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 real number such that 
Prove that
Let 
be real number . Prove that
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
be a triangle. The points 
and 
lie on the segments 
and 
respectively, such that the lines 
and 
intersect in a common point. Prove that it is possible to choose two of the triangles 
and 
whose inradii sum up to at least the inradius of the triangle .
and 
be two positive integer numbers such that 
.Prove that 
divides 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a point interior to triangle (with 
). The lines 
, 
and 
meet again its circumcircle 
at 
, 
, respectively . The tangent line at 
to meets the line 
at 
. Show that from 
follows 
.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of .
of positive integers such that 
is an integer.
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist.
, its circumcircle, and its incenter are drawn on the plane. Construct the circumcenter of using only a ruler.
such that:
is strictly increasing,
such that 
for all 
,
, there exists 
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
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
satisfying
Prove
\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
