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be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
with the following property: for 
but not 
consecutive positive integers, each one is a divisor of .
+1 w
be the incenter of triangle , and 
be the perpendicular bisector of 
. Suppose that is on the circumcircle of triangle , and line 
and intersect at point . Point 
is on such that 
.Suppose that line 
and the midsegment of that is parallel to 
intersect at . Show that 
holds![$$a\in[-2,10]\iff (a+2)(a-10)\le 0\iff a^2\le 8a+20.$$](http://latex.artofproblemsolving.com/2/1/0/210f528cf3ec1e68944430ffb10e49131148a4a2.png)
Therefore
Can there be equality? It holds iff 
numbers is equal to 
and 
numbers is equal to . It comes to
Hence no equality and finally
#1730
![$a_1,a_2,\cdots,a_n \in [A,B]$](http://latex.artofproblemsolving.com/1/c/b/1cb093fd8b5dbf38c36adbe49053e1142f5366ce.png)
, and 
,
.
for ![$x \in [-2,10]$](http://latex.artofproblemsolving.com/3/4/c/34c550e9e5f87530e063c7f96e73cb2f2636c5b3.png)
. Thus...
or for all 
. If there are 
's and 
's then. 
. But 
(mod 
). Contradiction.
![\[(a_i + 2)(a_i + 10) \leq 0\]](http://latex.artofproblemsolving.com/c/2/9/c29790445fa8d5397819ceaf7061d0e266348dcb.png)
![\[a_i(a_i - 8) \leq 20\]](http://latex.artofproblemsolving.com/8/9/e/89efc565b61edf94056b563b5481a84b66a1e15a.png)
Which is always true, and equality is achieved when 
, or 
.
times for each 
results in 
, but since the sum of all is 
, we cannot have equality, because not all can equal or , because there are terms.
Then summing we have,
We cannot have equality as there are terms.
for all 
. Summing among all yields the conclusion. (The inequality is strict as we cannot have all variables either or )
such that 
, we see that if 
is positive, 
Thus by smoothing (more like anti-smoothing) argument, by making the numbers farther apart, the sum of squares becomes greater. This operation cannot be applied further if all (or all except one) elements are at the limits (
). Thus, let our set have 
of 's, 
of 's, and one 
which gives a sum of square value of 
\qed
Expanding gives 
so 
After some experimentation, the equality case 
cannot be realized, so the inequality is strict. \qed
such that 
from the interval ![$[-2,10]$](http://latex.artofproblemsolving.com/d/6/8/d68464eed4c2fe38f134793ffca4a671614f8634.png)
such that their sum is 
real numbers, or else the maximum sum possible will be 
Hence no equality and finally
for all indices 
for all indices 
Inequality (1) rearranges to 
Summing this inequality over all indices 
with equality if and only 
for all indices 
However, equality cannot be achieved because there is no way for 
![$-2, 0]$](http://latex.artofproblemsolving.com/6/d/d/6dd3f53b736f5ac26a6a3d60ec4b9fa784f792c3.png)
and one is in the interval 
,
,
away from each other with constant sum also increases the sum of their squares.
and we can calculate the sum of all numbers and get the maximum, 
.
due to the interval constraint, and expanding yields 
.
,
to 
giving us 
.
tells us![\[0 \ge \sum (a_i+2)(a_i-10) = \left(\sum a_i^2\right) - 2020.\]](http://latex.artofproblemsolving.com/6/5/8/658ef0776081e439b9111fcbdc37a7d78d97c781.png)
,
.
.
which does not have an integer solution.
.
.
which is true because ![$a_x \in [-2,10]$](http://latex.artofproblemsolving.com/3/2/0/32067564c70bbbce04c9c034279f4d9cd2fa4b9e.png)
.
.