AoPS Academy
such that 
.
equations in real variables 
:![\[v_i = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} \qquad (i = 1, \ldots, 10).\]](http://latex.artofproblemsolving.com/a/7/5/a75a8d8b5d2f134372e154f12225d9e4ae820cbf.png)
Find all -tuples 
that are solutions of this system.
Prove that
, 
.Find the largest 
that satisfies:
be a sequence of reals such that 
and 
for all positive integers 
. As grows large, the value of 
approaches the real number 
. What is the greatest integer less than ?
![\[
\sqrt[3]{\sqrt[3]{x - \frac{3}{8}} - \frac{3}{8}} = x^3 + \frac{3}{8}
\]](http://latex.artofproblemsolving.com/b/b/9/bb99cd588b94b5716738e6e9ea01d15377b6efef.png)
has exactly two real positive solutions 
and 
. Compute 
.
be a regular nonadecagon. Lines 
and 
meet at 
. Compute 
.
die is laid down on an infinite grid of 
squares. The die starts perfectly aligned in one of the square of the grid with the 
facing up and the 
facing forward. The die can be "rolled" along an edge such that is moves exactly one unit up, down, left, or right and rotates 
correspondingly in that axis. facing up and facing forward).
and let 
and 
be permutations of 
. Suppose there exists a permutation 
such that 
for all 
in .
is the number of possible pairs of permutations 
, find the remainder when is divided by 1000.
and 
Prove that 
be real numbers such that 
. Find the minimum of 
and 
Prove that 
be all the divisors of . It is known, that 
are all divisors for some 
(except 
). Find all such 
be the divisors of 
is prime so 
is prime or equal to 4, therefore, 
or 
, 
, and by repeating the process, we will have that every integer is a divisor of and , contradiction!
and 
is prime, then 
, contradiction as 
, same contradiction
, but testing gives us no answer, therefore, 
![\[ a_1 \times a_n = a_2 \times a_{n-1} = \dots = n. \]](http://latex.artofproblemsolving.com/8/7/0/870f1a73954f088386ac9b7f665b14bc5cf03911.png)
Given that the divisors of 
are:![\[ a_1 + 1, a_2 + 1, \dots, a_k + 1, \]](http://latex.artofproblemsolving.com/0/d/0/0d0666c1a87538d4229f7978538c47c6a23bf8a9.png)
where:![\[ 1 < a_1 + 1 < a_2 + 1 < \dots < a_{k-1} + 1 < a_k + 1 = m. \]](http://latex.artofproblemsolving.com/8/8/8/888bc184a8c8d441ff5f4c4cbc36db25589ef74a.png)
![\[ (a_2 + 1)(a_{k-1} + 1) = m = (a_1 + 1)(a_k + 1). \]](http://latex.artofproblemsolving.com/9/3/0/9306aae4cad4e5b22d1c300e3e01404382757273.png)
2. Given 
and 
, we substitute:![\[ m = 2(n + 1). \]](http://latex.artofproblemsolving.com/1/0/3/1034ad55927ce5afa9f710a3198e5ac5c48e500e.png)
3. Equate the two expressions for :![\[ 2(a_2 \times a_{k-1}) + 2 = (a_2 + 1)(a_{k-1} + 1). \]](http://latex.artofproblemsolving.com/3/0/e/30edb1245b34e1078a288008bb795b86c4e38821.png)
4. Expand and simplify:![\[ 2a_2 a_{k-1} + 2 = a_2 a_{k-1} + a_2 + a_{k-1} + 1 \]](http://latex.artofproblemsolving.com/d/1/1/d113fba5803421b8273cc5780f8e291cc591ee74.png)
![\[ \implies a_2 a_{k-1} - a_2 - a_{k-1} + 1 = 0 \]](http://latex.artofproblemsolving.com/3/e/a/3ea23c41290f60e2c981e3e5a7198dba9fd2dd9a.png)
![\[ \implies (a_2 - 1)(a_{k-1} - 1) = 0. \]](http://latex.artofproblemsolving.com/9/3/5/935da4a73d8d0636ffa31b122fe91be31182783b.png)
5. Thus, either 
or 
., then 
violates 
., then 
contradicts the sequence order. exists under the given conditions.
then find the value of 
be the 19th-roots of unity with 
.![\[X = \frac{\omega^0 \omega^4 (\omega^2 + \omega^3) - \omega^2 \omega^3 (\omega^0 + \omega^4)}{\omega^0 \omega^4 - \omega^2 \omega^3}\]](http://latex.artofproblemsolving.com/3/b/5/3b535dd8a08df864eb062ebdb05d836ae1682cd0.png)
![\[= \frac{\omega^6 + \omega^7 - \omega^5 - \omega^9}{\omega^4 - \omega^5}\]](http://latex.artofproblemsolving.com/a/0/1/a015858628e275d220ebb89f4647fb2cdc00dfde.png)
![\[= \frac{(\omega^5 - \omega^3) - (\omega^2 - \omega)}{\omega - 1}\]](http://latex.artofproblemsolving.com/b/6/a/b6ae9d840aa9c9ed0e951a3b9a58368815192460.png)
![\[= \omega^4 + \omega^3 + \omega\]](http://latex.artofproblemsolving.com/1/4/4/144976ecce9cb7e1b2ef0b96eaf69e6d72dafc39.png)
Therefore, ![\[ \frac{A_7 - X}{A_5 - X} \]](http://latex.artofproblemsolving.com/d/e/7/de7218c41a3500b606805cebb420553e127d0c38.png)
![\[ = \frac{\omega^6 - \omega^4 - \omega^3 - \omega}{- \omega^3 - \omega} \]](http://latex.artofproblemsolving.com/a/0/4/a048a571a14118eac1da00fbf524685783ea5e3c.png)
![\[ = 1 - \omega^3 \]](http://latex.artofproblemsolving.com/8/a/4/8a49e13dd1b819e6d8e3776a5735012406636d50.png)
We plot this on the complex plane. We let 
be the origin, 
still be 
,
be 
.
is isosceles with 
.![\[\angle BOA_1 = \frac{\pi - \frac{6\pi}{19}}{2} = \boxed{\frac{13\pi}{38}}\]](http://latex.artofproblemsolving.com/8/0/9/809e94631304088982e4cf5e420e73f8cddbcb18.png)
