AoPS Academy
+1 w
, pairwise distinct points 
, 
, and 
are collinear iff 
.
are collinear iff the slope of the lines 
and 
are equal. That is, we need:
which simplifies as:
Now collecting terms and factoring out 
, we have . These steps are reversible, so the proof is complete.
, which would make 
and have 
collinear iff 
.
, let 
, Then for 
, we need that
Noticing that for 
,
For 
, let 
, then
Therefore are collinear if and only if .
; then 
, 
, 
are collinear iff the area of the triangle with these three points as vertices is 0, or by Shoelace![\[ 0 = \begin{vmatrix}1&a&\left(a-\frac{2014}{3}\right)^3\\1&b&\left(b-\frac{2014}{3}\right)^3\\1&c&\left(c-\frac{2014}{3}\right)^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014),\]](http://latex.artofproblemsolving.com/1/a/b/1ab2f2af01a0f6f1a8804fe4b6983c4e264c6774.png)
iff 
and 
are distinct, as desired.
![\[P_x=\left(x^2-2014x^2, x\right)\]](http://latex.artofproblemsolving.com/b/a/a/baa71c63d857307a0f0da1bbe86c4cc6f002ec39.png)
Then we have that 
are collinear if and only if![\[ 0 = \begin{vmatrix}a^3-2014a^2&a&1\\b^3-2014b^2&b&1\\c^3-2014c^2&c&1\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014)\]](http://latex.artofproblemsolving.com/c/e/1/ce198e5f3485ca287d559c7a5828cdb79d13cb67.png)
Since 
are distinct it follows that the 3 points are collinear iff QED.
be a projective elliptic curve with nonzero rank and let ![$O = [0: 1 : 0]$](http://latex.artofproblemsolving.com/3/c/f/3cf8e617f09958482e532c68a3bfc771777e395e.png)
be the identity in its group law. and let 
be a point on 
which is not in the torsion subgroup. Now, we define 
, whose points are all distinct due to the earlier condition.
,then it follows that![\[
P_a + P_b + P_c = (3a + 3b + 3c - 3 \cdot 2014) \cdot P = O.
\]](http://latex.artofproblemsolving.com/b/b/b/bbb4482eea87f566c1cb04a89e95b7c4c6b6010b.png)
and thus are collinear over projective space. Finally, since 
it follows that 
so all our points can just be put in 
.
instead of 
Have each 
Then by shoelace 
so our final answer is 
for each positive integer 
, we can replace the condition with 
.
works. Indeed, notice that the area determinant![\begin{align*}
[P_aP_bP_c] &= \begin{vmatrix} a & a^3-a^2 & 1 \\ b & b^3-b^2 & 1 \\ c & c^3-c^2 & 1\end{vmatrix} \\
&= \sum_{\mathrm{cyc}} a\left(b^3-b^2\right)-b\left(a^3-a^2\right) \\
&= (a-b)(b-c)(c-a)(a+b+c-1).
\end{align*}](http://latex.artofproblemsolving.com/3/e/9/3e9cdba1168e603ee3491ebe90ca456c2e89d41d.png)
Thus ![$[P_aP_bP_c]=0$](http://latex.artofproblemsolving.com/1/8/3/183f57efecc0ff3487d235f39500816bc8b55838.png)
, i.e. for distinct 
if and only if .
determinant, which we know must have these factors.
works. and , therefore the line they define is ![\[y=(x-a)(2014(a+b)-a^2-ab-b^2)-a^2(a-2014)\]](http://latex.artofproblemsolving.com/c/2/3/c234fdabffb4614bc7df44c221612bcb80555afa.png)
Plugging in 
, we get the equation ![\[0=c^3-2014c^2+c(2014(a+b)-a^2-ab-b^2)+ab(a+b-2014)=(c-a)(c-b)(c-(2014-a-b))\]](http://latex.artofproblemsolving.com/3/2/5/3257b644c0e777074f5535b6fda4d049af916ba5.png)
The conclusion follows.
, rather than only 
, hence the generalisation to 
fixed is also true.
be one of its generators, define the addition of points on it as usual so we need 
, then let 
this works as we get its 0 iff 
![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
where 
.
-
be Lincoln, NE, 
the North Pole, and 
any vertex of the Bermuda triangle. These points are noncollinear since they lie on the spherical Earth. We use barycentric coordinates with respect to triangle 
.
,
when 
,
,
are distinct. Back-construct the original matrix after expanding this:![\[ \left( \begin{array}{ccc} 1 & x & x^3 \\ 1 & y & y^3 \\ 1 & z & z^3 \end{array} \right). \]](http://latex.artofproblemsolving.com/0/b/1/0b1d8fe67fe0dfdefe9cc173bfcaaf2b12fd12e3.png)
,
.
.
are all roots of this, so if we want 
to be polynomials then we must have 
.
.
.
works. Then to finish retranslate everything 
units to get 
.
for any real number 
and that you will have the same property.
for polynomials 
and 
Now the thing is that the question "when is 
collinear" is a question that has an ALGEBRAIC answer. What does this mean? It means there will be some polynomial 
in 
So all you do here is pick 
be an elliptic curve in 
,
determine the whole set, it seems like Pappus/projective transformation should work, unless we run into other issues (I'm guessing "other issues" means it won't work)
problems in computational geometry" (available online 
be the line through 
Let the 
,
,
.
to be 
,
and 
.
Putting this into the slope equation for collinearity, we obtain that 
cancels out anyways, so we just take 
to be the coordinates (which can be checked to work).