AoPS Academy
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
be a convex hexagon, such that the triangles 
and 
are equilateral and the diagonals 
and 
are concurrent. Prove that 
or 
to the right; to the left.
squares north of the initial square after some finite number of moves if it is initially facing:
satisfying 
is a perfect square
and diameter 
. A line intersects at 
and the semicircle at 
and 
s.t. 
and 
. The circumcircles od the 
and 
intersect again at 
. Prove that 
.
so that we can angle chase pretty much anything. with . Well at least we know that the statement holds even in that case 
be the intersection of 
and 
.We will proceed proving that 
are collinear.
we have 
.Simillary in the quadrilateral 
we get 
..
and 
.
so the quadrilateral 
is cyclic.
.
hence the quadrilateral 
is cyclic.
intersect in the radical center.
intersect at so we get that are collinear.
where 
is the intersection of 
.
are altitudes in 
because is diameter,and 
is the median.
be the third altitude and let 
be the projection of on .
is cyclic because 
is cyclic 
.
contains the points 
.
are collinear so 
.
, 
; and 
having equal powers w.r.t. 
, 
, is on 
. Next, it is well known that for 3 points (let them be 
) on the sides 
of any 
, the circles 
, 
, 
have a common point, hence 
is cyclic, and easy to see, 
is its circumcircle diameter.
being cyclic, 
and, 
being median in 
, it is symmedian in 
, hence 
, 
are tangent to 
, or 
and 
, but 
and 
, or 
, hence 
, that is, 
is cyclic, but has equal powers w.r.t. 
and , or belongs to their radical axis, i.e. 
are collinear. As is a diameter of , it follows that 
, or 
.
, we note that 
maps to 
and 
maps to 
. Denoting the image of a point under that inversion 
, we have is the intersection of and 
. is mapped to itself and 
is mapped to 
. As is the intersection of and . 
is the intersection of 
and . But we note that 
is the nine point circle of 
, and as it intersects at and , where is the mid-point of 
, it follows that is the foot of altitude dropped from , that's 
, but By a property of inversion 
. So we are done.
, 
and 
are concurrent by Ceva. Let the radius of the circumcircle be 
, then ![\[ |AM'|=\frac{r^2}{|OA| \cdot |OM|} \cdot |AM| = \frac{r}{|OM|} \cdot |AM| \quad \text{and} \quad |M'B|=\frac{r}{|OM|} \cdot |BM|. \]](http://latex.artofproblemsolving.com/6/4/f/64f47dade2259fe10e8ba016ceeb75deb65c2f12.png)
We also have ![\[ |DK'| = \frac{r^2}{|OD| \cdot |OK|} \cdot |DK| = \frac{r}{|OK|} \cdot |DK| \quad \text{and} \quad |K'C| = \frac{r}{|OK|} \cdot |DK|. \]](http://latex.artofproblemsolving.com/1/e/2/1e25f2310dee516dea5d3bccd2ed3a956ed2dd39.png)
Both are just an application of the Inversion Distance Formula (which follows from simple similarity). Plugging those in yields ![\[ \frac{|AM'|}{|M'B|} \cdot \frac{|BD|}{|DK'|} \cdot \frac{|K'C|}{|CA|} = \frac{|AM|}{|BM|} \cdot \frac{|BD|}{|CA|}. \]](http://latex.artofproblemsolving.com/4/7/6/47681c4002b9781723baa262428440d8e2382b6f.png)
But that's equal to 
as 
by 
and 
as 
is cyclic. The similarity gives 
which is just what we wanted., and intersect in a point which we'll call . By Brocard's Theorem is the orthocenter of 
. Thus 
. Note that we needed that 
to be collinear for that. Thus ![\[ 90^{\circ} = \angle OM'K' = \angle OKM, \]](http://latex.artofproblemsolving.com/7/4/3/7436d47509352778893f229214e48d8511898549.png)
phew! 
and diameter . A line intersects at and the semicircle at and s.t. and . The circumcircles od the and intersect again at . Prove that .
denote image of 
under inversion.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/8/d/8/8d895c1b42173c5c9643a9ea5a58e5976f888d29.png)
with radius 
. We have the following as a result of this inversion :
is the Nine Point Circle of 
, is the foot of the perpendicular from to .
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Let denote the inversion image of .
Inverting about with radius 
,we see:
and 
,we get:
,
and is the midpoint of it follows that 
is the nine-point circle of 
, hence 
.
, - orthocenter of 
(also 
).
(Radical Center of 
).
- cyclic (by Miquel on 
). Since is also on this circle, 
. - orthocenter of 
, so 
. Hence proved.
.
collinear. By Power of Point on the semi-cricle, 
Radical Axis of 
The result follows.
we have that 
is the orthocenter of 
Notice that
also lies on the -median of 
, we have that is the -Humpty Point of this triangle. It is a well-known fact about the A- Humpty that 
and 
must concur, that 
are collinear. Thus, by (2), the result follows.
and 
be the centers of the circumcircles 
and 
of 
and 
. Let 
meet at 
, and let 
meet at .
as the line connecting two circle centers is perpendicular to their radical axis.
and 
, 
, so 
.
is a diameter of , 
, so 
and 
are tangent to circle .
, and 
and 
are tangent to circle . So 
and 
. meet at , and let 
meet at .
.
.,, and are collinear, so 
and 
.
and diameter . A line intersects at and the semicircle at and s.t. and . The circumcircles od the and intersect again at . Prove that .
becomes and 
becomes 
. Since the two circles intersect at 
, 
must be the intersection of and 
. Further, since 
are collinear, is the intersection of 
and 
. Since we want to show that is the diameter of 
it suffices to show that 
. But, observe 
. Since 
, we have that and 
are altitudes of . Hence, 
is the nine point circle, and must be the altitude from .
and diameter . A line intersects at and the semicircle at and s.t. and . The circumcircles od the and intersect again at . Prove that .![[asy]
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[/asy]](http://latex.artofproblemsolving.com/4/0/0/4009b9c3056e274eb70cb50ea2faee881628df39.png)
about the semicircle 
with center and diameter . The points 
are fixed under . Due to preservation of angles under inversion, if is the image of under and if is the image of under , then we need to prove that 
. It can be seen that due to inversion, and 
line 
circumcircle of 
. Now, we see that 
implies that 
and since is midpoint of segment , we see that circumcircle of is the nine-point circle of , now 
segment and nine point circle of and is not the midpoint of segment because is the midpoint of segment , therefore is the foot of the perpendicular from to segment or 
which is the desired result.
, the semicircle. Note that 
were free in the original diagram, and will be sent to themselves post inversion. Next, since 
, we have that 
.
, since (AOC) passes through the center of inversion, it will get sent to a line through . Similarly 
. Thus, we have 
.
, thus must be the foot of the altitude from so we have 
and we are done .
with diameter . The inverted diagram is given below.
and the result follows.
with 
. Note that 
, 
, and stay fixed. Since 
is cylic, , 
, 
are collinear after inversion. Similarly, , 
, 
are collinear, meaning is the intersection of lines and . After inversion 
, so it is enough to show that 
. Because is the diameter of the semicircle, 
, so and are the foot points of the altitudes on 
and 
. Since , , are collinear 
is cyclic. However, is the midpoint of , so 
is the nine point circle 
of 
. Hence, 
, meaning 
. Therefore, .
and diameter . Let and also lie on this semicircle, with closer to than is. Let be the intersection of and , and let be the intersection of 
with . Prove that 
.
and 
, and since is the midpoint of , is the nine-point circle of 
. This immediately implies the desired. 
, denoting images with 
. We have 
, 
, and we wish to prove that 
. is the midpoint of and 
, is the 9-point circle of , hence is the foot of the altitude from to , done. 
be the inversion of arbitrary radius centered at , and for any arbitrary point , let 
denote 
. Then we first observe: ![\[\Psi((AOC)) = AC; \phantom{c} \Psi((BOD)) = BD. \]](http://latex.artofproblemsolving.com/c/9/e/c9eec87e74bc0d9083e8e1678a210350cac4c1b9.png)
Now 
, and moreover we also procure that 
, so that it would suffice to show 
. However now observe that as is the nine point circle of , we must have 
, as was needed to show. 
is cyclic. So we want to show that 
.
.
and 
,
,
.
and 
.
.
and 
.
the intersection of 
around 
)
nine point circle of the triangle 
.
and notice that 
and 
Since 
then its easy to see that 
are inverses w.r.t. 
and now taking polars w.r.t. 
but by brokard we have 
thus 
thsu 
and since 
are colinear (by the inversion) we have 
as desired, thus we are done 
remain as it is after inversion. 
goes to 
goes to 
is the nine point circle of 
and we're done 
.
,
and 
invert to 
with 
,
and 
.
.
and 
is the polar of