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 non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
be a non-constant monic polynomial with integer coefficients and let be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence .
is the set of prime divisors of 
.
is the smallest positive integer such that 
and 
, where 
, then 
.
. We have 
and if 
is even, 
would give a contradiction. For odd , we get 
thus, 
but then 
where 
which results in a contradiction.
. We see that for 
, we have 
and 
for 
positive integer. Pick 
sufficiently large. If 
, then 
in order to get 
.
. Let 
for 
.![\[\nu_{p_i}(P(n)^{a_n}+1)\leq \nu_{p_i}((P(n)^{d_i})^{p_1^{\beta_1}\dots p_k^{\beta_k}m}+1)=\nu_{p_i}(P(n)^{d_i}+1)+\beta_i=\nu_{p_i}(P(1)^{d_i}+1)+\beta_i=\beta_i+\theta_i\]](http://latex.artofproblemsolving.com/0/e/9/0e9006891526f624edb7a43227857ce30e9c377d.png)
![\[P(n)^{p_1^{\beta}\dots p_k^{\beta_k}m}+1\leq p_1^{\beta_1+\theta_1}\dots p_k^{\beta_k+\theta_k}=(p_1^{\beta_1}\dots p_k^{\beta_k})(p_1^{\theta_1}\dots p_k^{\theta_k})\]](http://latex.artofproblemsolving.com/3/6/9/3691d3b066d0bf4883fba1fe852756485f6ccabf.png)
Let 
and note that 
is a constant. We have 
. Since is non-constant and monic, 
is unbounded above. 
exists hence 
does not hold. Hence we get contradiction as desired.
. We see that for , we have and for positive integer. Pick sufficiently large. If , then in order to get .
's exsist? What if 
or like the situation when 
? I'm confused. does not have to exist, but if 
for some 
, then there must be a minimum for .
is unbounded and when it is bounded.
such that 
Prove that 
Where 
Prove that 
which is contradiction, good problem.
,
is a prime such that 
,![\[ p^{\alpha} \leq (a^{p-1} - 1)p^{\nu_p(k)}. \]](http://latex.artofproblemsolving.com/3/b/a/3baa4f1c3cc466144094c5d7f3634e3987c118bd.png)
,
.
.
is the order of 
modulo 
,
,
.
,
,
is finite, and let 
be all such primes. From the lemma it follows that![\[
P(k)^{a_k} + 1 = \prod_{i=1}^s p_i^{\nu_{p_i}(P(k)^{a_k} + 1)} \leq \prod_{i=1}^s \left(P(k)^{p_i - 1} - 1\right)p_i^{\nu_{p_i}(a_k)} \leq a_k P(k)^{\sum_{i=1}^s p_i},
\]](http://latex.artofproblemsolving.com/3/2/a/32a5c2958d1e21c09a96bf9cc40efd02074b1888.png)
and thus 
.
for all sufficiently large 
,
.
.
for some 
,
,
.
modulo 
,
,
.
,
,
is bounded.
is not a power of 2 for 
because of modulo 4. Therefore, if 
for some 
,
dividing 
,
,
for all 
,
be such that 
for all 
be such that 
,
be sufficiently large. Then for![\[
k = \prod_{i=1}^s p_i^{2N} + t
\]](http://latex.artofproblemsolving.com/8/6/8/8685b6dcd274fb3ff18d35fcbee09b5b9955936d.png)
we have![\[
P(k)^{a_k} + 1 \equiv P(t)^{a_k} + 1 \pmod{p_1^N p_2^N \cdots p_s^N}.
\]](http://latex.artofproblemsolving.com/0/4/d/04de40f81ff5c3a0f55404b6a541255a33eef7fe.png)
Therefore, 
,![\[
P(k)^{a_k} + 1 \leq \prod_{i=1}^s p_i^N < \sqrt{k},
\]](http://latex.artofproblemsolving.com/7/5/d/75d24ffa7400081cdbe9b7109a9dbdc91e903e49.png)
which leads to a contradiction for sufficiently large 
.
for some 
,
,
such that each natural number appears in the sequence no more than 
are a finite number of nonconstant polynomials, and each of them attains a given value only finitely many times; here 
be all the primes with the required property. Write 
for 
.
for every 
,![\[
b_k < c^{k^{\varepsilon}} = 2^{k^{\varepsilon} \log_2 c}
\]](http://latex.artofproblemsolving.com/f/2/d/f2df46178ee2ec6378a6d2b828099c275af0e47b.png)
for large 
does not exceed 
.
does not exceed 
.
.![\[
m^{1 - s\varepsilon} \leq T (\log_2 c)^s.
\]](http://latex.artofproblemsolving.com/c/2/a/c2a77e687bc15caf815a04e79940d92eb365625d.png)
If we choose 
,
,