where 
are non-negative integers that are not necessarily distinct. Find the smallest positive integer 
such that no multiple of 
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
and 
meet at 
and 
meet at 
.
and 
be a positive integer. Claire writes 
in a row on a blackboard. In a 
William can erase a number 
and replace it with either 
or 
at the same location. His goal is to perform a sequence of moves such that after he is done, the number are strictly increasing from left to right.
independent of 
such that William can always reach his goal in at most 
moves.
independent of 
moves.
Y by Mango247, Mango247
Given 
with circumcenter 
. Let 
be a point satisfying 
and 
be the midpoint of 
. Suppose that 
intersect circle 
at another points 
, respectively. Let 
be a point on 
so that 
is tangent to circle . Prove that 
are concyclic.
with circumcenter 
. Let 
be a point satisfying 
and 
be the midpoint of 
. Suppose that 
intersect circle 
at another points 
, respectively. Let 
be a point on 
so that 
is tangent to circle . Prove that 
are concyclic.
This post has been edited 1 time. Last edited by parmenides51, Mar 21, 2020, 8:30 PM
lie on 
such that 
is a harmonic quadrilateral. An arbitrary circle through 
intersect 
again at 
intersect 
then 
bisects 
cuts 
then by Reim theorem 
so 
with 
resp. Let 
be the reflection of 
then 
is a isosceles trapezoid
to 
meets 
then by lemma 
passes through the midpoint of 
which is 
combine with 
meets 
so 
is tangent to 
are concyclic
combine with 
so 
is angle bisector of 
so 
.
is cyclic.
.
.
.
and 
.![\[ MJ \cdot MA = \text{Pow}_M (ABC) = MK \cdot MG \Rightarrow MJ = MG \]](http://latex.artofproblemsolving.com/5/f/b/5fbe536c4dd9b9ed98d1281e1eebe70c622c4771.png)
cyclic.
.![\[ \angle AMG = \angle KMD = 2 \angle KAD \equiv 2 \angle KAJ = 2 \angle AJG = 2 \angle ACG = \angle AOG \]](http://latex.artofproblemsolving.com/9/1/3/913f1b0d2165bd7629edc2e14e7eecb4c973a70b.png)
is a harmonic quadrilateral.![\[ -1 = (C,B;I,H) \overset{A}{=} (C,B;J,K) \overset{M}{=} (F,E;A,G) \]](http://latex.artofproblemsolving.com/e/8/1/e81f6a383bcd52bc62b38e0c599fe60abc6435d6.png)
and hence the conclusion.
is cyclic.
and 
.
and 
tangent to 
is cyclic. Since we already have 
are cyclic.
quite naturally. From here, we already get a "nearly" complete quadrilateral. Therefore, we would want to complete the picture, by adding point 
and point 
and 
.
,
being harmonic quadrilateral. Now, I rephrase the question into:
,
.
and 
,
.
are collinear. Therefore, we once again rephrase the question into an equivalent one:
are collinear.
cyclic instead of 
and 
are the same, or in other words, 
