AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a trapezoid with parallel sides 
. Points 
and 
lie on the line segments 
and 
, respectively, so that 
. Suppose that there are points 
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points , , 
and 
are concyclic.
, with 
, and the circle 
with center 
and radius 
. Denote by 
the midpoint of side . The line 
intersects the circle for the second time in 
. Let 
be a point on the circle such that 
and 
. Show that 
.
. Let and be the feet of the altitudes from to 
and from 
to respectively.
and 
as the reflections of across lines 
and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
![\[\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}\]](http://latex.artofproblemsolving.com/5/d/6/5d6881be29327bcdc3e7bd76c221968578c7f50b.png)
is isosceles.
be all the divisors of 
. It is known, that 
are all divisors for some 
(except 
). Find all such 
be a sequence of positive real numbers that satisfies 
and 
for every natural number . Prove that 
for every natural number .
be positive real numbers such that : 
. Prove that : 
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and .
(ans- 
)
satisfying 
Prove that 
When does equality hold?
, take 
and 
, notice that any two non-consecutive terms have a prime divisor in common, while any consecutive don't have one.
's inductively. Take 
distinct primes not chosen at some previous step 
and let ![\[a_i = p_{i,0}p_{i,1} \prod_{j=1}^{i-2}p_{j,i\pmod{2}} \]](http://latex.artofproblemsolving.com/1/7/8/178328faf144a644510b1c2e1b3d57c31a7e29d6.png)
Then, clearly, shares factors with all 
's 
![$j \in [1,i-2]$](http://latex.artofproblemsolving.com/a/a/4/aa4a2972f946a85fc6ecc6cc38d6bc0ce05e6a3e.png)
. However, 
and 
are coprime because primes chosen in and 
have opposite pairity indices and are therefore distinct.
, 
, 
, 
, ... and define![\[
a_{n}
=
p_nq_n \cdot
\begin{cases}
\prod_{k=1}^{n-2} p_k & n \text{ even} \\
\prod_{k=1}^{n-2} q_k & n \text{ odd}.
\end{cases}
\]](http://latex.artofproblemsolving.com/7/6/d/76dd28041b46ab9eff05c328ba1b600ecb8af26e.png)
you want to not be relatively prime (meaning 
) and throw in a prime. You can't do this by using a different prime for every pair (since each must be finite) and you can't use the same prime for a fixed 
, so you do the next best thing and alternate using even and odd and you're done.
y 
y sea 
. Luego, para 
, 
, donde representa el mayor primo que aparece en (
), y 
representa el menor primo que no divide a ninguno de los siguientes términos: 
, para 
, por ejemplo, 
. Probaremos que esta secuencia funciona. Si demuestro que para cualquier 
, se cumple que y son coprimos, y con no son coprimos, para todo 
, ya quedaría listo. En efecto, probaremos que para 
, cumple, notemos que la base ya la construimos, pues los 
primeros términos ya satisfacen, luego supongamos que todos los términos cumplen lo pedido, entonces como 
. Notemos que es coprimo con 
, con 
, y con . Por lo tanto, y son coprimos. Luego, es fácil ver que y 
no son coprimos, también vemos que como 
es el mayor de los primos en , entonces no aparece atrás, y por tanto, 
no es coprimos con , para
. Por lo tanto, no es coprimo con , para todo 
. Lo cual completa la inducción. tambíen es igual a , solo que usé otro sentido en la fórmula para que se vea la construcción.
, and define the other ´s recursively:
denotes the nth prime.
is prime. since if is prime, 
and 
must have a common factor, . Such that they cannot be coprime.
there exist a sequence 
satisfying the condition in the promp by induction on The base case is trivial. For the step let the statement be true for some Now we will construct our new sequence with 
members as follows:
be distinct primes greater than for each 
The sequence with members is 
Now it is easy to notice, that 
iff 
for 
and 
for all 
Therefore 
iff 
iff 
(By the induction hypothesis), we also have that 
and for all 
Thus, the problem is solved.
such that 
and 
are coprime if and only if 
?
if and only if 
Here's the construction: denote the th prime. Consider the sets 
Then, add alternating elements from previous sets, as shown:
Then the sets 
to be 
the th column works, for instance 
and 
the prime numbers.
with 
denote 
. with denote 
.
and 
for , and then down the line we are forced to pick 
for some , then 
. Then 
should be coprime to 
, but also not coprime with , which is impossible.
in some order. Let
where 
if is odd and 
otherwise. It is easy to check that this works, since 
and 
for 
. 
to chosen, we just need to pick suitable , which essentially just means that the sets of primes dividing satisfy 
are not subsets of 
. at this point we can just realize that if we selected 
to 
to contain really large arbitrary sets of primes (think of them as buffers) then this basically works, and we're done
, , 
. The construction is as follows
in , 
, 
, where by putting it in we simply mean divides these numbers
in , 
, 
, all terms 
with 
and 
having the same parity clearly share a common factor, namely or . Accesing the odd terms is also easy by the delay of some terms by 
indices instead of .
. Let 
denote a sequence of ordered pairs, such that 
for each positive integer (so 
, and so on). Define 
and 
. For odd , we let be the product of all the first elements in each of 
multiplied by the product of both elements in 
. For even , we let be the product of all the second elements in each of multiplied by the product of both elements in . The first few numbers of this sequence we ![\[2\cdot 3, 5\cdot 7, 2\cdot 11\cdot 13, 3\cdot 7\cdot 17 \cdot 19, \ldots, 2\cdot 5\cdot 11\cdot 23\cdot 29\]](http://latex.artofproblemsolving.com/4/d/0/4d09925d8261793492fae32811347d2a6465afa2.png)
, then divides both and , so they aren't relatively prime. If 
, then we can see the prime factors of and are disjoint ( and have different parity). Therefore, 
.
be the set of all prime numbers. Then, we define a sequence of sets 
which are subsets of 
and let be the product of all the primes in .
and 
. Then, we add to all sets where 
and to all sets 
where 
for all 
. Next, in the third round we add 
and 
to 
. Then, we add to all sets where 
(
) and 
to all sets where 
(
). Similarly, we keep continuing, where in the 
round we add 
and 
to 
. Then, we add to all sets where 
(
) and to all sets where 
(
).![\[S_i= \{p_1,p_4,p_5, \dots ,p_{2i-6},p_{2i-5},p_{2i-1},p_{2i}\} \text{ for all } i \equiv 1 \pmod{2} \text{ and } i>1\]](http://latex.artofproblemsolving.com/5/d/0/5d0c61672c181ffaaf0de24f7ebafbb2bf867d2a.png)
and![\[S_i = \{p_2,p_3,p_6,p_7, \dots ,p_{2i-6},p_{2i-5},p_{2i-1},p_{2i}\} \text{ for all } i \equiv 0 \pmod{2} \text{ and } i>2\]](http://latex.artofproblemsolving.com/7/6/9/7693ea3491a164dc94d62aa9a4e4c6e2aac124bd.png)
and do not have any common factor for consecutive 
(since 
, 
for any 
and 
and don't divide since they are added only to sets 
where 
by the nature of our construction). For all where 
, and must have one of or as a common factor depending on the parity of 
. Thus, this construction must work and we are done.
as a set of positive integers such that 
where represents the th prime.
and 
.
, let 
for even and let 
for odd .
, and split the primes into groups of 
. Then for 
, have divide if and only if is odd and at least 
. For 
, have divide if and only if is even and at least 
or is 
. For 
, have divide if and only if is even and at least 
. For 
, have 
divide if and only if is odd and at least 
or is . For simplicity, call 
the associated primes of 
.
be the primes in increasing order. Then for term of the sequence, we let
and divide immediately;
, 
divides if 
, and 
divides otherwise.
, one of and divides both and by construction;
and 
divide , and for each 
, if divides 
and not , divides and not , and vice versa.
when and 
. This is all we need to prove.
of numbers which satisfy the condition, let 
be the minimal primes needed for to satisfy 
for every (if there's more than one possible option, pick one randomly, it's only important to not include superfluous primes).
because all factors of are also factors of .
, given by ![\[\left\{\begin{array}{ll}
a_1=2\cdot 3 \\
a_2=5\cdot 7 \\
a_3=2\cdot 11 \\
a_4=3\cdot 5\cdot 13\\
a_{n+2}=d(a_n)\cdot p_{n-1} \cdot p_{n+2} & (n\geq 3)
\end{array}\right.\]](http://latex.artofproblemsolving.com/d/8/b/d8ba2cbc4c5514fef70d166504c13ca778a2c37d.png)
where is a prime we choose which hasn't come up before in the sequence (before ) satisfies.
always works regardless of the choice of : suppose it is true until , then clearly if 
, 
, so 
and 
share common factors by definition of 
.
. Now:
because 
is a new prime.
for 
is a new prime chosen after 
and 
so it doesn't divide them and 
is a prime which hadn't come up before , yet 
, since 
and 
don't share factors (so neither will and 
) and 
is chosen after .
, since is chosen after 
such that it is a new prime and so it will also not be equal to 
, the same way 
isn't equal to 
, proving our claim.
such that 
and 
.
?
.
denote the largest prime divisor of 
.
.
.
doesn't divide any of the smaller 
shares a prime factor with all 
.
share a factor with 
doesn't divide any of the smaller 
doesn't divide 
.
.
be the product of the primes 
and 
and 
where 
and 
.
.
,![\[ a_n = \begin{cases} p_1(p_4p_6p_8p_{10} \cdots p_{n-3}p_{n-1})p_{n+2} \text{ for } n \text{ odd} \\ p_2(p_3p_5p_7p_9 \cdots p_{n-3}p_{n-1})p_{n+2} \text{ for } n \text{ even}. \end{cases}\]](http://latex.artofproblemsolving.com/3/0/0/3003a02345d6790cd0f0cdaaa1f01081cb084ee2.png)
It is not hard to see that this works.
since there are infinitely many of them. Let 
,
and suppose we let![\[a_n = (p_nq_n)\cdot (p_1p_2\ldots p_{n-2})\]](http://latex.artofproblemsolving.com/1/f/3/1f33a7b3f923c1282b67648dc3fbf9e898588df6.png)
if ![\[a_n = (p_nq_n) \cdot (q_1q_2\ldots q_{n-2})\]](http://latex.artofproblemsolving.com/1/8/a/18af1131fe91872e22e9a0a502e79762b65295ea.png)
if 
denote the set of primes.
This should work,hence we are done 
From 
,
where p and q are primes
where if 
denotes the 
-
Now, we need to have 
if and only if 
We claim that if 
and
then 
works.
for 
will be nonempty, and so 
for 
is obvious, since by our definition 
Our second base case, 
is true, since 
and so since 
we have that 
and so at least one element in 
cannot be in 
For 
assume for the sake of contradiction that 
However, by the inductive hypothesis, since 
we know that this cannot be true (this is because the claim being made is equivalent to having 
for 
)
are the primes, our construction is as follows:
for 
.
in 
for 
'
for odd-even.