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.
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. 
is the orthocenter. 
is the midpoint of segment 
. Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from 
and 
intersect at 
. Prove that 
are colinear
be an integer and let 
be positive integers. In a group of 
people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following two conditions to hold at the same time:
.
with 
, there is someone who has played exactly 
games of chess.
be an acute triangle with orthocenter and circumcircle 
. Let be the midpoint of side . Point 
is chosen from the minor arc on 
such that 
. Let 
be a point on such that 
is perpendicular to 
, and 
be a point on line such that 
is perpendicular to . Lines 
and intersect at point 
, and point 
is the reflection point of with respect to .
.
satisfying the following equations: 
be positive real numbers such that 
. Prove that
.
be real numbers such that 
Prove that
Let be real numbers such that 
Prove that
Let be real numbers such that 
Prove that
be a real numbers such that 
Prove that
Where 
be a real numbers such that 
Prove that
Where
If 
be positive real numbers, show that:
be nonnegative positive integers.
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.
such that: 
.
.
![\[ S=\sum d_id_{i+1}<n^2\iff \sum\frac{1}{d_id_{i+1}}<1\]](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 
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 
.
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