AoPS Academy
have altitudes 
circumcenter 
and orthocenter 
. Let 
be a point on line 
. The points 
are on lines 
respectively such that 
and 
. Prove that 
is perpendicular to 
.
and 
be sequences of positive integers defined by the recurrence relations:![\[
a_{n+1} = n a_n + 1, \quad b_{n+1} = n b_n - 1
\]](http://latex.artofproblemsolving.com/e/8/f/e8fceef09debd22326dc85d36aa10a387650f797.png)
for all positive integers 
.
be the circumcenter and the orthocenter of an acute triangle 
Prove that the area of one of the triangles 
and 
is equal to the sum of the areas of the other two.
with any line through the centroid 
and I have no clue as to why that is true. Plz help
be positive integers, and define![\[ n = a + b + c + d + e + f. \]](http://latex.artofproblemsolving.com/9/d/d/9ddfecfa5800050bc15dcaa48d17d12e431fc358.png)
Suppose that divides both of the following expressions:![\[
abc + def \quad \text{and} \quad ab + bc + ca - de - ef - fd.
\]](http://latex.artofproblemsolving.com/c/5/8/c5817937f736a7b00c85e1f92b8e561ba26cf8b0.png)
Prove that is a composite number.
The sequence 
is defined by![\[a_1 =3 \ \text{and} \ 2a_{n+1} =a^2_n+1.\]](http://latex.artofproblemsolving.com/7/6/e/76e5b19b1c4baacda35ad138f5625d9911a10b04.png)
Prove that![\[\sum_{k=1}^{n}\frac{1}{a_k +1} < \frac{1}{2}.\]](http://latex.artofproblemsolving.com/4/f/6/4f6bfe179daeebe6f0aadddb5ed40798c90b8702.png)
![\[\sum_{k=1}^{n}\frac{1}{a_k +1}=\frac{1}{2}-\frac{1}{a_{n+1}-1}\]](http://latex.artofproblemsolving.com/e/0/a/e0adea7fe9cc8ec8c5834b50845a3946434c995a.png)
for all 
it's obvious that 
for all 
so 
is always positive, thus the proof of the claim will be sufficient to finish the solution. 
We will apply induction.
is trivial. Let us assume that![\[\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_n+1}=\frac{1}{2}-\frac{1}{a_{n+1}-1}\]](http://latex.artofproblemsolving.com/f/c/c/fcc745b76130ae085e4573c7b428925694cef2c3.png)
holds and try to prove it for 
Or we just need to prove the following equation:![\[\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_{n+1}+1}=\frac{1}{2}-\frac{1}{a_{n+2}-1}.\]](http://latex.artofproblemsolving.com/4/5/9/459a2defd4c3b8a315ac379ba053cb5bc1644696.png)
-----------------
![\[\frac{1}{2}-\frac{1}{a_{n+1}-1}+\frac{1}{a_{n+1}+1}=\frac{1}{2}-\frac{1}{a_{n+2}-1}.\]](http://latex.artofproblemsolving.com/e/8/2/e828da418e492bc8b296a155764a80f12bde31c4.png)
![\[\frac{1}{a_{n+1}+1}-\frac{1}{a_{n+1}-1}=-\frac{1}{a_{n+2}-1}.\]](http://latex.artofproblemsolving.com/7/a/c/7acb715a08ac9c6c3f85478686cae6b709b0b89d.png)
![\[\frac{2}{a^2_{n+1}-1}=\frac{1}{a_{n+2}-1}\]](http://latex.artofproblemsolving.com/7/f/b/7fb4d65ab158f790bf1097f35b1f90a5758d3818.png)
![\[2a_{n+2}-2=a^2_{n+1}-1\]](http://latex.artofproblemsolving.com/5/a/6/5a6f11d5d5595201796b16de8960c153158d63ec.png)
![\[2a_{n+2}=a^2_{n+1}+1,\]](http://latex.artofproblemsolving.com/9/3/6/936d9eaa3cf546129f2931360d3cc8d8869609ee.png)
which is true according to the given condition![\[2a_{n+1} =a^2_n+1\]](http://latex.artofproblemsolving.com/d/3/f/d3f3d9c9c26f34d73e8bfdbc1af53cbdd3141e10.png)
for all 
be a positive integer. Prove that there exists a positive integer 
such that exactly 
of the integers 
are squarefree.
be the number of squarefrees among 
. Notice that 
. Pick sufficiently large primes 
. By Chinese Remainder Theorem for 
for 
, we see that 
. If there exists 
then there is some 
. Suppose that 
for each positive integer 
. Since density of squarefree numbers is 
we get a contradiction.
, everything else is as bin_sherlo.
to divisible by 
for some prime 
is at most 
. Hence the amount of non-square free does not exceed 
by telescoping 
, so indeed the amount of squarefree is greater than 
.
