such that
for all real numbers 
satisfying the equation
?
has circumcenter 
.
and 
intersect at 
,
and 
intersect at 
and 
and 
intersect at 
.
are similar(with vertices in that order), than 
candidates and 
judges, where 
is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most 
candidates. Prove that ![\[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]](http://latex.artofproblemsolving.com/3/8/7/387c8824b28f6422306e3c25b34fccb0a841f4cd.png)
for Girls, PRISMS Online Math Meet for Girls, is happening this spring! 
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
Y by
It is considered a circle 
and a chord ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
of this circle. In one of the arcs determined by the chord two circles are inscribed 
and 
, 
, which are tangent to circle at points 
and 
, and at the chord in points 
and 
, respectively. Prove that points 
and lie on a circle.
and a chord ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
of this circle. In one of the arcs determined by the chord two circles are inscribed 
and 
, 
, which are tangent to circle at points 
and 
, and at the chord in points 
and 
, respectively. Prove that points 
and lie on a circle.This post has been edited 1 time. Last edited by parmenides51, Mar 17, 2021, 7:11 PM
let 
and 
respectively. Then 
is cyclic if and only if 
is parallel to the tangent to 
at 
be the tangent to ![\[ABCD\text{ cyclic} \iff \measuredangle ABP=\measuredangle (AP,\ell)=\measuredangle PDC\iff CD|| \ell\]](http://latex.artofproblemsolving.com/7/e/8/7e82b4746de73e82ceabfc49f1d096bc413b2af6.png)
so we're done. 
be the midpoint of the arc not containing the two inscribed circles. It's well-known that 
pass through 
with center 
and chord 
let 
be points on either arc 
Then if 
not containing 
let 
Prove that 
are concyclic.
so the desired result follows immediately from Reim. 
