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.
, let the incircle 
touch sides 
at 
, respectively. Let 
lie on the line through 
perpendicular to 
. Let 
be the intersections of 
with 
, respectively. Let 
be the projections of 
onto line . Let 
be the second intersections of 
with the incircle . Let 
be the intersection of 
and 
. Prove that the line 
bisects segment 
.
be the incentre of 
such that the incircle touches the sides at the points respectively. The perpendicular to at 
and 
meets at 
and 
respectively. 
and 
meet the incircle at and 
respectively. 
and 
intersect at 
Prove that 
bisects 
![$P\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/f/3/d/f3d763f4197792a2956273485eb1e94700a43044.png)
has degree 
having real roots in 
. Suppose 
is the leading coefficient of . Show that, 
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both and 
. Integers and are coprime if 
.
that satisfy 
are positive real numbers, then
such that 
is a perfect square for all integers and .
satisfied 
Find minimum and maximum of 
be a convex quadrilateral with 
. Define the points 
and 
. Prove that the triangles 
and 
have the same centroid if and only if 
and 
.
and 
follow on the lines 
in this order, the triangles 
do not have a common point and the proposition is trivial. Assume that the points 
and 
follow on the lines in this order. Assume that the lines 
are not parallel - they intersect at a point 
. WLOG, assume that the points 
and 
follow on the lines in this order. Let be the midpoints of the segments and 
. In a central projection of the line 
to the line 
, the cross-ratio of 4 points is preserved:
be the intersection the line of a line through the point parallel to the line 
. Obviously, 
. Since 
,
and the points 
follow on the segment in this order. Let 
be the intersections of the line 
with the lines , respectively. Project the point 
to infinity. The projected lines 
become parallel and the projected points 
become the midpoints of the projected segments . Again, the cross ratios of 4 points are preserved:
, 
and 
,
and the points 
follow on the segment in this order. Similarly, it can be shown that the points 
follow on the segment in this order. Thus the segments 
do not have a common point and consequently, the median segments 
of the triangles 
do not have a common point either. Hence, these 2 triangles cannot have a common centroid. are parallel and let be the intersections of the line with the lines . From similarity of the triangles 
, we have
of the triangle 
, i,e., is the midpoint of the segment . From similarity of the triangles , is also the midpoint of the segment . Thus the median lines 
of of the triangles 
are coincident. and parallel to the line 
intersect the lines 
at points . Since , from similarity of the triangles 
and 
we deduce that 
and 
. The cross ratio of 4 points is preserved in a central projection of the line 
to the line from the point . Thus
and 
, it follows that
be the centroids of the triangles 
. Since the centroids cuts the medians 
in the ratio 
, 
. For coincidence of the centroids , it is necessary and sufficient that
, we obtain a quadratic equation for the ratio 
, where 
is the golden ratio. From similarity of the triangles ,
be the origin of the vectorial system, i.e. for any 
we have the vectors 
. Thus,
.
such that
and 
have the same centroid 
and 
Prove that
then prove 
Prove that
Prove that
and 
Prove that
\overrightarrow{AB}
be a trapezoid 
.
and 
.
are the midpoints of 
respectively.
are collinear.
is harmonic (that is, 
) be an harmonic division with ratios 
.
is a point 
which divides the segments 
in the same ratio 
and 
. We will show that the triangles 
have the same centroid, for some 
we get that 
, because 
, where 
is the golden ratio. wich divides the segment 
in the ratio 
, divides the segment 
in the same ratio
.
, which means that the point is the centroid of 
and . and have the same centroid.
and 
be te common centroid. and the midpoint of are collinear.
belongs both to the lines 
, so the only possible result is that the points 
are collinear. of the triangle 
is also median for the triangle 
.
. So is a trapezoid., for aome in 
which divides the segments at the same ratio, and the ratio is 
is the one with this property, and the value of 
is 
, because 
is positive
, which yields
and the midpoint of are collinear. But . belongs both to the lines , so the only possible result is that the points are collinear.
, 
, 
, 
. So the line ![$BD=[-z: 0: x]$](http://latex.artofproblemsolving.com/f/2/9/f29c3344cf27b09a6348ea10f6702c7dd0ba76fc.png)
and the line ![$AC=[0: 1: 0]$](http://latex.artofproblemsolving.com/5/d/f/5dfab2e82e3837b7e971fb7fdb4d7d7cbcf11adc.png)
. This follows that 
. Similary we have that the line ![$BC=[1: 0: 0]$](http://latex.artofproblemsolving.com/d/0/a/d0ac7ebf9d8100ecfd9dc5f03cd5e6d24f2d037c.png)
and ![$AD=[0:-z: y]$](http://latex.artofproblemsolving.com/0/3/7/037f94224449580ab09669dde6aa7b57f76a8f0d.png)
. Thus 
.
and 
be the barycenters of 
and so 
and 
then if 
, it means that is in normalized coordinates.
, 
and 
.![$x=\frac{(1+\sqrt[]{5})}{2}$](http://latex.artofproblemsolving.com/9/3/4/934b227e92122e3ae22d7391f6a36aa12a4e0611.png)
, ![$y=-\frac{(1+\sqrt[]{5})}{2}$](http://latex.artofproblemsolving.com/8/2/d/82dbc92e361e879d50ce54380c67f46f14837eab.png)
, 
. This follows ![$\overrightarrow{CD}=(x: y: z-1)=(\frac{(1+\sqrt[]{5})}{2}:-\frac{(1+\sqrt[]{5})}{2}: 0)=-\frac{(1+\sqrt[]{5})}{2}(-1: 1: 0)=\frac{(1+\sqrt[]{5})}{2}\overrightarrow{AB}$](http://latex.artofproblemsolving.com/2/7/4/274bc0314609046b3875b55874f20dad0f7d859c.png)
. That's 
be ![$\frac{(1+\sqrt[]{5})}{2}$](http://latex.artofproblemsolving.com/6/e/4/6e4675c1d4f16a5ea4ed3a9f61ed9c0194dc7935.png)
, so 
, 
and 
and 
.
and 
so 
and with some betrayals we get 
and we get 
.
Capitán Mandarina
