AoPS Academy
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.
![$\frac{[APC]}{[APQ]}=\frac{b}{AQ}$](http://latex.artofproblemsolving.com/d/4/b/d4be417e605f9746919822a58d3e73906e3d3ed2.png)
and ![$\frac{[CPB]}{ACP}=\frac{PB}{AP}$](http://latex.artofproblemsolving.com/e/4/4/e44e3e0987aa54153108f2494545b7826f3c6666.png)
so we have ![$[CPB]=\frac{b\cdot PB}{AQ\cdot AP}\cdot [APQ]$](http://latex.artofproblemsolving.com/2/9/e/29e362af2b34bb2d8cb9dc02bf01d4a2a633d839.png)
so from ![$[APQ]=[BCQP]$](http://latex.artofproblemsolving.com/7/e/c/7ec264975a219baaae2859469a7ea80fdf652f3b.png)
we have 
and now from from 
applying in bary formula for colinearity we have:
so we have to prove these to are equivalent,ie. 
of these to are equal.
is the feet of angle bisector is redundant. It can be any point on the side 
as can be clearly seen in the following solution.
be a diameter of the circumcircle of 
. Observe that 
, so ![$[BDP]=[XDP]$](http://latex.artofproblemsolving.com/3/4/e/34ee74f1929156cea57ccc18a3076a56486ca5b0.png)
. Analogously, ![$[CDQ]=[XDQ]$](http://latex.artofproblemsolving.com/c/4/5/c450b8c6fdd100b33402158f16f8408f898e90ce.png)
. Thus ![$$[BCQP]=[BDP]+[PDQ]+[CDQ]=[XDP]+[BDP]+[XDQ]=[PXQ].$$](http://latex.artofproblemsolving.com/6/4/d/64d897a0577ab7f90a228a68464c8b7f8a39563b.png)
Therefore ![$[APQ]=[BCQP] \iff [APQ]=[PQX] \iff O \in PQ$](http://latex.artofproblemsolving.com/b/5/2/b5294853f57745bc722bde9c8a7b278792f27bf0.png)
, where 
is the midpoint of , i.e. the circumcenter of .
![$[APQ]=[BCQP] \iff [APQ]=\frac{[ABC]}{2} \iff \frac{AP \cdot AQ \cdot \sin \angle BAC}{2}=\frac{AB \cdot AC \cdot \sin \angle BAC}{4} \iff AP \cdot AQ=\frac{AB \cdot AC}{2}$](http://latex.artofproblemsolving.com/5/5/8/5588c82bedb8c6857df4ac06a536a38981018020.png)
. Consider transformation 
- composition of symetry wtr bisector of 
and inversion in 
with radius 
. Denote 
as midpoints of 
and let 
be a feet of -altitude. It is sufficient to show that 
are concyclic. Notice that 
.
are collinear, 
are concyclic. However, as 
and 
are isogonals wtr and 
, hence 
and 
. Points 
are concyclic, so we are done.
?
and 
are orthogonal projections of 
and 
.
.
