[/quote]
,
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that there exist positive integers 
for which the following holds 
for all 
.
is a perfect cube
of a prime number 
and a positive integer 
for which 
is an integer.
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
,
and 
at 
and 
,
and 
meet at 
,
and 
meet at 
Let 
be the point of intersection of the lines 
and 
be the point of intersection of the lines 
and 
Prove that 
is the circle that is tangent to the line segment 
,
.
such that if 
and 
are distinct rational numbers satisfying 
or 
,
?
and 
.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
and 
intersect at two points 
and 
.
be the line tangent to these circles at 
and 
,
be the line parallel to 
on 
on 
and 
meet at 
;
and 
;
and 
.
.
is such that 
for any positive integers 
.
for all positive integers 
.
.
satisfy that 
.
Y by ahmedosama, jishu2003, Adventure10, Mango247, and 1 other user
Let 
be a point in the interior of a triangle 
, and let 
be the point of intersection of the line 
and the side 
of the triangle, of the line 
and the side 
, and of the line 
and the side 
, respectively. Prove that the area of the triangle must be 
if the area of each of the triangles 
and 
is 
.
be a point in the interior of a triangle 
, and let 
be the point of intersection of the line 
and the side 
of the triangle, of the line 
and the side 
, and of the line 
and the side 
, respectively. Prove that the area of the triangle must be 
if the area of each of the triangles 
and 
is 
.This post has been edited 1 time. Last edited by syk0526, Apr 4, 2012, 6:43 AM
![$a = [PDC], b = [PAE], c = [BFP]$](http://latex.artofproblemsolving.com/8/4/2/84278d17e64c0cd430f938f0ad70686f1201a764.png)
.
is the only positive root except a root around 
of the cubic. You can bound the root between 
and 
and hence it is not actually a solution for 
.
as well and therefore ![$[ABC] = 6$](http://latex.artofproblemsolving.com/e/9/1/e9115efe128e2291f088c58a62044589d436fdbf.png)
.
,
,
,![\[b+b^2 = ab + 1\]](http://latex.artofproblemsolving.com/6/2/1/621132602f204abfe23ff064575f14b2cc04f729.png)
![\[c+ c^2 = bc+1\]](http://latex.artofproblemsolving.com/6/a/e/6aee0d84e697d79878ce1a0c85b919ff366482e9.png)
![\[a +a^2 = ac + 1\]](http://latex.artofproblemsolving.com/7/e/e/7ee439bd91d031e22d06b4061758713965b52ac2.png)
are also roots so I factored it then, so it sufficed to consider the cubic term. The cubic has three real roots (easy to verify) and there is only one positive root which happens to be between 0 and 1. Then guessing I found the root was less than 
,
and 
because of 
.
is a point inside triangle 
meet the opposite sides at 
respectively. If the areas of three of the six triangles so formed are equal then 
,
Its easy to see that 
.
.
.
.I feel sleepy so I will continue tommorow anyone wanting to modify my work is welcomed.
-
to solve the problem proposed by @synthetic_potato but.....
![$[PFA]=[PDB]$](http://latex.artofproblemsolving.com/e/4/e/e4ef9fdb76edfe16a0d47e144901fbf1bf28d11a.png)
,
and cyclic relation holds. Now we resort to barycentric coordinates w.r.t. 
.
.
and 
.
Cyclic relations hold so 
.
,
,
,
causes contradiction hence 
.
![$[PCD] = x, [PAE] = y, [PBF] = z$](http://latex.artofproblemsolving.com/1/8/1/18104c9a488c05405e1041511830c73377effd31.png)
.
.
,
(similarly, 
and 
)
,
.
and 
,
and 
.
,
,
as above. We have 
and similar equalities by taking ratio of areas. We also have 
by Ceva's theorem. Adding up all the similar equalities we obtain 
and then AM-GM gives 
![$x=[\triangle APE], y=[\triangle BPF], z=[\triangle CPD]$](http://latex.artofproblemsolving.com/1/c/c/1ccd8ff74cd226a0f10222b65108b2fa6008fa7c.png)
.
.![$\frac{[\triangle CPE]}{[\triangle CPB]}=\frac{[\triangle APE]}{[\triangle APB]}$](http://latex.artofproblemsolving.com/b/5/3/b53d5e33a197a5bb204f63036c03905d5605d617.png)
,
or 
.
is one, then the others are clearly one. Otherwise, by multiplying cyclically the last equation, we get 
,
then 
Divide the second equation by 
so that 
Let 
Then, this becomes 
By AM-GM, ![$$3=a+b+c\geq 3\sqrt[3]{abc}=3,$$](http://latex.artofproblemsolving.com/0/4/5/045205559a490415abf1e1d7536f804f5b023577.png)
so we must have 
![$$[APE]=x,[BPF]=y,[CPD]=z.$$](http://latex.artofproblemsolving.com/e/f/d/efd52db04bd0d002de8ce6a45fd3f121b03b4d72.png)
Then, 
etc. and Ceva. Furthermore, ![$$\frac{[ABD]}{[ACD]}=\frac{BD}{CD}=\frac{[BPD]}{[CPD]}.$$](http://latex.artofproblemsolving.com/0/1/a/01a6fbf8dc2da8a490255b9113264f7e766e59b9.png)
Hence, 
This rearranges to 
Similarly, 
Adding these three equations, we get that 
By our lemma, it follows that 
![$[BPF] = a, [PDC] = b, [PEA] = c$](http://latex.artofproblemsolving.com/6/5/f/65f6066d03eb393d7d8bcc4ba1fb983795d232c9.png)
.
Similarly, 
and 
Adding up, we see that 
But now 
,
.
.
.![$\frac{[AFP]}{[ABC]} = \frac{bc}{a+b}$](http://latex.artofproblemsolving.com/3/b/e/3bea4eeff5d6fd0c2734e20bdf6dcfa3383046f8.png)
.
so 
so 
so 
.
,
.
,
,
respectively. Use Ceva's theorem in 
and 
.
.
,
