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.
, 
, 
be non-negative real numbers, no two of which are zero. Such that 
Prove that :
be a convex quadrilateral and points 
and 
on sides 
such that![\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]](http://latex.artofproblemsolving.com/b/4/2/b42cd6305fab848a60c79689f17bdb55e1ba0d31.png)
is the area of 
show that 
and can teleport to other Cartesian coordinates with only 
of 
moves: 
, 
when 
, 
, and 
when 
.
be any polynomial with positive integer coefficients that passes through 
. Show that for all such , there exists a unique point on the curve where the wizard can land on., let be this unique point. Find the equation of the graph that contains all potential .
be a sequence of positive integers. Let 
be the number of 3-element subsequences 
with 
, such that 
and 
. Considering all such sequences 
, find the greatest value of .
, 
is the midpoint of 
. 
is an arbitrary point on 
. Let 
be the intersection of 
and 
. The line 
intersects with the circumcircle of 
, for the second time at 
. 
is the circumcenter of triangle 
. Prove that 
.
in the expansion of 
table. A cannon can fire at any 
cells of the board after that the tank will move to one of the adjacent cells (by side). Then the process is repeated. Find the smallest value of such that the cannon can definitely shoot the tank after some time.
is 
. Line 
meets line 
at point 
, and there is a point on 
such that 
. Line 
intersects at point . The perpendicular bisector of line segment intersects line 
at point 
, and line intersects 
at point 
.
is tangent to 
.
be positive real numbers. Prove that 
.
(doesn't have to be called 
), and we define the homogenized barycentric coordinate as![\[P=\left(\dfrac{[PBC]}{[ABC]},\dfrac{[PAC]}{[ABC]},\dfrac{[PAB]}{[ABC]}\right)\]](http://latex.artofproblemsolving.com/c/e/7/ce7719dabb2c95f0ccf1f4b54683338b8abde865.png)
Note that we can define 
in this system. Furthermore, if a point 
, then we have that ![$[PBC]=0$](http://latex.artofproblemsolving.com/2/b/9/2b9ba4a873941cb8c489868556f3f445dd6fbb74.png)
, so thus we easily have that![\[P=(0,\text{something},1-\text{something})\]](http://latex.artofproblemsolving.com/b/6/d/b6d4627b0b4722ef39e6ea62ca0c75c2caa95342.png)
What are the advantages of barycentric coordinates? Here are a few:
as![\[[ABC]\begin{vmatrix}p_1&q_1&r_1\\p_2&q_2&r_2\\p_3&q_3&r_3\end{vmatrix}\]](http://latex.artofproblemsolving.com/4/6/f/46fbe16a64f3097fd9d782d1ea40ff18adc263be.png)
where 
and 
are defined analogously.
triangles of equal areas, so![\[G=\left(\dfrac{[GBC]}{[ABC]},\dfrac{[GAC]}{[ABC]},\dfrac{[GAB]}{[ABC]}\right)=\left(\dfrac 13,\dfrac 13,\dfrac 13\right)\]](http://latex.artofproblemsolving.com/a/5/a/a5a2a534d28665b7a6bb750da827a7f82b0c3eec.png)
The circumcenter and orthocenter aren't as nice, but still are nice. Furthermore, we have that![\[[IBC]=\dfrac 12ar\]](http://latex.artofproblemsolving.com/7/4/9/7498b126bc2728e5e5d31ee2fafca134daefa781.png)
and analogous objects, so thus![\[I=\left(\dfrac{ar}{2rs},\dfrac{br}{2rs},\dfrac{cr}{2rs}\right)\]](http://latex.artofproblemsolving.com/b/b/a/bbab24e04e140133a5cd079549d2b263b70e811c.png)
where I used ![$[ABC]=rs$](http://latex.artofproblemsolving.com/5/f/e/5fe68c691b13681950a6542efe4b5ddd0bc41511.png)
. However, we can "cancel" the 
to get![\[I=\left(\dfrac{a}{2s},\dfrac{b}{2s},\dfrac{c}{2s}\right)\]](http://latex.artofproblemsolving.com/3/a/4/3a4b78f1c4b58686a9c62c44aca550f6ac19d855.png)
![\[P=(x:y:z)\]](http://latex.artofproblemsolving.com/4/1/e/41eca29375a7077fd6634d32225879cb390b7e12.png)
as long as 
are in the same ratio as ![$[PBC],[PAC],[PAB]$](http://latex.artofproblemsolving.com/7/7/0/770f726d7deb2ec1cfdf603700d98abe10120904.png)
. That allows us to remove the denominator in, for example, 
, so we get![\[I=(a:b:c)\]](http://latex.artofproblemsolving.com/8/b/7/8b78695d3df71a26c0c2c9c2be95ea51e1d36421.png)
![\[G=(1:1:1)\]](http://latex.artofproblemsolving.com/3/5/e/35e9fad690f3b21174131260bab24c6e7db399b2.png)
![\[\vec P=x\vec A+y\vec B+z\vec C\iff P=(x,y,z)\]](http://latex.artofproblemsolving.com/f/5/e/f5ef70b5d133a282c8785829b8eac8a48b51afbe.png)
That helps, for example, with finding the distance between two points, finding perpendicular lines, and also finding circles.
