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.
different positive integers , less than 
,
Y by Adventure10, HWenslawski, Mango247, and 1 other user
Let 
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides 
and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex 
is denoted by 
. Prove that 
.
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides 
and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex 
is denoted by 
. Prove that 
.
and 
the points where 
and 
the points on sides 
and 
,
and 
.
.
:
.
,
,
.
,
to 
.
;
can be found easily using the relation 
.
.
to points 
,
.
.
.
.
,
be its incircle. Denote by 
intersects segment 
.
. Therefore, 
.
. From the relations 
.
Denote 
.Prove easily that 
and
. In conclusion, 
, i.e. 
.
which is new for me, I just remarqued it from the diagram and proved it, and I think if we knew this result, the problem becomes easy.
touch side 
,
,
and 
,
.
that is cutted by the line 
,
then 
.
,
.
and 
thus 
,
,
,
call this result 
.
, 
and 
and 
and 
the area of 
.
,
,
and 
,
From the law of cosines, 
replacing on the previous result we get: 
from which we deduct that: 
,
we have 
then 
.
then 
from which follow that 
the point 
and hence 
It suffices to prove that 
and from here, I am not able to complete the proof. If someone knows the length of the nagel point from the vertex...
is the incenter of its antimedial triangle 
and orthogonal projections 
of 
on 
Let the incircle 
of 
at 
and let 
Since 
is the 
-
it follows that 
quadrilateral 
is a parallelogram, hence 
since 
Then 
If the projections of 
on 
as desired.
,
,
be 
.
and the semi-perimeter 
.
,
,
,
.
,
,
,
.
and of 
.
.
,
,
inside the triangle on the line 
.
of the way "up" the line segment from 
of the way "up".
-
.
,
,
,
.![$a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0$](http://latex.artofproblemsolving.com/f/e/e/fee79731f6129c50cd87d8355785c61d75acc13a.png)
,
.![$a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0$](http://latex.artofproblemsolving.com/e/c/5/ec575b0ecd5e13fd50d4faab534f13bf0f18b628.png)
.
,![$a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0$](http://latex.artofproblemsolving.com/2/2/9/229335fdf65dc82fbf88a5941509d193690f09f2.png)
.
.
![$2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0$](http://latex.artofproblemsolving.com/8/b/8/8b82183266bb2fc902da1ba9cde4d13afe90915b.png)
.
gives
,
,
,
(
.
bisects 
& 
(
s
(1)
:
bisects 
and 
so 
be such that 
and 
.
.
is the medial triangle of 
be the Nagel point of 
.
is a diameter of circle 
is a parallelogram and 
,
.
at 
and 
,
is a parallelogram. Thus 
.
,
,
,
as the area of 
is a diameter of 
,
is isotomic to 
such that 
,
and 
,
.
.
and 
,![\[ \frac{QD_2}{2\frac{K}{s}}=\frac{AD_2}{2\frac{K}{a}}\implies\frac{QD_2}{a}=\frac{AD_2}{s}=\frac{AD_2-QD_2}{s-a}=\frac{AQ}{s-a},\]](http://latex.artofproblemsolving.com/4/0/8/408335066d8839c9994472b7741d701026d28b6e.png)
.
.
and 
are tangents, we have 
.
and 
,![\[ \frac{D_2P}{\sin\angle{E_2BC}}=\frac{BD_2}{\sin\angle{D_2PB}}\text{ and }\frac{AP}{\sin\angle{AE_2B}}=\frac{AE_2}{\sin\angle{APE_2}},\]](http://latex.artofproblemsolving.com/0/4/c/04c3952d79169310934808cd8131552cd68fbfa4.png)
and 
,![\[ \frac{D_2P}{\sin\angle{E_2BC}}=\frac{AP}{\sin\angle{CE_2B}}.\]](http://latex.artofproblemsolving.com/e/7/2/e72ac32e98f175c7677c97dec88a933536c434f6.png)
,
,![\[\frac{D_2P}{CE_2}=\frac{AP}{a}=\frac{AP+D_2P}{a+CE_2}=\frac{AD_2}{a+CE_2}.\]](http://latex.artofproblemsolving.com/9/3/7/937e4289d801d78db6377efa3143c4dbfb361656.png)
.![\[ \frac{D_2P}{s-a}=\frac{AD_2}{a+s-a}\implies \frac{D_2P}{AD_2}=\frac{s-a}{s}.\]](http://latex.artofproblemsolving.com/b/a/6/ba6c059052e92251eb4a23f6687933428ead5702.png)
,
be the excircle of 
are collinear.
is the midpoint of 
, 
is the diameter and 
is the center of 
by using midpoint theorem in 
and 
(Nagel point ) 
(
we get that 
.
so 
,
.
Also, 
since 
Hence, 
,
.
is the Nagel point. Then, it suffices to show that 
,![\[\frac{-\frac12 a + \frac12 b + \frac12c}{\frac{a+b+c}{2}} + \frac{2a}{a+b+c} = 1 + 0\]](http://latex.artofproblemsolving.com/b/8/2/b821636612799cf43f10c09084852569623530ab.png)
which is true, so 
and we're done. 
Note that this implies 
Thus, 
This means that 
Also, since 
It is clear that 
since 
Therefore, 
as desired.
,
is the midpoint of 
and 
Thus, 
Also, 
are collinear with 
,
which implies result.
,
Some computation now gives 
,
gives 
,
,
,
by ratio lemma. Then we use ratio lemma again, get 
,
from 
to 
,
,
,
,
,
,
,
,
with respect to 
,
.
so the result follows.
.
formed by it and 
has similarity ratio 
from ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
,![$$k=\frac{AQ}{AD_2}=\sqrt{\frac{[\triangle]}{[ABC]}}=\sqrt{\frac{k(s-a)}{s}} \implies \frac{AQ}{AD_2}=\frac{s-a}{s}.$$](http://latex.artofproblemsolving.com/3/a/e/3ae5e08d39765b97afbdb4447b6d5d93b9b68b2a.png)
On the other hand, in barycentric coordinates is clear that 
and 
,
,
Also, by a homothety at 
is the projection of the 
,
which implies the conclusion.
the semiperimeter, inradius, and 
e
by 
respectively.
are the 
extouch points respectively (i.e. where the 
excentres are tangent to 
)
i
.
concur due to Ceva's theorem and the existence of the incentre. Therefore, from Van Aubel's theorem and some well-known lengths, we get 
so 
Also, it is well known that the homothety at 
where the final part follows by the exradius length formula given in EGMO Chapter 2.
and the result follows. 
and etc. Now, we know that 
and 
.
and 
.
since it is the intersection point of the cevians 
.
.
.![$[BQC]=2[BIC]$](http://latex.artofproblemsolving.com/3/c/6/3c6a8749102d9bd0b1b9d150c7c55ff74ac74223.png)
which means,
Thus, we conclude that 
.
,![$[ABQ] = [PBD_2]$](http://latex.artofproblemsolving.com/1/7/d/17dc387091b372198c17b01dcbaa55710e7fa8d3.png)
.![\[\begin{vmatrix}
0 &1 & 0 \\
1 & 0 & 0\\
\frac{2a^2}{a^2+ab+ac} & \frac{c^2 - (a-b)^2}{2(a^2+ab+ac)} & \frac{b^2 - (c-a)^2}{2(a^2+ab+ac)}
\end{vmatrix} = \frac{b^2 - (c-a)^2}{2a^2+ab+ac} = \frac{b^2 - (c-a)^2}{2a(a+b+c)}\]](http://latex.artofproblemsolving.com/a/3/7/a37c1016faa3a62941b6f082a6ff1686a1b41355.png)
and![\[\begin{vmatrix}
0 &1 & 0\\
0 & \frac{a-b+c}{2a} & \frac{a+b-c}{2a}\\
\frac{-a+b+c}{a+b+c} & \frac{a-b+c}{a+b+c} & \frac{a+b-c}{a+b+c}
\end{vmatrix} = \frac{(a+b-c)(-a+b+c)}{2a(a+b+c)} = \frac{b^2- (c-a)^2}{2a(a+b+c)}\]](http://latex.artofproblemsolving.com/6/4/5/64553a139558f9ba6c79a5495ab36425796e73d0.png)
so we are done.
.
denote the semiperimeter of the triangle.
gives:
,
,
and 
.
From 
and 
we can compute: 
.
and doing the calculations we get that the problem is equivalent to: 
which is true, as desired. 
.
