,
-
,
.
at 
.
parallel to 
respectively, so that 
meets 
.
is constant.
.
and from 
to 
and 
as the reflections of 
.
the second intersection of line 
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
and 
. Prove that
Where 
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the 
cars from the 
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Let 
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
This post has been edited 2 times. Last edited by djmathman, Jun 16, 2020, 4:02 AM
Reason: problem author
Reason: problem author
and 
is perpendicular to internal bisector of 
be points on smaller arc 
.
onto two equal parts. That means that 
(Archimed Lemma). Simialrly 
.
to cut the circumcircle at 
, also 
to cut the circumcircle at 
are cyclic (of course with center 
are also cyclics)
Given 
.
be the point on 
and let 
be the point such that 
.
.
intersects 
at 
.
so 
and 
.
.
.
so 
as desired.
on arc 
.
and 
congruent, so 
.
are parallel, 
and 
.
be the incenter of the triangle 
.
.
is cyclic, so 
,
.
,
,
,
,
and the circumcircle of 
we have![\[ \frac{AD}{\sin \angle AFD} = \frac{AF}{\sin \angle ADF} = \frac{AF}{\sin \angle BDF} = \frac{AF}{\sin \angle ABF} = 2R \]](http://latex.artofproblemsolving.com/4/7/c/47cba532afa31887d889d490fb02d9560c53f665.png)
![\[ \frac{AE}{\sin \angle AGE} = \frac{AG}{\sin \angle AFG} = \frac{AG}{\sin \angle GFC} = \frac{AG}{\sin \angle ACG} = 2R\]](http://latex.artofproblemsolving.com/a/d/9/ad991c91fe9220922d511ae068e2d94ad206c9d7.png)
and since 
.
,
.
.
and 
.
from the perpendicular bisector, we compute![\[ \angle APQ = \angle AFG + \angle BAF = \angle ACG + \angle BAF = \angle AFG + \angle BAF \]](http://latex.artofproblemsolving.com/9/7/8/9787520ab38141721fa6bcf46c6a981ae0660588.png)
![\[ = \angle GAF + \angle AGF + \angle BAF = x + y + z. \]](http://latex.artofproblemsolving.com/e/0/8/e08c5b4eb40141a421c0502338f2c814bd94b5ac.png)
Analogously 
.
,
and the result follows.
and 
,
.
,
to be the point on segment 
and define 
.
.
is 
.
,
.
,
,
,
.![\[\angle FBA=\angle FBC-b=(180-\angle FGC)-b=(\angle GYC+\angle GCA)-b=(90-\frac{a}{2})+x-b=\frac{-b+c}{2}+x,\]](http://latex.artofproblemsolving.com/f/b/f/fbf504da22d09bcc507be5184e8932100e114052.png)
which then gives that![\[\angle FAB= \frac{b+c}{2}-x,\]](http://latex.artofproblemsolving.com/a/c/1/ac1fd4f3d32e07d988b3efe4352ea3299c661265.png)
and![\[\angle CAG=b-x.\]](http://latex.artofproblemsolving.com/b/5/0/b5089f036c12227f1f9bf17ddf73ff3844c61f8d.png)
![\[CG=\sin \angle GAC=\sin (b-x),\]](http://latex.artofproblemsolving.com/3/6/7/3678d628180a4255f122b9b3f72a5636e1b51aab.png)
and since 
,![\[CE'=2\sin (b-x)\cos x,\]](http://latex.artofproblemsolving.com/4/9/c/49cd17c94f4ddb08643968895defe02ee8d88114.png)
which is equal to 
by Product-to-Sum formulas. This means that![\[AE'=AC-(\sin b+\sin (b-2x))=\sin b-(\sin b+\sin (b-2x))=-\sin(b-2x).\]](http://latex.artofproblemsolving.com/5/c/0/5c0ce6bf63dfc34df576c7fbc552869f6f4014bb.png)
Similarly, we get that![\[BD'=2\sin \left(\frac{b+c}{2}-x\right)\cos x,\]](http://latex.artofproblemsolving.com/1/6/d/16dcad67ed3711819daef3446e63c83e2a8dd6c9.png)
which is equal to 
by Product-to-Sum formulas. This means that![\[AD'=AB-(\sin c+\sin (b-2x))=\sin c-(\sin c+\sin (b-2x))=-\sin(b-2x),\]](http://latex.artofproblemsolving.com/1/6/6/166cf3aed36fcda007cd3c44de982c10a0b58365.png)
which is indeed equal to 
,
,
and 
)
or 
and 
also move linearlly. Hence its enough to prove that 
for three positions of the points 
,
and 
.
and 
be the tangents to 
at 
and 
be the angles that 
and 
,
will denote the derivative of the point 
has magnitude 
along 
as the direction of 
,
by the chain rule. Repeating the same argument with 
.
so 
as well.
as functions of 
as when 
are collinear and both angles can be chased to be equal to 
.
with the initial condition 
.
is "nice" enough, the solution to this differential equation is unique, so indeed we have that 
and 
are parallel, so 
are points on 
then we get that:
so 
so 
are collinear.
so 
is cyclic. We are motivated that 
is isosceles. We make the following claim:
and 
are both isosceles.![\[\angle GXA = \angle GCE = \angle GEC = \angle AEX. \phantom{c} \square\]](http://latex.artofproblemsolving.com/1/c/d/1cdb9de727f3b7b8c20d61c76ccf8cea05611564.png)
,
là một hình diều. Tương tự, 
l
.
l
t
l
v
.
so that 
is a parallelogram, and construct 
.
and 
.
and 
.
,
becomes a trapezoid. Therefore 
.
and 
are similar. Hence with the parallelism stated they are necessarily homothetic : the parallelism that was unknown to us is especially 
,
.
and 
are isosceles trapezoids. Then, since 
over the line through 
and 
are both parallelograms. Thus, 
,
is an isosceles trapezoid. Additionally, letting 
and 
.
,
,
,
are all equal, which now gives that 
is an isosceles trapezoid. Thus, 
,
.![\[\angle(DE,BC) = B - \angle{ADE} = B - \left(90 - \frac{A}{2}\right) = \frac{B-C}{2}.\]](http://latex.artofproblemsolving.com/f/b/8/fb85bbfb4236f4b8866f4de8fc0f5a522697cbf0.png)
We will show that this value is equal to 
.![\[\angle(MN,BC) = \angle{NBC} - \angle{MNB} = \frac{B}{2}-\frac{C}{2},\]](http://latex.artofproblemsolving.com/7/7/a/77a75920c1490898b54fee71bf109c61162fa6d2.png)
as desired.
,
.
,
,
gives the condition as 
,
on both sides this becomes 
,
,
gives 
as the equivalent condition.
is self conjugating, which is equivalent to 
,
,
,
,
,
meet. Let 
is a rhombus.
,
.
therefore by sliding principal we have 
.
.
,
and 
are symmetric about 
is a cyclic isosceles trapezoid with 
It is well known that the arcmidpoints of ![[asy]
size(7cm);
pair A = dir(110);
pair B = dir(205);
pair C = dir(335);
pair D = A+0.4dir(degrees(B-A));
pair E = A+0.4dir(degrees(C-A));
pair X = 0.5(B+D);
pair F = intersectionpoints(X+10*dir(degrees(A-B)-90)--X-10*dir(degrees(A-B)-90), circumcircle(A, B, C))[1];
pair Y = 0.5(C+E);
pair G = intersectionpoints(Y+10*dir(degrees(A-C)-90)--Y-10*dir(degrees(A-C)-90), circumcircle(A, B, C))[0];
draw(extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06+dir(degrees(A-dir(-90)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(A-dir(-90)))*0.06, red+linewidth(0.5));
draw(A--B--C--cycle);
draw(D--E^^C--G--F--B);
draw(circumcircle(A, B, C));
draw(A--dir(-90), dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(160));
dot("$E'$", E, dir(40));
dot("$F$", F, dir(F));
dot("$G'$", G, dir(G));
dot("$P$", extension(A, B, F, G), dir(125));
dot("$Q$", extension(A, C, F, G), dir(60));
[/asy]](http://latex.artofproblemsolving.com/0/c/d/0cd32b83b651cd2f4a007840d7f75bc2a6aed50f.png)
where 
is the angle bisector of 
such that 
.
and 
.
.
:
which is equal to 
.
.
Doing the same thing on the other side, 
.
,
.
