AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and colors are given. We will say that a set of 
points on a plane is 
, if it contains exactly 2 points of each color and if lines connecting every two points of the same color are pairwise distinct. Find, in terms of the least integer 
such that: in every set of 
points of a plane, no three of which are collinear, consisting of 
points of every color there exists a subset.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
, let 
be the set of all positive integers not exceeding that are relatively prime to . Consider the polynomial![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
such that 
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
integers are such that 
is prime and 
. Prove that there exists a positive integer 
such that 
.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
, where 
, then the number of integers in the range of 
is __
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
+1 w
be reals such that 
and 
.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
satisfying 
be the diameter of a semicircle which cuts 
are the feet of the perpendiculars from 
and 
intersect at 
.
Let side BC of triangle ABC be the diameter of a semicircle which
cuts AB and AC at D and E respectively. F and G are the feet of the
perpendiculars from D and E to BC respectively. DG and EF intersect
at M. Prove that AM is perpendicular to BC.Dear Orlando,
In Hyacinthos message #9531, you wrote:
>> Let side BC of triangle ABC be the diameter
>> of a semicircle which cuts AB and AC at D and
>> E respectively. F and G are the feet of the
>> perpendiculars from D and E to BC respectively.
>> DG and EF intersect at M. Prove that AM is
>> perpendicular to BC.
Well, D and E are just the feet of the altitudes
from C and B, respectively.
Let S be the foot of the altitude from A, and let
H be the orthocenter of triangle ABC. Let the
line DG meet the line AS at M', and let the
line EF meet the line AS at M". We will show that
the points M' and M" coincide.
Since the point M' lies on the lines AS and DG,
the Menelaos theorem applied to triangle ABS
yields
AM' SG BD
--- * -- * -- = -1,
M'S GB DA
and therefore
AM' DA GB AD GB
--- = - -- * -- = -- * --.
M'S BD SG BD SG
Since EG || HS, we have GB / SG = EB / HE, so
that
AM' AD EB AD DH EB AE
--- = -- * -- = -- * -- * -- * --.
M'S BD HE DH BD AE HE
The triangles BDH and BEA are similar, so
DH AE DH EB
-- = --, and -- * -- = 1,
BD EB BD AE
so that our equation above simplifies to
AM' AD AE AD AE
--- = -- * -- = - -- * --.
M'S DH HE DH EH
Now, this expression is symmetric in B and C;
hence, by analogy,
AM" AD AE
--- = - -- * --,
M"S DH EH
and the points M' and M" coincide. Thus, the
lines DG, EF and AS meet at one point M' = M" = M,
and AM is perpendicular to BC.
There should be a simpler proof.
Thanks for this and all the other problems, many
of them are still too difficult or non-standard
for me.
Friendly,
Darij GrinbergDear Darij,
here is a shorter solution
to Hyacinthos message #9531 --
> >> Let side BC of triangle ABC be the diameter
> >> of a semicircle which cuts AB and AC at D and
> >> E respectively. F and G are the feet of the
> >> perpendiculars from D and E to BC respectively.
> >> DG and EF intersect at M. Prove that AM is
> >> perpendicular to BC.
Let D', E' be reflections of D, E in BC, N the intersection of DE' and D'E.
Obviously, N lies on BC and MN is parallel to DD' and EE', or perpendicular
to BC (because M and N divide DG and DE', resp., in the same ratio
DF:EG=DD':EE'). So it remains to show that, in your notation, N=S, the foot
of the altitude from A, or, equivalently, that angle NEC = angle ABC..
But angle NEC = angle D'EC = angle DE'C = angle DBC. QED
Best,
Vladimir Dubrovsky
, intersects 
at .quadrilateral 
is perfect,, to circle..suppose 
intersects the circle at points 
are parallel , then quadrilateral 
is a trapezoid, and
also meet , at ., is also perfect, and the polar of , to circle , goes
.
, it is clear that it lies on ., to circle, is line 
, which is. 
and 
, on the two
is a trapezoid,and because of the symmetry of the
, turn into each other with a respect to line , and it means that
is perpendicular to . 
: is perpendicular to . And is also perpendicular to .
are collinear, and 
is perpendicular to .
and 
are the altitudes of the triangle 
so let them intersect at the orthocentre 
of 
Now join 
Let 
and 
Then 
this leads to 
implying 
so that we have
so that 
But, 
which, from Thale's theorem, gives 
Since 
we note that 
ie 
lies on 
So is the -altitude of 
is cyclic, 
and is on the polar of (Brocard) perpendicular to and let it cut , at , 
, that is, (,
,)
(
,
,)
is the polar of , 
is the polar of and 
,although it noway explians why result is true,it admits inelegent but short co ordinate sol,take A-0,1,B-b,oand C-c,o.