[/quote]
,
be a tree on 
vertices with exactly 
leaves. Suppose that there exists a subset of at least 
vertices of 
be an acute scalene triangle. Denote by 
the circle with diameter 
,
be the contact points of the tangents from 
to 
lie on opposite sides of 
lie on opposite sides of 
be the circle with diameter 
,
,
the circle with diameter 
.
are concurrent.
be a triangle with circumcircle 
intersects the line 
at 
and 
is the midpoint of 
.
and 
intersect on 
![\[x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d355433e896e8b97f28c2e4235140fd2348e77.png)
correct to four decimal places.
be an integer, 
a real number, and 
be positive real numbers such that 
.![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
be positive real numbers. Prove that there exists a cyclic permutation 
of 
such that for all positive real numbers 
the following holds:![\[
\frac{a}{x\,a + y\,b + z\,c}
\;+\;
\frac{b}{x\,b + y\,c + z\,a}
\;+\;
\frac{c}{x\,c + y\,a + z\,b}
\;\ge\;
\frac{3}{x + y + z}.
\]](http://latex.artofproblemsolving.com/1/5/2/15261b3f712c2cc8ffbeb92a9711e06ad0992b5d.png)
be positive integers such that 
,
,
are perfect squares. Prove that 
is also a perfect square.
be distinct primes and let 
be nonnegative integers. Define![\[
m \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;+\;\sum_{i=1}^k(p_i-1)\Bigr),
\]](http://latex.artofproblemsolving.com/7/9/a/79af85375957986298d37c00d288b6d402f40106.png)
![\[
n \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;-\;\sum_{i=1}^k(p_i-1)\Bigr).
\]](http://latex.artofproblemsolving.com/e/2/3/e23799ff03400d5eade3731bb83456fcd702a5e1.png)
Prove that![\[
p^2-1 \;\bigm|\; p\,m \;-\; n.
\]](http://latex.artofproblemsolving.com/3/7/e/37e4535b59ef183908730946213429e9ebc05dd7.png)
,
be positive real numbers. Prove that![\[\frac{a^4 + 3}{b} + \frac{b^4 + 3}{c} + \frac{c^4 + 3}{a} \ge 12.\]](http://latex.artofproblemsolving.com/1/f/d/1fdfd64c9aafa8741ad1594db46cc16a403b1a99.png)
When does equality hold?
with 
.
be the midpoint of side 
,
and 
be points on segments 
and 
,
.
be the circumcircle of 
,
the circumcircle of 
.
to 
,
and 
,
intersect 
,
intersect 
.
is tangent to both 
islands of the Arctic Ocean. Every bear sometimes swims from one island to another. It turned out that every bear made at least one swim in a year, but no two bears made equal swams. At the same time, exactly one swim was made between each two islands 
and 
such that 
.
Y by ImSh95, Adventure10, Rounak_iitr
Let 
be a trapezium inscribed in a circle 
with diameter 
. Let 
be the intersection point of the diagonals 
and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.
Greece - Silouanos Brazitikos
be a trapezium inscribed in a circle 
with diameter 
. Let 
be the intersection point of the diagonals 
and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.Greece - Silouanos Brazitikos
it is enough to prove that 
,
.
so 
.
.
.
.![\[ \cos{\widehat{KBL}}=1-\dfrac{KL^2}{2KB^2}=1-\dfrac{KL^2}{2BE^2}=1-\dfrac{2(AB^2-BE^2)}{AB^2}=2\dfrac{BE^2}{AB^2}-1 \]](http://latex.artofproblemsolving.com/c/4/7/c472344faf3c8081631f8208dcbce85fd7d3d337.png)
,
and with the basic trigonometry we get that both expression equal to 
.
and 
.
.
is center of 
.
is cyclic and tangents to 
.
we get that 
are concurrent at one point 
.
and 
.
it gives 
and 
.
.
and 
It is enough to prove that 
.
.
parallel to 
.
;
.
.
.
.
is parallelogram.. 
.
.
.
.
as well as 
.
.
.
.
,
.
.
and 
are equivalent triangles. Thus 
so 
and 
.
.
be the circle with center 
intersect 
respectively.
are inverses with respect to 
.
,
,
be the circle with centre 
,
,
and 
.
,
so 
as they share 
and both have a right angle.
.
,
,
and 
so that 
.
is isosceles, so that as 
,
is isosceles; moreover, 
,
.
.
,
as required.
.
lies on the radical axis of the two circles,or in other words 
are collinear.From this it immediately follows that 
(
is cyclic.Thus 
or in other words 
.
not containing 
,
we see that 
at 
,
be the reflection of 
the orthocenter of 
are collinear and 
is a parallelogram. From 
and 
,
.
be the midpoint of 
,
is a rhombus so 
so 
,
and 
are concurrent at a point 
and 
;
.
;
is an isosceles trapezoid; hence, line 
,
,
.
it is well known that 
are concurrent...and its easy from here as shown by posts above 
.
and 
are reflections of each other in 
.
around 
and 
.
is tangent to 
lies on 
.
,
.
.
.
with 
.
hence 
,
.
in 
.
and 
hence 
is a diameter in 
unit circle...so that
Let us firstly note that
Since 
On the other hand 
Equalizing 
and 
yields
.
.
.
be 
and 
be 
.
is 
and so 
is 
.
is 
and so 
,
at 
is equal to 
, or equivalently, 
is isosceles. Since 
is the perpendicular bisector of 
.
.
=
=
=
is cyclic. We are done. 
and 
is tangent to 
.
is tangent to 
.
are concyclic , we get 
are concyclic
.
,
and 
,
.
as required. 
we deduce that 
and 
.
,
,
and 
so 
collinear. Note that the radical axis must be a line perpendicular to 
,
are reflections across 
