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.
+1 w
be a triangle and let 
be a point in its interior. Lines 
, 
, 
intersect sides 
, 
, 
at 
, 
, 
, respectively. Prove that![\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC] \]](http://latex.artofproblemsolving.com/b/6/f/b6f492b82d7b43d9a5a0a7f9e81205cb8e732bc9.png)
lies on at least one of the medians of triangle . (Here ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
denotes the area of triangle 
.)
, such that:
.
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
are nonzero real numbers such that![\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]](http://latex.artofproblemsolving.com/6/d/9/6d9aabd8130b99d7128de9a3f063eb4c7124febf.png)
let 
be the maximum value of the expression![\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]](http://latex.artofproblemsolving.com/a/b/2/ab28769126365574d4c0108500c370826082de6b.png)
Determine the sum of the numerator and denominator of the simplified fraction representing .
be a circle and 
a line in its plane. Let 
be a point on , located outside of . Let 
and 
be the tangents from to , where 
and 
are distinct points on . Let and 
be two points on . Lines and intersect line in two points 
and respectively 
. Lines 
and 
intersect the second time circle in points 
and . Prove that the tangents from and to are concurrent on line .
and 
be the roots of the equation 
where 
is an odd integer. Let 
is an integer
) = 1 .
denote the sum of all integers 
such that 
and exactly one of 
and 
is divisible by 
. Compute the remainder when is divided by 
.
is given in the plane. The incircle of triangle has center 
and is tangent to diagonal 
at 
. Let 
be the center of parallelogram . Show that 
and 
are parallel.
is valid while 
is not)
of positive integers such that 
is an integer.
are positive integers such that for every natural number the set of prime factors of 
are same. prove that 
.
be circumcenter of triangle and let angle bisector at cut circumcircle of triangle at 
.
and are concyclic with 
it is enought to prove that bisect 
.
this follows. 
we have that 
. Let be the point in such that 
and 
.
and 
are angle bisectors of angles 
and 
, respectively, and triangles 
and 
are isosceles we have that and are the perpendicular bisectors of 
and 
, respectively. Hence is the circumcenter of triangle 
. lies on the perpendicular bisector of 
and on the angle bisector of 
we have that 
is cyclic.
is cyclic, where 
are the points of perpendiculars from to and respectively. So we need to prove is 
. From the condition 
we get that 
. Thus 
. So we get that , from which follows that is cyclic.
. i.e.
where 
and like wise 
.
which is nothing but 
.
]
as desired.
as the reference triangle. We have 
. Consider the circumcircle of 
. We must show that also lies on this circle. Plugging in each of the points 
into the circle formula, we get 
. The equation of 
is 
Now, plugging in 
, we have
which is clear from the problem statement, so we are done. 
, the midpoints of 
in addition to the circumcenter 
and are concyclic with diameter 
Thus, it suffices to show that lies on .
for some constants 
. To find the constants which determine 
, we can plug in each of in the general form and solve for . Note that the equation of a circle is homogenous so we can use homogenized coordinates here.
gives 
.
gives 
.
gives 
. has equation 
Hence, using Barycentric Power of a Point, we can compute 
since 
. But this is 
since 
. Done.
.
and 
be midpoints of 
are concyclic.
be the point inside 
,
isosceles (such point exists because 
)
we find that 
is also an isosceles triangle. Since 
angle bisector and 
.
,
,
hence 