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.
and 
such that
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
satisfy 
. Determine the minimal and the maximal value of the expression 
be an odd prime congruent to 2 modulo 3. Prove that at most 
members of the set 
are divisible by .
be a random point on the smaller arc 
of the circumcircle of square 
, and let 
be the intersection point of segments 
and 
. The feet of the tangents from point 
to the circumcircle of the triangle 
are 
and 
, where 
is the center of the square. Prove that points 
, 
, and lie on a single circle.
with 
, let its incircle be tangent to sides 
, 
, and at , 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of .
greater than 
is divisible by 
be a convex quadrilateral with pairwise distinct side lengths such that 
. Let 
be the circumcenters of 
, respectively. Show that 
, the Euler line of 
and the Euler line of 
are concurrent.
such that 
is the square of an integer.
be the circumcenter and 
the orthocenter of an acute triangle . Show that there exist points , , and on sides , , and respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](http://latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
be finite.
satisfying the following condition for all real numbers 
:![\[
f(x - 1 - f(x)) = f(x) - x - 1.
\]](http://latex.artofproblemsolving.com/3/a/5/3a51479b61a5c025ed81f539e964ce96b5857f0f.png)
of positive integers such that ![\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]](http://latex.artofproblemsolving.com/e/c/7/ec778edf97100f3b19350d8a817317ebf1f391bd.png)
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.
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 
,
,
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 
positive divisors, we must have 
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 
varies from 
.
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