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 fixed integer. Find the least constant 
such the inequality![\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]](http://latex.artofproblemsolving.com/d/d/e/dde2d1dd0c412d8ba640f1d4c044e2fa2dcccc82.png)
(the sum on the left consists of 
summands). For this constant , characterize the instances of equality.
, 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 
.
such that 
. Find the minimum value of:
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 
.
, 
, 
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 
work. Note, that 
doesn't work, because, the sequence, stays constant at 
is even. is even, we have that 
which is an integer. is odd.
so 
and now we are done. (note we subtracted 1 because of the denominator)
. Clearly 
does not have an integer term as every term is 
.
. We would eventually want it to equal 
since that would be equivalent to being even, which is a vacuous case.
,![\[a_1 = M \left(M+\frac{1}{2} \right) = \frac{2M^2+M-1}{2} + \frac{1}{2}.\]](http://latex.artofproblemsolving.com/e/0/b/e0be141d994d6c2431869675b90829b9dc018608.png)
, we have that![\[\nu_2(N-1) = \nu_2\left(\frac{2M^2+M-3}{2}\right) = \nu_2((2M+3)(M-1))-1 = \nu_2(M-1)-1.\]](http://latex.artofproblemsolving.com/2/9/4/294dce46a63af7fd9105f62d6cbcf80e6a933278.png)
, at which point the term will be an integer.
work. If is even, 
. Otherwise assume 
where 
is maximal and so is odd.
.
. For 
, 
where is odd. Thus 
, 
which is an integer as is odd. and prove it for 
.
, 
.
is an integer itself. Now note that after each move, the power of two in 
decreases at most by due to the 
in the denominator after multiplication. So after moves, the power of remaining in it is at least 
. But note that after moves, the 
turns into an integer due to induction hypothesis. Thus we are done. 
. It is clear that 
is invalid, as all terms are and thus will never become integer., we have 
for all nonnegative integers , such that 
is an integer. Then all are odd, otherwise an integer term would exist. Then
Now taking the difference from 
yields
Since 
, all 
are distinct, so 
is an odd integer for all nonnegative integers 
, implying that 
decreases as we progress through the sequence. At some point, we reach a sufficient 
where 
must be odd, a contradiction as all are odd. Our proof is complete.
doesn't work. This is clear, since 
so 
and the sequence is always a constant non-integer. work.
for 
odd. Then 
is an integer.
The base case 
is obvious since is even and thus 
is an integer.
we will use it to show that the claim holds for 
Then
Note that 
is odd since is odd. Applying the inductive hypothesis on as the starting term gives us that 
is an integer. The induction is complete. 
is an integer, so all work. These are the only solutions, and we are done.
if M is even we have that 
and since 
, we have that is an integer. So obviously every even M works. Now assume M is odd. For M = 1, it doesn't work. Now we will prove by induction that each M works.
, we will always get an integer.. If it is 0, then it becomes an integer in one step and if not, then the next term will be 
if 
, we have that 
, so the value reduces and the inductive hypothesis will end.
. fails because 
for all 
. work because 
. Therefore, assume odd. is odd, we can let 
. Then, 
.
is even when is odd, an odd works if 
is odd. is not odd, so let 
. Then 
.
is odd when 
is odd, an odd works if 
is odd. is not odd, so let 
. Then, 
.
is odd when is odd, an odd works if 
is odd.
and assume that an odd works when 
is odd for 
. Then, 
.
is odd when is odd, an odd works if 
is odd. works when 
is odd for some integer 
. This is clearly true for all odd , as desired.
. fails, note that 
and 
, so the sequence is constant. works. The denominator of the sequence is always either or (trivial by induction).Now we rephrase the problem in terms of the number on the numerator while the denominator is constant, consider the sequence where 
. Then if 
, is the last element of the sequence, otherwise we have 
, so 
. Now we show that 
decreases by for each increment in . At some point this forces and we are done. Let 
, then we have 
, so we know 
which is 
, so 
as desired.
. Clearly does not satisfy the desired condition, and is an integer for all even .
for any index and odd 
. Then 
, so that 
. Hence, for some large index 
we have 
, implying that 
is an integer. Setting 
for any odd 
finishes. 
such that no term in the sequence is an integer.
for all . The proof is by induction. is odd, else 
.. Then note that shifting the sequence as 
, we must have 
.
. obviously does not work. The idea is casework on 
. If 
for odd, then I claim is the desired integer. The most straightforward way to do so is induction. The 
case is immediate from 
. If we suppose that it is true for 
, we wish to prove it is so for 
. For 
, then 
which when viewing 
has 
. Using our inductive hypothesis finishes.
is even is an integer. If 
then write 
then 
which has an even integer part so is an integer.
and 
For the sake of a contradiction, assume that none of the terms will be integers. Then clearly all 
Write 
we have that 
Hence is even. Let 
Then we can write 
Similar to above, for 
we must have that 
is even, so continuing this logic eventually yields a contradiction as is finite.
work. Clearly, does not work as 
for all 
so are the only solutions.
.
, which is a half-integer.
.
is a half-integer and 
is even, then 
is an integer.
**![\[a_0 = \frac{3}{2}, \quad \left\lfloor a_0 \right\rfloor = 1\]](http://latex.artofproblemsolving.com/2/6/a/26a9deb93793311ea31c07775985be0207e8c107.png)
and for all 
,![\[a_{k+1} = a_k \times 1 = a_k\]](http://latex.artofproblemsolving.com/c/2/5/c25895e2e769834f82f8998161fea5ec030ed998.png)
forever, never reaching an integer.**![\[a_0 = M + \frac{1}{2}, \quad \left\lfloor a_0 \right\rfloor = M\]](http://latex.artofproblemsolving.com/c/f/a/cfac57dd585417cd9eab8d2480ee34653804c20b.png)
![\[a_1 = \left( M + \frac{1}{2} \right) \times M\]](http://latex.artofproblemsolving.com/1/8/a/18a7349bdf87c04725dc0a40b4037a698520890c.png)
is even, then 
is an integer, so 
is an integer. is odd, then is a half-integer, but multiplying it by (which is at least 3) in the next steps ensures that eventually we will multiply a half-integer by an even integer, making the product an integer.
, the product's denominator (which starts as 2) will eventually be canceled once an even factor appears., either is even (done in one step) or we multiply a half-integer by increasing integers, and since after at most one multiplication by an even number the product becomes an integer, the sequence must eventually reach an integer.
Now when we take gif of the function "
" term vanishes and rest of the term becomes odd, So it does not cancel out with factor in denominator of the next term.
but only thing remains :
now notice that that the
is reducing by a factor of every time.
and observe that the 
would all be reduced to which is odd and hence we are done at 
where the even term in numerator would get cancelled with the factor of in denominator .
defined by ![\[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]](http://latex.artofproblemsolving.com/2/1/b/21b2a1f93b11a94b84e7b55f4b4f679aa20e36c6.png)
contains at least one integer term.
,
is an integer as desired. If 
,
,
,
.
.
So 
is good.
,
is even, so 
,
is good. Then notice that 
which is in the form of 
for some suitable 
,
and then 
for all 
. So now if we have a bad number 
is divizible with every power of 
,
.
.
.
,
I guess?
,
doesn't contain any integer term.In this case it's obvious that all the terms of the sequence will be of the form 
.
,
i.e. 
.
i.e. 
or 
i.e. 
.
or 
,
(
,
,
.
is an integer.
for some positive integer 
and odd natural 
,
is of the form 
for odd natural 
.
are not integers. Since 
and 
only multiplies the previous number of the sequence with a integer, all numbers 
where 
.
.![\[ a_{k+1}=a_k \cdot \left(a_0^{k+1}-\frac12 a_0^k-\frac12 a_0^{k-1}-\dots-\frac12 \right). \]](http://latex.artofproblemsolving.com/f/a/0/fa081f2dd64afe1a12def2a15f8fc076191fb011.png)
Let ![\[ b_k = a_0^k-\frac12 a_0^{k-1}-\frac12 a_0^{k-2}-\dots-\frac12 \quad \text{for } k \geq 1. \]](http://latex.artofproblemsolving.com/b/8/e/b8e621d798bf624d30fcc9561092c70bfa592fd7.png)
It then suffices to show that there is a number in the sequence 
that is even to get the desired contradiction, as it then follows that 
is an integer. Again, assume the contrary, in specific that all number 
are uneven. Then 
is even for all positive integers 
It thus suffices to show that there is a ![\[ v_2 \left((2M+1)^{n+1}-3 \cdot (2M+1)^n \right) = n+1 \]](http://latex.artofproblemsolving.com/f/3/0/f30f92f5b47a1da2bc26726be74cefa24be6dc62.png)
as then ![\[ v_2 \left((2M+1)^{n+1}-3 \cdot (2M+1)^n \right) = v_2 \left((2M+1)^{n}(M-1) \cdot 2 \right) = v_2 (M-1) +1 \]](http://latex.artofproblemsolving.com/0/4/e/04eee913fbb9cc195d88a84f6453d2ef5fcfc8ac.png)
For 
and will hence have found the desired contradiction.
=
.
=
+
:
.
:
=
)
,
:
)
)
=
is an integer.
,
for all non-negative integers 
is not an integer for any non-negative integer 
is easy to see.
;
for some integer 
,
for some integer 
,
,