AoPS Academy
has incentre 
and incircle touching 
at 
. Let 
be the antipode of 
in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
x square board has the row and column numbered from 
to , square in 
row and 
column get symbolized by square 
. Subset of squares on the board is called "good" subset if two random squares 
belong to satisfy that the squares 
with 
or 
are not belong to . Find the minimum number of "good" distinct subsets such that each square on the board belongs to only one subset
be a triangle and let and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from ; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
. +1 w
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 
, define![\[
b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}.
\]](http://latex.artofproblemsolving.com/a/9/3/a937b379140df5aa51d7ede1a685d263c3229e96.png)
Compute the limit 
, with help of 
and Moivre's formula.
. Easy to guess that the answer is , but I haven't found the reason yet. :(
houses on a straight street are one or more boys. At what point should all the boys meet so that the sum of the distances that they walk is as small as possible?
![\[
\text{Find: }
10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z},
\]](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)
with 
and 
given as 
Find the best (real) bounds 
and 
such that ![$f(M) = \{f(a,b,c): (a,b,c)\in M\}\subseteq [\alpha,\beta]$](http://latex.artofproblemsolving.com/c/f/e/cfe67211f334f6d6220566cc229565070e641ee9.png)
and determine whether any of them is achievable.
. Note that 
is bounded and closed, meaning it is a compact set. Hence 
has minimum/maximum value in constraint 
. Let's say 
gets its minimum/maximum value at the boundary. Therefore we have either 
or 
. Since 
is undefined, we can still calculate the value by limits.It is not hard to see that we have to check 
,
case. 
case is clearly impossible. therefore for second case which is 
we can see 
we have to check 
and 
. The first case is same as above.The second case gives . Therefore we have finished for this case.
lies on open subset only, we can bash this with lagrange multipliers since all the functions are differentiable in set . The equation 
gives obvious answer 
. By Graphing the function estimately, we can see there is no such 
satisfying . For case 
we can let 
and analyze new 1 variable function, which again gives 
.
![$f(M) = (8,9]$](http://latex.artofproblemsolving.com/8/5/a/85a4d17216e35ae50eca65faabda932dd35b0721.png)
. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable.
for all 
to transform
. What we want to prove is
, using to homogenize the inequality yields
. Which is true by Schur's Inequality. As desired.
for all 
What we want to prove is
. Homogenizing the inequality and cancelling yields (same as before)
. Note that are the sides of a triangle (
, and similarly for others), let 
where 
. We have,
. While
. So,. As desired.
![$(8,9]$](http://latex.artofproblemsolving.com/d/d/6/dd6d852bbeb2fb74be0f14b9344c0d7c9bbb86d5.png)
is attainable to get 
. Next, pick 
and 
where 
, it obeys the restriction 
. Then,
. Define 
where , then
. So 
is strictly decreasing and clearly continuous at 
. We have 
and 
. Hence every value in 
is attainable by , and our job is finished 
.
and 
where is an infimum while is a maximum.
. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable.
for all to transform
. What we want to prove is, using to homogenize the inequality yields. Which is true by Schur's Inequality. As desired.
for all What we want to prove is. Homogenizing the inequality and cancelling yields (same as before). Note that are the sides of a triangle (, and similarly for others), let where . We have,. While. So,. As desired.
is attainable to get . Next, pick and where , it obeys the restriction . Then,. Define where , then. So is strictly decreasing and clearly continuous at . We have and . Hence every value in is attainable by , and our job is finished .
and where is an infimum while is a maximum.'s continuity? i think you should prove that the limit exists
![\[
I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du
\]](http://latex.artofproblemsolving.com/b/c/5/bc5d60ff6679870b47776c401658e1ba62af75aa.png)
-
![\[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]](http://latex.artofproblemsolving.com/b/7/2/b72b5e302cecdf2649c43a078aea3aa9a88ef2cc.png)
find 
and 
.
,
.
But the last inequality is Schur's inequality. Thus equality is achieved if and only if 
Note that 
,
and in a triangle 
