such that 
Prove that 
Where 
be an isosceles triangle with 
.
,
and 
,
at 
.
is tangent to 
at 
.
and 
be the midpoints of segments 
,
intersects 
and 
.
be the incenter of the 
intersects the circumscribed circumference 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 GeoKing, Rounak_iitr, SatisfiedMagma
Let 
be an acute angled scalene triangle with circumcircle 
. Let 
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and , respectively, such that
Prove that 
and 
intersect on the circumcircle of 
.
Proposed by Antareep Nath
be an acute angled scalene triangle with circumcircle 
. Let 
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and , respectively, such that
Prove that 
and 
intersect on the circumcircle of 
.Proposed by Antareep Nath
This post has been edited 2 times. Last edited by taptya17, Dec 17, 2024, 6:08 AM
so that 
lies on 
lies on 
and 
are similar, we observe that ![\[\frac{BF}{CE} = \frac{BH}{HC}.\]](http://latex.artofproblemsolving.com/a/5/4/a540476b20d8c7ab9c063e3c6f2af249cb04aa87.png)
Since ![\[\frac{BQ}{QC} = \frac{BH}{HC}\]](http://latex.artofproblemsolving.com/a/3/4/a34da1a19aa4eca3c3a00424e3fb8c6cc7aa0394.png)
as well, which leads to 
cyclic from spiral-sim.
again at 
.
are reflections over 
.
,
is the point where 
.
.
and note that 
and thus 
we determine this isogonal conjugate to be 
,
which finishes aong with the previous paragraph. 
be the feet of the altitudes from 
to 
respectively. Note that 
q
.
are concyclic.
and 
.![\[\frac{EL.EA}{EB.EA}\overset{?}{=}\frac{FK.FA}{FC.FA}\iff \frac{HL.HK}{HB.HC}=\frac{EA.\frac{LH}{HC}}{EA.\frac{HB}{HK}}=\frac{EL}{EB}\overset{?}{=}\frac{FK}{FC}=\frac{FA.\frac{HK}{HB}}{FA.\frac{HC}{HL}}=\frac{HK.HL}{HB.HC}\]](http://latex.artofproblemsolving.com/7/9/9/79950946e13081bdd87b94dd65b3c8564b69da52.png)
Which proves the result
.![\[\overline{e}=\frac{1}{a}+\frac{1}{b}-\frac{e}{ab}, \ \ \frac{h-b}{h-e}=-\frac{\overline{h}-\overline{b}}{\overline{h}-\overline{e}}\implies \frac{1}{e-a-b-c}=\frac{1}{a+\frac{ec}{b}}\]](http://latex.artofproblemsolving.com/8/4/e/84eac33eb0e110fb5ee92cd5cd4994270cacabef.png)
Thus, 
.
.![\[e.\overline{e}=\frac{(2a+b+c)b}{b-c}.\frac{\overline{(2a+b+c)}.\frac{1}{b}}{c-b}=\frac{(2a+b+c)c}{c-b}.\frac{\overline{(2a+b+c)}.\frac{1}{c}}{b-c}=f.\overline{f}\]](http://latex.artofproblemsolving.com/3/e/7/3e7110f64541032f6d79d6234e6304a341915250.png)
So 
.
thus, 
.
and 
are concyclic and 
hence the perpendicular bisectors of 
intersect at another point than circumcenter of 
.
is an isosceles trapezoid and 
,
must hold. 
which proves the collinearity of 
as desired
.
and 
intersect on 
Indeed cool problem
,
and 
.
.
As 
,
.
we get 
is cyclic trapezium.
is point on 
.
where ![\[\Box AEHF \sim \Box CHBA'\]](http://latex.artofproblemsolving.com/4/c/8/4c8a1f5dba52eab47318ddc7215a994f39eae593.png)
then assuming 
we get 
Thus 
Hence 
lie on circle 
cyclic and to chase 
.
,
.
are tangents to 
.
and 
are similar and thus an angle chase tells us 
.
and 
then by symmetry 
must bisect 
.
.
be the feet from 
to 
respectively. By similar triangles we have 
,
.
.
is clearly a parallelogram, 
is the center of 
lies on the perpendicular bisector of 
cannot be 
.
intersect 
.
.
and 
This implies 
,
be the common midpoint of 
.
. 
, then we have 
.
we obtain 
collinear. Thus 
.
.
and 
meet 
which implies 
is an isosceles trapezoid.
.
are collinear.
so 
.
.
and 
is foot of 
foot of 
but this is because of 
and similarly defined altitude.
.![\[\frac{EE'}{EB}=\frac{EE'\times EB}{EB^2}=\frac{EH^2}{EB^2}=\sin^2{90^\circ-\angle A}=\cos^2{\angle A}\]](http://latex.artofproblemsolving.com/8/6/9/86924ac259c052f98ebe9f6762fc5c8fe088ce6a.png)
Now use ratio lemma on 
so![\[\frac{GH}{GH'}=\frac{QH}{QH'}\times\frac{\sin HQO}{\sin H'QO}\]](http://latex.artofproblemsolving.com/4/c/a/4ca815c645252e4918c13ff004f79732ce7c16ef.png)
now 
(because angle at centre theorem), and 
.
.
,![\[\cos{\angle A}=\frac{QH}{QH'}\]](http://latex.artofproblemsolving.com/d/5/9/d5972a06bf7c527bd87cb3f8f88c60b2add0db3c.png)
Now, doing more trig,![\[\frac{QH}{QH'}=\frac{AH \sin \angle QAH}{\sin\angle QAH}\times \frac{\sin \angle AQH'}{AH'}=\frac{AH}{AH'}\times \sin\angle AQH'\]](http://latex.artofproblemsolving.com/8/0/4/80470be4458bfee16177920731010053c938b5ee.png)
hence we just need ![\[\frac{AH}{AH'}=\frac{\sin \angle ACH}{\sin \angle ACH'}\]](http://latex.artofproblemsolving.com/3/e/3/3e339540f730c5167acd1469bd5f448641e8f7e3.png)
This is a direct consequence of ratio lemma on 
and the well known fact 
.
be orthic triangle of 
.![\[\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))}=\frac{\text{Pow}(F,(AH))}{\text{Pow}(F,(ABC))} \iff \frac{EZ}{EB}=\frac{FY}{FC}\]](http://latex.artofproblemsolving.com/e/e/2/ee2ba2980615681cfc987bb874b9f86067455915.png)
Which is true since 
because see that 
.
.
and so all we need to prove is that 
in 
,
and it is obvious that this must be ![\[\angle OFE=\angle HFC=90 ^{\circ}-\angle ACH=\angle BAC=\angle EAF\]](http://latex.artofproblemsolving.com/4/2/5/4255aa95a8cf5d47d1257dc3ef5b2f68907e9fac.png)
And so 
is tangent to 
and done.
,
,
and analogously 
are both tangents to the circumcircle of 
.
is the symmedian in 
.
.
where 
to 
.
and 
,
can be expressed in trigonometric ratios in 
,
which we can calculate analogously, thus the spiral similarity also sends 
invert then the whole problem turns into a very famous 
,
config which is very easy to prove.
