such that 
be the orthocenter of an acute angled triangle ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
intersect at point 
,
be the midpoint of the small arc 
of the circumcircle of the triangle 
the midpoint of the large arc 
Prove that points 
are concyclic.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
such that for any integer 
there exists an integer 
for which 
is divisible by 
and 
are marked, draw the line connecting 
and 
are drawn, mark the intersection of 
is drawn, draw a line parallel to 
Y by
Initially, three non-collinear points, 
, 
, and 
, are marked on the plane. You have a pencil and a double-edged ruler of width 
. Using them, you may perform the following operations:
using these operations.
, 
, and 
, are marked on the plane. You have a pencil and a double-edged ruler of width 
. Using them, you may perform the following operations:
- Mark an arbitrary point in the plane.
- Mark an arbitrary point on an already drawn line.
- If two points
and 
are marked, draw the line connecting and . - If two non-parallel lines
and 
are drawn, mark the intersection of and . - If a line
is drawn, draw a line parallel to that is at distance away from (note that two such lines may be drawn).
using these operations.This post has been edited 1 time. Last edited by Quantum-Phantom, Mar 8, 2024, 4:22 PM
,
at 
.
,
to get the median. This way we can construct the midpoint of any segment. Now note that by Ceva-Menelaus, given collinear 
we can always construct 
such that 
.
,
.
on the ray 
,
.
.
and 
.
and the internal angle bisector of 
intersect on the perpendicular bisector of 
.
now evidently gives us the centroid. We are now done by applying operation 
.
and 
take 
and draw a line parallel to 
and 
at 
,
lies on the median by Ceva.
over 
,
.
,
.
is the desired point since 
so 
.
,
and 
which is well known to be the midpoint of the 
.
and 
,
bisects 
.
.
.
and 
and 
.
is the centroid of the wanted triangle 
)
is constructed easily.
where 
and the centroid 
.
be the reflection of 
be a point at a distance further than 
and 
at 
is a median of 
,
at a distance 
which is also parallel to 
is a parallelogram. Now, construct the midpoint of 
and let the line through this point and 
.
.
and 
.
is parallel to 
and 
.
,
,
with distance 1 and suppose
=
.so we have M as midpoint.
=1 , then in 
we can find its centroid ,as desired .
and suppose 
=W then clearly 
.
on BC such that 
=1 with 
lie between M and C.
=1 and 
,
and 
.
median of 
.
we can mark the angle bisectors of 
.
median at 
is the 
,
is a rhombus, and so 
is the angle bisector.
,
at 
respectively. Then, let 
intersect at 
is the median in 
,
,
of 
on 
.
intersect 
at 
intersects 
.
where the vertices are the excenters of 
.
.
to reflect 
,
along 
and 
bisects 
.
.
.
.
.
and 
which are a distance of one apart where 
where 
and taking its intersections with the sides 
and the 
.
.
Consider 
,
we can construct its bisectors. 
parallel to 
form a rhombus. Taking 
is the internal bisector from symmetry. The external bisector is the same but with the rhombus on the outside.
.
.
intersect 
at 
for any point 
intersects the parallel line to 
,
is an isosceles trapezium.
with 
and 
.![[asy]
pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(unitcircle, dotted+blue); draw(A--B--C--cycle, blue); pair N = dir(270); pair M = midpoint(B--C); pair Q = 0.6*C+0.4*A; pair P = extension(B, Q, A, M); pair R = extension(B, P, A, A+B-C); draw(B--N--C, grey); draw(B--R, red); pair S = extension(A, R, M, N); pair L = midpoint(A--M); pair D = foot(A, B, C); draw(A--D, dashed+deepgreen); draw(A--M, blue); draw(A--R, blue); draw(S--N, blue); draw(D--S, deepgreen+dashed);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$N$", N, dir(N)); dot("$M$", M, dir(45)); dot("$Q$", Q, dir(85)); dot("$P$", P, dir(250)); dot("$R$", R, dir(R)); dot("$S$", S, dir(S)); dot("$L$", L, dir(L)); dot(D);
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 130 B = dir 210 C = dir 330 unitcircle / dotted blue A--B--C--cycle / blue N = dir 270 M 45 = midpoint B--C Q 85 = 0.6*C+0.4*A P 250 = extension B Q A M R = extension B P A A+B-C B--N--C / grey B--R / red S = extension A R M N L = midpoint A--M D .= foot A B C A--D / dashed deepgreen A--M / blue A--R / blue S--N / blue D--S / deepgreen dashed */
[/asy]](http://latex.artofproblemsolving.com/1/d/2/1d2c7e5a2c6465264583473a7b9ff14c644d7556.png)
,
which is equidistant from 
is the perpendicular bisector of side 
through 
at 
at 
of 
.
,
is the point at infinity along line 
.
.
meet the perpendicular bisector of 
.
passes through the foot of the 
on 
to bisect 
to be 
gives us the relfection of 
and using the above Claim two times.
,
as the two vertices and 
as the point of intersection of the diagonals. Get the line through the midpoint where the circumcentre must lie, repeat the same to get the circumcentre.
through 
and the incenter 
.
to be 
.
to be 
can easily be checked to be the foot of the altitude from 
