AoPS Academy
, 
and 
be points on circle 
divided into three equal parts. Construct three equal circles 
, 
, and 
tangent to internally at points , , and respectively. Let 
be any point on arc 
, and draw tangents 
, 
, and 
to circles , , and respectively. Prove that 
.
, 
, 
, 
be points on circle divided into 
equal parts. Construct equal circles , , , 
tangent to internally at , , , . Let be any point on circle , and draw tangents 
, 
, , 
to circles , , , . If the sum of 
of , , , equals the sum of the remaining 
(where 
), find all such .
denote the set of positive integers
such that for all 
it holds that 
Note: consider 
and 
, ∀x∈[0,1].
and 
and there exists a positive integer 
such that 
. Let 
be minimal polynomial 
over 
. Prove that 
be a probability distribution on the set of nonnegative integers. Select a number according to this distribution and repeat the selection independently until either a zero or an already selected number is obtained. Write the selected numbers in a row in order of selection without the last one. Below this line, write the numbers again in increasing order. Let 
denote the event that the number 
has been selected and that it is in the same place in both lines. Prove that the events 
are mutually independent, and 
.
. Calculate the first and second derivative of the function at the point 
.
kg sky diver that jumps out of a plane with no initial velocity and an air resistance of 
. For this example assume that the positive direction is downward.
satisfying ![\[|z -a | + |z + a| = 2|c|\]](http://latex.artofproblemsolving.com/4/b/d/4bdafe46f8d3ed1c8945a1553fc3bc0d23fe11e2.png)
if and only if 
If this condition is fulfilled, what are the smallest and largest values of 
with k and u continuous in the unit square and unit interval can have only the trivial solution. Prove this in detail. Here k(x,y)=sin(xy)
, done.
and 
are given on side 
of convex quadrilateral 
(with closer than to 
). It is known that 
and 
. Prove that 
.
+ 
..... (1)
- 
....(2)
= 180
=
is cyclic, which implies 
. This also implies that 
. However, we note that 
which implies is cyclic. To finish, we note that 
which implies 
. 
is cyclic because 
.
. We then proceed to angle chase. 
because is cyclic and opposite angles in a cyclic quadrilateral add to 
.
by supplementary angles and 
because the angles in a triangle add up to .
, which shows that 
. Finally, because you just subtract equal angles 
and 
. 
and 
, if we add those equalities to the desired conclusion, we then need to prove that is cyclic., we know that is cyclic, so let 
and 
. Let 
. We then have 
, and by sums of angles in triangle we have 
, so is cyclic as desired. 
, 
is cyclic, wherefore
So is cyclic as well, yielding 
or![\[\angle FAC=\angle BAC-\angle BAE-\angle EAF=\angle CDB-\angle CDF-\angle FDE=\angle EDB,\]](http://latex.artofproblemsolving.com/d/0/1/d01bda978ce395ccda1aa19db70f77153ed72658.png)
as desired. 
is cyclic. Because of this, we can set 
and 
. Now, setting 
and 
, we can angle chase to get that 
and 
. Clearly, 
, so is cyclic. Finally, we have 
, which implies that . We are done. 
be the intersection of 
and 
Since 
is cyclic. As such, 
and 
We also have 
so 
As such, 
As such, 
so 
and is cyclic. As such, 
so 
as desired. 
, we know that 
is cyclic.
, we just need to prove that or that is cyclic
, is cyclic, as desired.
, is a cyclic quadrilateral. It follows that 
However, we also have 
. Thus, 
and 
, so it follows that 
or is cyclic. Then, 
, but we have 
, so 
as desired.
if and only if is cyclic.
and 
is cyclic. This means that 
Therefore, 
which means that 
proving that is cyclic, as desired.
. So, quadrilateral is cyclic.
.
. is cyclic. Hence 
.
is a cyclic quadrilateral. Take point 
on line 
such that 
. Then by angle chasing,
which implies quadrilateral being cyclic. Thus
as desired.
. Since is cyclic, 
. 
and 
also equal to so 
and 
. 
so is cyclic. Hence so 
, we know that is cyclic. This also means that 
, implying 
.
. If we substitute in angles (
from earlier and 
from the given), we find that 
. Thus, is cyclic, as the opposite angles in the quadrilateral are supplementary.
by the cyclic condition, and subtracting off some equal angles yields 
, as desired.
.
.
.
,
,
is cyclic, so 
.
.
,
Hence, 
,
Q.E.D.
so, 
,
to intersect 
and 
at 
and 
,
are cyclic.
is cyclic. Now note that 
.
,
.
after simplifying, so quadrilateral 
,
as required. This was a lot easier in hindsight.
.
.![\[ \measuredangle ABC = \measuredangle ABE = \measuredangle BAE + \measuredangle AEB = \measuredangle BAE + \angle AEF \]](http://latex.artofproblemsolving.com/0/e/8/0e8883554573b4877b050977e489a1b68d3e27fd.png)
and that ![\[ \measuredangle ADC = \angle ADF + \angle FDC \]](http://latex.artofproblemsolving.com/b/3/6/b36a40fce2548385508623223ab4710492d843c6.png)
Finally, ![\[ \measuredangle ABC = \measuredangle BAE + \measuredangle AEF = \measuredangle BAE + \measuredangle ADF = \measuredangle FDC + \measuredangle ADF = \measuredangle ADC \]](http://latex.artofproblemsolving.com/6/6/4/664d33dd786dd8481cb5785135a0014d483438fd.png)
![[asy]
pair a, b, c, d, e, f;
a = dir(140); d = dir(220); f = dir(300); e = dir(30); b = 1.3 * e - 0.3 * f; c = intersectionpoints((100*e-99*f)--(100*f-99*e),circumcircle(a,d,b))[1];
draw(a--e--f--d--cycle,orange); draw(d--c--f,blue); draw(e--b--a,blue); draw(circumcircle(a, b, c),blue); draw(circle((0,0),1),orange);
dot("$A$", a, dir(120));
dot("$D$", d, dir(240));
dot("$F$", f, dir(330));
dot("$E$", e, dir(30));
dot("$B$", b, dir(30));
dot("$C$", c, dir(270));
[/asy]](http://latex.artofproblemsolving.com/c/f/a/cfab0fa694eb45c43d633409951a99b4f16775d8.png)
is cyclic. Because 
implies that 
