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 cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and , and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly., and are concurrent on line .
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
, and an odd prime 
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some 
(which may depend on 
) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly 
.
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all such that 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer such that 
for all 
. be a right triangle with 
. Point lies on the line 
such that is between and . Let 
be the midpoint of 
and let be the seconf intersection point of the circumcircle of 
and the circumcircle of 
. Prove that as varies, the line 
passes through a fixed point.
and 
intersect each other at 
and 
. The common tangent to two circles nearer to touch and at 
and 
respectively. Let 
and 
be the reflection of and respectively with respect to . The circumcircle of the triangle 
intersect circles and respectively at points 
and 
(both distinct from ). Show that the line 
is the second tangent to and .
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
with 
and the circle 
of radius centered at 
Let be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let be the intersection of 
and Prove that 
cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from 
to 
is a sequence of roads such that one can move from to along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.
and be two distinct cities in Anisotropy. Let 
denote the maximal number of paths in a diverse collection of paths from to . Similarly, let 
denote the maximal number of paths in a diverse collection of paths from to . Prove that the equality 
holds if and only if the number of roads going out from is the same as the number of roads going out from .
decide whether there exist pairwise distinct positive integers 
such that for every 
, 
divides 
.
and 
and 
![$b_n<=[ta_n]+b_1$](http://latex.artofproblemsolving.com/d/7/0/d708574866027662bc903d1c725535e98b7d44d2.png)
,![$n-b_1>[ta_n]$](http://latex.artofproblemsolving.com/2/e/5/2e55f611360c6a0f9cceaf77c5acdaa5c77839c6.png)
}
of C
and 
decide whether there exist pairwise distinct positive integers such that for every , divides .
Consider any prime 
Then the problem gives
Now we have the following key claim which can be proven by simple induction:
denote the th harmonic number. Then
are pairwise distinct. The claim gives 
In particular 
is eventually constant. So ignore primes for which 
This means we only have a finite set of prime factors to worry about for the sequence now.
for some constant and all primes Now it is not too hard to see that we can find arbitrarily large intervals over which 
is constant. Since the set of prime factors of 
is finite now, hence by pigeonhole two terms 
will have the same 
for all primes 
hence will be equal, a contradiction. 
. For convenience, define 
for every integer . Furthermore, define 
. Finally, let 
. The pith of the problem lies in the following claim:
.
, there exists an arbitrary long sub-sequence of terms 
such that 
for sufficiently large 
and 
. Note that for primes in but not , 
, so there are only a finite number of possible terms with equal 
for all primes 
. Therefore, for sufficiently large there will be two equal terms, the end.
decide whether there exist pairwise distinct positive integers such that for every , divides .
is bounded and when is sufficiently large we will get 
, contradiction.
there exists such a sequence. I'll start with the following claim:
for every prime .
is constant, but note that 
can take only values which are divisors of 
where 
. Now at some point 
is going to be smaller than the length of these intervals because 
grows very slowly, so the 's can't be distinct over these intervals.
's differently without the harmonic number).
, such a sequence does not exist. Assume contrary. Fix and such a sequence. Call a prime nice if it divides some . Note all nice primes must divide 
, in particular number of nice primes is finite. Fix a 
such that 
for all primes . Fix any nice prime . Let 
. Then we know that
Intuitively, we will show that 
's grow slowly. Call an 
a peak if 
. Let 
be all the peaks.
we have
. Note 
is equivalent to
Observe that for any that
Now note 
. So it follows that
Plugging 
in 
and using above implies our claim. 
an 
such that 
.
and Claim 1 to prove this. Observe 
, since all of them cannot be (by Claim 1 for 
). Now choose a very small such that:
We show this 
works. Suppose for some 
it holds that 
. We will show 
(note this would imply our claim by induction). 
in 
gives
So it suffices to show
Indeed,
This proves our claim. 
, all of 
are at most 
.
. As 
, so it follows 
Now 
, thus all of 
are 
, and 
for large . This proves our claim. 
be all the nice primes, and 
be any numbers for which Claim 3 is true. Let 
. For a large , look at the numbers
By Claim 3 we know that for any 
adic valuation of any of these numbers is 
. It follows number of distinct numbers between them is at most
As all 's are distinct, so this forces
But this is a contradiction as exponential functions grow faster than polynomial functions. This completes the proof. 
's are not given to be distinct, so we somehow had to prove that. So we basically had to prove the 's don't grow fast. Now was only interesting for peaks. So it was natural to consider peaks. Now after getting Claim 1, I was sure we only have to use Claim 1 and ignore all other conditions on peaks, which along with 
would give us the sequence 
doesn't grow fast. So we only had to prove Claim 2. Basically, we conjecture 
for all 
(for some ). We then find the sufficient conditions on for which we can prove this by induction. The two equations in proof of Claim 2 were precisely those. Now we had some problems like it might happen 
, but that was easy to fix by showing for large .
. Let be some prime, and let 
, 
for 
. Viewing the divisibility condition in -adic terms only, it is equivalent to
Let 
denote the sequence of harmonic numbers. The crux of the problem is the following:
.
doesn't modify the truth of the condition, so WLOG let 
. We now use strong induction:
so 
as desired.
is zero if 
, 
if 
but 
, and 
otherwise. Suppose there are 
distinct primes dividing and 
distinct primes dividing but not . Then there are 
choices for the value of 
—we have options for if and , and options for if 
. But dominates 
, so by Pigeonhole for sufficiently large there must exist two non-distinct elements of the sequence: contradiction. 
.
. Similarly and because of , we get 
, for every prime dividing both sides of the equations. Now we calculate 
, which is equivalent after some boring basics in arithmetic to ![$(k+1) \cdot [v_p(a_{k+2}) - v_p(a_{k+1})] \le v_p(C)$](http://latex.artofproblemsolving.com/4/1/0/410d8c6350272a5077dd707bda3a997f2477fda2.png)
. But this inequality is not true for a large , hence we get a contradiction and we are done 
k![$, hence we get a contradiction and we are done :ninja:[/quote]
Why ?If {a_{k+1}})^{k+1} \le C$](http://latex.artofproblemsolving.com/f/f/3/ff37743383c4040ec1d59b5686fa83c927706050.png)
. isn't very big ,how can explain the numer is larger than C?
's grow? Define 
as the set of prime divisors of 
. Note that the given sequence has no prime factors outside of . Let prime 
be arbitrary. Denote 
for all positive integers . The given condition rewrites as:![\[kb_{k + 1} \leq k\nu_p(C) + b_1 + b_2 + \cdots + b_k.\]](http://latex.artofproblemsolving.com/1/d/6/1d66093f7ebf6d1ba0c96d362e955f3c0d0445ca.png)
Define 
as the average of the first elements of our new sequence. Adding 
to both sides of this inequality and dividing through by 
lends us a useful inequality.![\[\frac{b_{k+1}+\cdots + b_1}{k+1} \leq \frac{1}{k+1} \nu_p(C) + \frac{b_k + \cdots + b_1}{k}.\]](http://latex.artofproblemsolving.com/8/a/6/8a6774aac39a3ad2b6aa036c53728bbe38b8d2e5.png)
Applying this inequality repeatedly gives us the following inequality.![\[s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{2}\right)\nu_p(C) + s_1.\]](http://latex.artofproblemsolving.com/9/e/0/9e07f26eba45d02468d7da95be4ff4eee3515d87.png)
Plugging back into the original inequality and bounding the harmonic series gives us:![\[b_{k+1} \leq \nu_p(C) + s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{1}\right)\nu_p(C) + b_1 \leq \ln(k+1) + b_1.\]](http://latex.artofproblemsolving.com/7/0/1/701ea61c8b028a2bdb9ee2dd620b49d464db65b8.png)
Something is seriously wrong, and we are very happy about it. Since the above is true for all primes , there are 
different possibilities for the values of 
. Since![\[O((\ln k)^{|S|}) < k\]](http://latex.artofproblemsolving.com/8/e/c/8ec6f533782cba4334846e23482fde1af075a755.png)
as approaches infinity, there must exist two terms in the sequence that are equal.
and is constant, so the set of primes dividing an element in 
is finite. Let be a arbitrary prime dividing one of 
. Then the divisibility condition becomes 
. Now consider the following claim:
, we have 
.. Base case is clear. For the induction step, 
. is finite, let be a number of elements in the set of primes dividing an element in . Then taking large , we see that there are most 
distinct values in 
, a contradiction. Therefore there are no such . 
is finite as any prime dividing for 
must divide 
by induction due to the given divisibility condition. Now, we prove our key bound.
and positive integers ,![\[\nu_p(a_k) \le \nu_p(a_1)+\nu_p(C)\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k-1}\right)\]](http://latex.artofproblemsolving.com/c/f/2/cf2568f80dbd15ce37d50745c2865d2f3f2a60d6.png)
,![\[(k-1)\nu_p(a_k)\le (k-1)\nu_p(C) + \nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-1})\]](http://latex.artofproblemsolving.com/f/a/8/fa8dd57b5c4ae3fceaf4d985466ec2048b6e8c50.png)
Now replacing 
inductively we obtain,
which proves the claim.
for all sufficiently large . This is because if 
for all 
then for all 
, 
but this is a set of distinct positive integers bounded above by 
which is a clear contradiction. the number of distinct positive integers that are possible for 
is![\[ \prod_{i=1}^r \left(\left \lfloor \nu_{p_i}(a_1) + \nu_{p_i}(C)H_{k-1}\right \rfloor +1\right) = O\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k}\right)^{r} = O(\ln k)^r < k\]](http://latex.artofproblemsolving.com/8/f/0/8f0e294590fe17f09d102ab0c35abd19c900c1a7.png)
for all sufficiently large where 
is the set of all primes dividing . But this is a contradiction to what we noted above so the desired is impossible for all and we are done.
.
; hence we can write each as 
for fixed primes 
, 
, 
( is finite).
for each 
.![\[\nu_p(a_n) \le \nu_p(C)+\frac{\nu_p(a_1)+\dots+\nu_p(a_{n-1})}{n-1}\]](http://latex.artofproblemsolving.com/2/f/6/2f6ddb28f8b2b3f081a72c4ec304b8e6c40e965b.png)
By simple induction we have that 
As required. 
to each . Now fix some large and look at the number of distinct sequences among , , 
which is ![\[O(\log N)^t \ll N \]](http://latex.artofproblemsolving.com/b/7/a/b7a0a4f22fcee2b8a26624e3befd333ce35f0ceb.png)
And hence by PHP, two the sequences must coincide and hence their corresponding numbers are same which is a contradiction.
,
,
,
.
.
,
.
is finite. Let 
.
denote the sum of exponents of prime numbers appeared in the canonical form of 
.
that 
For induction step, suppose we want to prove for the case when 
.
,
.
.
for all 
.
that 
is at most 
.
by simple bound.
denote the 
Then
Proof: The proof is just strong induction on 
Firstly, put 
in Equation 1 to get
which serves as the base case since 
Now assume the result till some 
in Equation 1, we find
and hence
and the induction is complete. 
to be equal for all primes to get a contradiction. The claim gives 
(basically 
for 
then 
for 
Also, if we can show 
then by Pigeonhole Principle, we would find two 
such that 
So if the interval ![$[n,m]$](http://latex.artofproblemsolving.com/9/3/3/933804a455b543d85e536a3fc82eae1be1e27ab2.png)
is large enough, then this will hold for all primes 
So we prove the following:
there exists an interval 
of length 
Next,
So even if we pick 
for large enough 
lie between the same two integers. 
