Y by GeoKing, buratinogigle, Deadline
Let points 
, 
and 
lie on the circle 
in a counter-clockwise order, and let 
be a point in the same plane. For 
, let 
denote the counter-clockwise rotation of the plane centred at 
, where the angle of rotation is equial to the angle at vertex in 
. Further, define 
to be the point 
, where the indices are taken modulo 
(i.e., 
and 
).
Prove that the radius of the circumcircle of
is at most the radius of .
Proposed by Anant Mudgal
, 
and 
lie on the circle 
in a counter-clockwise order, and let 
be a point in the same plane. For 
, let 
denote the counter-clockwise rotation of the plane centred at 
, where the angle of rotation is equial to the angle at vertex in 
. Further, define 
to be the point 
, where the indices are taken modulo 
(i.e., 
and 
).Prove that the radius of the circumcircle of
is at most the radius of .Proposed by Anant Mudgal
This post has been edited 2 times. Last edited by kamatadu, Jan 21, 2024, 12:53 PM
is reflection about 
,
.
will have radius 
(from a homothety centred at 
will get the bound.
is a spiral sim itself, and we can observe easily (by following the rotations) that 
to the reflections of 
points, we are done. This finishes the problem as well. 
.
half turns after the triple composition, I realized what the fixed points of this should be, and bam, we had this problem setup. Many versions were suggested about the conclusion, including the much more awful: "among all six points obtained by composing the three rotations in any order, show that there exists three for which the circumradius is smaller than that of 
.
is equal to 
where 
is the unique point so that 
and 
(where angles are taken with sign).
where 
is the incentre of 
where 
i.e. Halfturns about 
,
where 
and 
are the foot of perpendicular from 
and 
.
to 
.
applied to 
This problem destroyed me
denote rotation of point 
(counter-clockwise).
where 
.
and let its incircle touch 
at 
respectively.
denote the angle of vertex 
Thus, 
.
by Euler's Geometric Inequality : 
.
represent the points 
in the complex plane. The centre of 
unit by trivializing. We take the subscripts with variable 
instead of 
,
denote points 
on the complex plane.
representing points 
,
is the reflection of 
by an angle ![\[(z-y)=(x-y)e^{i\theta},\]](http://latex.artofproblemsolving.com/2/a/f/2afc6785426c398d8f4d42d0753487f80f125aff.png)
where 
is Euler's constant.
in independent of choice of ![\[\tau_{k+1}(p)=pe^{iA_{k+1}}-a_{k+1}(e^{iA_{k+1}}-1)\]](http://latex.artofproblemsolving.com/6/d/c/6dc88b66127082b71d9ab04d3bfa1c57e3425de3.png)
![\[\tau_{k}(\tau_{k+1}(p))=pe^{iA_{k+1}}e^{iA_k}-a_{k+1}(e^{iA_{k+1}}-1)e^{iA_k}-a_{k}(e^{iA_{k}}-1)\]](http://latex.artofproblemsolving.com/b/2/0/b20caae75dac2c8ad19a0c898ed0a3a248a9c007.png)
![\[\tau_{k+2}(\tau_{k+1}(\tau_k(p)))=pe^{iA_{k+1}}e^{iA_k}e^{iA_{k+2}}\]](http://latex.artofproblemsolving.com/3/0/2/30276a241ec90945508223720f2d0fd9fc42858d.png)
![\[-a_{k+1}(e^{iA_{k+1}}-1)e^{iA_k}e^{iA_{k+2}}-a_{k}(e^{iA_{k}}-1)e^{iA_{k+2}}-a_{k+2}(e^{iA_{k+2}}-1)\]](http://latex.artofproblemsolving.com/8/c/0/8c0e427cfc03a7ca9eb32191b597497425e136a7.png)
We know that 
and that 
.
,
is completely independent of 
.
are the in-touch points of triangle 
,
on 
on 
,
on 
,
,
,
denotes the complex number for 
.
.
and 
.
,
is the reflection of 
is reflection of 
is reflection of 
,
,
is it's inradius. Hence, circumradius of 
.
.
.
for any two points 
.
is a parallelogram. Therefore, 
.
and 
and thus triangles 
are congruent. So it suffices to show this result for one point 
(possibly 
)
and it is easy to show the problem in this case lol.
counter clockwise. How will it ever reach (even close to) 
?
denotes a rotation of angle 
,
,
