[/quote]
,
-
on the planes 
respectively. Let 
be the intersection point of the line 
and the 
-
.
moves on the perimeter of the circle with center 
, radius 1 on the plane 
,
and 
on the plane 
.
moves on the perimeter of the circle with center 
, radius 1 on the plane 
with 
,
,
,
at 
,
,
,
intersects lines 
and 
at 
and 
,
and 
be distinct points on side 
.
be the circumcircle of 
.
.
and 
such that 
divides 
.
and 
;
and 
.
.
wrt 
.
Prove that 
is the angle bisector of 
.
.
represented by tokens to players 
to 
so that they all have the same number of tokens in their urns.
beats 
when, when 
.
What is 
such that the following equations are both true:
be reals such that 
Prove that
of the triangle 
intersect 
and denote the excentre of 
by 
.
and 
be the midpoints of 
respectively.
and 
are concyclic.
.
.
cut 
at 
be the centroid of 
meets 
and 
lie on a common circle.
be 
.
and 
are two moving points on the circle 
:
,
.
.
.
such that there exists a non-negative integer 
for which the set 
is fragrant?
and 
Show that
Where 
Where 
is a positive integer, then a user is an 
-
-
Y by Adventure10, Mango247, Rounak_iitr
Let 
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.This post has been edited 1 time. Last edited by fattypiggy123, Mar 13, 2017, 1:34 AM
Reason: Typo
Reason: Typo
are pairs of an involution on 
to all three circles by harmonics, so the perpendicular bisector of 
must exist, the center of inversion is still the midpoint of these two points, so this is another way to patch the original solution. One still must take a certain amount of care to work out the details, though, when working in the complex projective line.
,
,
,
be four circles. Define 
to be the exsimilicenter of 
for 
.
are coaxal.
belong to a line 
;
similarly. If these lines are pairwise distinct, the claim follows from 
;
centered at 
,
are the points 
,
are the exsimilicenters of 
.
must be swapped under some inversion with center 
Since 
However, we know that the centers of the circles are collinear, so 
must be coaxial; clearly, the former is impossible, so we're done. 
are pairs swapped under some involution. If this is a reflection, the problem is done. Otherwise, suppose that the involution is an inversion, say with center 
meaning that 
and 
Now if you draw the line perpendicular to 
circles, as desired. Note: 
(lengths are signed).
there exists a unique 
such that 
.
respectively, thus 
-
and any other conic through 
(which can be expressed as 
for some reals 
)
is the same as that of 
and 
,
)
we have (in signed distances) 
,
:
which is perpendicular to 
then 
so 
which is absurd. 
![[asy]
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Converted from GeoGebra by User:Azjps using Evan's magic cleaner
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pair X = (-180.85092,-305.52787);
pair Y = (-72.06909,-80.72397);
pair Zp = (118.67646,313.46266);
pair Z = (-28.24650,9.83792);
pair Xp = (25.82900,121.58806);
pair Yp = (54.56188,180.96621);
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draw(A--B, linewidth(0.5));
draw(B--C, linewidth(0.5));
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draw(B--Zp, linewidth(0.5));
draw(Zp--X, linewidth(0.5) + red);
draw(X--B, linewidth(0.5));
draw(D--Yp, linewidth(0.5));
draw(A--C, linewidth(0.5));
draw(circle((-77.51096,-91.96990), 237.24703), linewidth(0.5) + blue);
draw(circle((-8.75360,50.12111), 145.35917), linewidth(0.5) + blue);
draw(circle((45.21497,161.65029), 168.65226), linewidth(0.5) + blue);
dot("$A$", A, NW);
dot("$B$", B, W);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$X$", X, SW);
dot("$Y$", Y, 2*dir(270));
dot("$Z'$", Zp, NE);
dot("$Z$", Z, 2*dir(270));
dot("$X'$", Xp, NE);
dot("$Y'$", Yp, 2*E);
[/asy]](http://latex.artofproblemsolving.com/f/6/9/f69c388758f51ac2bcc44af87538ced9b1a3a703.png)
,
and 
.
and 
share the same midpoint and thus the order of the points would be 
or 
.
with 
as given in the problem statement.
.
and 
too. Now consider the radical axis of 
)
share a common radical axis. Similarly, 
also share the common radical axis which means that the circles 
are coaxial and we are done. 
but this is just inversion for some point 
,
and hence the circles are coaxial.
![[asy]
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pen dotstyle = black; /* point style */
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[/asy]](http://latex.artofproblemsolving.com/5/1/5/515d1883395ec90a80745949df3363426325ac56.png)
,
and 
be the circles 
,
and 
.
,
are reciprocal pairs for a unique involution a unique involution along 
Then 
and 
,
are pairs of points swapped by an involution. Also, the involution can be defined as inversion from a point 
.
by inversion. Now this means that 
and 
are swapped by some involution. Since involutions are inversions/reflections, the center 
so 
and 
the circles with diameters 
and 
and 
.![\[PX\cdot PX' = PY \cdot PY' = PZ \cdot PZ'\]](http://latex.artofproblemsolving.com/9/7/b/97bd37e348024f222164cc354229b83f0c613191.png)
Thus, the pairwise radical axes of 
, 
and 
where 
is moving. 
is 
To prove the three functions are linearly dependent, we only need
However we may write 
where 
thus the above equality is 
with degree 3. Therefore we only need to plug in 4 special 
and using 
it follows that the radius of inversion squared is 
Thus, 
which implies that they are coaxial.
