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.
such that 
.
students, each pair of students have watched a movie together. Let the set of movies watched by a student be his movie collection. If every student has watched every movie at most once, at least how many different movie collections can these students have?
be non-negative real numbers such that 
Prove that
and 
be real numbers. Let 
be the matrix with entries ![\[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\]](http://latex.artofproblemsolving.com/c/0/8/c0856fe7b2b113180659952c4d216a01d8849f64.png)
Suppose that 
is an 
matrix with entries 
, 
such that the sum of the elements in each row and each column of is equal to the corresponding sum for the matrix 
. Prove that 
.
be an acute triangle and let 
be an interior point for it such that 
. Prove that
When does equality hold?
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be real numbers such that 
.![\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]](http://latex.artofproblemsolving.com/5/a/8/5a83eaa7470bcca6546599ff4e296e4efec69b16.png)
and determine all the cases when the equality occurs.
have the same incenter 
.Prove that 
is concyclic iff 
is concurrent
be the nonnegative integers in 
where 
. Define 
. Prove that 
.
such that 
where 
, and the statement is still true when 
? such that 
where 
, and the statement is still true when ?.
be a scalene triangle with circumcircle 
and circumcircle 
. Denote 
and 
as the midpoints of 
and 
, respectively. The altitude from 
to intersects again at 
. The line 
meets a second time at 
, and 
intersects at 
. The perpendicular bisector of intersects 
at 
and 
at 
. Let the circumcircle of 
intersect again at 
, and let the line 
intersect at 
.
is the reflection of over , prove that the intersection of 
and 
lies on the circumcircle of 
. be real numbers such that 
and 
and determine the equality cases.
with orthocenter 
, and 
the feet of altitudes for 
, we have on 
such as that 
. 
is on segment 
so 
. prove 
is a parallelogram. 
is perpendicular to 
, deduce: 
is perpendicular to 
i.e is parallel to 
. Using butterfly theorem, it's easy to realize 
so 
the midpoint of , the midpoint of 
, 
the midpoint of 
, 
the intersection of 
and 
, 
the ,intersection of 
and circumcircle, 
midpoint of 
is perpendicular to and is parallel to hence is perpendicular to and because is centre of circle passing through 
hence R is the intersection of 
so 
lie on a circle so 
also 
so 
hence 
and that means 
lie on a circle so
queue point as 
, midpoint of 
as , 
, 
midpoint , 
, 
is the nine point center ,
antipode of in 
, 
and 
, midpoint of 
and midpoint of 
being midpoint of 
.
by butterfly theorem in 
we have 
are collinear and is circumcircle of 
so by butterfly theorem in 
for 
we get 
be the midpoint of 
and 
be the intersection of circle 
and 
and 
be the intersection of 
and be the intersection of 
and 
is foot of perpendicular of and 
be the intersection of 
and 
be the intersection of 
. in is ( facing diameter of ) so we have 
we have 
are congruent .
is cyclic 
is radical axis of 
, 
. Then 
is the orthic axis of , and it is well known that this is perpendicular to the Euler line. Let 
and 
. Then![\[(K,L;P,P_\infty)\stackrel{U}{=}(A,B;F,V)=-1\]](http://latex.artofproblemsolving.com/a/f/3/af38e4cc99d3cf9f962f0e08602b8b34bcb25f32.png)
which shows that 
. We also have![\[(K,H;P,L)\stackrel{U}{=}(A,\overline{UH}\cap\overline{AB};F,B)=-1\]](http://latex.artofproblemsolving.com/2/7/2/27213c1ccf4fead704d3342d60c27033fd98a3a4.png)
which shows that 
. Therefore, 
. As both 
and 
are perpendicular to 
, we have![\[\frac{PH}{PK}=\frac{QH}{QA}=\frac{1}{3}\]](http://latex.artofproblemsolving.com/c/5/2/c520223b04c81107329cec092fe18cc4aaac3cc7.png)
as desired.
and 
.
where 
is a real number.
,
and
.
Thus![\begin{align*}
0=(b^2-c^2)(-a^2+b^2+c^2)t+\left[a^2\left( 1-\frac{a^2S_a}{S^2} \right)+b^2\left( -\frac{S_B}{b^2-c^2}-\frac{b^2S_B}{S^2} \right)+c^2\left( \frac{S_C}{b^2-c^2}-\frac{c^2S_C}{S^2} \right) \right]
\end{align*}](http://latex.artofproblemsolving.com/8/7/f/87f6301df20eff66b1d6771def5016f59b7e8ec3.png)
It is clear that 
.
.
.
Hence, 
which implies that 
.
such that 
,
.
.
is 
and 
.
where 
and 
.
and the last two equalities hold since 
.
Since 
,
.
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/* end of picture */[/asy]](http://latex.artofproblemsolving.com/1/6/0/1607c17527951e095b702a6f0216d430bc5aef63.png)
is the midpoint of 
is cyclic now since 
is cyclic we can see that:![\[\angle OND=\angle DMH=90^{\circ} - \angle HMD=\angle HQO_N \quad \textbf{So} \quad QO_N||ON\]](http://latex.artofproblemsolving.com/2/3/6/236665a546e25b17fe8a3f29d65b72afe4760cbb.png)
as desired