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 an even integer, and let 
denote the number of (positive) divisors of . does not divide 
.
+1 w
such that 
for all 
a regular hexagon inscribed in a circle 
. Let 
be a point inside and 
its polar reflection wrt . The rays 
meet again at 
. Call 
the polygon formed by the vertices . Similarly construct the polygon 
using instead. and are congruent.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
with an area of 
, with 
. If 
, Hence find the value of 
.
such that for all 
we have:
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
Let 
be positive reals such that 
.Prove the inequality 
pieces of equal weights. Can Vova prevent Pasha's win?
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
, with 
, let 
and 
denote its circumcenter and orthocenter, respectively. Let 
be the reflection of with respect to 
. Prove that 
and 
are collinear if and only if 
.
, with , let and denote its circumcenter and orthocenter, respectively. Let be the reflection of with respect to . Prove that and are collinear if and only if .
.
.Hence 
is 
-isosceles. Again a little angle chase proves that 
is the foot of perpendicular from ).But 
From midpoint theorem 
as well. 
are collinear.
are collinear
be the midpoint of 
.It is well known that 
.Let 
intersect 
at 
.Since 
and 
is the midpoint of 
as well 
is a paralellogram. A little angle chasing shows 
.But since 
we have 
as desired
are co-linear.
and 
. be the mid point of and intersect at 
.
and 
.
, as desired.
.
, find out 
.
.
as the origin.From it you can easily find the equations of the perpendicular bisectors and thus you can find the co-ordinates of the orthocentre and the circumcentre.Let the co-ordinate of be 
and that of point of be 
.Then,little calculation will yield co-ordinate of the orthocentre is 
.As the point is the reflection of the orthocentre,then the origin is the midpoint of the line segmeny joining the orthocentre and the point .So co-ordinate of the point is 
Similarly,you can find the co-ordinate of the orthocentre.Now lies the important part.Determine the equation of the straight line which joins the point and the vertex. Now as the points , and are collinear ,so the co-ordinates of the point will satisfy the equation.Put the values and you will have the relation 
.Now this relation is eqivalent to 
.And for the rest go the reverse way and we are done.
be a diameter of the circumcircle (
cyclic).
(easy angle chasing).
; with 
and 
we got 
parallelogram, hence 
., with 
and 
we got a parallelogram, i.e. 
, which gives 
, hence we are done.
let 
be the orthocenter and circumcenter respectively. Let the reflection of 
wrt 
be 
and reflection of wrt 
, the midpoint of 
be 
. Then 
are collinear points.
sends to . Since, 
(well known) and 
, 
and are corresponding points of the homothety and hence are collinear.
. Let be the reflection of in the midpoint of . Then 
are collinear. So 
passes thru 
iff 
i.e, iff 
i.e, iff is the circumcenter of 
, i.e, iff 
i.e, iff 
.
and 
to get the lengths and 
.Now its just trigo....(I think calculations are not too long but boring...)
and with scale 
. is sent to , the nine-point center of 
, goes to 
, the midpoint of line segment 
. goes to . Thus 
are collinear. But it is well-known that lies on the nine-point circle of , and the midpoint 
of 
is its antipode. Thus 
are collinear, that is, lies on line segment . This in turn means that 
. Consider all angles to be directed. Thus if 
, then 
, and so 
and 
Thus
as required.
such that:
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
Prove that
Let 
Prove that
Prove that
and then 
.
and 
.
are collinear 
.
is the reflection of 
is a line. Then note assuming collinearity we have 
where the latter is parallel to 
so done.
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be inscribed in the unit circle. Let the images of the points in the complex plane be denoted by their lowercase letters. We have 
Note that 
and 
.
Now, 
are collinear iff![\[\dfrac{a-b-c}{c} \in \mathbb{R} \iff \dfrac{a-b}{c-o} \in \mathbb{R} \iff OC \parallel AB \\
\iff \angle OCB = \angle CBA \iff \angle A - 90^{\circ} = \angle B \iff \angle A - \angle B = 90^{\circ},\]](http://latex.artofproblemsolving.com/9/2/1/921887fee5fca12f64de72214247ba96c4125a8c.png)
