.
be a triangle with circumcenter 
and let the angle bisector of 
intersect 
.
is such that 
and 
.
and 
intersect at
.
be reals such that 
.Prove that
Where 
be real numbers . Prove that
girls and 
players named 
is to be sent for batting. If 
and 
Then find the total number of batting orders if other players could go in any order.
that cannot be expressed as 
for some positive real number 
.
of cardboard, with 5 rectangular faces instead of 6 and there is no leftover cardboard. What is the largest volume this box can hold given that 
of cardboard is needed to cover the box?
Subject to the constraints:
Prove that
. Prove that
can be written in the form![\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]](http://latex.artofproblemsolving.com/9/e/9/9e96c07ea3b0750af8bed344df6e8a86fcf0d772.png)
where 
and 
are positive integers. Find the value of 
.
Y by
Suppose that two circles 
with centers 
, respectively , intersect orthogonally at 
,
. Let 
be a diameter of 
that is exterior to 
. Let 
be points on such that 
are tangent to , with 
on one side of 
and 
on the other side of . Let 
be the intersection of 
and 
be the intersection of 
. Prove that 
is parallel to and is tangent to
with centers 
, respectively , intersect orthogonally at 
,
. Let 
be a diameter of 
that is exterior to 
. Let 
be points on such that 
are tangent to , with 
on one side of 
and 
on the other side of . Let 
be the intersection of 
and 
be the intersection of 
. Prove that 
is parallel to and is tangent to 
![[asy]
import graph; size(11.3cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.29, xmax = 5.24, ymin = -1.65, ymax = 2.88; /* image dimensions */
pen uququq = rgb(0.25,0.25,0.25); pen xdxdff = rgb(0.49,0.49,1);
pair P = (0,0), A = (0.76,0.66), B = (0.88,-0.48), Q = (1.21,0.13), C = (1.26,0.82), D = (1.16,-0.56), F = (0.43,-0.9), T = (0.98,-0.29), G = (-1,0.08), H = (0.83,-0.06), I = (1,-0.08), J = (-0.81,2.46);
/* draw figures */
draw(circle(P, 1));
draw(circle(Q, 0.69));
draw(D--F);
draw(C--(0.15,0.99));
draw((1.03,0.34)--T);
draw(D--(0.15,0.99));
draw(C--F);
draw(A--Q);
draw(B--Q);
draw(A--P);
draw(P--B);
draw(C--D);
draw(D--G);
draw(G--C);
draw(A--B);
draw(D--A);
draw(C--B);
draw(G--I);
draw(J--A);
draw(J--G);
draw((0.15,0.99)--J);
/* dots and labels */
dot(P,uququq);
label("$P$", (0.02,0.03), NE * labelscalefactor,uququq);
dot(A,xdxdff);
label("$A$", (0.77,0.68), NE * labelscalefactor,xdxdff);
dot(B,xdxdff);
label("$B$", (0.9,-0.44), NE * labelscalefactor,xdxdff);
dot(Q,uququq);
label("$Q$", (1.23,0.16), NE * labelscalefactor,uququq);
dot(C,xdxdff);
label("$C$", (1.28,0.85), NE * labelscalefactor,xdxdff);
dot(D,xdxdff);
label("$D$", (1.17,-0.53), NE * labelscalefactor,xdxdff);
dot((0.15,0.99),uququq);
label("$E$", (0.17,1.02), NE * labelscalefactor,uququq);
dot(F,uququq);
label("$F$", (0.45,-0.87), NE * labelscalefactor,uququq);
dot((1.03,0.34),uququq);
label("$S$", (1.05,0.37), NE * labelscalefactor,uququq);
dot(T,uququq);
label("$T$", (0.96,-0.42), NE * labelscalefactor,uququq);
dot(G,uququq);
label("$G$", (-0.98,0.11), NE * labelscalefactor,uququq);
dot(H,uququq);
label("$H$", (0.72,-0.13), NE * labelscalefactor,uququq);
dot(I,uququq);
label("$I$", (1.06,-0.08), NE * labelscalefactor,uququq);
dot(J,uququq);
label("$J$", (-0.79,2.49), NE * labelscalefactor,uququq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/0/2/f028b37fc09aa35b39aeddbcab801a4a8668bda0.png)
and 
so we get that 
is the orthocenter of 
so it also lies on the circle 
at 
.
are collinear so 
.
is perpendicular to both 
and 
.
and 
theorem but i wasn't able to complete that idea, unfortunately.
be a cyclic quadrilateral and let 
where by 
we mean the tangent to the circle at point 
are collinear.
and 
we get 
and 
are collinear so 
are collinear.
the polar line of 
with respect to 
because the midpoint of 
is on the radical axis of 
(because 
)
.
.
lies on 
.
is perpendicular to 
.
so because 
,
is tangent to 
and the magazine you are refering is from 2021.
.
but we have 
and 
but 
is cyclic so we are done.
collinear.
is right angled and 
.
.
then 
and thus 
.
collinear.
cyclic.
,
cyclic. Also, 
and 
collinear. Thus, 
cyclic. Thus, the claim follows.
and similarly 
.
for the parallel condition.
.
.
.
is diameter of 
.
.
(
)
and 
and we are done.
