s.t 
be a parallelogram with 
A point 
is chosen on the extension of ray 
past 
The circumcircle of 
meets the segment 
again at 
The circumcircle of triangle 
meets the segment 
at 
Prove that lines 
are concurrent.
numbers 
are written on the blackboard . An allowed operation consists of choosing two different numbers 
and 
of the blackboard such that 
,
. Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
- Greatest common divisor of numbers 
be a closed map with the property that the inverse image of each point in 
is a compact subset of 
.
is compact whenever 
is compact in 
and 
.
.
,
is a finite sequence of positive integers 
,
;
,
.
-
;
;
;
.
is a finite sequence of positive integers 
such that:
;
;
each 
is in the sequence 
defined by 
,
,
;
if 
,
is a subsequence. For example, there are five 
;
;
;
.
the number of type 
and 
.
such that
Prove that
derivable on 
with 
.
such that 
is uniformly continuous on 
.
satisfied
for all 
and all 
.
be defined by 
(a) Show that 
has exactly one point of discontinuity.
Y by Adventure10
Q1) Let 
be 3 distinct mumbers from {1,2,3,4,,5,6}. Show that 7 divides 
.
Q2) Show that there is a natural number
such that the number 
ends exacly in 2009 zeros.
Q3) Let a,b,c be positive integers with no common factor and satisfy the conditions
. Prove that a+b is a square.
Q4) Suppose that
, where 
. Prove that a is divisible by 23 for any positive integer n.
Q5) Prove that
is divisible by 42 for any positive integer m.
Q6) Suppose that 4 real numbers a,b,c,d satisfy the conditions
Find the set of all possible values the number M=ab+cd can take
Q7) Let a,b,c,d be positive integers such that a+b+c+d=99.Find the smallest and the greatest values of the following product P=abcd.
Q8) Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is three times less than 1004.
Q9) Give an acute-angled triangle ABC with area S, let points A',B',C' be located as follows: A' is the point where altitude from A on BC meets the outwards facing semicirle drawn on BX as diameter.Points B',C' are located similarly. Evaluate the sum
.
Q10) Prove that
,where d is diameter of the inscribed cricle of 
.
Q11) Let A={1,2, . . . ,100} and B is a subset of A having 48 elements. Show that B has two distint elements
and 
whose sum is divisible by 11.
be 3 distinct mumbers from {1,2,3,4,,5,6}. Show that 7 divides 
.Q2) Show that there is a natural number
such that the number 
ends exacly in 2009 zeros.Q3) Let a,b,c be positive integers with no common factor and satisfy the conditions
. Prove that a+b is a square.Q4) Suppose that
, where 
. Prove that a is divisible by 23 for any positive integer n.Q5) Prove that
is divisible by 42 for any positive integer m.Q6) Suppose that 4 real numbers a,b,c,d satisfy the conditions
Find the set of all possible values the number M=ab+cd can take
Q7) Let a,b,c,d be positive integers such that a+b+c+d=99.Find the smallest and the greatest values of the following product P=abcd.
Q8) Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is three times less than 1004.
Q9) Give an acute-angled triangle ABC with area S, let points A',B',C' be located as follows: A' is the point where altitude from A on BC meets the outwards facing semicirle drawn on BX as diameter.Points B',C' are located similarly. Evaluate the sum
.Q10) Prove that
,where d is diameter of the inscribed cricle of 
.Q11) Let A={1,2, . . . ,100} and B is a subset of A having 48 elements. Show that B has two distint elements
and 
whose sum is divisible by 11.
with the area 
. Let 
be located as follows: 
is the point where altitude
on 
meets the outwards facing semicirle drawn on ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
as diameter. Points 
and 
are located similarly.![$T=[BCA_1]^2+[CAB_1]^2+[ABC_1]^2$](http://latex.artofproblemsolving.com/c/b/f/cbf63b73acd4959c04538c0d447e051516a3fd56.png)
, where denoted ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of ![$4\cdot \sum\sin A\cos B\cos C=2\cdot \sum\sin A\left[\cos (B+C)+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/f/4/6/f46411773081f30cabc9df6ebfff1f5c5964f90c.png)
![$2\cdot \sum\sin A\left[-\cos A+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/1/2/0/120d35382ab946fbceb8202916b3acb82ecb0e77.png)
![$\sum \left[-\sin 2A+2\sin (B+C)\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/e/3/0/e30f439cadb9cc95cbebff8352fb5dbd2355a834.png)
.
. So 
![$[BCA_1]^2=\left(\frac 12\cdot BC\cdot DA_1\right)^2=$](http://latex.artofproblemsolving.com/2/0/e/20e4047abe7bd7a93df385a945d8f2848eca6c8f.png)
![$\sum [BCA_1]^2=\frac {abc}{4}\cdot\sum a\cdot\cos B\cos C=$](http://latex.artofproblemsolving.com/7/c/d/7cdb7c672f76e1fbbd15a386d968b96c3117321c.png)
![$\boxed{\sum [BCA_1]^2=2R^2S\cdot \frac {S}{2R^2}=S^2=[ABC]^2}$](http://latex.artofproblemsolving.com/7/5/f/75fb9c79561c2b67d6c6992342c57e7671b06d76.png)
. Otherwise, 
.
,
.
has exactly 2008 zeros and the number 
has exactly 2010 zeros, and since the function 
,
,
But since 
,
,
,
,
,
,
.
![$\sqrt[4]{abcd} \leq \frac{99}{4} \Leftrightarrow abcd \leq \frac{99^4}{16}$](http://latex.artofproblemsolving.com/6/a/e/6aef3fe61b96c49f5afdac20505556b4a869feb3.png)
but this is assuming 
,
and so maximum will be at 
,
(I offer no proof, it is something very intuitive, I know that wouldn't probably suffice at olympiad. It is easy to prove for 
variables), and so maximum is 
.
,
and so minimum is 
.
?
and 
divids both 
divides all three of 
and 
s
(
rather than 
from the way the problem was stated.
and 
the orthocenter of ![$\frac{[A'BC]^2}{[ABC]^2}=\frac{A'D^2}{AD^2}$](http://latex.artofproblemsolving.com/e/3/b/e3b81f3113755c887c912afa389dc1e865933205.png)
,
,
,![$\frac{[A'BC]^2}{[ABC]^2}=\frac{DH\cdot AD}{AD^2}=\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$](http://latex.artofproblemsolving.com/5/4/4/544aa54683c42e9d35c7b5143752383e6c3efa62.png)
.![$\frac{\Sigma{[BCH]}}{[ABC]}=\frac{[ABC]}{[ABC]}=1$](http://latex.artofproblemsolving.com/f/1/1/f11cd513b58867ecb83cf48ccdbd5fed00f19ffd.png)
.
.
and 
because of factor 
.
,
,
are clear again from factor 
,
.
,
.
,
,
and so 
.
.
,
in order for the sum 
to be divisible by 
.
elements of 
which give a remainder of 
or 
when divisible by 
when divisible by 
,![$area [ABC]^{2}$](http://latex.artofproblemsolving.com/9/c/0/9c060357345d784b54636ffebf466e2541828657.png)
,
and 
satisfy given system of equations.
