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 
,
.
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Points 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
and Z is outside 
Though it's not competition-oriented, I believe it's around the same audience.
be the midpoint of 
.
is cyclic. Let 
Then, 
so 
colinear.
,
.
,
,
By Power of a Point and cyclic quadrilaterals, 
So 
are on the same circle, which implies our desired conclusion.
is similar to 
,
is similar to 
.
.
is a median
be the point at infinity along lines 
and 
,
.
is harmonic,![\[-1 = (A,C ;B , D) \stackrel{E}{=} (A, C; M, P_{\infty}) \]](http://latex.artofproblemsolving.com/3/b/a/3bac455ec07839e6b78f06a8fd19884b103e48b3.png)
so 
is the midpoint of 
,
.
cyclic. Now let 
.
and clearly 
where 
is the center of 
,
and thus 
.
is perpendicular to 
is cyclic.
and the tangent criterion, we have ![\[\angle BMP = \angle BMD = \angle DBP = \angle BOP\]](http://latex.artofproblemsolving.com/a/3/6/a363b3afbc14fcf16302abdb30b3f845fadecc28.png)
as desired.
,
and 
let the center of the circle be 
and let 
so we have 
we have 
from which it follows that 
so we are done. 
and 
and 
respectively. Then, by the formula for the intersection of tangents, we get 
Then, since 
and 
must have the same measures, so it follows that 
Then, since 
lie on the unit circle while 
we have
Therefore, the midpoint of 
It now suffices to show that 
and 
are collinear. By the complex number collinearity condition, this is equivalent to
concluding the proof. 
,
.
are all concyclic.![\[ \angle{BMP} = \angle{MED} = \angle{BDP}, \]](http://latex.artofproblemsolving.com/2/d/d/2dddfc10aac77de9281ee88e4f3cf7a47c46cf03.png)
where the last part comes from the so-called Tangent Lemma. In particular, 
is a cyclic quadrilateral, so 
.
is clearly cyclic 
)
tells us that 
,
is a symmedian. Also, 
is cyclic with 
,
implies 
,
.
,
is cyclic. But since 
has diameter 
,
and hence 
.
;
is isosceles.
is cyclic. We can do so by proving 
:![\[\angle{BXO} = \angle{BED} = \frac{\overarc{BD}}{2} = \frac{\angle{BOD}}{2} = \angle{BOP}\]](http://latex.artofproblemsolving.com/f/a/7/fa7332189b642ee5ad9f46b558a4b8e5aae1b7d4.png)
,
(this can be shown by proving 
,
,
(as 
by the tangent condition). Since 
and then project through 
onto 
is cyclic and by the 
criterion, 
.
is cyclic. This means 
,
and we're done. 
,
