[/quote]
,
,
be the orthocenter of 
at 
at 
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
be positive real numbers satisfying 
.![\[
3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.
\]](http://latex.artofproblemsolving.com/f/3/c/f3cd114f909f388eacee142d3f90d07f6b149131.png)
be the maximal number of the vertices of a convex lattice polygon with exactly 
lattice points in the interior. Show that:
for 
.
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)
be reals such that 
and 
.
and a set 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
be real numbers such that 
.
Let 
.
and 
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be a convex quadrilateral. Suppose that the circles with diameters 
intersect at points 
.
and 
.
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
.
?
Y by anantmudgal09, canhhoang30011999, Adventure10, Mango247
Consider a triangle 
and a point 
on its side 
. Let 
be a point inside 
on the angle bisector of 
. The second intersections of lines 
and 
with circle 
are 
and 
, respectively. Similarly, the second intersection of lines 
and with circle 
are 
and 
, respectively. Show that if 
and 
, then lines 
, 
and 
pass through a point or are parallel.
and a point 
on its side 
. Let 
be a point inside 
on the angle bisector of 
. The second intersections of lines 
and 
with circle 
are 
and 
, respectively. Similarly, the second intersection of lines 
and with circle 
are 
and 
, respectively. Show that if 
and 
, then lines 
, 
and 
pass through a point or are parallel.
intersects 
intersects 
at 
and 
.
.
are collinear.
are collinear.
,
which leads to 
concurrent.
be the intersection of the tangent at 
Since 
,
lie on a circle
be the internal bisector of angle 
with 
.
and vary 
.
remain fixed but 
vary and so 
are projectivities. Define 
and 
.
and 
are also projectivities, we need to show thus for exactly three positions of 
the result is obvious. Our third choice is 
.
now and vary 
parallel to 
respectively. Note that 
and 
.
are affine maps and same for 
.
where 
lie on 
.
this solution ? I'm really new to this projective thingy
and animate 
.
is a projective map, and simiarly 
is projective. To show that these points coincide, we need to check for three points, viz:
,
.
.
.
,
,
![[asy]
unitsize(2inches);
pair C=dir(100);
pair A=dir(140);
pair B=dir(40);
pair X=extension(C,incenter(A,B,C),A,B);
pair D=0.7*B+0.3*A;
pair U=extension(C,X,A,(A-C)*(D-A)/(D-C)+A);
pair V=extension(C,X,B,(B-C)*(D-B)/(D-C)+B);
pair Q=2*foot(circumcenter(A,C,D),C,X)-C;
pair S=2*foot(circumcenter(B,C,D),C,X)-C;
pair I=0.4*C+0.6*X;
pair P=2*foot(circumcenter(A,C,D),A,I)-A;
pair R=2*foot(circumcenter(B,C,D),B,I)-B;
draw(A--B--C--cycle);
draw(S--Q);
draw(A--U);
draw(B--V);
draw(A--P);
draw(B--I);
draw(circumcircle(A,C,D));
draw(circumcircle(B,C,D));
dot("$A$",A,dir(A));
dot("$B$",B,dir(55));
dot("$C$",C,dir(50));
dot("$X$",X,dir(230));
dot("$D$",D,dir(240));
dot("$U$",U,dir(0));
dot("$V$",V,dir(190));
dot("$Q$",Q,dir(270));
dot("$S$",S,dir(120));
dot("$P$",P,dir(260));
dot("$R$",R,dir(270));
dot("$I$",I,dir(130));
[/asy]](http://latex.artofproblemsolving.com/a/0/9/a09d05ee3c134f75224619f74bd5b5af2d8873ad.png)
,
to be the signed ratio 
.
,
.
,
.
,
,
,
are collinear, so 
,
,
,
and 
concur on ![\[(PA;CD)\stackrel{P}{=}(PP\cap CD,AI\cap CD;C,D)\]](http://latex.artofproblemsolving.com/c/6/b/c6b8f61300f3a4106c873817958c2815bfdeb720.png)
and![\[(RB;CD)\stackrel{Q}{=}(RR\cap CD,BI\cap CD;C,D).\]](http://latex.artofproblemsolving.com/7/8/3/7834adfb70eac36cbe401eb9c353db78834b13d7.png)
Thus,![\[\frac{(PA;CD)}{(RB;CD)}=\frac{(PP\cap CD;C,D)}{(RR\cap CD;C,D)}\cdot(BI\cap CD,AI\cap CD;C,D)\stackrel{I}{=}\frac{(PP\cap CD;C,D)}{(RR\cap CD;C,D)}(BA;XD).\]](http://latex.artofproblemsolving.com/a/3/a/a3aa0de2f111dcb214c48ec923731ada7951ba4e.png)
However, we also have![\[(PA;CD)\stackrel{A}{=}(I,AA\cap CX;C,X)\]](http://latex.artofproblemsolving.com/f/e/8/fe8dadbc092c847580087d6395d3a0a7b270f56d.png)
and![\[(RB;CD)\stackrel{B}{=}(I,BB\cap CX;C,X),\]](http://latex.artofproblemsolving.com/0/b/7/0b74b507eba2fd109988f0f1aa87cfc68cb404d3.png)
so by similar logic,![\[\frac{(PA;CD)}{(RB;CD)}=(VU;CX).\]](http://latex.artofproblemsolving.com/5/c/2/5c2291f49075f3e6547eeb63b0fd82cd3ea80758.png)
Our goal is to show that 
,
,
and point 
,![\[(V;CX)=\frac{CB}{XB}\cdot\frac{\sin\angle CBV}{\sin\angle XBV}=\frac{CB}{XB}\cdot\frac{\sin\angle CDB}{\sin\angle DCB},\]](http://latex.artofproblemsolving.com/6/1/6/61641ce3923c4a708e18df515b7b5081ee378de2.png)
and similarly![\[(U;CX)=\frac{CA}{XA}\cdot\frac{\sin\angle CDA}{\sin\angle DCA}.\]](http://latex.artofproblemsolving.com/7/b/1/7b12fedcaf6a2e94bf48d06d58e3192590b168d0.png)
Since 
by the angle bisector theorem, we have that![\[(VU;CX)=\frac{\sin\angle DCA}{\sin\angle DCB}=\frac{AD}{DB}\div\frac{AC}{BC}=\frac{AD}{DB}\div\frac{AX}{BX}=(BA;XD),\]](http://latex.artofproblemsolving.com/0/b/b/0bb6a1ce5c54fd73c0aa1121be6c8c6a30f1337b.png)
as desired.
and animate 
bisector (which we denote by 
)
are fixed and 
move as a function of 
Denote 
and 
is projective, and 
is also projective. It suffices to check these maps coincide for three positions of 
is obvious.
is obvious.
in the direction of 
Note we have 
and 
In particular, we have 
is fixed, so the map 
where the last map was induced by projection about 
Similarly, 
is fixed, so we have 
where the last map was induced by projection about 
It suffices to check these maps concur for three positions of 
But this isn't hard to do. Take 
and 
