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 positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be a scalene triangle with orthocenter 
and circumcenter 
. Denote by 
, 
the midpoints of 
, 
. Suppose the circle 
with diameter meets the circumcircle of at 
, and meets line 
at a point 
. The tangent to at 
meets line 
at 
. Show that the circumcircles of 
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
be a triangle with circumcircle 
and incenter 
and let be the midpoint of . The points 
, 
, 
are selected on sides , 
, 
such that 
, 
, and 
. Suppose that the circumcircle of 
intersects at a point 
other than 
. Prove that lines 
and 
meet on .
, 
, 
, 
, 
, 
initially contains one coin. The following operations are allowed
, 
, remove one coin from and add two coins to 
;
, 
, remove one coin from and swap the contents (maybe empty) of the boxes 
and 
., , , , become empty, while box contains exactly 
coins. be a triangle, its orthocenter, its circumcenter, and 
its circumradius. Let be the reflection of the point across the line 
, let be the reflection of the point 
across the line 
, and let be the reflection of the point 
across the line 
. Prove that the points , and are collinear if and only if 
.
be a real number and let 
be arbitrary real numbers in 
. The sequences 
and 
are then defined recursively by
. Prove that 
for all sufficiently large 
.
is a prime number
is inscribed in circle 
. Let denote its incenter and 
its -excenter. Let denote the midpoint of arc 
. Line 
meets a second time at . The perpendicular to 
at meets sides 
and at and respectively. The circumcircle of 
meets 
a second time at , and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
be a triangle with circumcircle and incentre . A line 
intersects the lines , 
, and 
at points , , and , respectively, distinct from the points , , , and . The perpendicular bisectors 
, 
, and 
of the segments 
, 
, and 
, respectively determine a triangle 
. Show that the circumcircle of the triangle is tangent to .
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
?
and let 
be the distinct odd primes, not 
or 
, strictly less than . Then, we can choose by the Chestnut Ramification Theorem![\[m\equiv 1\pmod 2\]](http://latex.artofproblemsolving.com/1/e/6/1e6c1ce5218d440ffef0b4c5bfbf43b24a372cec.png)
![\[m\equiv 0\pmod p_j\qquad 1\leq j\leq \ell\]](http://latex.artofproblemsolving.com/b/1/6/b16dc84ba1f5b5b118e14c62656afda9f7b0dffc.png)
![\[m\equiv -2\pmod p\]](http://latex.artofproblemsolving.com/d/a/7/da776b64bba663c1654978c2835555fbed207b3e.png)
![\[m\equiv -4\pmod q\]](http://latex.artofproblemsolving.com/4/6/b/46b9e805bcfaa50e893976e5438e2401d0108964.png)
Now, if we take 
, we can see it is odd, not divisible by 
for 
, and is divisible by . Thus, 
. By similar logic, 
.
, and consecutive instead of 
.
and does not exceed 2 (
). Prove that there are two consecutive positive integers such that the largest prime divisor of one of them is and that of the other is .
, and consecutive instead of .
and does not exceed 2 (). Prove that there are two consecutive positive integers such that the largest prime divisor of one of them is and that of the other is .
where is the smallest prime divisor of and 
. Clearly, . Also, 
. We want 
. For this we wish to show that such exists. It is well known that the solution to 
exists. As we want that should be the smallest prime dividing 
, we define a sequence of primes 
such that:![\[2<x_1<x_2 \ldots < x_m<q\]](http://latex.artofproblemsolving.com/a/3/9/a3938c77c0d77e0cd7945720f2fedc1c238e622f.png)
and 
where 
. Now we don't want the following set of congruences to hold true:
From here on, I think it is possible to pick such which doesn't satisfy any of the above congruences and is a solution to . As soon as we pick such , we are done. 
is so easy? Basically, is my solution correct? I personally think there should be more details, but I am not able to figure it out. Can I get some help? 
exists is wrong,otherwise what is left in the problem 
exclude .
to get .
, is never congruent to 
.
doesnt contain a prime smaller than as none of the divide it.Also choose odd.
and let be the distinct odd primes, not or , strictly less than . Then, we can choose by the Chestnut Ramification TheoremNow, if we take , we can see it is odd, not divisible by for , and is divisible by . Thus, . By similar logic, .
and let be the distinct odd primes, not or , strictly less than . Then, we can choose by the Chestnut Ramification TheoremNow, if we take , we can see it is odd, not divisible by for , and is divisible by . Thus, . By similar logic, .
and let be the distinct odd primes, not or , strictly less than . Then, we can choose by the Chestnut Ramification TheoremNow, if we take , we can see it is odd, not divisible by for , and is divisible by . Thus, . By similar logic, .
be the set of primes less than and distinct to 
. such that 
, 
and 
for every 
. and 
. 
is prime number.
- all prime numbers products less then .
exist prime great .
Prove that
where 
is the smallest prime factor of 
and 
.
and
,
.
,
modulo all other primes less than 
except 
.