then
where 
Also, 
and 
is largest real root of the equation
then 
Equality holds iff 
and any cyclic permutations.
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of 
denote the circle passing through 
.
denote the circle passing through 
at 
.
)
.
and 
,
[list=a]
and 
,
satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.
that are each continuous and last on a certain time interval. Each event has a weight 
However, one can only choose to attend activities that do not overlap with each other. The goal is to maximize the sum of weights of all activities attended. Prove or disprove that the following algorithm allows for an optimal selection:
consider 
the sum of 
over all 
such that 
and 
If there are multiple such events, delete the event with the minimal weight.
Y by Adventure10, Adventure10, Mango247
Let 
be a cyclic quadrilateral, and 
a point in its interior such that 
. Let 
, 
and 
be the orthogonal projections of on the sides 
, 
and 
, respectively. Show that triangles 
and 
are similar.
Alternative formulation. A point is in the interior of the cyclic quadrilateral and has the property 
. The feet of the perpendiculars from to , 
and 
are respectively denoted , and . Show that 
and 
are similar.
be a cyclic quadrilateral, and 
a point in its interior such that 
. Let 
, 
and 
be the orthogonal projections of on the sides 
, 
and 
, respectively. Show that triangles 
and 
are similar.Alternative formulation. A point
is in the interior of the cyclic quadrilateral and has the property 
. The feet of the perpendiculars from to , 
and 
are respectively denoted , and . Show that 
and 
are similar.
and 
.
.
seems to be not so easy.
,
.
and the circle (L) at points 
,
and the circle (L) at points 
,
respectively. Since X is the internal homothety center of the circles (K), (L), 
Combining, 
,
are both cyclic and generally not similar. Since (X) is the common P-Apollonius circle of the triangles 
and the common Q-Apollonius circle of the triangles 
,
and 
,
(inversion preserves angles). The quadrilateral vertices A, B, C, D are inverted into the intersections A', B', C', D' of the lines 
are similar. Hence, 
,
,
are parallel. Thus A'B'C'D' is a cyclic trapezoid, isosceles, and B'C' = A'D'. Consequently, and the line x bisecting the angle formed by its diagonals 
is also the perpendicular bisector of the sides A'B', C'D', so that the triangles 
are congruent. As a result, the triangles 
are similar. But the pedal triangle 
of the triangle 
with respect to the point P is similar to the image 
of this triangle in the inversion in the circle (P), so that the triangles 
are similar as well.
)
is a trapezoid (using the condition given in the problem). It must also be cyclic since 
), so it must be an isosceles trapezoid and 
.
and 
,
and 
So we're done. 
such that 
.
and by the converse of the Alternate Segment Theorem, 
is the common tangent of the circles through 
and 
.
.
.
is cyclic so 
.
is cyclic so 
.
(This follows from 
)
by the Alternate Segment Theorem, and 
,
.
) but I didn't manage to complete the proof. In fact, I tried considering the second intersection of the circumcircles of 
and 
as well, but that didn't really work - probably because there is no second intersection 
Anyways, the idea of that construction is not thát unnatural...
and the line 
,
of the quadrilateral 
,
for the triangles 
are cyclically and 
,
for which 
and 
.
is the common tangent in the point 
and 
,
.
.
Therefore, 
and 
and two fixed points ![$[AB]\cap d =\emptyset$](http://latex.artofproblemsolving.com/e/4/9/e49820d949d4a6c8d442d622b9074165f8bd3c44.png)
and 
.
,
for which the circumcircles of the triangles 
,
are tangent to the line 
is maximum belongs to the set 
.
