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.
and 
Show that
Where 
and 
Show that
Where
, where 
and 
are (not necessarily distinct) positive integers. The integer 
is written as the sum of 
Tiranian integers. What is the smallest possible value of ?
be non-nagative real numbers such that 
.
.
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
be a triangle with circumcircle 
. Arbitrary points 
on 
respectively. Circumcircle 
of triangle 
intersects at 
. 
intersects 
at 
. 
cuts and at 
respectively. Construct parallelogram 
. Prove that 
are collinear.
be an isosceles triangle with 
, and let be its incenter. Incircle touches sides 
at 
, respectively. Foot of altitudes from 
to 
are 
, respectively. Rays 
intersect 
at 
, respectively. Prove that 
touches incircle at 
.
with two altitudes and .Let be the point on the line such that 
.The lines 
and 
intersect at point 
, and 
is the point on segment 
such that 
.Suppose that bisects .Prove that triangle 
is isosceles.
-plane given points 
on the planes 
respectively. Let 
be the intersection point of the line 
and the 
-plane.
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane 
,.
and 
on the plane 
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane and the point moves on the perimeter of the square 
, draw the domain swept by the point on the -plane, then find the area.
with 
, let its incircle be tangent to sides , 
, and 
at , 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at and , respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of . with their incenter . Point lies on on the same side as 
wrt 
. Point lies on the shorter arc of the circumcircle . It is given that 
Prove that 
is the angle bisector of 
.
of the triangle touch its side at . Let the line intersect at 
and denote the excentre of opposite to 
by 
. Let 
and 
be the midpoints of and 
respectively.
and 
are concyclic.
be a triangle with 
. is the midpoint of 
. Let the angle bisector of 
cut 
at and 
be the centroid of . 
meets at 
and is the projection of onto . Prove that 
lie on a common circle.
lies in the interior of 
. If it isn't, it is on the segments 
or in the interior of one of the triangles 
. If it is on one of the segments (say on the segment 
) then ![$[CYPX]=[CYX]=\frac{1}{3}[ABC]$](http://latex.artofproblemsolving.com/c/f/6/cf6a170deecab15b846b23ffc7786cbbd9067392.png)
. But lies on the opposite side of the segment 
to hence ![$[AZPY]=[AZY]+[ZPY]=\frac{1}{3}[ABC]+[ZPY]>\frac{1}{3}[ABC]>[CYPX]$](http://latex.artofproblemsolving.com/f/2/c/f2c449cf226ac0d939627e2ef7bcbce4b6a4c49e.png)
, a contradiction. If lies in the interior of one of the triangles (again suppose it is 
) then use the same argument, except ![$[AZPY]>\frac{1}{3}[ABC]>[CYPX]$](http://latex.artofproblemsolving.com/8/b/e/8beb97be61cae8d2d6f13f50541b8c7ac5f892e1.png)
to reach another contradiction. lies in the interior of triangle 
. Hence ![$[AZPY]=[BXPZ]=[CYPX]$](http://latex.artofproblemsolving.com/6/d/6/6d6f91e7354483c90ae3fd57c3139a3ea6072bbf.png)
is equivalent to ![$[PXY]=[PYZ]=[PZX]$](http://latex.artofproblemsolving.com/e/a/5/ea5e188abb5de33fca624d8385a4fa83fe3eed97.png)
. Areal coordinates now turn this into algebra. If has areal coordinates 
then we find that 
which gives 
so is the centroid of triangle and hence the centroid of triangle . Conversely if 
then the areas of the quadrilaterals are clearly equal.
![$[AZY]=[BZX]=[CXY]= \frac 14 [ABC]$](http://latex.artofproblemsolving.com/5/c/a/5ca61a613b61d33bba451133518c8fc2e84b1e40.png)
so the equality in the statement implies ![$[PZY]=[PZX]=[PXY]$](http://latex.artofproblemsolving.com/3/0/d/30d3a42d96a493d25980869f1c62fdc7d30d515c.png)
and thus is the centroid of . But centroid of and is the same, which implies is the centroid of . is inside " is necessary-)![$ [AZPY]=\ [BXPZ] =\ [CYPX] = \frac 13[ABC] > [AZY],[BZX],[CXY]$](http://latex.artofproblemsolving.com/e/6/8/e6863b0570bae00ffcbece0cd7b7aeb13038471a.png)
so all the areas ![$[PXY],[PYZ],[PZX]$](http://latex.artofproblemsolving.com/e/f/4/ef410d8d12d5f73d6036deafddc71090863e3bcc.png)
are positive.
and 
be two points inside triangle .Let the distance between and is 
and between and is 
.Also let the distance between and is 
and between and is 
.If 
and 
then 
.
must be interior to (because the areas of the quadrilaterals are 
of the area of , while the areas of the triangles 
, 
, 
are 
of the area of ), hence the areas of the triangles 
, 
, 
must be equal, notice that such a point is uniquely determined, since it has to lie on parallels at determined distances from the sides of . Since the centroid satisfies, it must be that coincides with .
.
and 
be the midpoints of sides 
,
,
have equal areas if, and only if, 
.
denotes area of 
.
,
,
and similarly 
,
.
.
.
.
and 
.