[/quote]
,
satisfy,
Where 
}
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
and 
.
![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
be real numbers . Prove that
They fill in real numbers to empty spaces (the coefficients of the terms other than 
and the constant term) in turn. If the resulting polynomial has no real root, Mr. Fat wins; otherwise, Mr. Taf wins. If Mr. Fat goes first, who has a winning strategy?
be a positive integer. Consider a 
array of square unit cells. Two corner cells that lie on the same longest diagonal are colored black, and the rest of the array is white. A move consists of choosing a row or a column and changing the color of every cell in the chosen row or column.
,
,
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 
such that 
for all 
.
are non-negative real numbers such that 
Prove that![\[\sqrt{a^2+6ab}+\sqrt{b^2+6bc}+\sqrt{c^2+6ca} \ge 5\sqrt{1 + \frac{153abc}{50(a+b+c)}}.\]](http://latex.artofproblemsolving.com/4/2/9/4295e20975f0a7c9902a73695bebb665385f01b3.png)
or 
of real numbers such that![\[ a_{m^2 + m + n} = a_{m}^2 + a_m + a_n\]](http://latex.artofproblemsolving.com/b/b/9/bb97053e86e7733fe5c26f7859c87566eb444b52.png)
for all positive integers 
and 
.
be a convex quadrilateral with 
and 
.
and 
be points on segments 
and 
,
intersects lines 
and 
at points 
and 
.
be 
and 
lie on a circle.
Y by aopser123, Davi-8191, Executioner230607, Adventure10, Mango247, Rounak_iitr
An equilateral pentagon 
is inscribed in triangle 
such that 
, 
, and 
. Let 
be the intersection of 
and 
. Denote by 
the angle bisector of 
.
Prove that
is parallel to , where 
is the circumcenter of triangle , and 
is the incenter of triangle .
is inscribed in triangle 
such that 
, 
, and 
. Let 
be the intersection of 
and 
. Denote by 
the angle bisector of 
.Prove that
is parallel to , where 
is the circumcenter of triangle , and 
is the incenter of triangle .This post has been edited 1 time. Last edited by v_Enhance, Apr 20, 2016, 9:33 PM
Reason: Change \ell to l
Reason: Change \ell to l
and 
denote 
and 
.
.
.
is trivial, although technically the lines coincide and so they are not parallel, oops.)
Also, by Law of Sines, we have 
and similarly on triangle 
,![\[ \sin \epsilon - \sin \delta = \sin B - \sin C. \]](http://latex.artofproblemsolving.com/7/4/b/74bffc6f9e4125912be3373ce60be345e53026e3.png)
Using sum-to-product formulas on our relations implies that ![\[ \tan \left( \frac{\epsilon-\delta}{2} \right) = \frac{\sin B - \sin C}{\cos B - \cos C + 1}. \]](http://latex.artofproblemsolving.com/0/5/7/057aae18078279eb7f6b6827a48a7fab0b2c723d.png)
Now note that 
with line 
intersects line 
then ![\[ \tan\varphi = \frac{r - R \cos A}{\frac12(b-c)}. \]](http://latex.artofproblemsolving.com/6/c/e/6ce1d11a00dfb8ffbae514168784507fb401b678.png)
So in order to prove the result, we only need to check that ![\[ \frac{r - R \cos A}{\frac12(b-c)} = \frac{\cos B - \cos C + 1}{\sin B - \sin C}. \]](http://latex.artofproblemsolving.com/4/9/2/4924f47d69c37714970dc81571ab5ccd558a89ed.png)
Using the fact that 
,
,
,
and 
.
,
,
,
,
be the complex numbers corresponding to the arc midpoints of 
,
is the incenter of 
.
be the side length of 
and 
,![\[ m - a = i \cdot sx. \]](http://latex.artofproblemsolving.com/0/1/b/01b7eabf0c118f8fcb55310b1d5e186bb6af6193.png)
Similarly, 
and 
,![\[ i \cdot s(x+y+z) = (p-q) + (m-n) = (m-n) - (q-p). \]](http://latex.artofproblemsolving.com/1/d/8/1d8ac470b2e5f78ab5f2896d8adf99b42ee47248.png)
Since 
,
is along the external angle bisector of the angle formed, which is perpendicular to 
,
is fixed, is a line which is perpendicular to the line connecting the incenter and the circumcenter of 
be a point varies on the external bisector of 
From Lemma 
it suffices to prove 
(oriented distance) is constant. Using oriented areas we get ![$$ [TNP]+[TQA]+[TAM]=[AMNPQ]-[TMN]-[TPQ]=[AMNPQ]=\text{constant}, $$](http://latex.artofproblemsolving.com/3/f/2/3f229489d00455d3af093d0dc2607bb1f2959c26.png)
so combining 
we conclude that 
be the midpoint of 
and let 
be the midpoint of 
.
.![\[
d(N,AB)+d(N,AC)+d(Q,BC)=d(P,AB)+d(P,AC)+d(M,BC)
\]](http://latex.artofproblemsolving.com/5/9/d/59d7dc859fc1fee44f47f07510216292d59c0e5d.png)
![\[
[NAQ]+[NAM]+[QPN]=[PAM]+[PAQ]+[MNP]
\]](http://latex.artofproblemsolving.com/0/d/6/0d64f6d034b275def0b2c2251f74e252338dde61.png)
are sometimes the same line, and thus are not parallel. 
and 
intersect at the point of infinity and hence the angle bisector is essentially a line also parallel to 
but I imagine in that case we can just take limits and continuity or something and be like gg)
):
and let 
denote the midpoints of 
respectively. We will show that 
,
that 
.
.
.
.
,
are homothetic and consequently 
.
,
.
and we are done.
lie on 
respectively such that 
Denote by 
the respective midpoints of 
and let 
Then 
is a parallelogram and let 
be the reflections of 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/4/b/4/4b4d7dab7a46bdebff791d1e74769c763873b39e.png)
is a parallelogram since its diagonals bisect one another. Hence, 
and 
Using 
,
and 
Because 
,
is isosceles. Hence, the bisector 
of 
is perpendicular to 
But since 
are homothetic, 
,
,![\[\vec{NM}-\vec{PQ}=\vec{M}-\vec{N}-\vec{Q}+\vec{P}=\vec{AM}+\vec{NP}+\vec{QA}.\]](http://latex.artofproblemsolving.com/2/2/3/223eb3fdf47198e6283859b180fb487c839359e1.png)
Note that the final expression is parallel to the sum of the unit vectors along the sides of 
,![\[\vec{OL_A}+\vec{OL_B}+\vec{OL_C}\]](http://latex.artofproblemsolving.com/b/8/6/b86a8aac4a26d4fe7210e59af8d615ab45a9a71a.png)
where 
are the arc-midpoints on 
.
,
is a rhombus. Then, the angle bisector of 
is parallel to 
,
.
is homothetic to 
,
and 
on 
.
is perpendicular to 
,
.
,
as desired.
,
and (using points as vectors) 
are perpendicular.
is perpendicular to 
,
and 
.
is a parallelogram. Note that by these parallelisms, 
,
is isosceles, so it suffices to show 
.
again at point 
.
and 
,
.
.
.
and by definition 
.
and 
,
,
.![\[\frac{MQ}{DQ} = \frac{MQ}{QA} = \frac{BK}{KO} = \frac{IK}{KO} \]](http://latex.artofproblemsolving.com/3/e/f/3effb95008e4b9d4ff2e24c87ade0e478c729406.png)
as desired. 
.
and 
be the circumcenters of 
and 
,
the midpoint of arc 
and 
,
.
.
is the external bisector of 
,
.
,
are parallel. Finally, since 
and 
,
be the Miquel Point of quadrilateral 
.
and 
,
and 
are isosceles, so 
and 
.
,
be points on the angle bisector of 
,
and 
.
,
with the last equality following from the incenter-excenter lemma. Therefore, 
as desired. 
,
.
the side length of 
be pairwise different points. Then 
if and only if ![\[\frac{d-c}{b-a}\in i\mathbb{R}.\]](http://latex.artofproblemsolving.com/1/e/0/1e038a3f9239c569f548d6bd877789cf771c9d3f.png)
,
and 
.
and the radius of the circle is ![\[\frac{m-a}{m_{a}}=\frac{p-n}{m_{b}}=\frac{a-q}{m_{c}}=it\]](http://latex.artofproblemsolving.com/0/d/0/0d0890ac6eb3c8f176d5c0942f60b6c602455cb2.png)
Since all three of these are the same value, ![\[\frac{m-a+p-n+a-q}{m_{a}+m_{b}+m_{c}}=\frac{m-n+p-q}{m_{a}+m_{b}+m_{c}}=it,\]](http://latex.artofproblemsolving.com/b/c/8/bc81e0b505da56a4e83d9b1538ff4331ec208c56.png)
implying that the argument of 
is perpendicular to that of 
and 
,
and 
,
is perpendicular to 
This is a relatively simple computation in terms of the side lengths of triangle 
(as NM=QP), we have![\[\ell\perp(\overrightarrow{NM}+\overrightarrow{QP})=(\overrightarrow{AM}+\overrightarrow{NP}+\overrightarrow{QA})\perp OI\quad\implies\quad\ell\parallel OI.\]](http://latex.artofproblemsolving.com/7/1/1/7115cb1955e5872b24946662934af0c94a08b517.png)
and 
.
and 
a = 1 + BN + CP 
.
,
.
we have that 
.
.
.
and 
.
.
.
and 
,
,
and we are ready.
;
as 
