AoPS Academy
positive real numbers such that 
. Prove that :
, points 
and 
are taken on the legs 
and 
, respectively, so that 
and 
. Find the acute angle between line segments 
and 
.
be real numbers satisfying 
. Find the maximum value of![\[P = \dfrac{a(b+c)}{a^2+2bc+3} + \dfrac{b(a+c) }{b^2+2ca +3 } + \dfrac{c(a+b)}{c^2+2ab+3}.\]](http://latex.artofproblemsolving.com/7/b/2/7b2ec0a87ca05d3cb29da888a4dfa73e10f238f8.png)
points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least 
of the other points in its interior.
points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least of the other points in its interior.
on the convex hull. between all other points 
, there exists a unique one such that 
is minimal (because no four are collinear). it is clear now that 
is nice.
are points such that 
such that is not an edge of the convex hull and there exists a such that 
is nice, then there is another point 
on the other side of such that 
is also nice., divide the polygon 
into two parts 
such that 
and call the circumcircle of 
, 
. it is clear that except the rest of 
lies strictly inside . now continuously change such that the side not containing shrinks and the side containing grows. we know it still contains 
. do this until touches at a third point. if we choose this point as it is clear that 
is nice.(and of course is vertex and not in the middle of an edge) so the lemma is proven. as long as the polygon isn't fully triangulated, we can add another triangle to it. note that because we are adding triangles in the parts of the polygon that were previously untouched, this actually does create a triangulation.
vertices, there exists an edge in the triangulation such that the endpoints are at least 
points apart on the perimeter both clockwise and counter-clockwise.-gon on the vertices of a regular -gon (such that the order of the points on the convex hull remains the same) so its enough to prove it for a regular polygon. which can be done by taking the longest edge in the triangulation. if it's ends are less than 
apart, then one of the two triangles on each side of it has a longer edge, a contradiction (look on the perimeter of its circumcircle) .
points! so we are done.
？ 
？ 
be real numbers. Prove that
of 
contains all the other points of ![[asy]
unitsize(2cm);
pair A = (Cos(100),Sin(100));
pair B = (Cos(220),Sin(220));
pair C = (Cos(320), Sin(320));
pair X1 = (-0.44728,-0.74601);
pair X2 = (-0.16890,-0.77087);
pair X3 = (0.20890,-0.75098);
pair X4 = (0.50717,-0.71121);
pair Y1 = (0.79549,-0.08983);
pair Y2 = (0.53699,0.69064);
pair Z1 = (-0.58647,0.75029);
dot(A);dot(B);dot(C);dot(X1);dot(X2);dot(X3);dot(X4);dot(Y1);dot(Y2);dot(Z1);
label("$A$",1.15*A);label("$B$",1.15*B);label("$C$",1.15*C);
draw(A--Z1--B--X1--X2--X3--X4--C--Y1--Y2--cycle);
draw(circle((0,0),1));
[/asy]](http://latex.artofproblemsolving.com/4/5/5/45547ca1996d95c7562d293d7205cb53fc337fd7.png)
and significance 
.
of smallest area which contains all of 
is a large triple), or there exist two distinct elements 
of 
which minimizes the undirected angle 
.
is a large triple).![[asy]
unitsize(2cm);
real x = 1.0/15;
fill(circle((20*x,0),25*x),gray);
clip(currentpicture,circle((-20*x,0),25*x),invisible);
draw(circle((0,0),1));
draw(circle((20*x,0),25*x),dashed);
pair A = (0,1); dot(A); label("$A$",1.15*A);
pair B = (0,-1); dot(B); label("$B$",1.15*B);
pair D = (-4*x,-7*x); dot(D); label("$D$",D,(-0.45,0));
label("$\mathcal{C}$",(-1.15,0));
dot((-0.1,0.6)); dot((-0.14,0.49)); dot((0.18,0.45)); dot((-0.21,0.37)); dot((0.04,0.35)); dot((0.23,0.20)); dot((-0.04,0.18)); dot((-0.17,-0.03));
label("$S$",(0,-0.3));
[/asy]](http://latex.artofproblemsolving.com/e/b/9/eb9e25b28074629cc0f2d2eaeccbb58b16a2e4d6.png)
and 
,
first).
.
not containing 
.
.
and 
,
is minimized. Then 
is also a large triple, and the old arc has been split into two new arcs.![[asy]
unitsize(2cm);
pair A = (Cos(100),Sin(100));
pair B = (Cos(220),Sin(220));
pair C = (Cos(320), Sin(320));
pair X1 = (-0.44728,-0.74601);
pair X2 = (-0.16890,-0.77087);
pair X3 = (0.20890,-0.75098);
pair X4 = (0.50717,-0.71121);
pair Y1 = (0.79549,-0.08983);
pair Y2 = (0.53699,0.69064);
pair Z1 = (-0.58647,0.75029);
dot(A);dot(B);dot(C);dot(X1);dot(X2);dot(X3);dot(X4);dot(Y1);dot(Y2);dot(Z1);
label("$A$",1.15*A);label("$B$",1.15*B);label("$C$",1.15*C);
draw(A--Z1--B--X1--X2--X3--X4--C--Y1--Y2--cycle);
draw(circle((0,0),1));
draw(circle((0,1.17905),1.976),dashed);
label("$D$",0.85*X1);
[/asy]](http://latex.artofproblemsolving.com/6/5/a/65adfd4cde038673a5b6524c62356035f62ef6c4.png)
and 
,
.
,
.
.
be a large triple with significance at most 
.
not containing 
,
.
as the two desired points in the problem statement. Indeed, any circle through 
points of 
,
which are all good triangles
,
is the smallest
is a good triangle
,
be a diagonal of 
on the other side of 
wrt 
is the smallest
,
,
points.