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.
such that 
is a positive integer.
be positive real numbers such that 
. Prove that
.
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
such that 
for every 
.
and 
be polynomials with integer coefficients. Suppose that the integers 
and 
are roots of , and that 
. Prove that the equation 
has no integer solutions.
. Find .
be three positive real numbers with 
. Find the minimum value of the expression 
is strictly increasing : 
such that:
such that for all 
. 
.
satisfying the following conditions: divides 
,
divides 
,
.
+1 w
be positive integers and let 
be a non-increasing list of positive real numbers. Prove that there exists 
sets 
which partition the set 
such that 
be the minimum of the RHS such that 
for all 
.
proves itself. The case 
is also trivial as the sets are just the elements and 
then we can make a set with just 
and then one realizes that the rest of the inequalities are the same as the case 
and 
so we can induct down on until we have that all 
are smaller than .
to 
, respectively. We then preform the greedy algorithm, assigning the remaining elements until each set has a sum of at least . The amount of 'waste' in this process is at most 
. So it suffices to show that 
However we know that 
so it suffices to show that 
If this holds then we are done so assume it does not. If 
then we have a contradiction by assumption. Notice that we have that 
and so we can still come up with a partition if 
by pairing elements which clearly must hold.
.
, then we can group them into groups so that each sums to 
. In this case, we will be done, simply with the 
term on the RHS.
. Then, 
. We now put by itself in the partition. Since , this means that it is definitely not the smallest amongst 
. Hence, we can effectively remove it from the LHS of the inequality. It suffices to show the exact same inequality with replaced by 
and replaced by 
, and so we are done by induction on .
be :
,
blocks must be 
,and we must try for the other groups to have "even" number of elements.
we exclude elements 
from 
such that :
and 
we take the excluded elements from 
such that:
by excluding and including elements which are following certain basic rules.
observe that:
,hence it means we can exclude those elements also from the last such that its sum <
, such that sum <
and exclude them also.and continue our process. set then it may have size problem so we can do these steps but just fill from the 
set.
.
Let 
.
and 
,
.
.
(note 
)
However, it is also, by definition, 
,
or 
,
,
are similar.