[/quote]
,
is the foot of the altitude from 
of triangle 
.
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles 
.
.
.
.![\[\sum ab +\sum ab^2 \le 6\]](http://latex.artofproblemsolving.com/9/f/6/9f632ab1e6df5029bcd4416eb06623123ad2cd8c.png)
be a triangle with side‐lengths 
,
,
.
the area of 
be the areas of triangles 
,
,
,![\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]](http://latex.artofproblemsolving.com/1/3/9/139a3e15f08afb78ca0b8301f24d30db8f202207.png)
the circle with diameter 
,
be the contact points of the tangents from 
and 
lie on opposite sides of 
and 
lie on opposite sides of 
be the circle with diameter 
,
,
the circle with diameter 
.
are concurrent.
and 
are prime and satisfy![\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]](http://latex.artofproblemsolving.com/7/3/5/735ec9416a654f40121021bdd1233ffa86e42495.png)
.
.
P![$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$](http://latex.artofproblemsolving.com/6/b/6/6b6011a32b103fe94652974a521eeae479cba1d8.png)
with orthocenter 
and a point 
lying on the Euler line of 
Prove that the midpoint of 
lies on the Thomson cubic of the pedal triangle of 
.
is the inverse of function 
and thus, they must intersect that 
axis (as its clearly not self-inverse function). Thus: 
is the only solution.
. Prove that
W here 
be an acute triangle with 
,
be the center of its circumcircle. Let 
be the reflection of 
with respect to 
.
at 
,
intersects the line through 
.
be a point on the ray 
,
.
.
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least 
of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly 
leaders present. The chairperson must seat them in a square-shaped conference hall of size 
,
cell. It is known that exactly 
of these 
neighbors and will vote "FOR" only if both of their neighbors voted "FOR."
be a quadrilateral with 
.
meet at 
and 
denote the circumcircle and the circumcenter of triangle 
.
and 
denote the circumcircle and circumcenter of triangle 
.
and 
(other than 
and 
be the midpoints of minor arcs 
(not including 
(not including 
.
Y by Adventure10, Mango247, Mango247, Mango247
Let 
be one of the two points of intersection of two circles with centers 
respectively.The tangents at to the two circles meet the circles again at 
. Let a point 
be located so that 
is a parallelogram. Show that is also the circumcenter of triangle 
.
be one of the two points of intersection of two circles with centers 
respectively.The tangents at to the two circles meet the circles again at 
. Let a point 
be located so that 
is a parallelogram. Show that is also the circumcenter of triangle 
.
.
the second intersection point of 
.
is perpendicular to 
and 
respectively.(because AB,AC,AQ are the radical axes of 
,
.
is perpendicular at 
,
.
(because of the parallelogram).So it suffices to prove that 
which is true because these angles have their lines perpendicular.
.
then 
lies on perpendicular bisector of ![$ [AC]$](http://latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
,
.
,
,
is the perpendicular bisector of 
is also perpendicular bisector of 
,
, let 
and 
we get 
is parallelogram we have 
, hence we get 
