.
and 
are chosen on segments 
and 
such that 
.
and 
is the midpoint of the arc 
,
.
squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of 
squares.
can Bob ensure a win no matter how Alice plays?
with incenter 
= intersection of 
and 
= intersection of 
and 
concycle
and 
intersect the altitude from 
to 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
such that 
is a perfect square.
and 
,
and 
.
.
be real numbers such that 
.
and 
.
,
be 
lies on arc 
,
intersects the circumcircle of 
.
and 
intersects 
at 
respectively, prove that 
are concurrent.
such that
Y by Funcshun840, Kizaruno, Rounak_iitr, GA34-261
Let 
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.This post has been edited 1 time. Last edited by ErTeeEs06, Apr 26, 2025, 11:15 AM
Reason: latex
Reason: latex
has orthocenter 
then 
is a parallelogram, hence 
bisects 
,
,
,
.
,
,
,
,
,
.
with orthocenter 
,
,
,
,
,
,
,
,
![\[
\frac{2yz+xy+xz}{\frac{2yz+xy+xz}{2}}=2 \in \mathbb{R}
\]](http://latex.artofproblemsolving.com/9/5/0/950896a7f4064d8c0398647447803b03dc8c30bd.png)
be the unit circle
