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such that
for all real numbers 
satisfying the equation
has circumcenter 
. The lines 
and 
intersect at 
, while 
and 
intersect at 
and 
and 
intersect at 
. Show that if the triangles and 
are similar(with vertices in that order), than is equilateral.
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! https://www.prismsus.org/events/prom/home/index and be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
. So we need to show 
.
. But by Power of a point, 
. So we need to show 
(angle bisector theorem). But then that would mean CP=CM, where's my mistake?
and radius 
be 
+
=
=
are concyclic
=
are concyclic
=
and radius be 
+
== are concyclic= are concyclic=+=.
)
meet 
at 
.
implies that 
= 
.
= 
,
= 
, = . = ,
= 
.
is cyclic.
= 
,
= 
,
is cyclic.
= 
.
be the circumcircle of 
. Let 
and 
be the point on , that lies on angle bisector of 
. We ultimately want to show that 
are collinear.
but by Power of a Point we have that
Thus, 
takes the following form:
We also have that 
, since 
and 
, thus 
By 
and 
, we conclude that 
, which means that 
is the angle bisector of 
. Also, since is the defined as intersection of and angle bisector of , we have that 
is isosceles, from here we have that 
is the angle bisector of . We conclude that are collinear.
. By the angle bisector theorem, we know that
.(1)
, by (1), we get 
. is ciclic 
is ciclic, therefore 
, QED
intersect at 
.
. By angle bisector theorem, we get that 
.
, implying that 
is cyclic by converse of PoP.
, implying that 
is cyclic, and 
to meet 
at 
. Then 
, and from here some angle chasing shows us that 
is cyclic as desired.
. Then by the given condition, we have 
.
then 
is the angle bisector of 
. By internal angle bisector theorem, we have 
and 
. Thus, 
and 
. Now note that 
implying 
is cyclic. Thus, 
. Thus is also cyclic implying 
as desired. (
)
?
of a convex quadrilateral 
intersect at point 
.
meets the ray 
at 
.
.
(
according to(1)
,
(
is cyclic,then 
is cyclic and 
I have corrected it right now.
be the point where the circumcircle of 
intersects line 
at a point other than 
.
It follows that 
which means that 
.
.
by properties of exterior angles and the fact that 
.
, which implies that quadrilateral 
is cyclic. The result follows immediately.
and let 
be the point on ray 
such that 
.
is cyclic, and by PoP, 
.
so 
as desired.
intersect 
,
at 
.![\[MF \cdot MC = MB \cdot MD = MA \cdot MC + MA \cdot CD\]](http://latex.artofproblemsolving.com/a/d/0/ad089ac20a23105f47c46bce304961befb19e3e0.png)
implies![\[MA \cdot CD = MC \cdot (MF - MA) = MC \cdot AF\]](http://latex.artofproblemsolving.com/c/8/a/c8a83cefc927a0d0abeee8a35a1d0e3dc54f1bc5.png)
By this and the angle bisector theorem,![\[\frac{AF}{AM} = \frac{CD}{CM} = \frac{ED}{EM}\]](http://latex.artofproblemsolving.com/a/d/1/ad1a6b226948c28aeaf8002142d54d00e39b1710.png)
so 
.![\[\angle AEB = \angle FDB = \angle ACB\]](http://latex.artofproblemsolving.com/8/9/4/8940e049744767a14f52f30e8cda1fb72ef2d3d2.png)
so 
,
,![\[\angle BKC = \angle BEC - \angle KBE = \angle BEC - \angle MCE = \angle BEC - \angle DCE = \angle CDB\]](http://latex.artofproblemsolving.com/a/1/6/a161261eba2290189aa0e412793b2b159dd7c1d8.png)
and 
.
is cyclic.
is cyclic so 
.