93f7bb08558fc005b6ff690431fdc6b8b0a8dea0.png)
![\[s_7 = a_1 + a_3 + a_4 + a_7.\]](http://latex.artofproblemsolving.com/e/3/4/e34ae718b22b292751c6e4252b3e6a2bc3c31d99.png)
![\[s_1 + s_2 + \cdots + s_7 = 4N.\]](http://latex.artofproblemsolving.com/5/2/8/528009eb9e0aa620a30f4e011a409e3511e5ec46.png)
be 
.
,
,![\[\frac 17 \cdot \frac 47 + \frac 37 \cdot \frac 47 + \frac 37 \cdot \frac 47 = \frac 47,\]](http://latex.artofproblemsolving.com/1/2/e/12e1de41c76b01262653b03bcb8b970e0fcf34b1.png)
