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.
is satisfied. Point 
is taken on the circumcircle of 
such that 
.A line parallel to 
which passes from intersects 
and 
respectively at 
and 
.Show that circumcircle of 
passes from circumcenter of 
.
rectangles that can be covered without gaps and without overlaps with hooks such that
is the foot of the altitude from 
of triangle . On the lines and points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.![\[{x^3}{y^2}\left( {2y - x} \right) = {x^2}{y^4} - 36\]](http://latex.artofproblemsolving.com/d/5/6/d5668b2a486f5264e1cf2faeb363ac69bf91e3f0.png)
has centroid 
. A line parallel to passing through intersects the circumcircle of at a point . Let lines 
and intersect at . Suppose a point 
is chosen on such that the tangent of the circumcircle of 
at , the tangent of the circumcircle of at and concur. Prove that 
.
be an acute triangle with 
, and let 
be the center of its circumcircle. Let 
be the reflection of 
with respect to 
. The line through parallel to intersects 
at 
, and the tangent at to the circle 
intersects the line through parallel to at point 
. Let 
be a point on the ray 
, starting at , such that 
. lies on the circle with diameter 
.
be an even integer, and let 
be an arbitrary set of 
distinct integers. Define![\[
E(V,n)
\;=\;
\bigl\{(u,v)\in V\times V : u < v,\ u+v = n^k\text{ for some }k\in\mathbb{N}\bigr\}.
\]](http://latex.artofproblemsolving.com/4/b/6/4b65bed83b027596f154741763fb518e2415863d.png)
For each even , determine the maximum possible size of the set 
.
be a triangle with side‐lengths 
, incenter 
, and circumradius 
. Denote by the area of , and let 
be the areas of triangles 
, 
, and 
, respectively. Prove that![\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]](http://latex.artofproblemsolving.com/1/3/9/139a3e15f08afb78ca0b8301f24d30db8f202207.png)
such that
divides 
for every 
, and
we have![\[
f\bigl(f(a)+kb\bigr)\;=\;f\bigl(a + k\,f(b)\bigr).
\]](http://latex.artofproblemsolving.com/c/f/f/cff41352422c8765978b8fe6dc966ba934a9df49.png)
be positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
. Initially there is a single lotus leaf somewhere in the lake, sufficiently far from the shore. Each day, from every existing leaf a new leaf may grow at distance 
(measured between centers), provided it does not overlap any other leaf. If the lake is large enough that edge effects never interfere, what is the least number of days required to have 
leaves in the lake?
lie the following points: 
and 
on 
, 
on 
, and 
on 
. Let![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet at 
. Prove that the points 
, , and 
are collinear.
such that 
for all 
.
be a natural number, and let be the square of side length 
subdivided into 
unit squares. Determine for which values of it is possible to dissect into connected regions of equal area using only the diagonals of those unit squares, subject to the condition that from each unit square at most one of its diagonals is used (some unit squares may have neither diagonal).
is an acute angle triangle such that 
and 
. Let's denote by 
the center of the circumscribed circle of the triangle and the intersection of altitudes of this triangle. Line 
intersects in point and in point 
. Find the value of the ration 
.
is an acute angle triangle such that and . Let's denote by the center of the circumscribed circle of the triangle and the intersection of altitudes of this triangle. Line intersects in point and in point . Find the value of the ration .
is equlilateral. Now the nine point centre lies on the bisector of 
. Because if 
are the middle points of 
and 
the nine point centre, the quadrilateral 
is concyclic 
,So 
are symmetric points about the bisector of 
, So 
(since 
is an altitude)
(since 
is inscribed in the circle with center )
(since 
is isoceles since 
)
be the perpendicular from to . In any triangle, 
to meet the circumcircle at , then draw 
. 
since it's inscribed in a semicircle, so 
, and since 
. Now we have 
, and 
. Thus, 
, so 
is a parallelogram, so 
. be the perpendicular from to . Then 
so 
is a 
right triangle, so 
. Thus, since 
and 
, 
so 
.
of the line 
( is incenter) with the circumcircle 
of . Since 
obtain that the points , , belong to the circle 
and the quadrilateral 
is a rhombus, 
a.s.o.
.
.
and 
are congruent, and so 
.
and so 
is cyclic and so by simple angle chasing we get that 
and 
claims we get that 
