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 
. Prove that
Where 
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are 
vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
be a triangle inscribed in a circle 
with orthocenter 
and altitude 
. Let 
be the midpoint of 
. Line 
meets again at 
. Line 
meets again at 
. Let 
be the orthogonal projection of 
on the line 
. Line 
meets again at 
. Prove that 
.
satisfy 
. Prove that
Also, find the values of for which the equality happens.
such that![\[
x+f(xy)=f(1+f(y))x
\]](http://latex.artofproblemsolving.com/4/3/3/433f95836dad1224754d480aff8432ef033a37cb.png)
for all 
.
cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from 
to . In a move, Madina is allowed to choose any 
cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?
is such that number 
has exactly 
positive divisors. Prove that GCD 
consecutive even numbers, such that the sum of their squares divides the square of their product.
satisfying 
, find the biggest value the following equation could acquire:
and 
on 
and 
respectively in 
such that 
then draw line 
through 
such that 
let 
and 
reflection of 
and to respectively prove that 
and are congruence.
be a set of 
distinct integers all of whose prime factors are at most 
Prove that contains distinct integers such that their product is a perfect cube.
, 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.
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.
![[asy]
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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)
