AoPS Academy
chips, numbered 
, 
, 
, 
, and . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds . What is the probability that draws are required?
be a convex quadrilateral with 
and 
Suppose that the centroids of 
and 
form the vertices of an equilateral triangle. What is the maximum possible value of the area of ?
be real positive number satisfy that: 
. Prove that: 
, where 
integers and 
is a positive integer. Suppose that the integers
satisfy: 
are divisible by k. Prove that:
is divisible by k
such that 
and
.
, let 
denote the number of elements in . FInd the number of ordered pairs 
such that 
and 
are (not necessarily distinct) subsets of 
that satisfy
or 
then subtract 
. Can someone please explain how to get 
?
and realize that either 
or :
possibilities of where 
Times that by then add in the possibilities where 
to get 
, so 
(use Vieta). This implies either 
or 
. has three possibilities: in both and , in but not , or in neither nor . This gives 
possibilities in that case. Analogously, possibilities in the second case.
cases where 
gives a final answer of 
.
the number of elements in but not , 
the number of elements in but not , and 
the number of elements and . Then we get 
so 
.
, then each of the remaining numbers must be in or or neither, meaning ways.
, it is also ways.
, each number can either be in or not in , so 
.
.
all the time, and if 
and 
then 
so either 
or 
is equal to 
and then proceed as others said.
, if the products are equal as well, then by Vieta formula it follows that and are roots of same polynomial as and 
, hence the main claim.
. Can someone please explain how to get ?
. Can someone please explain how to get ?
.
is -4.
is incorrect?
is incorrect? and dont have to be nonempty
is empty.
. has one element.
. is not empty, then 
. Any such works. So there are 
ways. is empty, then there is way.
. has elements.
. contains both and , then all such work, giving 
. doesn't contain or , then is empty, giving . contains but not , then 
. Thus, there is way. contains but not , then there is way.
. has elements.
.
, then is empty. way.
, then 
. ways.
, then 
. ways.
, then there are total ways, all work.
.
.
., then is empty. way., then 
. ways., then 
. 
ways., then 
. ways.
, then there are ways.
.
.
. works, giving a total of .
.
with 
in my previous solution. or and everything of this form works..
ways..
ways.. ways.
.
. Hence![\[ |A|\cdot |B| = |A\cap B|\cdot (|A| + |B| - |A\cap B|).\]](http://latex.artofproblemsolving.com/8/4/7/847b6fd785f4fdc245483377a0cd5984d18976aa.png)
, 
, and 
. Then![\[ ab = ai +bi - i^2\]](http://latex.artofproblemsolving.com/8/0/f/80f5031b554443374f97ee97d6f148c66a5bab26.png)
![\[ a(b-i) = bi - i^2 = i(b-i).\]](http://latex.artofproblemsolving.com/6/e/9/6e9fb450b6f9445705a52483d952a67ee296d77c.png)
or 
. The only way for this to happen is either if or . WLOG, let (we will deal with some complications later). If has 
elements, there are 
ways to choose . Then there are 
ways to choose . The total number of ways is![\[ \sum_{j = 0}^{5}2^j\binom{5}{j} = 1 + 10 + 40 + 80 + 80 + 32 = 243.\]](http://latex.artofproblemsolving.com/d/5/3/d5310b63400afc67e1166e30499c83c3042b59fc.png)
ways, but we overcounted the case once, which has ways of happening. The answer is 
.
, let denote the number of elements in . FInd the number of ordered pairs such that and are (not necessarily distinct) subsets of that satisfy
Do a bit of casework to get 
Multiply by because 
We've counted twice here because it doesn't replicate; subtract by the number of times once ( for correction), and we're done.
or is a subset of the other. How I found this was letting 
. Then denoting , and 
we have,
and hence either 
, 
or 
. Now its just a simple counting problem.
then we must have . This can be done in 
ways. is a subset of . Then do casework on the size of .
then 
, we've already counted this.
we have choice for adding to our count.
we have 
choices for giving 
to our count.
we have 
choices for giving 
to our count.
we have 
choices for giving 
to our count.
we have 
choices for giving 
to our count. we have 
in this case.
.
where 
is a positive integer.
. The equation provided in the problem is equivalent to
Suppose . This implies that 
. Observe that for every element in , it is either in , the set when 
is removed, or in neither. Hence, we obtain 
possibilities. Similarly, gives . Now we subtract off the duplicates. If 
, , so there are 
ways to choose the common subset. We obtain our final answer of 
. (It can easily be checked that 
yields , the answer to the original problem.)
.
or
, we get that . So, number of pairs 
. pairs. such pairs where .
.
holds inequality 
and 
Prove that
and 
Prove that![$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$](http://latex.artofproblemsolving.com/5/6/1/561a990e5885863c77d454ffa3f1973e1bbf60b4.png)
![$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$](http://latex.artofproblemsolving.com/4/b/1/4b14911672688250c909571d788aaf10724462eb.png)
Where 
which can be obtained by a casework bash on 
(I heard there's a slick solution but I haven't figured it out yet.) Unfortunately during the test I got 454 but changed my answer later 
sets where 
,
or 
,
,
.
,
.
and 
.
is a permutation of 
.
or not.
.
then by Viète’s Formulas 
and 
.
and 
or 
and 
i.e. either 
or the other way around. Let 
.
where 
for 
where the first part comes from choosing 
.
so 
but we counted the case where 
times. The answer is 
.
.
and 
.
,
,
,
,
is a permutation of 
.
,![\[ |{A \cup B}| \geq |A|, |B| \geq |{A \cap B}| \]](http://latex.artofproblemsolving.com/5/1/9/5196df29ef4fac420e0d8a641028d94bfe2909ec.png)
for inequality reasons we are forced to have 
meaning either 
so it follows that either 
This implies that either 
If 
then each of the numbers 
Hence, the answer is
