such that 
.
be positive real numbers such that 
and 
,
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
and the distribution function is 
then
and find the expression of 
be a function whose derivative is continuous on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
.![$$\lim_{n \to \infty} \sum^n_{k = 1}\left[f\left(\frac{k}{n}\right) - f\left(\frac{2k - 1}{2n}\right)\right] = \frac{f(1) - f(0)}{2}.$$](http://latex.artofproblemsolving.com/3/1/4/314a1597acae030381f980e6e26c12432f31e069.png)
be a positive integer greater than 
.
where each of 
is either 
.
with integer coefficients, such that 
and 
for some distinct primes 
and 
.
be a finite subset of 
such that 
.
be functions satisfying
for all 
.
such that
for all 
,
for all 
.
be an integer and let 
be the set of all 
invertible matrices in which their entries are 
or 
be the number of 
.
.![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
.
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
,
for all 
.
If G has a normal subgroup of order 
then prove that 
is abelian without using Sylow Theorems.
is differentiable and 
for all 
,
.
Y by Adventure10, Mango247
Let 
be a circle with center 
and 
be a diameter. Point 
lies on segment 
and point 
is the midpoint of one of the arcs 
of . Construct point 
such that 
is a parallelogram. Lines 
and 
meet at 
. Line 
meets the minor arc 
at 
. Prove that 
bisects 
.
be a circle with center 
and 
be a diameter. Point 
lies on segment 
and point 
is the midpoint of one of the arcs 
of . Construct point 
such that 
is a parallelogram. Lines 
and 
meet at 
. Line 
meets the minor arc 
at 
. Prove that 
bisects 
.
are cyclic. Let 
.
and 
.
is isosceles 
so 
so 
,
and 
cyclic and then we just angle chase again.I think the problems in Saudi Arabia TST are too easy.
