)
.
passes through 
and 
.
and circumcircle of 
and 
be the midpoint of 
be the midpoint of 
.
,
and 
in the array can be changed into either 
or 
so that the row-sums and column-sums remain unchanged. (Note that 
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points 
a non-negative real number 
,
,
.
such that 
and 
are connected by an edge for every 
.
is not fixed and the first selected vertex always has to be 
.![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
,
for which Bobby has a winning strategy.
,
,
,
,
,
be positive integers and let 
.
divides 
and 
.
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
and 
,
and 
meet at 
and 
meet at 
Let 
be the point of intersection of the lines 
and 
be the point of intersection of the lines 
and 
Prove that 
is the circle that is tangent to the line segment 
,
be positive reals such that 
.
such that:
and 
: Prove that :
Y by baopbc, anantmudgal09, mymathboy, Mathuzb, AlexGr, Adventure10, Mango247, Rounak_iitr, Tastymooncake2
In the non-isosceles triangle 
,
is the midpoint of side 
,
is the midpoint of side 
,
is the midpoint of side 
.The line(different from line ) that is tangent to the inscribed circle of triangle and passing through point intersect line 
at 
.Define 
similarly.Prove that 
are collinear.
,
is the midpoint of side 
,
is the midpoint of side 
,
is the midpoint of side 
.The line(different from line ) that is tangent to the inscribed circle of triangle and passing through point intersect line 
at 
.Define 
similarly.Prove that 
are collinear.
....
,
the points of tangency of 
and 
and 
,
(
and 
hence 
is tangent to 
and define 
,
,
,
.
with line 
we get: 
On the other hand: 
,
we get 
similarly 
and 
which is the converse of Menelao's theorem in 
with line 
,
are collinear.
in 
.
be the midpoints of 
and let 
.
be the touch point of 
.
so is 
and so 
and hence as 
is the A-Nagell cevian in the medial triangle so is 
in 
.
,
is 
and so on the radical axis of point 
.
be the intouch triangle of 
and let 
be the intersection of 
By Newton's theorem we get 
are concurrent, so 
Let 
be the Nagel point of 
be the intersection of 
It's well-known that the reflection 
of 
in 
lies on 
respectively, so 
hence we conclude that 
lies on the trilinear polar of 
be the medial triangle of 
to the incircle 
and 
lie on 
respectively. Evidently, 
It is sufficient to show that 
are collinear. Let 
be the antipode of 
.
and observe that 
in standard notations. Since 
we have 
Finally, we have 
Taking the cyclic product, the result follows by the converse of Menelaus' theorem.
This is true since 
and 
where 
is the 
be the triangle formed by the three tangents. By Desargues' Theorem, it suffices to prove that 
,
concur. We claim that they all pass through the Nagel point 
of 
and incenter 
be the reflection of 
,
,
intersect the incircle again at 
,
,
.
,
,
concur at 
,
,
are tangents to the incircle. Note that the tangents to the incircle at 
and 
intersect at 
.
.
,
is the centroid of 
as 
in the ratio 
.
must be the midpoint of 
and we are done.
be the orthocenter of 
,
,
similarly. Suppose 
,
;
is concurrent with its cyclic variations. To prove this, since 
and cyclic variations are obviously concurrent at 
where the first equality follows from Power of a Point. Call this concurrency point 
,
and 
be the reflection of 
respectively. Denote by 
and 
are collinear and 
from which we conclude that 
and 
be the reflection of 
Notice that![\[\measuredangle XVU=\measuredangle DP_2U=\measuredangle P_2UD=\measuredangle VUX \implies XV=XU. \]](http://latex.artofproblemsolving.com/7/f/f/7ffd6ce3f64951bfeed255cfd8937ce1b798ffa3.png)
Hence 
which amounts to 
is the angle bisector of 
Therefore 
The statement then follows by Menelaus' theorem in ![[asy]
size(8cm); defaultpen(fontsize(9pt)); pathpen=black; pointpen=black;
pair A=dir(55);
pair B=dir(-140);
pair C=dir(-40);
pair D=midpoint(B--C);
pair E=midpoint(A--C);
pair F=midpoint(A--B);
pair I=incenter(A,B,C);
pair P=foot(I,B,C);
pair P2=2*D-P;
pair P1=2*I-P;
pair U=OP(incircle(A,B,C),P1--P2);
pair V=IP(E--F,P1--U);
pair X=extension(E,F,D,U);
pair X1=2*X-V;
D(A--B--C--cycle,purple+linewidth(1.1));
D(incircle(A,B,C),dotted+red);
D(A--P2,cyan);
D(D--X,dashed);
D(X1--E,red+linewidth(1.2));
D(X1--P--P1);
D(E--U--F);
D(rightanglemark(P,U,P1,2),deepgreen);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$I$",I,dir(I));
dot("$P$",P,dir(P));
dot("$P_1$",P1,dir(120));
dot("$P_2$",P2,dir(P2));
dot("$U$",U,dir(-160));
dot("$V$",V,dir(135));
dot("$X$",X,dir(90));
dot("$X^*$",X1,dir(90));
[/asy]](http://latex.artofproblemsolving.com/c/5/7/c57e6310b446d10a37775e65ef2cff3ed3e7868e.png)
are not on the incircle, so be aware of the fact that the property 
does not hold.
.
similary.
are concurrent
concur.
be the incircle of 
be the touchpoints of 
,
be the antipode of 
be the reflection of 
.
and 
are homothetic with homothety center 
sends 
,
and 
are also homothetic with homothety center 
.
,
)
are collinear. Now let 
.
.
and so 
is the touchpoint of the tangent 
.
are collinear.
be the 
be the antipode of 
,
.
.
is a parallelogram, we have that 
lies on the radical axis of the incircles of 
the tangency points and 
the second tangency points from 
difine 
similary
are collinear is equivalent to 
are concurrent which is equivalent to 
are concurrent
similary 
denote the orthocenters of 
,
.
and 
and 
respectively. Note that it is well known that 
is the pole of 
are concurrent. Note that 
are collinear along the polar of 
are concurrent at 
are concyclic, hence we conclude that 
.
![[asy]
size(10cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A,B,C);
pair A1 = foot(I,B,C);
pair E = (A+C)/2;
pair F = (A+B)/2;
pair D = (B+C)/2;
path inc = incircle(A,B,C);
pair A4 = B+C-A1;
pair A3 = foot(A1,A,A4);
pair A2 = 2*I-A1;
pair X = extension(D,A3,F,E);
draw(A--B--C--A--cycle);
draw(inc);
draw(E--F);
draw(F--X);
draw(D--X);
draw(A--A4);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$I$",I,dir(I));
dot("$X$",X,dir(X));
dot("$A_1$",A1,dir(A1));
dot("$A_2$",A2,dir(A2));
dot("$A_3$",A3,dir(A3));
dot("$A_4$",A4,dir(A4));
[/asy]](http://latex.artofproblemsolving.com/b/3/d/b3d28ddb8bb255903344cdd81ed1ef5cd002b656.png)
be the point of tangency of the incircle with side 
be a diameter of the incircle. Let the other tangent from 
.
.
meets the side 
.
Note that the displacement vector 
is given by
be a displacement vector. Then a displacement vector perpendicular to 
In this case, a displacement vector perpendicular to 
is given by
Simplifying, this is equal to
Also, 
.
is
Substituting this into line 
,
From this,
Similarly,
From this, we get that 
is
Since 
,
Similarly,
It remains to verify that
But this is true as the sum of the three rows is the zero row.
respectively, the aforementioned tangents from 
touch the incircle at 
respectively, 
be the orthocenter of 
,
,
.
is the polar of 
,
passes through 
and 
,
.
,
be the orthocenters of 
respectively, then analogous reasoning implies 
is the polar of 
and 
is the polar of 
.
is the polar of 
lies on the polar of 
,
are the polars of 
.
concur.
are collinear.
is the polar of 
and 
are also sets of collinear points.
concur at 
,
be the antipode of 
.
intersects 
,
,
touches 
is a harmonic quadrilateral, so 
Since 
,
,
,
is the inradius of 
,
,
.
.
and 
.
so we are done by Menelaus' Theorem on triangle 
WRT line 
,
,
to be incircle touchpoints, 
,
to be second tangency points to incircle from the midpoints. The pole of 
,
,
,
.
is the same as 
.
.
,
are concurrent but this is true by Carnot's theorem.
respectively. If 
concur then so does 
concur
,
,
concur. If we manage to prove this, we can do the same thing as the original proof.
.
,
are concurrent, we must have 
,
and 
collinear, hence the generalisation is proved.
.
be the antipodes of 
meet the incircle at 
for second time. Note that 
are tangent to incircle by homothety. In order to use menelaus,![\[\Pi{(\frac{XN}{XP})}=\Pi{(\frac{\frac{\sin NMT_D.MN}{\sin MXN}}{\frac{\sin PMT_D.MP}{\sin MXN}})}=\Pi{(\frac{\sin MNT_D}{\sin PMT_D})}.\Pi{(\frac{MN}{MP})}=1\]](http://latex.artofproblemsolving.com/d/d/b/ddb375044afccf946306f2fd6e9d947a459f058c.png)
We observe that 
.![\[\Pi{(\frac{\sin MNT_D}{\sin PMT_D})}=\Pi{(\frac{\sin FIT_D}{\sin EIT_D})}=\Pi{(\frac{T_DF}{T_DE})}=\Pi{(\frac{\sin FD'A}{\sin AD'E})}=\Pi{(\frac{AF.\frac{\sin \frac{B}{2}}{AD'}}{AE.\frac{\sin \frac{C}{2}}{AD'}})}=1\]](http://latex.artofproblemsolving.com/f/0/3/f03abedb5206e28bb324052e13e3262dc21fb1d5.png)
As desired
.
and 
.
is tangent to 
,
.
and similar.
.
and 
for lack of better names. Then by Iran Lemma, we have 
.
and 
is the 
and hence 
.
and 
.
,
,
concurrent and now see that 
Now we are done by converse of Ceva. Taking the dual of this, we get 
