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 
connect the points 
and 
These arcs lie in the same half-plane defined by line 
in such a way that arc 
lies between the arcs 
and 
Point 
lies on the segment 
Let 
, and 
be three rays starting at 
lying in the same half-plane, 
being between 
and 
For 
denote by 
the point of intersection of 
and 
(see the Figure below). Denote by 
the curved quadrilateral, whose sides are the segments 
and arcs 
and 
We say that this quadrilateral is 
if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals 
are circumscribed, then the curved quadrilateral 
is circumscribed, too.
be integers, let 
be a set with 
elements, and let 
be pairwise distinct non-empty, not necessary disjoint subset of . A function 
is called nice if there exists an index 
such that ![\[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]](http://latex.artofproblemsolving.com/3/d/a/3da5ad1bcade98ee08caa35763de226027bcb2c6.png)
Prove that the number of nice functions is at least 
.
, , and 
are fixed on a line in this order. Let 
be a circle passing through and whose center does not lie on the line . Denote by 
the intersection of the tangents to at and . Suppose meets the segment 
at 
. Prove that the intersection of the bisector of 
and the line does not depend on the choice of .
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is
+1 w
. Prove that for any given positive integer , the product
has at most distinct prime divisors.
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
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
Prove that
Prove that
Where 
satisfying
Prove
