AoPS Academy
be the set of real numbers. Determine all functions 
such that 
and for any real numbers 
and 
,
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
be a triangle. Find all points 
on segment 
satisfying the following property: If 
and 
are the intersections of line 
with the common external tangent lines of the circumcircles of triangles 
and 
, then ![\[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]](http://latex.artofproblemsolving.com/0/8/0/080a689aea002f1b997294cb6aa48d3334881be7.png)
be positive real numbers. Prove that 
.
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
and 
, there exists 
such[/center]
is directly similar to either 
or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to , 
to , and 
to 
.
. Adding to its area the area of 
we get:![$[CRB]+[CRD]=[BCD]=R^2\ (\ 2\ )$](http://latex.artofproblemsolving.com/9/d/0/9d09cf84b632d93d68eb266f0c47d1d75612d4b9.png)
. Adding to ![$[RDS]$](http://latex.artofproblemsolving.com/2/7/e/27e6a50c0abfcae32df9fcf2b2a8d0925998515b.png)
the same ![$[CDR]$](http://latex.artofproblemsolving.com/6/4/6/6462e2fd63e60e317abb0d4831f987ba2da5dbff.png)
we get ![$[RDS]+[CDR]=[CDSR]$](http://latex.artofproblemsolving.com/3/c/6/3c68ba02871db35c8fdf828286901404cd364a77.png)
. 
having perpendicular diagonals, its area is ![$[CDSR]=\frac{CS\cdot DR}{2}\ (\ 3\ )$](http://latex.artofproblemsolving.com/6/3/9/6396ef4e79fe87988d731e007023e2e4f68aff81.png)
, so we need 
. The last relation we get from 
, done.
is the circumradius the square.
be the center of 
and let 
be the point at infinity of 
(also point at infinity of the tangent of 
at D).
![$\frac{OB}{OD} \cdot \frac{RD}{RB}=\frac{OA}{OS} \Longrightarrow OS \cdot RD=OA \cdot RB \Longrightarrow [DSR]=[ARB].$](http://latex.artofproblemsolving.com/1/1/0/1109c08461d9b93da58cc7ded47e73461e3fddfa.png)
, then 
too. Also let 
be the length of the radius of the circle.
we have 
. Consequently 
is: ![$$[ARB] = \frac{1}{2} \cdot AB \cdot BR \cdot \sin(\angle ABR)=\frac{1}{2} r\sqrt{2}\cdot r(1-tan\phi) \frac{\sqrt{2}}{2}=\frac{1}{2}r^2(1-tan\phi)$$](http://latex.artofproblemsolving.com/a/e/2/ae251524382847a6702441fef1efe00c1eb2ffe8.png)
On the other hand, in the right-angled triangle 
we have: 
, so 
is: ![$$[DSR]=\frac{1}{2}\cdot OS \cdot DR=\frac{1}{2}rtan(45^\circ-\phi)r(1+tan\phi) = \frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)$$](http://latex.artofproblemsolving.com/8/4/c/84cce5c5a32ebf85639c1bae46edc6dd2cc39fb6.png)
Finally, using the trigonometrical formula:
So the area of the triangle is:![$$[DSR]=\frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)=\frac{1}{2}r^2\frac{1-tan\phi}{1+tan\phi}(1+tan\phi)=\frac{1}{2}r^2(1-tan\phi)$$](http://latex.artofproblemsolving.com/9/d/2/9d21642f91837863d2e8444737cf6b80395a72fa.png)
Thus the areas of and are equal.
be the intersection point of lines 
and and let 
be the projection of onto line . be the circumcenter of circle . Now 
, assume that side length of square = 1.
= 
= 
= 
, 
is cyclic.
= 
= 
, implying that 
.
area of 
, 
is an isosceles right triangle. So 
= 
and 
above, ![$[DSR]=[ARB].$](http://latex.artofproblemsolving.com/a/1/c/a1cf1807f2fdd7a23d58ae91bc83e1b38dd7d710.png)
satisfying
Prove
,such that 
P
and 
.
,
.
.![\[\frac {DA}{DS}=\frac {\sin \angle ASD}{\sin \angle SAD}=\frac {\sin \angle OSD}{\sin \angle SAD}=\frac {\sin \angle PBR}{\sin \angle BPR}=\frac {RP}{RB}\,.\]](http://latex.artofproblemsolving.com/4/0/5/405fb8d8f8b8eb64a33716fbe5c92b863c68420b.png)
Calculating the power of 
.
and 
.
,
and 
have equal areas. But 
.
is tangent to 
.
the claim follows.
which is true by the claim, we are done.![[asy]
size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta;
pair O,A,B,C,D,P,R,S;
O=(0,0);B=dir(45);A=dir(135);D=dir(225);C=dir(315);path w=circumcircle(A,B,C);P=dir(100);R=extension(P,C,D,B);S=extension(P,D,A,C);
draw(A--B--C--D--cycle,deep);draw(w,deep);draw(A--C,deep);draw(B--D,deep);draw(P--D,deep);draw(P--C,deep);draw(D--S--R--cycle,org);draw(A--R--B--cycle,light);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$P$",P,dir(P));
dot("$R$",R,dir(R));
dot("$S$",S,dir(S));
[/asy]](http://latex.artofproblemsolving.com/d/e/6/de65c126817b1946dcb5e31f18111a80e915f020.png)
.
.![$[PAB]=[PSR]$](http://latex.artofproblemsolving.com/e/7/f/e7f7b0ce26638596ff2035e06f35a3d64441509c.png)
.![$[DSRC]=[PDC]-[PSR]=[PDC]-[PAB]=[DBC]$](http://latex.artofproblemsolving.com/8/7/4/874118f361646a6781f453351857dc5cd25c0ac5.png)
.![$[DSR]=[CRB]$](http://latex.artofproblemsolving.com/2/2/e/22e13fa7d195107bf8b935959ea5147e32c4e0b5.png)
,