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.
a list of 
real numbers 
is said to be area definite for 
if the inequality ![\[\sum_{1\le i<j<k\le m}a_{ijk}\cdot\text{Area}(\triangle A_iA_jA_k)\ge0\]](http://latex.artofproblemsolving.com/e/7/7/e778445aa49547e4a227d16cc8685370a5d1a7d9.png)
holds for every choice of 
points 
in 
For example, the list of four numbers 
is area definite for 
Prove that if a list of numbers is area definite for 
then it is area definite for 
![$f,g \in C^1([0,1])$](http://latex.artofproblemsolving.com/6/1/7/617c5be50035205b21074842d7266911aa1dc4d8.png)
.
?
be a ring with unity such that for every 
there exist 
such that 
. Prove that
then 
such that 
then the result from a) may no longer hold.
there exists an -digit number divisible by 
all of whose digits are odd.
, 
, 
, 
be distinct positive integers, 
. Prove that there exist distinct indices 
and 
such that 
does not divide any of the numbers 
, 
, , 
.
, show that![\[
\sum_{i=1}^{n} \tan^{-1} r_i = \tan^{-1} \frac{a_1 - a_3 + a_5 - \cdots}{a_0 - a_2 + a_4 - \cdots}
\]](http://latex.artofproblemsolving.com/8/1/2/812c7c690b2237f45bc7453ff2b4160fb72c7b0d.png)
and![\[
\sum_{i=1}^{n} \tanh^{-1} r_i = \tanh^{-1} \frac{a_1 + a_3 + a_5 + \cdots}{a_0 + a_2 + a_4 + \cdots}.
\]](http://latex.artofproblemsolving.com/6/1/9/61962993a2c34e1c0ce961d7057127ae75876ac2.png)
, 
, 
be fixed positive integers. There are 
ducks sitting in a ducks ducks picking paper, and ducks picking scissors., , and , the maximum number of moves which could take
be the circumcenter of triangle 
. Line 
intersects the altitude from at point 
. Let 
be the midpoints of 
, 
respectively. If 
intersects 
at 
, and the circumcircle of triangle 
meets 
at 
, prove that 
is cyclic. be a triangle with incentre 
and circumcentre . Let 
be the touchpoints of the incircle with , 
, respectively. Prove that 
is the Euler line of 
.
with its incircle 
. Let 
and 
be the points of tangency of with and respectively. On the edges 
and 
, take points 
and 
such that 
is tangent to at 
. Suppose 
is the intersection point of the lines 
and 
. Prove that two lines 
and 
are parallel or coincide.
is a finite set, let denote the number of elements in . Call an ordered pair 
of subsets of 
if 
for each 
, and 
for each 
. How many admissible ordered pairs of subsets 
are there? Prove your answer.
is a finite set, let denote the number of elements in . Call an ordered pair of subsets of if for each , and for each . How many admissible ordered pairs of subsets are there? Prove your answer.
where 
denotes the Fibonacci sequence, with 
and 
. For some , by summing over 
and 
, it is clear that there are
ways to choose and 
. Thus consider the formal series
Then, we have
where 
and 
are the roots of 
. By the geometric series formula it follows that 
.
and 
, so we have
as desired. 
We sum this by parts according to 
. Eaach of these parts is equal to 
Here 
and 
. This is just the 
coefficient in 
Expanding out the series for this, the desired coefficient is 
. Hence, the answer is 
and are 
and 
respectively, we find we need to choose numbers from 
numbers to fill set and numbers from 
numbers to fill set . Thus we have 
choices for and 
choices for . Note we have the added restriction that 
, else the binomial coefficients are undefined.
Then we will use snake oil. The idea is to let 
, and consider the generating function,
Now by substituting 
we find,
Now we find,
This then simplifies to,
Which in turn becomes,
where we take 
and 
. Then we can use geometric series to find,
Now we want to find the coefficient of 
. Thus the coefficient of is,
, 
.
for each 
, is a subset of 
(
if 
).
These are 
questions in total. 2's and 
1's.
These are 
questions in total. 2's and 1's.
2's and 
1's, which sum to 
.. must have an odd number of 1's, and the middle 1 is the separator. and can be determined, and the two sets can be uniquely recovered.
by using partial fractions and then comparing with the formula of Fibonacci numbers.
, where 
denotes the th Fibonacci number.![\[S_{n} = \sum_{A \geq 0}\sum_{B \geq 0} \binom{n - B}{A} \binom{n - A}{B}\]](http://latex.artofproblemsolving.com/d/c/9/dc9df6659ce077dccdaadd74111c9b6b724bb839.png)
and 
, deriving a closed form for this given and , and sum over all possible pairs of and . Note that for valid elements of set , we must choose elements that are NOT in the range 
to ; thus, there are 
combinations. The same logic holds for choosing elements from set . Since these selections are independent, we multiply these to obtain the result![\[\binom{n - B}{A}\binom{n - A}{B}\]](http://latex.artofproblemsolving.com/6/e/e/6eeca201ea388918aa768fc59c27fa73a0583e5d.png)
, we use generating functions. Define 
to be![\[F(x) = \sum_{n \geq 0} S_{n} x^n \]](http://latex.artofproblemsolving.com/b/e/0/be0f26d9cd03fb67ea7132dcc54ebe3d0222ffe8.png)
![\[F(x) = \sum_{n \geq 0} \sum_{A \geq 0}\sum_{B \geq 0} \binom{n - B}{A} \binom{n - A}{B} x^n\]](http://latex.artofproblemsolving.com/b/1/7/b17eb5eb9dc516df6b0f397e0aea35d0935acd55.png)
to obtain![\[F(x) = \sum_{A \geq 0} \sum_{B \geq 0}\sum_{n \geq 0} \binom{n + A}{A} \binom{n + B}{B} x^{A + B + n}\]](http://latex.artofproblemsolving.com/b/0/7/b075be07acfa23f55afce3df46c81f8f9fe0a1c5.png)
![\[F(x) = \sum_{n \geq 0} x^{n} \sum_{B \geq 0} \binom{n + B}{B} x^{B} \sum_{A \geq 0} \binom{n + A}{A} x^{A}\]](http://latex.artofproblemsolving.com/3/7/e/37ed4f6ebe5630501f7034235309a73e60424720.png)
![\[F(x) = \sum_{n \geq 0} x^{n} \left( \sum_{A \geq 0} \binom{n + A}{A} x^{A} \right)^2 \]](http://latex.artofproblemsolving.com/d/9/6/d9624ea79f8fff1b9d9252acaf910d7a55522b77.png)
![\[F(x) = \sum_{n \geq 0} x^{n} \left( \frac{1}{(1 - x)^{n + 1}} \right)^2 \]](http://latex.artofproblemsolving.com/b/6/2/b6210bad86dd7dc403873c0daa2d130ad147c6cb.png)
![\[F(x) = \frac{1}{(1 - x)^2} \sum_{n \geq 0} \left( \frac{x}{(1 - x)^{2}} \right)^n \]](http://latex.artofproblemsolving.com/d/3/4/d3477946708b6bc5ac711c146c035ab67ed7aa1e.png)
![\[F(x) = \frac{1}{(1 - x)^2} \cdot \frac{1}{1 - \frac{x}{(1 - x)^2}} \]](http://latex.artofproblemsolving.com/d/a/7/da7ded8d3aab234743d1dee319bc3cb5d722cf6f.png)
![\[F(x) = \frac{1}{(1 - x)^2 -x} = \frac{1}{x^2 - 3x + 1} \]](http://latex.artofproblemsolving.com/9/f/9/9f942c864a93f4f803134bdc5759ae74ff065fad.png)
and 
; these are the roots of the quadratic in the denominator of . Then can be written as![\[F(x) = \frac{1}{\sqrt{5}} \left(\frac{1}{x - \alpha} - \frac{1}{x - \beta}\right)\]](http://latex.artofproblemsolving.com/9/e/b/9eb4d684d23dce1c6288e01620e2edf6dbb807ac.png)
![\[F(x) = \sum_{n \geq 0} \frac{1}{x\sqrt{5}} \left(\frac{1}{1 - \frac{\alpha}{x}} - \frac{1}{1 - \frac{\beta}{x}}\right) x^n\]](http://latex.artofproblemsolving.com/0/7/a/07ab5c3952716a7dd40f46f554a216f3d60c5725.png)
with 
:![\[F(x) = \sum_{n \geq 0} \frac{1}{x\sqrt{5}} \left(\frac{1}{1 - x \alpha} - \frac{1}{1 - x \beta}\right) x^n\]](http://latex.artofproblemsolving.com/c/9/c/c9cdf6fa3cbf4b0b998faa45ef9a0d8fe9eea6c0.png)
![\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \alpha^{n + 1} + \beta^{n + 1} \right) x^n\]](http://latex.artofproblemsolving.com/5/8/4/584a953a5cb09fcc2c7506925518c9cb51fb31d6.png)
![\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \left(\left(\frac{1 + \sqrt{5}}{2}\right)^2\right)^{n + 1} + \left(\left(\frac{1 - \sqrt{5}}{2}\right)^2\right)^{n + 1} \right) x^n\]](http://latex.artofproblemsolving.com/b/d/c/bdccf0ab67073202d5c8953735810393788190b3.png)
![\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \left(\frac{1 + \sqrt{5}}{2}\right)^{2n + 2} + \left(\frac{1 - \sqrt{5}}{2}\right)^{2n + 2} \right) x^n\]](http://latex.artofproblemsolving.com/b/1/f/b1fa42910348346a2c4be36678bd3545274e74ee.png)
![\[F(x) = \sum_{n \geq 0} F_{2n + 2} x^n\]](http://latex.artofproblemsolving.com/2/e/a/2eaec20ab925157b9eb80d1ad5cde1db2f11991b.png)
, as desired.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
and 
.
,
is the number of admissible pairs. This function can also be expressed as
![\[\sum_{a\ge 0}\sum_{b\ge 0}\binom{n-b}a\binom{n-a}b=\sum_{a\ge 0}\sum_{b\ge 0}\binom{n-b}{n-a-b}\binom{n-a}{n-a-b}=\sum_{k\ge 0}\sum_{a+b=k}\binom{n-b}{n-k}\binom{n-a}{n-k}.\]](http://latex.artofproblemsolving.com/9/c/f/9cf2399fd9b8b54fa4789aeeaec18329b0df1e8a.png)
Now we claim ![\[\sum_{a+b=k}\binom am\binom bn=\binom{k+1}{m+n+1}.\]](http://latex.artofproblemsolving.com/2/c/8/2c82c78c07367c516bb11d7103598c374642f8a1.png)
This follows from ![\[\sum_{k\ge 0}\sum_{a+b=k}\binom am\binom bnx^k=\left(\sum_{a\ge 0}\binom amx^a\right)\left(\sum_{b\ge 0}\binom bnx^b\right)=\frac{x^m}{(1-x)^{m+1}}\cdot\frac{x^n}{(1-x)^{n+1}}=\frac{x^{m+n}}{(1-x)^{m+n+2}}=\sum_{c\ge 0}\binom{c+1}{m+n+1}x^c.\]](http://latex.artofproblemsolving.com/d/b/1/db1be8017865c4042b40acc576329778fa99e6d1.png)
Thus we want ![\[\sum_{k\ge 0}\binom{2n-k+1}{2n-2k+1}=\sum_{k\ge 0}\binom{2n-k+1}k.\]](http://latex.artofproblemsolving.com/7/c/c/7cc7466ff7a5f6918d756e6af5dd75b39ea55cc2.png)
Now we claim ![\[\sum_{k\ge 0}\binom{a-k}k=F_{a+1}.\]](http://latex.artofproblemsolving.com/b/1/f/b1ff51ac1dcc4bfc512423a7f36dfe2304b76058.png)
This is because ![\[\sum_{a\ge 0}\sum_{k\ge 0}\binom{a-k}kx^a=\sum_{k\ge 0}\sum_{a\ge 0}\binom{a-k}kx^a=\sum_{k\ge 0}\frac{x^{2k}}{(1-x)^{k+1}}=\frac1{1-x}\left(\frac1{1-\frac{x^2}{1-x}}\right)=\frac1{1-x-x^2}=\sum_{a\ge 0}F_{a+1}x^a.\]](http://latex.artofproblemsolving.com/a/6/f/a6ff036eb071ae126452827e9fbbd9ed47db6a08.png)
Thus the answer is 
then 
and 
Thus, it follows that the number of admissible pairs is 
We wish to separate 
such that 
We can let 
both of the binomial coefficients will have a value of 
It follows that our expression becomes 
so if we let 
be the number of admissible pairs of subsets from the set 
then we get
Partially-decomposing this yields
where 
so it follows that 
Manipulating this using Binet's Formula (motivated by the fact that 
and 
)
