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, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
be the set of all lines in the plane and 
be a set of 
points in the plane. For a line 
let 
be the number of distinct points among the orthogonal projections of the points from on 
. Find the maximum possible number of distinct values of (this quantity is computed for any line ) as varies.
, where 
, the internal angle bisectors of angles 
, 
, and 
meet the sides , 
, and 
at points , 
, and 
, respectively. Let 
be the incenter of triangle 
, and let 
be the foot of the perpendicular from to line 
. Prove that if the quadrilateral 
is cyclic, then the center of its circumcircle lies on segment 
.
, 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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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)
