with diameter 
is given. The point 
lies in the interior of the circle, but not on 
intersects 
and 
.
.
lies on 
is tangent to 
.
,
colours (so that adjacent vertices have different colours).
edges can be removed from the graph so that it remains connected.
be the incenter and the circumcenter of triangle 
,
be the circumcenters of triangle 
.
be the midpoints of segments 
.
,
.
Prove that
,
.
satisfies the quadratic equation 
and deduce that the value of 
.
.
,
is 
.
![$f\in C^1[a,b]$](http://latex.artofproblemsolving.com/8/4/8/848da12534d564560fe1cb016142159234be98b4.png)
,
and suppose that 
,
,
for all ![$x\in [a,b]$](http://latex.artofproblemsolving.com/7/8/d/78d9dcd969a2c45f7cc9234aa1377b6ec1c2d5d3.png)
.
for all 
![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
is called perfect if the sum of its positive divisors 
is twice 
.
is a perfect number since the sum of its positive divisors is 
,![\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]](http://latex.artofproblemsolving.com/0/c/0/0c08f527e87f7aae4ca5226f19bd084249551fe3.png)
where the sums are taken over all prime divisors 
of 
![$f\in C[0,b]$](http://latex.artofproblemsolving.com/8/9/8/898aec0300888e562794b7a551be34f5c2676236.png)
,
and let 
be periodic with period 
.
has a limit as 
and
be an even positive integer. Let 
be a monic, real polynomial of degree 
;
for some real coefficients 
.
for all integers 
such that 
.
for which 
.
where 
,
are nonnegative integers.
,
,
and 
for 
.
and give an example where 
.
Y by Adventure10, Mathlover_1, OronSH, aidan0626, Blue_banana4, PikaPika999, and 7 other users
Given are 
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
scalene triangles.
.
to be the number of bases of these triangles (we count three bases for each equilateral and one for each nonequilateral isosceles triangle). Clearly 
.
isosceles triangles, yielding that there are at least 
scalene triangles. 
points of the same color and we apply the lemma.
points on the perpendicular bisector). Hence there are at most 
isosceles triangles, so at lesat 
scalene triangles.
points of that color.
isosceles triangles. At the same time there are 
triangles in total, making 130 non-isosceles AKA scalene triangles. 
triangles that can be formed.
be the number of isosceles triangles in the 
to 
is the maximum number of isosceles triangles. Then the minimum number of scalene triangles is 
,
as a base, since in order for that to happen, 
isosceles triangles, thus at least 
scalene triangles formed from those points.
isosceles triangles and total triangles equal 
so at least 
triangles.
![\[\binom{x}{3}-2\binom{x}2\]](http://latex.artofproblemsolving.com/e/5/e/e5ebd5bd3558da44fe6d1d25dbe24441be359efa.png)
scalene triangles formed by it.
points and the number of scalene triangles it form are atleast ![\[\binom{13}3-2\binom{13}2=130\]](http://latex.artofproblemsolving.com/3/e/4/3e4c8ec22face221b0219029db3c7ffd7171a46d.png)
as desired.
scalene triangles with same colored vertices.
triangles. For any line segment connecting any two of these points, at most 2 isosceles triangles with this segment as the base may be constructed since we are given that no three points are collinear as any vertex for which the resulting triangle is isosceles must lie on the perpendicular bisector of this segment. Hence we have atleast,![\[S\ge \binom{x}{3} -2\binom{x}{2} = \frac{x(x-1)(x-2)}{6} - x(x-1) = \frac{x(x-1)(x-8)}{6}\]](http://latex.artofproblemsolving.com/1/9/c/19c4147458ac2c22d61220b9987f03337dcabf73.png)
scalene triangles all of whose vertices are of this same color. However since we have 50 points and only 4 colors, there exists some color with atleast, 
points of that color. But then,![\[S \ge \frac{x(x-1)(x-8)}{6} \ge \frac{13\cdot 12 \cdot 5}{6} = 130\]](http://latex.artofproblemsolving.com/8/d/1/8d1eddfe9112dfd38d84ef100c77f4c241484d17.png)
scalene triangles with all vertices of the same color, as desired.
