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.
is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let 
be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .
be a right angled triangle with 
. Point 
is the midpoint of side 
And 
be an arbitrary point on 
. The reflection of 
over 
intersects lines 
and at and 
, respectively. The circumcircles of 
and intersect again at 
. Prove that the center of the circumcircle of 
lies on 
.
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.
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
positive integers 
such that 
for any 
, 
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
. For 
, 
is the smallest number different from 
which doesn't divide 
. Proof that every number large enough appears in the sequence.
be a triangle for which there exists an interior point such that 
. Let the lines 
and 
meet the sides and at 
and 
respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
, the point is the foot of the altitude from , and is a point on the segment 
. The lines through parallel to and meet at and , respectively. Points 
and 
lie on the circles 
and 
, respectively, such that 
and 
.
and 
are concyclic.
and the recursive relation
for all natural number . Find the general form of .
.
be a positive integer, with divisors 
. Prove that 
is always less than 
, and determine when it is a divisor of .
=
=
= is a divisor of 
when is a prime number:=
isn't a prime we have at least 3 divisors of = the maximum that can be: 
=
is a divisor of only if is a prime!
=== is a divisor of when is a prime number:= isn't a prime we have at least 3 divisors of = the maximum that can be: = is a divisor of only if is a prime!
and 
we will have
.
, we will have that
, it is enough to prove that
, we will have
, if 
and 
,
and so the above is true.
. If is not prime then 
and so 
, where 
is the smallest prime divisor of (which exists because 
). And so if 
then there exists a 
such that 
. Notice that 
by the first part and so 
. Since 
, then 
and we get 
which is false because 
. must be prime, and it is easy to see 
.
if and only if 
for every i
so clearly this inequality holds
) is 
and equality holds only when n is a prime so thats the only posible case
we know 
and 
where 
is prime.
, so![\[ d_id_{i+1} = \frac{n^2}{d_{k+1-i}d_{k-i}} = \frac{n^2}{d_{k+1-i}-d_{k-i}}\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right) \le n^2\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right), \]](http://latex.artofproblemsolving.com/8/4/8/8487980413d9f0f96c57231700b84b2759de0570.png)
since 
. Summing, we get![\[ \sum_{i=1}^n d_id_{i+1} \le n^2 \sum_{i=1}^n \frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}} = n^2\left(\frac{1}{d_1}-\frac{1}{d_k}\right)=n^2(1-1/n) < n^2.\]](http://latex.artofproblemsolving.com/3/4/7/347a5f5e7be57ba575889f92ba4c5c9daad0b8e5.png)
(b) Let 
. First consider the case where is composite. We know is the smallest prime divisor of . Let 
. Since 
, so 
. But is also the smallest prime divisor of , so 
is the largest proper divisor of . Therefore, 
, which is impossible, since it is at most 
. Hence, no composite works. Now consider the case where is prime. Then 
, so 
, which divides . Therefore, 
if and only if is prime.
This proves the first part. is prime (in which case we get 
). Now assume is not prime (meaning 
) and let be the smallest prime dividing . Then ![\[ d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 > d_k d_{k-1} = \frac{n^2}{p} \]](http://latex.artofproblemsolving.com/9/f/9/9f94e65ff5fdd58bc6d8981d02ace08eb0a704d9.png)
exceeds the largest proper divisor of , but is less than , so does not divide .
be a positive integer, with divisors . Prove that is always less than , and determine when it is a divisor of .
so the first assertion is true. Observe that 
where is the smallest prime factor of ; so if 
then 
contradicting that this quotient is a divisor of . So, 
and 
. For all prime numbers , the assertion holds; hence these are the only solutions.
![\[S:=d_1d_2+\cdots+d_{k-1}d_k=n^2\left(\frac{1}{d_1d_2}+\cdots+\frac{1}{d_{k-1}d_k}\right)\le n^2\left(\left(\frac{1}{d_1}-\frac{1}{d_2}\right)+\cdots+\left(\frac{1}{d_{k-1}}-\frac{1}{d_k}\right)\right),\]](http://latex.artofproblemsolving.com/0/6/a/06a48f00face21b335a14b5e17e8b9f64993b0d7.png)
so![\[S\le n^2(1-1/n)=n^2-n<n^2,\]](http://latex.artofproblemsolving.com/9/5/6/9568b632cb19c955f87e6bc2041bf1c74bbe9823.png)
as desired.. Any divisor of can be written as a product of two divisors of , so suppose 
where 
. We'll show that this isn't possible for size reasons. To do so, we have the following ``smoothing'' lemma.
, then 
.![$[x,y]$](http://latex.artofproblemsolving.com/3/d/3/3d3692c09209718827eeb3f7fcd218f7c92ad98e.png)
, the extrema of the linear function 
will be at the edges, so![\[x^2+xy\le xz+zy\le y^2+xy,\]](http://latex.artofproblemsolving.com/e/2/a/e2a6f29f65e8763896ce4b67f47d61aea9924d73.png)
so certainly . 
Using the lemma, we see that![\[d_ad_b<d_ad_{a+1}+d_{a+1}d_{a+2}+\cdots+d_{b-1}d_b,\]](http://latex.artofproblemsolving.com/3/d/9/3d990e926776bc5a102ce6471dc18098c11cf502.png)
so 
. If 
, then the only way 
is if has only two factors, so is prime. Primes clearly work as 
. Thus, the answer is 
.
and so on.
.
part over. that'll work should be prime - once we realise this, it basically becomes a one-liner: consider non-prime with smallest divisor . Notice that the largest divisor of 
must be 
, and as our sum 
but 
if there are more than 
factors, 
can't divide . Verifying, we see that each prime indeed works, hence the only that work are primes.
and 
Prove that
satisfying
Prove
![\[ S=\sum d_id_{i+1}<n^2\iff \sum\frac{1}{d_id_{i+1}}<1\]](http://latex.artofproblemsolving.com/3/b/2/3b211e0d1666d5dacc2f354c78b781c7e3954002.png)
![\[ \sum\frac{1}{d_id_{i+1}}\leq\sum\frac{d_{i+1}-d_i}{d_{i+1}d_i}=\sum\left(\frac{1}{d_i}-\frac{1}{d_{i+1}}\right)=1-\frac{1}{n}<1\]](http://latex.artofproblemsolving.com/e/8/1/e81e4f453c0ac5df88f69e112877b9a562e1e9b2.png)
.
is a divisor of 
also. Since the smallest non trivial divisor of 
which is 
,
.
and therefore 
,
.
.
.
,
,
,
,
,
.
.
The summation telescopes into 
,
,
is 
and 
.
However, if 
,
for all 
.
has![\[ D = n^2 (\frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}) \]](http://latex.artofproblemsolving.com/1/9/8/1984aaa1fbc73c91590a7f765d4bbf418f1c1c8c.png)
Since 
,
for all ![\[ \frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}
\leq \sum_{i = 1}^{k-1}\frac{1}{(i)(i+1)}
= \sum_{i = 1}^{k-1}\frac{1}{i} - \frac{1}{i+1}\]](http://latex.artofproblemsolving.com/c/3/3/c33e1fd69034ccf3266c447013e526893e123f07.png)
By telescoping, the above sum is equal to 
,
.
if and only if 
,
.
for some positive integer 
the smallest non-one divisor of 
.
.
is the only term in 
,
,
.
we get it suffices to prove
Now note that
For the other part, note that
But 
are the bigger and the second bigger divisors of 
.
which divides 
where 
if 
Thus, we just need to show that 
Indeed, we know that 
,
,
,
and the desired inequality follows.
divides 
,
.
.
for some positive integer 
which is equivalent to 
.
as then we only have 
,
,
so 
For the second part, we claim the answer is when 
,
,
.
.
so 
,
,
.
Where 
where 
,
,
,
,
.
Now, since 
,
,
,
,
,
Notice that: 
where 
is the largest divisor of 
inorder to have 
.
The second part is only true when 
for some prime 
.
,
is less than 
.
is less than or equal t
.
which is less than 
is a prime, 
and 
are all the divisors of ![\[d_1d_2=n \nmid n^2\]](http://latex.artofproblemsolving.com/0/b/4/0b490114cf04897f20cc482101d77a603519c478.png)
so the inequality holds with ![\[d_1d_2+d_2d_3 + \dots + d_{k-1}d_k = \frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1}\]](http://latex.artofproblemsolving.com/7/b/8/7b8a8604f3ec81d99f6a1fb34e4c15c27e5f24c6.png)
First note,![\[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} > \frac{n^2}{d_2d_1} = \frac{n^2}{d_2}\]](http://latex.artofproblemsolving.com/f/2/f/f2f5be8a9d28f5daf8889e7f1eefbd1610791ee4.png)
Further, since clearly 
for all 
,![\[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} \le \frac{n^2}{k(k-1)}+\frac{n^2}{(k-1)(k-2)}+\dots + \frac{n^2}{2 \cdot 1} = n^2 \left(1- \frac{1}{k}\right) < n^2\]](http://latex.artofproblemsolving.com/6/6/0/660f3eb3d9a1ea4c87281cafa62a66edb163c619.png)
Since ![\[\frac{n^2}{d_2}<\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} < n^2\]](http://latex.artofproblemsolving.com/8/d/a/8da6bfb1748b55e1c34f9d1b7511ffb690f44c35.png)
it cannot be a divisor of 
Now if 
so 
so this cannot possibly divide 
But if 
for all 
Hence, 
For equality, observe that 
Hence, 
but it also can't be