Unfortunately I was being silly and wrote my solution in terms of barycentric coordinates (since I was trying to exploit $3$-symmetry). So the opening lines of my solution were: "Let $X$ be Lincoln, NE, $Y$ the North Pole, and $Z$ 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 $XYZ$."

How'd you find this construction?

@v_Enhance What was the motivation for the construction?

I turned in "Let $C$ be an elliptic curve in $\mathbb{R}^2$. Clearly some subset of $C$ satisfies the problem." 1 pt?

That's hilarious, did you actually know how that worked or did you just write that down at random?

Well the problem reminded me of the group law, but I didn't really look at #3 very much so I just wrote that down before running out of time.

I came up with the same construction that everyone else did (namely, P(a) = (a - 2014/3, (a - 2014/3)^3) and proved that it worked. However, for reasons that escape me, I included some random (and false) information that every line intersecting the cubic intersected it in either 1 or 3 points (forgot about tangent lines). This in no way harmed the proof. Do you think I will still get a 7, or at least a 6?

Ignore this

@pythag011 Its crazy to claim that anyone bellow black mop doesnt know real math. There are plenty of people who never took the AMC or didnt even qualify for usamo who know alot of theory.

But those people aren't exactly "below" black MOP are they?

Imo they are. And even if it isnt being used in the same exact sense it still implies that being less accomplished at competition math implies less knowledge of theory which is just stupid.

I think he was more implying that the very best at competition math aren't simply pulling from a bundle of tricks but rather have a deeper understanding of the theory behind these problems. I'm not exactly sure if this generalization really does apply in all cases, but it isn't as if it doesn't make any sense at all.

I think we should stop feeding the troll and refocus the thread. On that note , a couple of posts above i asked a grading question about this problem. Any opinions?

I think if you don't get a 7, you'll get a 6.

I think that at most you'd be docked 1 point; if your proof all in all is sound, you shouldn't lose much. I mean, saying incorrect stuff is never good, but I can't see why a full proof with a little (incorrect) fluff would lose much.

HOLLLLYYYYY I SOLVED A PROBLEM THREE P_(x-671)=(x-1/3,(x-1/3)^3) WORKS!!! ELIPTIC CURVES FTW

We claim that the construction \[P_n=(n,n^3-2014n^2)\]works. To show this, we derive from the slope formula \[\frac{a^3-2014a^2-b^3+2014b^2}{a-b}=\frac{b^3-2014b^2+c^3-2014c^2}{b-c}\]which gives, after factoring, \[(a-c)(a+b+c-2014)=0.\]Since if $a=c$, $P_a$ and $P_c$ would be the same point, we have that $a+b+c=2014$ for collinearity, as desired. The motivation for this solution comes after thinking about Vieta's formulas; then, with the slope intercept, we hope that our result factors into something like \[(a+b+c-2014)(...)=0.\]

I see many people talk about elliptic curves, but what exactly are they? I can't seem to find a good handout online.

it's like advanced theory (group theory concept)

Claim: For any cubic polynomial $P(x)=a_3x^3+a_2x^2+a_1x+a_0$, pairwise distinct points $Q_a=(a,P(a))$, $Q_b=(b,P(b))$, and $Q_c=(c,P(c))$ are collinear iff $a+b+c=-\frac{a_2}{a_3}$. The points $Q_a,Q_b,Q_c$ are collinear iff the slope of the lines $\overline{Q_aQ_b}$ and $\overline{Q_bQ_c}$ are equal. That is, we need: $$\frac{P(b)-P(a)}{b-a}=\frac{P(c)-P(b)}{c-b},$$which simplifies as: $$a_3(a^2+ab+b^2)+a_2(a+b)+a_1=a_3(b^2+bc+c^2)+a_2(b+c)+a_1.$$Now collecting terms and factoring out $a-c$, we have $a+b+c=-\frac{a_2}{a_3}$. These steps are reversible, so the proof is complete. This claim clearly suffices by taking, say, $P(x)=x^3-2014x^2$, which would make $P_a=(a,a^3-2014a^2)$ and have $P_a,P_b,P_c$ collinear iff $a+b+c=2014$.

For $n\in\mathbb Z$, let $P_n=(x(n),y(n))$, Then for $\forall a,b,c\in\mathbb Z$, we need that $$\begin{vmatrix}1&x(a)&y(a)\\1&x(b)&y(b)\\1&x(c)&y(c)\end{vmatrix}\Leftrightarrow a+b+c=2014.$$Noticing that for $\forall a,b,c\in\mathbb R$, $$\begin{vmatrix}1&a&a\\1&b&b\\1&c&c\end{vmatrix}=0;$$$$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(a-b)(b-c)(c-a);$$$$\begin{vmatrix}1&a&a^3\\1&b&b^3\\1&c&c^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c).$$For $\forall n\in\mathbb Z$, let $x(n)=x,y(n)=n^3-2014n^2$, then $$\begin{aligned}\begin{vmatrix}1&x(a)&y(a)\\1&x(b)&y(b)\\1&x(c)&y(c)\end{vmatrix} &=\begin{vmatrix}1&a&a^3-2014a^2\\1&b&b^3-2014b^2\\1&c&c^3-2014c^2\end{vmatrix}\\ &=\begin{vmatrix}1&a&a^3\\1&b&b^3\\1&c&c^3\end{vmatrix}-2014\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}\\ &=(a-b)(b-c)(c-a)(a+b+c)-2014(a-b)(b-c)(c-a)\\ &=(a-b)(b-c)(c-a)(a+b+c-2014).\end{aligned}$$Therefore $P_a,P_b,P_c$ are collinear if and only if $a+b+c=2014$.

pi e so xor Take $P_x = \left(x, \left(x-\frac{2014}{3}\right)^3\right)$; then $P_a$, $P_b$, $P_c$ are collinear iff the area of the triangle with these three points as vertices is 0, or by Shoelace \[ 0 = \begin{vmatrix}1&a&\left(a-\frac{2014}{3}\right)^3\\1&b&\left(b-\frac{2014}{3}\right)^3\\1&c&\left(c-\frac{2014}{3}\right)^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014),\] which is indeed $0$ iff $a+b+c = 2014$ and $P_a, P_b, P_c$ are distinct, as desired. (i thought this problem was very easy but i also guessed the construction very quickly, so i dont think im qualified to comment)

Take \[P_x=\left(x^2-2014x^2, x\right)\]Then we have that $P_a,P_b, P_c$ are collinear if and only if \[ 0 = \begin{vmatrix}a^3-2014a^2&a&1\\b^3-2014b^2&b&1\\c^3-2014c^2&c&1\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014)\]Since $a,b,c$ are distinct it follows that the 3 points are collinear iff $a+b+c=2014$ QED.

$P_{-3},P_{-2},P_{-1},P_0,P_1$ 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)

I think "group law" just works. I am not an expert on elliptic curves so there might be technical issues but I digress. Let $\gamma = E(\mathbb{Q})$ be a projective elliptic curve with nonzero rank and let $O = [0: 1 : 0]$ be the identity in its group law. and let $P$ be a point on $\gamma$ which is not in the torsion subgroup. Now, we define $P_i = (3i - 2014) \cdot P$, whose points are all distinct due to the earlier condition. If $a + b + c = 2014$,then it follows that \[ P_a + P_b + P_c = (3a + 3b + 3c - 3 \cdot 2014) \cdot P = O. \]and thus $P_a, P_b, P_c$ are collinear over projective space. Finally, since $3 \nmid 2014$ it follows that $3i - 2014 \ne 0$ so all our points can just be put in $\mathbb{R}^2$.

Shift to when the sum is $0$ instead of $2014.$ Have each $P_a=(a,f(a)).$ Then by shoelace $f(a)=a^3,$ so our final answer is $$P_n=(n-\frac{2014}{3}, (n-\frac{2014}{3})^3).$$

The idea in the solution to this problem also appears in the 1995 paper "On a class of $O(n^2)$ problems in computational geometry" (available online here and here for example), in Theorem 4.1 and Figure 2.

solved with NTguy spotted EC fast enough, clowned on trivial nonsense

By considering the set $Q_n = P_{n+671}$ for each positive integer $n$, we can replace the condition with $a+b+c=1$. Then, I claim that the construction $P_n = \left(n, n^3-n^2\right)$ 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*}Thus $[P_aP_bP_c]=0$, i.e. $P_a, P_b, P_c$ for distinct $a, b, c$ if and only if $a+b+c=1$. Remark: This construction is almost completely reverse-engineered from the $(a-b)(b-c)(c-a)(a+b+c-1)$ determinant, which we know must have these factors.

pythag011 wrote: But if you do understand all these things, then olympiads are trivial. go on, perf imo then