be a point in the plane, and 
a line not passing through 
is a cyclic quadrilateral with 
.
and 
be the feet of perpendiculars from 
to 
respectively. Prove that 
.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
.
and 
Prove that
be positive integers. If for each positive integer 
Prove that 
.
cats consisting of five different breeds. He records the number of cats of each breed and after multiplying these five numbers he obtains the number 
.
satisfying 
Prove that 
,
is fantabulous, then all the elements are fantabulous. Does it follow that the number 
is fantabulous?
,
:
,
Y by khan.academy, cubres, Mathdreams
Consider a triangle 
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .
(Shining Sun, USA)
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .(Shining Sun, USA)
This post has been edited 6 times. Last edited by Bata325, Apr 13, 2025, 2:54 AM
Reason: italics
Reason: italics
denote the circumcenter of 
.
.
denote the foot of the altitude from ![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - 2\angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/e/2/d/e2d9375826c14c5e7197102e8c1d14f2dad41068.png)
Hence, the points 
are collinear. Similarly 
are all collinear.
,
be the midpoint of 
hence 
are the centres of the circles 
respectively
and 
such that 
and 
and 
are the perpendicular bisectors of the radial axes 
and 
hence 
and 
so 
and 
and the required claim is proved
![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - \angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/7/d/9/7d972f3079ea48f9b57a8f171777f9dabada44e0.png)
Hence, the points 
hence 
where ![\[\measuredangle BAQ = \measuredangle CXQ = 90 + \measuredangle BCA\]](http://latex.artofproblemsolving.com/9/4/2/9429375d4f0c28c6a6939e228bc971645d77e8ec.png)
which implies that ![\[\measuredangle CAQ = \measuredangle CXP = 90 + \measuredangle CBA\]](http://latex.artofproblemsolving.com/0/0/a/00ab53c14582f60a8519bd3daef06f3048629879.png)
which implies that 
