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.
be the midpoint of 
on triangle 
. Point 
lies on segment 
such that 
and 
on line 
such that 
. Prove that 
.
be a convex quadrilateral. The diagonals 
and 
intersect at 
, with 
. Let 
and 
be the midpoints of sides 
, and 
, respectively. Find the greatest positive real number 
such that![\[
EG + 3FH \geq k \cdot d + (1 - k) \cdot s,
\]](http://latex.artofproblemsolving.com/b/e/f/bef24b0d482f95d2e6ef22b3797a432c69d98af1.png)
where 
is the semiperimeter of and 
is the sum of the lengths of its diagonals (i.e., 
). Determine when equality holds.
such that the equation
and 
passes through the center of the other one. Triangle is inscribed into , and lines 
touch . Prove that 
.
and 
be positive real numbers such that 
. Prove that
When equality holds?
+1 w
is the circumcircle of 
wwith diameter 
. The tangents of at points 
and 
intersect at 
. The line perpendicular to 
at point 
intersects the extension of at point 
. Extend 
at length 
. If 
, find the measure of 
.
. Prove that
set of numbers:
, ..., 
is k-degree elementary symmetric polonomial of 
+1 w
, 
, 
, 
are given. For each 
, 
, define![\[ d_{i} = \max \{ a_{j}\mid 1 \leq j \leq i \} - \min \{ a_{j}\mid i \leq j \leq n \}
\]](http://latex.artofproblemsolving.com/6/4/f/64f48f259b0d44340c5162f704420635d79404a8.png)
.
,![\[ \max \{ |x_{i} - a_{i}| \mid 1 \leq i \leq n \}\geq \frac {d}{2}. \quad \quad (*)
\]](http://latex.artofproblemsolving.com/1/9/9/19950d29994fe166d1ce80d193e79ec24de8bda4.png)
such that the equality holds in (*).
be a positive integer which is not a perfect square, and consider the equation ![\[k = \frac{x^2-a}{x^2-y^2}.\]](http://latex.artofproblemsolving.com/c/2/2/c223ccbc272d07c19632c5b8883571023e47a395.png)
Let be the set of positive integers 
for which the equation admits a solution in 
with 
, and let be the set of positive integers for which the equation admits a solution in with 
. Show that 
.
we can find a set of distinct integers greater than 1, such that the average of the elements of any subset of is a square, cube or higher power.
is 
. Line 
meets line at point , and there is a point 
on 
such that 
. Line 
intersects at point 
. The perpendicular bisector of line segment intersects line at point 
, and line intersects 
at point 
.
is tangent to 
.
is a triangle. 
is a point on the line . 
is the midpoint of . The line through parallel to 
meets the line 
at . The line through parallel to meets the line at . The line through parallel to meets the line at 
. The line through parallel to meets the line at . The lines 
and 
meet at 
is the point on the line such that is the midpoint of 
. Show that 
and are collinear.
is a triangle. is a point on the line . is the midpoint of . The line through parallel to meets the line at . The line through parallel to meets the line at . The line through parallel to meets the line at . The line through parallel to meets the line at . The lines and meet at is the point on the line such that is the midpoint of . Show that and are collinear.
is undefined, tried figure with geogebra with no success, as there were two points (
) undefined
parallel to meets the line at .
such that: 
và 
.
