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.
with 
, let its incircle be tangent to sides 
, 
, and 
at 
, 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of .
greater than 
is divisible by 
be a convex quadrilateral with pairwise distinct side lengths such that 
. Let 
be the circumcenters of 
, respectively. Show that 
, the Euler line of 
and the Euler line of 
are concurrent.
such that 
is the square of an integer.
+1 w
be the circumcenter and 
the orthocenter of an acute triangle . Show that there exist points , , and on sides , , and respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](http://latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
be finite.
satisfying the following condition for all real numbers 
:![\[
f(x - 1 - f(x)) = f(x) - x - 1.
\]](http://latex.artofproblemsolving.com/3/a/5/3a51479b61a5c025ed81f539e964ce96b5857f0f.png)
of positive integers such that ![\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]](http://latex.artofproblemsolving.com/e/c/7/ec778edf97100f3b19350d8a817317ebf1f391bd.png)
be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
. be a right angled triangle with 
.Let 
be the midpoint of , and 
be an arbitrary point on 
. The reflection of 
over intersects lines 
and at and 
, respectively. The circumcircles of 
and intersect again at . Prove that the center of the circumcircle of 
lies on 
.
and be on segment of an acute triangle such that 
and 
. Let and 
be the points on 
and 
, respectively, such that is the midpoint of and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
The last line only since I like these variables better, heh. Let 
[which is called the Liouville Function] then we can see that 
is multiplicative, as 
. But by Dirichlet Convolution ![\[ \sum_{d \mid n} \lambda(d) = \lambda \ast \textbf{1} \]](http://latex.artofproblemsolving.com/8/3/f/83f7786937968849d11e513011352975cbddf8ea.png)
we can see that 
is also multiplicative. So it suffices to look at prime powers 
. It's now easy to conclude that ![\[ \sum_{d \mid n} \lambda(d)= \begin{cases} 1 \text{, if } n \text{ is a perfect square}
\\ 0 \text{, else.} \end{cases}\]](http://latex.artofproblemsolving.com/6/9/a/69ac3eb71aac8e945da61a5fc46e0917a77ea671.png)
So it suffices to count the number of perfect squares smaller than 
and well 
, hence the answer is 
.
is multiplicative. We have that![\[\displaystyle\sum_{n=1}^{1989}{(-1)^{\Omega(n)}\lfloor\frac{1989}{n}\rfloor} = \displaystyle\sum_{n=1}^{1989}{\displaystyle\sum_{n\mid m, m\leq 1989}{(-1)^{\Omega(n)}}}\]](http://latex.artofproblemsolving.com/1/d/0/1d08d012ecc2a91a2274fd6d81ff99dda96e42e3.png)
We can switch the order of the summation to make it![\[\displaystyle\sum_{m=1}^{1989}{\displaystyle\sum_{n\mid m}{(-1)^{\Omega(n)}}} = \displaystyle\sum_{m=1}^{1989}{\textbf{1}*(-1)^{\Omega(n)}}\]](http://latex.artofproblemsolving.com/6/0/3/603147c02250d726a594c2f7acb538c0d9704ef6.png)
Because is multiplicative, 
must also be multiplicative. For a prime power 
, note 
is if 
is even, and 
if is odd. Thus, we have that is if is a square and otherwise.![\[\displaystyle\sum_{m=1}^{1989}{\textbf{1}*(-1)^{\Omega(n)}} = \lfloor\sqrt{1989}\rfloor = \boxed{44}\]](http://latex.artofproblemsolving.com/4/b/7/4b7ce3a04966a30cd2da183013ebe57c19dd4110.png)
Now let 
where 
are distinct primes and 
are positive integers. Since is mutiplicative, we have
If at least one of the is odd, the sum is . On the other hand, if all are even, the sum is . Then the final sum counts the number of perfect squares between and inclusive, which is .
.![\[
\sum_{n=1}^{1989} (-1)^{\Omega(n)}\left\lfloor \frac{1989}{n} \right\rfloor = \sum_{n=1}^{1989} \sum_{m \mid n} (-1)^{\Omega(m)}.
\]](http://latex.artofproblemsolving.com/6/f/1/6f1df800004014b1683a01bb1b237f3268575286.png)
We will now show that![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 1
\]](http://latex.artofproblemsolving.com/9/2/2/9224382e59164762af692df149ec6d0c16cd8151.png)
if is a square and 
if is not a square. Notice that if is not a square, it has an even number of divisors, so we can partition the divisors of into groups 
. Notice that each group has sum because one of them has one more prime divisor than the other. If is a square, then we can do the same thing but with 
for some 
. Then 
, so the sum is 
. So summing from 
to 
is just the number of squares less than , which is .
Now see that the inner function is basically 
and both are multiplicative functions and hence their Dirichlet convulation is also a multiplicative function; and now we can check easily that ![\[\sum_{n \mid m} (-1)^{\Omega(n)}= \begin{cases} 1 \text{ if $m$ perfect square} \\ 0 \text{ otherise} \end{cases}\]](http://latex.artofproblemsolving.com/a/1/9/a19cebd6063504d98b3e0c8c93084b113117d03f.png)
This basically gives us that the answer is 
.
I claim that 
unless 
is a perfect square, which makes it 1. This can be done by pairing 
and 
Thus, we are done. The answer is 
satisfying
Prove
denote the number of prime factors of ![\[ \sum_{n=1}^{1989} (-1)^{\Omega(n)}\left\lfloor \frac{1989}{n} \right \rfloor. \]](http://latex.artofproblemsolving.com/2/0/4/2043dca9f2ff2adfd3c8d49fa4dd278871968db0.png)
It is often useful to rewrite the floor function in order to swap the order of summation, so let us do exactly that here:
Now, we will prove the following lemma that we obtained from checking small cases:![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 1 \iff n\text{ is a perfect square;}
\]](http://latex.artofproblemsolving.com/e/b/0/eb0f4b6a7767af51d40578ac00534ae8032681de.png)
![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 0 \text{ otherwise.}
\]](http://latex.artofproblemsolving.com/4/9/1/4915b23e26fb5c5e1ccf55e5614bca5eb64d8579.png)
Proof: 
- then we can easily see that it is multiplicative. Therefore, it suffices to check the function for prime powers 
if 
is even and 
if it is odd. Next, observe that since![\[ \sum_{d \mid n} f(d) = f \ast \textbf{1},\]](http://latex.artofproblemsolving.com/e/7/6/e7654598a9425ac392bee97751577393a5b1d2e0.png)
the function 
is also multiplicative. However, if a prime number 
has its 
odd, then we would have a zero - otherwise, we have a 
e
so the lemma is proven.
From here, since 
,
.
Now, if we let 
then we have
If at least one of the 
implying that the whole product will be 
Otherwise, all of the summations are 
so the whole product will be 
will be 
which is 
