[/quote]
,
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
passes through the midpoint of 
.
& 
.
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.
Y by Adventure10, Mango247
Hi guys, so I'm struggling with this seemingly simple BMO1 problem:
Let ABC be an acute-angled triangle, and let O be its circumcentre. The circle through C,O and B is called S.
The lines AC and AB meet the circle S again at P and Q respectively. Prove that the lines AO and PQ are
perpendicular.
What I've done so far:
___________________________
Let angle QPC=x and angle PQB=y.
PCBQ all lie on a circle, and thus form a cyclic quadrilateral. Therefore, angle QBC=180-x and angle PCQ=180-y. As points P, C, and A are collinear, angle BCA=y. Similarly, angle CBA=x
Because the line AO goes through the circumcentre, it meets line CB at 90 degrees.
___________________________
To achieve the desired result, I want to prove that the line CB is parallel to the line PQ.
I think I might be close, but I am terrible at geometry questions so I'm not sure.
Please give me some hints / advice, I'd prefer that to a complete solution. Thank you kindly
Let ABC be an acute-angled triangle, and let O be its circumcentre. The circle through C,O and B is called S.
The lines AC and AB meet the circle S again at P and Q respectively. Prove that the lines AO and PQ are
perpendicular.
What I've done so far:
___________________________
Let angle QPC=x and angle PQB=y.
PCBQ all lie on a circle, and thus form a cyclic quadrilateral. Therefore, angle QBC=180-x and angle PCQ=180-y. As points P, C, and A are collinear, angle BCA=y. Similarly, angle CBA=x
Because the line AO goes through the circumcentre, it meets line CB at 90 degrees.
___________________________
To achieve the desired result, I want to prove that the line CB is parallel to the line PQ.
I think I might be close, but I am terrible at geometry questions so I'm not sure.
Please give me some hints / advice, I'd prefer that to a complete solution. Thank you kindly
is not perpendicular to 
.
be isogonal conjugate of 
.
meet 
at 
, respectively .
~ 
~ 
