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, let 
be a point on the segment 
and let 
be the intersection point of 
with the diagonal 
. The line perpendicular to the segment through intersects the side 
at point 
. Let 
be a point on the segment 
such that 
. Let 
be the intersection point of the diagonal 
and the segment 
.
and 
are concyclic.
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
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 
, but we also have that MO = MA = MB. We have therefore that
. But this also follows from the triangle inequality if one out of triangle ABC, make a parallelogram with center M.
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