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 real numbers such that 
for all 
.
.
and 
be touch points of the incenter of 
at 
and 
, respectively. Let 
and 
be the circumcenter of triangles 
and 
, respectively. Show that 
and 
concurrent.
of one variable that fullfill the following for all real numbers 
and 
:
.
be a nonempty set of positive integers such that:
if 
then 
. for any prime 
, there exists 
such that 
. is finite.
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer 
such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture and replaced with same amount of cocoa powder to form a new mixture 
If the ratio of coffee and cocoa in the mixture 
is 
then the ratio of cocoa in mixture to that in mixture is
which satisfy the following equality for all 
![\[f(x)f(y)-f(x-1)-f(y+1)=f(xy)+2x-2y-4.\]](http://latex.artofproblemsolving.com/2/0/a/20ae7491750bac4c94deaed32d5b9aa66b26491f.png)
Proposed by Dániel Dobák, Budapest
such that 
for all 
and 
and coefficients in the set 
. Prove that the probability for this polynomial to be special is between 
and 
, where a polynomial 
is called special if for every 
in the sequence 
there are infinitely many numbers relatively prime with 
.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let 
be the midpoint of the side 
. Let 
be a point on the side and 
a point on the side 
such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.
Then there are two cases either 
or 
.When we are admitting the case we cannot divide by 
. So, in the case we apply divison by then is a solid prerequisite to do so.Thus, 
from 
we must take the assumption in hand that . For example take the extension of the same problem in 
then the same problem reads 
,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in 
. Thankfully, we are lucky enough that we are in the field of 
.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of .
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ..
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ...The assumption whenever we get 
else we get the case .I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.
Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the from each of the terms on the left hand side and get which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of and and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. 
is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution is not actually lost. The reason we are getting but not is beacuse when we are dividing by we making an assumption that and since we are making this assumption the solution is lost. For example when we divide by 
the solution is lost why 
and we are assuming 
which is equivalent to 
. Now divison by zero is not possible which is not at all very easy to explain. Now and is not possible at the same time. Thus either or . and 
can be same when we are in a field of 
. If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that 
is possible in finite fields of charecteristic .
. Prove that
