AoPS Academy
be an isosceles triangle with 
. The circle 
, passing through 
and 
, intersects segment 
at 
. The circle 
is tangent to 
at and passes through 
. Let 
and 
be the midpoints of segments and 
, respectively. The line 
intersects and at points 
and 
, respectively, where and are the intersections closer to . Prove that 
.
be the incenter of the triangle. The circumference that passes through and has center
intersects the circumscribed circumference of the triangle at points and. Prove that the line is tangent to the inscribed circle of the triangle.
be defined by 
(a) Show that 
has exactly one point of discontinuity. at its point of discontinuity.
![$f \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/7/a/a/7aa8f551ba6422af33ab978aaf842329b6c2aced.png)
such that its Fourier series diverges at 
. There is nothing special about the point , it was just for convenience. The same proof showed that for every ![$t \in [-\pi,\pi]$](http://latex.artofproblemsolving.com/d/1/3/d1331fab5626e64ec75c6c40edc6379bc591c803.png)
, there is ![$f_t \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/e/1/e/e1ef8ac797a9743ebd9e71bf1682b0246b221086.png)
such that the Fourier series of 
diverges at 
, not to show.
be a Banach space and for every 
we have a normed space 
. Suppose that for every there is 
and 
such that![\[
\sup_{k \geq 1} \| T_{n,k}x_n \|_{Y_n} = \infty.
\]](http://latex.artofproblemsolving.com/a/1/b/a1b7a2453200f95f61c2451fc91a8fc3a81fd8fd.png)
Show that![\[
B = \left\{ x \in X : \sup_{k \geq 1} \| T_{n,k}x \|_{Y_n} = \infty \ \forall n \geq 1 \right\}
\]](http://latex.artofproblemsolving.com/4/1/a/41a671b6c40492f454bf79d1a3205cd5592a94e6.png)
is of second category. (I am given the hint to write 
as![\[
A = \bigcup_{n \geq 1} A_n = \bigcup_{n \geq 1} \left\{ x \in X : \sup_{k \geq 1} \| T_{n,k}x \|_{Y_n} < \infty \right\}
\]](http://latex.artofproblemsolving.com/b/9/1/b9174ec68dd2a503c6d68deaf281c5e110637cbd.png)
and show that 
is of first category.)
. Show that there is ![$f_D \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/8/a/b/8ab83356343e4c5e30778770377027d0b2d57735.png)
such that the Fourier series of 
diverges at 
for all . (I'm given the hint to use part a) with![\[
T_{n,k} : (C_{\text{per}}([-\pi,\pi]; \mathbb{C}), \| \cdot \|_\infty) \to \mathbb{C}, \quad f \mapsto S_k(f)(t_n),
\]](http://latex.artofproblemsolving.com/2/2/5/225a5eec7a9582981841918fc28de96379b31404.png)
where![\[
S_k(f)(x) = \sum_{|j| \leq k} c_j(f) e^{ijx}.
\]](http://latex.artofproblemsolving.com/a/c/3/ac37ad818424867f4504e1472b2abe0883b6284b.png)
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solve
be a 
-dimensional inner product space of column vectors, where for 
and 
, the inner product of 
and 
is defined as ![\[\langle v, w \rangle = \sum_{i=1}^{10} v_i w_i.\]](http://latex.artofproblemsolving.com/9/f/9/9f9b7ff2fb19fee7e935c1b18dea3ec80547b9f9.png)
For 
, define a linear transformation 
on as follows:![\[ P_u : V \to V, \quad x \mapsto x - \frac{2\langle x, u \rangle u}{\langle u, u \rangle} \]](http://latex.artofproblemsolving.com/9/6/4/9646290e2b5e540b7f85d0d71e4d8992ccd8f542.png)
Given 
satisfying![\[ 0 < \langle v, w \rangle < \sqrt{\langle v, v \rangle \langle w, w \rangle} \]](http://latex.artofproblemsolving.com/d/8/0/d800fe48958cfee9ac01aa8bf11fe431ef45f8ef.png)
let 
. Then the dimension of the linear space formed by all linear transformations 
satisfying 
is 
.
be a triangle with 
. 
is bisector of 
. A line, which passes through 
intersect side 
at 
. Then 
is bisector of 
.
.
in triangle 
for bisector 
, line 
and obtain that 
is bisector of angle 
. By analogy 
is bisector of angle 
.
.
, 
. Let the bisectors of angles 
, 
and 
, 
and respectively. Prove that 
. 
, 
and 
. Observe that the ray 
is the bisector of 
and 
is harmonically. Is well-known that the division 
is harmonically. Thus, 
, i.e. the ray 
. Analogously prove that the ray 
is the bisector of 
. In conclusion, 
. See and here .
.
, then 
is harmonical pencil, i.e. 
is harmonical division 
.If denote the second intersection 
of 
with the circumcircle of , then 
is equilateral.
(Pompeiu's theorem). But 
( 
). In conclusion, 
. equilateral triangles a.s.o.
bisects the exterior angle 
, 
must be the -excenter of 
. By symmetry, 
must be the -excenter of 
. Thus, 
bisects 
, while 
bisects 
, from which we can find that 
. The desired conclusion follows immediately.
such that 
Prove that 
Where 
Prove that 
.
a
meet the opposite sides at 
an
passes through 
and 
.
be incenter. Then quadrilateral 
is cyclic (because 
)
and 
,
,
so 
Hence, 
and similarly 
Thus, 
