AoPS Academy
data values 
are all equal to 
. What is the value of ?
plot 
equally spaced points around a circle. Label one of them 
, and place a marker at . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of 
distinct moves available; two from each point. Let 
count the number of ways to advance around the circle exactly twice, beginning and ending at , without repeating a move. Prove that 
for all 
.
be the real solution of the system of equations 
, 
, 
. The greatest possible value of 
can be written in the form 
, where 
and 
are relatively prime positive integers. Find 
.
![\[
a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 + a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 + a_1 \cdot 3^1 + a_0 \cdot 3^0
\]](http://latex.artofproblemsolving.com/0/c/e/0ce2c259010b678238f11d35fc1328ea76f945f6.png)
where 
for 
?
sides and 
sides, respectively. The total number of sides is 33, and the total number of diagonals is 243. What are the values of and ?
be a polynomial with integer coefficients such that 
and 
for some integer 
. Prove that there exists no integer 
such that 
.
for some real number and ![\[ a_n=na_{n-1}+(n-1)(n!(n-1)!-1) \]](http://latex.artofproblemsolving.com/4/1/f/41f0cc7b041a8bba056882b5c794503195c374fd.png)
for integers 
. Given that 
, and given that 
, where 
and 
are positive integers whose greatest common divisor is 
, compute 
![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
, points 
, 
, and 
are collinear if and only if 
.
-symmetry). So the opening lines of my solution were: "Let 
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 
."
be an elliptic curve in 
. Clearly some subset of satisfies the problem." 1 pt? 
satisfying:
and 
Then 
in terms of 
where 
.
,
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 
such that 
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 
so that 
,
,
,
,
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 
,
,
and have 
collinear iff 
,
,
,
Noticing that for 
,
For 
,
,
Therefore 
;![\[ 0 = \begin{vmatrix}1&a&\left(a-\frac{2014}{3}\right)^3\\1&b&\left(b-\frac{2014}{3}\right)^3\\1&c&\left(c-\frac{2014}{3}\right)^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014),\]](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 
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)
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 
,
,![\[
P_a + P_b + P_c = (3a + 3b + 3c - 3 \cdot 2014) \cdot P = O.
\]](http://latex.artofproblemsolving.com/b/b/b/bbb4482eea87f566c1cb04a89e95b7c4c6b6010b.png)
and thus 
it follows that 
so all our points can just be put in 
Have each 
Then by shoelace 
so our final answer is 
problems in computational geometry" (available online 
,
.
be the line through 
Let the 
,
,
.
,
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).
for each positive integer 
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)
,
if and only if 
determinant, which we know must have these factors.
works.![\[y=(x-a)(2014(a+b)-a^2-ab-b^2)-a^2(a-2014)\]](http://latex.artofproblemsolving.com/c/2/3/c234fdabffb4614bc7df44c221612bcb80555afa.png)
Plugging in 
,![\[0=c^3-2014c^2+c(2014(a+b)-a^2-ab-b^2)+ab(a+b-2014)=(c-a)(c-b)(c-(2014-a-b))\]](http://latex.artofproblemsolving.com/3/2/5/3257b644c0e777074f5535b6fda4d049af916ba5.png)
The conclusion follows.
,
fixed is also true.