has incentre 
and incircle touching 
at 
.
be the antipode of 
in the circumcircle of 
is chosen on the internal angle bisector of 
such that 
.
be the midpoint of arc 
,
be the midpoint of 
.
x
to 
row and 
column get symbolized by square 
. Subset 
belong to 
with 
or 
are not belong to 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle 
and 
are defined similarly. The circles 
distinct from 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
defined as follows:![\[
a_1 = \frac{\sqrt{6}}{3},
\quad
a_{n+1} = a_n + \frac{1}{3a_n},
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/8/9/2/892a2676214a3c837d6b18a10a4f6f83bc099f74.png)
![\[
0 \le a_n \sqrt{6} - 2\sqrt{n}
\le
\frac{1}{4\sqrt{n}}
\Bigl(
1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
\Bigr),
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/f/0/8/f0824edd96ddfeaa107d729466134812ad7ea777.png)
b,For each 
,![\[
b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}.
\]](http://latex.artofproblemsolving.com/a/9/3/a937b379140df5aa51d7ede1a685d263c3229e96.png)
Compute the limit 
,
and Moivre's formula.
.
![\[
I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du
\]](http://latex.artofproblemsolving.com/b/c/5/bc5d60ff6679870b47776c401658e1ba62af75aa.png)
![\[
\text{Find: }
10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z},
\]](http://latex.artofproblemsolving.com/f/3/3/f3360b85ce092804a9de263cac49aed025ea8cee.png)
![\[
11. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{(-1)^{n+m}}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/9/4/5/945aaa2340a4fb314e5eb218ba5286eb054c1b6f.png)
![\[12.
\int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos(2x) + 2} \tan^{-1}\left(\frac{\cos x \cot(2x)}{\sqrt{2}}\right) dx
= \frac{5\pi^2}{48\sqrt{2}} - \frac{\pi}{4\sqrt{2}} \cos^{-1}\left(\frac{1}{3}\right).
\]](http://latex.artofproblemsolving.com/b/1/3/b13a5b1c537926c8323170d3d533b0bc71bb06a1.png)
-
![\[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]](http://latex.artofproblemsolving.com/b/7/2/b72b5e302cecdf2649c43a078aea3aa9a88ef2cc.png)
find 
and 
Y by cubres
Given 
, orthocenter 
, 
is the center of Euler's circle of . 
is the midpoint of 
. From draw the tangent of 
, cut a line through perpendicular to sides 
at 
.Prove that 
concurrent
, orthocenter 
, 
is the center of Euler's circle of . 
is the midpoint of 
. From draw the tangent of 
, cut a line through perpendicular to sides 
at 
.Prove that 
concurrentThis post has been edited 2 times. Last edited by Giahuytls2326, Feb 23, 2025, 1:20 AM
Reason: wrong source
Reason: wrong source
be 
,
.
is cyclic. Now,
However, observe that 
is actually the ratio of the distances from 
to 
and 
to 
.
as reference.
