and 
. Prove that
Where 
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
,
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.
and 
. Prove that
Where 
probability of coming up heads and 
probability of coming up tails. The expected number of flips necessary to first see the sequence 
in that consecutive order can be written as 
for relatively prime positive integers 
,
be four distinct points in the plane.
and 
are distinct and parallel, then
is equal to the oriented angle DCB.
lies on the segment 
,
plus the oriented angle 
e
.
equals 0°, then
a
and 
Let 
be a point on that plane such that 
Prove that ![$$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$](http://latex.artofproblemsolving.com/7/8/e/78e428cbc99ce4e9448f5c4cec86f50567123812.png)
be reals such that 
. Prove that
specifies the greatest common divisor of 
and 
.
and 
Prove that
Let 
Prove that
Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user
Let be given a triangle 
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
be midpoint of 
.
,
,
.
then:
,
,
,
(a.a), 
,
.
(thanks to 
,
(Combine 
)
is a harmonic quadrilateral. Therefore, 
,
is the circumcenter of 
.
the midpoint of 
.
is cyclic (let 
its circumcenter) . Then 
are collinear . So if the circumcircle of 
, then 
. So by polars 
is tangent to 
is the 
meets the circumcircle of 
, then 
. But also because 
we have that 
is the incenter of 
so 
and because 
we also have 
. This shows that 
is cyclic and so 
and we are done .
cuts 
at 
are incenter and B-excenter of 
is a circle with center 
The cross ratio 
is harmonic and 
is radical axis of 
Pairwise radical axes 
of 
meet at their radical center 
and 
is cyclic. Its circumcircle is the B-Apollonius circle of 
at 
. Denote the following points : the middlepoint ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
;
for which 
; the point 
; the 
-
,
; the point 
. Then 
, i.e. the quadrilateral 
is cyclically.
- the tangent in 
to 
. Observe that 
and 
.
and the line 
is the 
.
is harmonically. Since 
results 
.
(a.a.) . Thus, 
, i.e. 
.
,
, i.e. 
.
obtain 
and in consequence, 
. Since 
obtain 
, i.e. the quadrilateral 
.
.
the midpoint of segment 
meets 
,we have 
are collinear.
is cyclic and 
is cylic ,we have :
.
.
.
is harmonically,so 
are collinear.
and denote by 
the intersection of two tangent to 
,respectively.
lie on the polar of 
wrt 
,we obtain 
are collinear.
(where 
at a point 
such that 
.![\[ \angle{MBC} = \angle{MTA} = \angle{ABF}.
\]](http://latex.artofproblemsolving.com/2/3/9/239532c7002a5d2f2d9ff6e8e73e2630d0ff7ad5.png)
inversion. Since this swaps 
and 
we have 
are swapped, so the circle with diameter 
is fixed, and 
gets sent to the foot from 
to 
is a symmedian.
.
themselves and same with 
and 
and the circle with diameter 
,
and obviously is the midpoint. 
be the midpoint of 
be the top point of 
,
collinear.
,
is cyclic.![\[\measuredangle MAD = \measuredangle MND = \measuredangle DAF. \quad \blacksquare\]](http://latex.artofproblemsolving.com/b/7/1/b7149517ba1ca4659451f0df62c429fecc7d338d.png)
which means that 
is fixed under this inversion. Now 
and thus, 
.
is simply the midpoint of 
be 
.
must lie on line 
,
;
;
is the midpoint of arc 
is nothing but the midpoint of 
is the reflection of the median about the angle bisector, i.e. the symmedian. 
be the antipode of 
,
.
,
,
.
denote the circle with diameter 
.
at 
and 
.
,
.
and 
with center 
passes through 
Take 
so 
and 
collinear. DDIT on 
means that 
and 
are isogonal from existing isogonal pairs 
and 
.
concur clearly. This is as 
is projected to 
.
is harmonic conj of 
are collinear and 
is cyclic. Note that 
so 
are also collinear and we get:![\[
\angle FAD=\angle FAE = \angle FNE = \angle DNM=\angle DAM
\]](http://latex.artofproblemsolving.com/0/a/b/0ab0704c1156f3fab3189527e5b838287d79b6d2.png)
and we're done since 
