be a point on side 
.
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle 
distinct from point 
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
and 
on the plane such that 
is an equilateral . Let 
and 
at 
and 
at 
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
over 
.
and 
Prove that
![\[\lfloor x \rfloor\]](http://latex.artofproblemsolving.com/d/3/f/d3f8facf62413e1f17c89b9c183a8735c2defe8b.png)
denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which ![\[ x^2 - 20\lfloor x \rfloor + 19 = 0 \]](http://latex.artofproblemsolving.com/f/a/9/fa96f2d6217a16029e5f03499e0a83deffdfa28a.png)
Prove that
of the pizza, where 
is a positive integer. Find 
![\[\mathcal{P}(x) = x^{108}+x^{102}+x^{96}+2x^{54}+3x^{36}+4x^{24}+5x^{18}+6\]](http://latex.artofproblemsolving.com/e/b/2/eb28918359848dd07784a3c2c6a34b934743c016.png)
be 
.![\[\dfrac{r_1^6+r_2^6+\dots+r_{108}^6}{r_1^6r_2^6+r_1^6r_3^6+\dots+r_{107}^6r_{108}^6}\]](http://latex.artofproblemsolving.com/3/1/5/315341a7adf35bc04e40ee0685a011c9f7b76115.png)
without Newton Sums.
Is there an elegant way to prove this?
Y by
A circle with center 
is inscribed in an angle. Let 
be the reflection of across one side of the angle. Tangents to the circle from intersect the other side of the angle at points 
and 
. Prove that the circumcenter of triangle 
lies on the bisector of the original angle.
(I.Sharygin)
is inscribed in an angle. Let 
be the reflection of across one side of the angle. Tangents to the circle from intersect the other side of the angle at points 
and 
. Prove that the circumcenter of triangle 
lies on the bisector of the original angle.(I.Sharygin)
![[asy]
import graph; size(20cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -22.052250553809948, xmax = 55.602503588611654, ymin = -14.6157368833881, ymax = 32.66078702540154; /* image dimensions */
pen wwqqcc = rgb(0.4,0,0.8); pen qqwwzz = rgb(0,0.4,0.6);
/* draw figures */
draw(circle((5,5), 5), linewidth(1));
draw((-14.898670900541106,0)--(xmax, 0*xmax + 0), linewidth(1) + wwqqcc); /* ray */
draw((-14.898670900541106,0)--(xmax, 0.5364143375878395*xmax + 7.9918606820529785), linewidth(1) + wwqqcc); /* ray */
draw(circle((11.199791397931394,6.557840578629505), 13.115681157259), linewidth(1));
draw((18.109279252949484,17.70593771671708)--(22.558304468054573,0), linewidth(1));
draw((18.109279252949484,17.70593771671708)--(-0.15872167219178485,0), linewidth(1));
draw((0.27299320287298867,13.812230520130251)--(5,5), linewidth(1));
draw((0.27299320287298867,13.812230520130251)--(-0.15872167219178485,0), linewidth(1));
draw((0.27299320287298867,13.812230520130251)--(22.558304468054573,0), linewidth(1));
draw((-1.909487855018086,6.967584019171428)--(22.558304468054573,0), linewidth(1));
draw((-14.898670900541106,0)--(xmax, 0.25127306366296226*xmax + 3.743634681685189), linewidth(1) + qqwwzz); /* ray */
draw((-1.909487855018086,6.967584019171428)--(-0.15872167219178485,0), linewidth(1));
/* dots and labels */
dot((5,5),dotstyle);
label("$O$", (5.230109082605615,5.636416605506779), N * labelscalefactor);
dot((-14.898670900541106,0),dotstyle);
label("$T$", (-14.635060581734795,0.6701241894402005), NE * labelscalefactor);
dot((0.27299320287298867,13.812230520130251),dotstyle);
label("$A$", (-1.219621327894518,15.053023004801851), NE * labelscalefactor);
dot((-0.15872167219178485,0),linewidth(5pt) + dotstyle);
label("$B$", (-2.058086281259535,-2.1032598870645125), NE * labelscalefactor);
dot((22.558304468054573,0),linewidth(5pt) + dotstyle);
label("$C$", (21.805916237590957,-2.2967517993787947), NE * labelscalefactor);
dot((11.199791397931394,6.557840578629505),linewidth(4pt) + dotstyle);
label("$P$", (11.615342189000748,4.346470523411564), NE * labelscalefactor);
dot((18.109279252949484,17.70593771671708),linewidth(4pt) + dotstyle);
label("$R$", (18.387559120025887,18.213390905935128), NE * labelscalefactor);
dot((-1.909487855018086,6.967584019171428),linewidth(4pt) + dotstyle);
label("$Q$", (-4.379989229039583,7.3133465122305585), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/5/e/4/5e429f053088966d8d159febf5990dee1d1de513.png)
intersect the side of the angle 
Because 
bisects 
it follows from the incenter-excenter lemma that 
are the midpoints of arc 
or arc 
form an isosceles trapezoid and line 
coincides with the perpendicular bisector of the bases of this trapezoid, implying the result. 
