[/quote]
,
be a fixed integer. Find the least constant 
such the inequality![\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]](http://latex.artofproblemsolving.com/d/d/e/dde2d1dd0c412d8ba640f1d4c044e2fa2dcccc82.png)
(the sum on the left consists of 
summands). For this constant 
,
be the least prime that does not divide 
,
to be the product of all primes less than 
For 
,
.
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
.
such that 
.
such that 
.
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
and 
and 
meet at 
.
and 
and orthocenter 
.
denote the circumcircle of triangle 
the midpoint of 
.
and 
,
.
and 
similarly.
such that for all positive integers 
and ![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
,
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of 
(which may depend on 
)
.
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
for all 
.
.
such that 
and the circumcircle of 
.
passes through a fixed point.
and 
intersect each other at 
and 
.
and 
respectively. Let 
and 
be the reflection of 
intersect circles 
and 
(both distinct from 
is the second tangent to 
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
,
,
,
be the 
intersects 
.
is perpendicular to 
,
is the incenter of 
and the circle 
of radius 
Let 
The line 
intersects 
Let 
Let 
and 
Y by ilovemath121, Davi-8191, Adventure10, Mango247, Rounak_iitr, and 4 other users
In the triangle 
, 
is biggest. On the circumcircle of 
, let 
be the midpoint of 
and 
be the midpoint of 
. The circle 
passes through 
and is tangent to 
at 
, the circle 
passes through 
and is tangent 
at . and intersect at and 
. Prove that 
bisects 
.
, 
is biggest. On the circumcircle of 
, let 
be the midpoint of 
and 
be the midpoint of 
. The circle 
passes through 
and is tangent to 
at 
, the circle 
passes through 
and is tangent 
at . and intersect at and 
. Prove that 
bisects 
.![[asy]
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draw(circle((-1.32,1.36), 2.98));
draw(circle((3.56,1.53), 3.18));
draw((0.92,3.31)--(-2.72,-1.27));
draw(circle((0.08,0.25), 3.18));
draw((-2.72,-1.27)--(3.13,-0.65));
draw((3.13,-0.65)--(0.92,3.31));
draw((0.92,3.31)--(2.71,-1.54));
draw((-2.41,-1.74)--(0.92,3.31));
draw((0.92,3.31)--(1.05,-0.43));
/* dots and labels */
dot((-1.32,1.36),dotstyle);
dot((0.92,3.31),dotstyle);
label("$A$", (0.81,3.72), NE * labelscalefactor);
label("$c_1$", (-2.81,3.53), NE * labelscalefactor);
dot((3.56,1.53),dotstyle);
label("$c_2$", (3.43,3.98), NE * labelscalefactor);
dot((1.05,-0.43),dotstyle);
label("$P$", (0.5,-0.43), NE * labelscalefactor);
dot((-2.72,-1.27),dotstyle);
label("$B$", (-3.02,-1.57), NE * labelscalefactor);
dot((2.71,-1.54),dotstyle);
label("$E$", (2.71,-1.86), NE * labelscalefactor);
dot((3.13,-0.65),dotstyle);
label("$C$", (3.39,-0.9), NE * labelscalefactor);
dot((-2.41,-1.74),dotstyle);
label("$D$", (-2.78,-2.07), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/e/f/4/ef437106f2f9dfc796fb26c8861005194064a52a.png)
This post has been edited 3 times. Last edited by WakeUp, Dec 15, 2013, 8:21 PM
Reason: Asymptote diagram added.
Reason: Asymptote diagram added.
be an inversion with center 
denote the image of 
under 
and radius 
,
is the point on ray 
satisfying 
.
satisfying 
.
is the intersection of the line through 
parallel to 
and the line through 
parallel to 
.
.
and 
and 
hence 
so 
is isoceles and 
.
is isoceles and so 
as desired.
passes 
passes 
.
?
,
are collinear.
and let the center of 
.![$[EO_1]$](http://latex.artofproblemsolving.com/b/5/5/b551a4f560dec7d8ead229a3b03952a75b69a54a.png)
such that 
,
.
bisects angle 
which implies that 
and hence that 
.
of 
.
.
.
which implies the result.
, by angle chasing , we have 
, so 
are concyclic .
and 
. At the same time , 
is exactly the orthocentre of 
is perpendicular to 
.
be the midpoints of 
Let the internal bisector of 
Note that 
leads to the fact that 
is tangent to 
We will now show that 
is tangent to 
Hence we get 
Furthermore, 
which leads to 
and 
we get the equation 
by sine law we get 
which means the triangle 
is isosceles. The rest is just angle calculation.
and so on.
.![\[\sin \angle{ABP}:\sin \angle{AEP}\]](http://latex.artofproblemsolving.com/8/0/6/80658552a980801fb5966d29bf950b3100d59866.png)
![\[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2C}}{\sin {2A}}\]](http://latex.artofproblemsolving.com/2/6/4/264188c88b6855ff7e0c4a8b0ca1b99377aca34e.png)
![\[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2A+2B}}{\sin {2B+2C}}\]](http://latex.artofproblemsolving.com/9/1/1/9119db810db454f58a03f979b1777f84fef62800.png)
![\[=\sin A:\sin C\]](http://latex.artofproblemsolving.com/6/4/9/649879995105fa6c999749bc8eead2bbf2de00a5.png)
,
and the function 
in increasing when 
.
and let 
.
.
.
.
since they are formed by the intercepts of pairs of parallel lines.Also 
since 
and so 
so we are done.
,
(they are just the midpoints of arcs.) Let 
,
.
,
is tangent to 
.
and this is trivial, so we are done.
is actually complex numbers. This lets us compute 
by angle chasing,we have
cyclic,and 
cyclic
then we find 
isosceles.
from 
and using the Sine Law in 
we can find 
Then we can find easily 
where using the Sine Law in 
As desired.
be the ẽ-bisector of triangle 
and 
are collinear. Let 
cut 
at R, we have 
.
are collinear so we have 
are concyclic so 
are the external angle bisectors of 
and 
.
so that it meets 
at 
and extend 
so that it meets 
at 
and 
by Reim's Theorem. So 
are collinear.![\begin{align*}
\angle IJE&=\angle AJE\qquad[\text{Since }A,I,J\text{ are collinear}]\\
&=\angle DAE\\
&=\angle DBE,
\end{align*}](http://latex.artofproblemsolving.com/e/8/3/e83437c2bc3d64e9fd9c384ce6777f8a484c6725.png)
giving us that 
are concyclic.
,
,
to be concurrent, giving us what we need. ![\[\measuredangle ABP' = \measuredangle P'AB = \measuredangle CAP' \]](http://latex.artofproblemsolving.com/e/b/3/eb331e3def1999c410a83f6fedb2da047453e17d.png)
which implies that 
, i.e ![\[\measuredangle P'AD = \measuredangle MAD = \measuredangle ICA = \measuredangle P'EA\]](http://latex.artofproblemsolving.com/b/f/5/bf53e492b2b0ccef204ae58a04be464aadac1066.png)
which implies that 
i.e 
and thus, 
which implies that 
b
