[/quote]
,
prove that for each 
there exists 
such that either 
or 
is multiple of 
such that 
.
and 
and the line passing through 
,
at 
.
)
is the orthocenter and point 
is the midpoint of the segment 
intersects the circumcircle of triangle 
.
intersects the perpendicular bisector of 
and the circumcircle of the triangle 
.
lies on circle 
,
and 
is a trapezoid 
)
and 
meet on 
![\[
S = \left\{\, x \in \mathbb{Q} : x \text{ has a finite decimal expansion} \,\right\}.
\]](http://latex.artofproblemsolving.com/1/c/e/1ce81426f5d169a9b4c254ef91a1fe34250504b4.png)
For each 
,![\[
d(x) = \text{the number of digits after the decimal point in the (reduced) decimal form of } x.
\]](http://latex.artofproblemsolving.com/8/1/9/819d07d26274a051a776fcf9f6d7c668ab8d0dab.png)
Find all functions 
such that, whenever both 
and 
,![\[
f(x) + f(y) = f(x+y) + d(x y).
\]](http://latex.artofproblemsolving.com/5/c/a/5ca142ba21a23dc21db6e3c712bc0a76d561647c.png)
be a finite set of points in the plane such that no three of them are on a line. For each convex polygon 
whose vertices are in 
be the number of vertices of 
be the number of points of 
,
,
vertices respectively. Prove that for every real number 
![\[\sum_{P}{x^{a(P)}(1 - x)^{b(P)}} = 1,\]](http://latex.artofproblemsolving.com/3/2/7/3279b2917f21c8f3c2b819e39a32528ad18e4b4b.png)
where the sum is taken over all convex polygons with vertices in 
be a finite point set in the plane and no three points are collinear. A subset 
of 
g
For each round subset let 
be the number of points from 
and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset ![\[ P_A(x) = x^{|A|}(1 - x)^{r(A)}.
\]](http://latex.artofproblemsolving.com/e/b/c/ebc6f8fa2cc303e1062c5a95d4f65c9d2691d0ca.png)
be an odd prime. Define 
over the positive integers as follows:
where 
such that 
.
such that 
and 
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
grid. Initially, 
stones are placed on some of the unit squares of this grid.
,
and 
meet at the orthocenter 
from 
and 
meet at a point 
meets 
at 
is perpendicular to the angle bisector of 
![\[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\]](http://latex.artofproblemsolving.com/d/6/2/d622094b95a1f317063cf2f02a8b66b31751638f.png)
Show that 
,
.
letters, with exactly 
and 
.
such that 
swaps.
such that one can find a number of triangles in the Cartesian plane, the centroid of each triangle is a lattice point, the union of these triangles is a square of side length 
with no obtuse angles.![$$\frac1{\sqrt3}\cdot\left(\cot A+\cot B+\cot C\right)+\left(2-\sqrt 3\right)\sqrt[3]{\cot A\cot B\cot C}\geq\frac2{\sqrt3}.$$](http://latex.artofproblemsolving.com/d/1/4/d14154281d7ccf60a2dcfe1d937b974ae1d74f97.png)
Y by Mango247
In the triangle 
denote 
the center of the inscribed circle, 
the midpoint of the side 
, 
is the point of contact of the 
-exscribed circle with the side , 
is the midpoint of the arc 
of the circumscribed circle triangle. Line 
intersects for the second time the circle circumscribed around the triangle 
, at the point 
. Prove that line 
divides the segment 
in half.
(D. Hilko)
denote 
the center of the inscribed circle, 
the midpoint of the side 
, 
is the point of contact of the 
-exscribed circle with the side , 
is the midpoint of the arc 
of the circumscribed circle triangle. Line 
intersects for the second time the circle circumscribed around the triangle 
, at the point 
. Prove that line 
divides the segment 
in half.(D. Hilko)
This post has been edited 1 time. Last edited by parmenides51, Jun 26, 2021, 1:22 PM
Reason: incircle to A-excircle, see #3
Reason: incircle to A-excircle, see #3
be the 
be the reflection of 
.
bisects 
is the 
because 
is tangent to 
and 
is also this symmedian by isogonality, so 
as desired.
