such that 
Prove that 
Where 
be an isosceles triangle with 
.
,
,
.
is tangent to 
at 
.
be the midpoints of segments 
,
intersects 
and 
,
.
be the incenter of the 
Prove that 
be defined by 
(a) Show that 
has exactly one 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 
.![$t \in [-\pi,\pi]$](http://latex.artofproblemsolving.com/d/1/3/d1331fab5626e64ec75c6c40edc6379bc591c803.png)
,![$f_t \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/e/1/e/e1ef8ac797a9743ebd9e71bf1682b0246b221086.png)
such that the Fourier series of 
diverges at 
,
be a Banach space and for every 
we have a normed space 
.
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.)
.![$f_D \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/8/a/b/8ab83356343e4c5e30778770377027d0b2d57735.png)
such that the Fourier series of 
diverges at 
for all ![\[
T_{n,k} : (C_{\text{per}}([-\pi,\pi]; \mathbb{C}), \| \cdot \|_\infty) \to \mathbb{C}, \quad f \mapsto S_k(f)(t_n),
\]](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 solv
be a 
-
and 
,
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 
,
on ![\[ P_u : V \to V, \quad x \mapsto x - \frac{2\langle x, u \rangle u}{\langle u, u \rangle} \]](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 
.
satisfying 
is 
Y by Physicsknight, WizardMath, amar_04, Geometry013, Purple_Planet
Let 
and 
be two distinct points in the plane. Let 
be the midpoint of the segment 
, and let 
be a circle that goes through and . Let 
be a point on such that the line 
is tangent to . Let 
be a point (other than ) on the line such that 
, and let 
be the foot of the perpendicular from onto the line .
Prove that the lines
and 
are parallel.
Navneel Singhal, India
and 
be two distinct points in the plane. Let 
be the midpoint of the segment 
, and let 
be a circle that goes through and . Let 
be a point on such that the line 
is tangent to . Let 
be a point (other than ) on the line such that 
, and let 
be the foot of the perpendicular from onto the line .Prove that the lines
and 
are parallel.Navneel Singhal, India
,
is isosceles and 
hence 
is cyclic so 
the lines 
be the foot of 
is cyclic , so 
=
. We get that 
=
(
. After squaring and using pop it reduces to (1).
.
,
where the first equality holds because 
and the last equality holds because 
.
where the last equality holds since 
and 
,
to the segment 
,
and 
,
therefore, this homothety also sends 
and 
,
.
.
,
where the last equality holds due to the tangency of 
,
,
the point 
therefore, 
where the last equality holds due to the tangency of 
is equivalent to 
.
,
which by looking at the power of 
being concyclic.
,
lie on a circle with midpoint 
and therefore 
.
,
.
.
and 
.
.
.
.
we have 
.
which implies 
.
Since 
and 
which implies 
![[asy]
size(7cm);
pair A=(-8.86,2.91);
pair B=(1.8499163754792471,2.9844830809920997);
pair M=(-3.505041812260376,2.9472415404960497);
pair T=(-2.0504048164735122,9.47612561960746);
pair X=(-5.860059676194846,2.9308633560035102);
pair Y=(-0.957891074132805,7.657760327427519);
filldraw(A--M--T--cycle,cyan,blue);
draw(circumcircle(A,M,T),green);
draw(B--T,purple);
draw(B--M,purple);
draw(circumcircle(T,M,X),orange+dashed);
draw(A--Y,purple);
draw(X--Y,purple);
draw(Y--M,purple);
draw(T--X,purple);
dot("$A$",A,S);
dot("$M$",M,S);
dot("$T$",T,N);
dot("$Y$",Y,NE);
dot("$X$",X,S);
dot("$B$",B,S);
[/asy]](http://latex.artofproblemsolving.com/8/9/4/8942629ae1b3da3a7c82798f2769df4e5d600929.png)
is cyclic
hence 
.
and we are done 
and we are done 
.
,
.
( since 
)
is tangent to 
,
.
.
which implies that 
and we are done.
cyclic
is cyclic.
which implies the conclusion.
- cyclic. So by Reim 
is cyclic.
then 
we find that 
which gives that 
so 
This gives 
as desired.![[asy]
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import graph; size(7cm);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.596925851065945, xmax = 9.250987235646235, ymin = -2.347242959423775, ymax = 7.058189585802859; /* image dimensions */
pen ffqqtt = rgb(1,0,0.2); pen dbwrru = rgb(0.8588235294117646,0.3803921568627451,0.0784313725490196); pen ffttww = rgb(1,0.2,0.4); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961);
/* draw figures */
draw((-4,-0.45)--(-0.5584965546539985,-0.416629514997293), linewidth(0.4) + ffqqtt);
draw(circle((-3.1738984537761103,3.09305821011176), 3.6380908791444244), linewidth(0.4) + dbwrru);
draw((0.25494455728530113,1.8770233214637557)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-4,-0.45)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-4,-0.45)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffqqtt);
draw((-0.9535063381410938,-1.5304351697975988)--(1.112709278027912,-0.4004246934529864), linewidth(0.4) + ffqqtt);
draw((1.112709278027912,-0.4004246934529864)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffttww);
draw((-2.279248277326999,-0.4333147574986465)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-2.279248277326999,-0.4333147574986465)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffqqtt);
draw((-0.5584965546539985,-0.416629514997293)--(1.112709278027912,-0.4004246934529864), linewidth(0.4) + ffqqtt);
draw(circle((-0.5898191880272617,0.25860068246196543), 1.8256280080640814), linewidth(0.4) + linetype("2 2") + dtsfsf);
/* dots and labels */
dot((-4,-0.45),dotstyle);
label("$A$", (-3.9420814276242977,-0.2942805256941039), NE * labelscalefactor);
dot((-0.5584965546539985,-0.416629514997293),dotstyle);
label("$B$", (-0.48864849646663094,-0.2624516507525586), NE * labelscalefactor);
dot((-2.279248277326999,-0.4333147574986465),linewidth(4pt) + dotstyle);
label("$M$", (-2.2233221807808507,-0.31019496316487654), NE * labelscalefactor);
dot((0.25494455728530113,1.8770233214637557),linewidth(4pt) + dotstyle);
label("$T$", (0.4503033143089558,1.8541685328602033), NE * labelscalefactor);
dot((-0.9535063381410938,-1.5304351697975988),linewidth(4pt) + dotstyle);
label("$Y$", (-1.2525414950637186,-1.8538953978298233), NE * labelscalefactor);
dot((1.112709278027912,-0.4004246934529864),dotstyle);
label("$X$", (1.182367437964498,-0.246537213281786), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/b/2/7/b272d7a7d80a8b2bb76bf46b7db2cb972054707e.png)
,
and 
.
is cyclic.
,
