[/quote]
,
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
and puts 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
.
at 
,
at 
,
at 
are concurrent.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
Y by Adventure10, Mango247
In acute-angled triangle 
, 
is an altitude and 
is a median. Points 
and 
on lines 
and 
respectively are such that 
and 
. Prove that the midpoint of 
is equidistant from 
and 
.
Proposed by Danylo Khilko
, 
is an altitude and 
is a median. Points 
and 
on lines 
and 
respectively are such that 
and 
. Prove that the midpoint of 
is equidistant from 
and 
.Proposed by Danylo Khilko
This post has been edited 2 times. Last edited by Mindstormer, Jul 22, 2018, 8:49 PM
Reason: typo
Reason: typo
and 
- midpoints of 
a
so 
- midpoint of common hypotenuse of 
so 
and 
.
- midpoint of 
and so 
.
is the unit circle, then we know that 
So, the midpoint of 
For computing 
we know 
(unit circle chord) and 
(because 
Using symmetry, 
Then the midpoint of 
,
We only need to show that the foot from 
is 
And hence we're done!
