[/quote]
,
be the foot of altitude dropped for 
.
intersects with the said circles at 
and 
,
in the plane is given so that 
.
and 
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers 
such that all unit squares on the board can be made into all black after a finite number of moves.
and 
Show that
Let 
Show that
be a square of center 
and let 
with respect to point 
and 
,
and 
.
is the angle bisector of 
.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these 
?
Prove that
vertices, each edge has one of the colors 
,
,
.
,
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
such that, for all 
,
such that for all integer 
and a subset ![$[n],$](http://latex.artofproblemsolving.com/d/2/7/d27d2715f4f95c24a28b7c6371a006d4c803b2cf.png)
satisfy 
for all 
then 
be the set of positive integers. Find all functions 
such that for all 
,![\[m^2f(m) + n^2f(n) + 3mn(m + n)\]](http://latex.artofproblemsolving.com/0/f/f/0ff783fa217b7be4bda5006e417f1471ea3a6fca.png)
is a perfect cube.
and 
.
![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
be the set of all polynomials with real coefficients. Find all functions ![$f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/3/3/033be3ab043bd47e537f01791c813739ad5c7da5.png)
satisfying the following conditions:
maps the zero polynomial to itself,![$P \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/f/7/c/f7c4c970908aa5570a9afcc45fc16c1d36cc3d5d.png)
,
,![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
,
and 
have the same set of real roots.
Y by mathematicsy, samrocksnature, HamstPan38825, jhu08, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with incenter 
, and suppose the incircle is tangent to 
and 
at 
and 
. Denote by 
and 
the reflections of and over . Let 
be the intersection of 
with 
, and let 
be the midpoint of . Prove that 
and 
are perpendicular.
be a triangle with incenter 
, and suppose the incircle is tangent to 
and 
at 
and 
. Denote by 
and 
the reflections of and over . Let 
be the intersection of 
with 
, and let 
be the midpoint of . Prove that 
and 
are perpendicular.
denote the reflection of 
the polar of 
,
and 
on 
is the Nagel point of 
be the point on the incircle so that the tangent at 
.
is the pole of 
.
(just extend 
be the reflect point 
intersect the incircle at 
now we have 
so we have that 
is tangent to the incircle where 
is the reflect of 
so we are done 
where 
equals ![$[(s-b)(s-2c)/(s-c)(s-2b) -1]^{-1}$](http://latex.artofproblemsolving.com/d/5/d/d5d2a6ed42b32188430065191a5593b471b1be06.png)
.
= 
we are done.
and let the incircle be a unit circle.
,
and 
,
,
,
and
p
A
wrt incircle , 
M
,
M
, 
.
equal to inradius of 
.
map to 
and 
map to 
.Consider a homothety of ratio 
sending 
and 
.Note that 
and 
denote the image of 
.
.
is the radical axis of 
.
,
.But this is obvious from the fact that 
are isotomic conjugates wrt 
.
be the intersection of tangents from 
wrt the incircle.Obviously 
are collinear.To prove this extend the before mentioned tangents to cut the sides 
,
is a rhombus thus 
is the angle bisector of 
so we need to prove that 
is the angle bisector in 
which is not hard (length chasing by parallel sides)
![[asy]
unitsize(3cm);
pointpen=pathpen=black;
pointfontpen=fontsize(10pt);
pair A=D("A",dir(110),NW), B=D("B",dir(200),SW), C=D("C",dir(-20),SE), I=D("I",incenter(A,B,C),NE);
path uc=D(incircle(A,B,C));
pair D=D("D",IP(uc,D(B--C))), E=D("E",IP(uc,D(C--A)),NE), F=D("F",IP(uc,D(A--B)),NW), G=D("G",2*I-E), H=D("H",2*I-F), Q=D("Q",extension(B,C,G,H),W), M=D("M",(B+C)/2);
D(B--Q--I--M);
[/asy]](http://latex.artofproblemsolving.com/c/6/e/c6e6172d2db35d81a4530b03bcc64c92d42b94de.png)
and 
,
Now use the chord intersection formula to find 
so 
and 
so 
,
.
be the midpoint of 
is cyclic 
-(i) Also, angle chasing tells that: 
-(ii) 
from (i, ii) we have: 
,
and 
meet 
at 
and 
,
lies on a circle with center 
meet 
is a rectangle we have that 
,
be the midpoint of arc 
intersect the circumcircle of 
intersect 
,
circle, we have that 
is perpendicular to 
.
and 
are spirally similar - this is sufficient. Note that since 
,
is cyclic and thus 
.
.
.
and 
.
Thus, 
and we are done.
so that 
(the third tangency point of the incircle and 
.
so that 
because then 
concur and thus 
is tangent to 
is a kite so 
.
.
.
is collinear with 
,
is harmonic. Letting 
is harmonic. However, 
is harmonic by definition! So 
and we're done.
by construction, 
,
.
ratio homothety at 
ratio homothety at 
lie on the unit circle, so 
and 
.
Further, 
,
It suffices to show that
is pure imaginary, but this is clear since its conjugate is its negative:
is tangent to the incircle. So 
is the polar of 
be the reflection of 
is a diameter.
is a tangent, and since 
.
,
,
be this point of tangency, so that 
.
,
,
to be the reflection of 
.
,
.
,
,
and thus 
,
.
,
.
be the second intersection of 
and 
be the unique point on 
is tangent to 
,
are all tangents to 
,
and 
are both harmonic, forcing us to have 
.
are both tangent to 
.
,
to 
,
to 
.
,
.
,
,
.
.
hit 
and 
,
,
e
but 
too which makes it clear that 
.
which implies 
thus 
.
is the polar of 
.
but as 
is tangent to the incircle therefore 
.
is the polar of 
.
and 
we have 
a parallelogram. Thus 
which implies 
.
was parallel to 
.
and 
we get 
is polar of 
with incircle then as 
we get 
which give us 
is 
and by La Hire's, 
and 
,
Hence 
is the negative of its conjugate and thus imaginary, as needed.
.
be the polar of 
finishes. 
clearly 
.
by homothety at 
is the polar of 
.
because 
.
is tangent to the incircle. Now since 
and therefore we are done.
be unit circle on Argand plane and let 
,
.
.
.
,
:
