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 a triangle with centroid 
. Points 
and 
are chosen on rays 
and 
, respectively, such that![\[ \angle ABS=\angle ACR=180^\circ-\angle BGC.\]](http://latex.artofproblemsolving.com/0/a/d/0ad01e29194358ea0d3e10cb74b0769d1bd3a656.png)
Prove that 
.
and 
for which![\[ \left \lfloor \frac{a^2}{b} \right \rfloor + \left \lfloor \frac{b^2}{a} \right \rfloor = \left \lfloor \frac{a^2 + b^2}{ab} \right \rfloor + ab.\]](http://latex.artofproblemsolving.com/c/3/6/c3682397d21443cb720e56251c260b984392a84a.png)
be a triangle with 
, incenter 
, circumcenter 
, and circumcircle 
. The line 
meets 
at 
, and meets again at 
. Let 
be the reflection of over 
. Let the line through perpendicular to meet at 
. 
meets 
and 
at 
and 
, respectively. Suppose that the circumcircle of 
meets again at 
. Prove that 
, , 
, are concurrent if and only if the sides of 
form an arithmetic progression.
+1 w
such that 
for all 
.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
with 
right and 
, 
. On the ray 
we take two points and 
so that 
and 
. Consider the squares 
and 
containing inside 
and 
, respectively. If 
and 
, 
and 
, prove that 
and that 
, and are collinear.
inside a square 
we raise a segment 
perpendicular to the plane of the square. Show that the projections of on the planes 
, 
, 
and 
are coplanar if and only if ![$O\in [AC]\cup [BD]$](http://latex.artofproblemsolving.com/6/6/6/666df8f2afa87f15c79f9c96d9d1cea17a2a7af0.png)
.
. Proof that![\[
(2^x - 1)(5^x - 1) = y^n
\]](http://latex.artofproblemsolving.com/f/8/f/f8f013a96863bc5f39aa23edb98eed9aec54a17f.png)
have no positive integer solution 
.
(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.
