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 
and 
be two similar cyclic quadrilaterals (with the vertices denoted counterclockwise). Their circumcircles intersect again at point 
. Prove that lies on line 
.
, or 
, so 
, done.
