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![$f \in C^1([0,1]), g \in C^2(f([0;1]))$](http://latex.artofproblemsolving.com/f/5/2/f52369171ef8571734950120c43aacac10154fcd.png)
.
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![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
, where 
are positive integers and 
is odd.
be the least positive integer for which 
is divisible by 
. Find the number of positive divisors of .
on a plane with centers 
, respectively. 
is feet of the perpendicular from 
to 
and 
and 
are intersections of line 
with circle 
. Points 
and 
can move around the circle 
and 
, respectively. Find the maximum possible value of 
.
and 
.
with multiplication of matrixes operation is making an isomorphic-group structure with 
.
.
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
. Let 
, 
, 
, which gives us our problem statement. Hence, it factors to![\[(x + y - 1)(x^2 + y^2 + 1 - xy + x + y) = 0\]](http://latex.artofproblemsolving.com/c/6/1/c6191986e52ad12b8f1841d63063b9ca729b1adc.png)
so either 
or 
. Hence the locus of all points satisfying the problem is the line and the point 
. The other two vertices of this triangle are clearly unique so the triangle is unique. to can be calculated to be 
, which also happens to be the height of the unique equilateral triangle. Hence, the equilateral triangle has area 
.
so the graph of the curve is the union of 
and 
. From here the result is immediate. 
: 
, when 
in a neighborhood of the point 
.
give the singular points of .
, there are no singular points; for 
, we obtain the sole point which is an isolated point of .
small , has no points close to .
, the curve presents a kind of small ellipse close to .
the complex set contains two conjugate curves that don't intersect; yet, they intersect -when - in the isolated singular real point .
: 
, where 
. If 
, then contains the conjugate complex curves
that don't intersect.
, these 
curves intersect in the isolated point 
.
small, we obtain a little circle close to .
It follows that the curve consists of a point 
and the line 
We need one of the vertices to be for the triangle with three points on this curve to be formed. Moreover, the distance from to 
is 
, in fact this is the altitude from 
So the area of the triangle is 
factors as 
and we may notice that the second factor is zero only at 
since it is the average of 
and 
Since the curve is a point and a line we see that one of 
must be and the other two are on Their midpoint must be the foot from to 
and now it is easy to finish. The triangle has height 
so it has area 
so the given equation is equivalent to 
. Notice that this implies that is a solution. Since this is a line, an equilateral triangle can be made from any point such that 
and points on . Therefore, it suffices to prove that there is only one solution to such that 
. Since we've already considered , we can divide by 
to get 
or 
. We can rewrite this as 
. Clearly, is a solution to this. Plugging in 
and 
into the original equation gives , and we have already considered , so we multiply both sides by 
to get that 
so 
, which we've already shown only has solution . Now, we calculate the area of this unique equilateral triangle to be 
.
to the line and there are two unique points on the line making 
with the perp and hence we are done 
. Now it suffices to twiddle with the given equation.
. So either 
or . So the three vertices of the equilateral triangle are 
It can be shown that the 2nd factor is never 0 unless 
From here, the problem is done, since one vertex must be at 
and the other two points lie on the line.
contains only one set of three distinct points, 
and 
which are the vertices of an equilateral triangle.
,
.
over to the right, then adding in the middle terms needed to complete the cube:
with 
![$= \frac12(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$](http://latex.artofproblemsolving.com/7/b/d/7bdec712f5c3716891b1ad459a81f64c395dd730.png)
and 
.
from which it is easy to see that there is only one set of three points, which form an equilateral triangle, whose area is 
.
the area is 
as the area. Thanks!
from the line 
times that, which is 
.
![\[(x+y-1)(x^2+y^2+x+y-xy+1) = 0\]](http://latex.artofproblemsolving.com/d/1/2/d12d75f847f3a18cd7c1baa1a8cdc3ee97c8361e.png)
So either ![\[x^2+y^2+x+y-xy+1 = 0\]](http://latex.artofproblemsolving.com/9/2/6/92609a3e2b5f1385ae5ec1c89a0bb0806d9bb44a.png)
Since 
is real, taking the discriminant gives![\[-3(y+1)^2 \ge 0 \implies y = -1 \implies x = -1\]](http://latex.artofproblemsolving.com/0/c/9/0c91d4e4cbac7da43d53f8c2bd71f022d47bd332.png)
Thus we have 
.![\[\boxed{\frac{3\sqrt{3}}{2}}\]](http://latex.artofproblemsolving.com/8/b/a/8ba187c3ac58aa120ede027a46f7f2e016e052d5.png)