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-plane given points 
on the planes 
respectively. Let 
be the intersection point of the line 
and the 
-plane.
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane 
,.
and 
on the plane 
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane and the point moves on the perimeter of the square 
, draw the domain swept by the point on the -plane, then find the area.
with 
, let its incircle be tangent to sides 
, 
, and 
at 
, 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of .
and 
be positive integers with 
such that 
divides 
. Prove that
and 
; or
and 
.
and every 
, and also for 
and 
, and then it suffices to notice that satisfies the recurrence relation 
: there are two types of groups of size 
in 
, namely those that contain 
and those which do not. The average of the smallest element over those which do not is 
and there are 
of them, while the average of the smallest element over those which do contain is 
and there are 
such groups, hence 
.
element subsets of the given 
element set is 
by the given statement.
which gives us an idea to consider an 
element set containing elements from the given set. The additional element can be chosen as 
as we are dealing with the smallest number of each element subset here.
with the given set through some function. Consider the 
element subsets of the set and strike off the smallest number from each of the element subsets. Now, it is obvious that each element subset of the given set 
occurs 
times where is the least element in each of the element subsets. Thus we can instead count the sum of the element subsets of the given set as 
.
.
is the average of the 
-th least elements of all the -element subsets of 
is 
Consider a regular -gon inscribed in a circle. We take 
of the points, which split the circle into parts 
. By symmetry, ![$\mathbb{E} \left [|\rho_i| \right]=\tfrac{n+1}{r+1}$](http://latex.artofproblemsolving.com/a/d/a/adac96a74c3b194df47e60c958470de4126aa4e4.png)
. Cut the circle just after the 
-th chosen point, and straighten the circle into a line segment, which has length , so we have chosen points among 
. Then the expected value of the minimal distance of a point to the endpoint of the segment (the origin) is . Again by symmetry, the distance of the -th chosen point to the origin is 
is the average of the -th least elements of all the -element subsets of is Consider a regular -gon inscribed in a circle. We take of the points, which split the circle into parts . By symmetry, . Cut the circle just after the -th chosen point, and straighten the circle into a line segment, which has length , so we have chosen points among . Then the expected value of the minimal distance of a point to the endpoint of the segment (the origin) is . Again by symmetry, the distance of the -th chosen point to the origin is 
is the average of the -th least elements of all the -element subsets of is 
is the average of the -th least elements of all the -element subsets of is 
element subsets of ![$[{0,...,n}] $](http://latex.artofproblemsolving.com/7/8/4/784d95fbad7c087b9d54ff9c72e6d53d2af32cb8.png)
,the white vertices are the 
-element subsets of ![$[{1,2,...,n}] $](http://latex.artofproblemsolving.com/6/f/7/6f75f94e7b8f1ccbc5231ab1752c3c9050acf295.png)
and a black vertex 
is adjacent to the white vertex 
by deleting the smallest element from .
black vertices, 
white vertices and 
edges.
of a white vertex.
element subsets of the given element set is by the given statement. which gives us an idea to consider an element set containing elements from the given set. The additional element can be chosen as as we are dealing with the smallest number of each element subset here. with the given set through some function. Consider the element subsets of the set and strike off the smallest number from each of the element subsets. Now, it is obvious that each element subset of the given set occurs times where is the least element in each of the element subsets. Thus we can instead count the sum of the element subsets of the given set as ..
is the smallest element, we can choose the remaining 
elements from 
is the smallest element, we can choose the remaining elements from 
Therefore, we can write 
Now using the Hockey Stick Identity we can write -
Use the Hockey Stick Identity again to get 
By the same Identity again we can write - 
reduces to 
Since it is the same as taking the smallest one in each turn in that way.![\[\frac{1\binom{n-1}{r-1} + 2\binom{n-2}{r-1} + \dots + n-r+1\binom{r-1}{r-1}}{\binom{n}{r}}\]](http://latex.artofproblemsolving.com/6/e/5/6e5bfe0f1e390196493bc32e592d54cb3d58cfda.png)
.![\[1\binom{n-1}{r-1} + 2\binom{n-2}{r-1} + \dots + n-r+1\binom{r-1}{r-1}=\binom{n+1}{r+1}\]](http://latex.artofproblemsolving.com/5/4/9/5496b0771acab0f942b01a0a03e0c17937b9bc99.png)
Now by simple algebraic manipulations we can arrive to ![\[\frac{\binom{n+1}{r+1}}{\binom{n}{r}}=\frac{n+1}{r+1}\]](http://latex.artofproblemsolving.com/9/1/b/91b0147992442d9deae88ea0d07874ee3900368c.png)
will be included 
times, will be included 
times, and so on until 
will be included 
times. So we have ![\[F(n,r)=\frac{1\cdot \binom{n-1}{r-1}+2\cdot \binom{n-2}{r-1} +\cdots+(n-r+1)\binom{r-1}{r-1}}{\binom{n}{r}}\]](http://latex.artofproblemsolving.com/6/f/0/6f09157a7939a61cab752b8dbe472a97f1a3d03b.png)
.
. When we divide, the 
's cancel out. So we are left with 
, as desired.
![\[\frac{\binom{n+1}{r+1}}{\binom{n}{r}} = \frac{(n+1)! r! (n-r)!}{(r+1)! (n-r)! n!} = \frac{n+1}{r+1}\]](http://latex.artofproblemsolving.com/e/6/0/e60fabce5ae6c3926282bb3cbec9e3faba6f7b16.png)
and we're done. 
is chosen as the minimum element then the number of ways forming subsets such that is the least element, is 
students at MOP. The director wants to interview one of students, but he does not know who is who so he keeps asking different people until he find one of the students. Prove that the expected number of students he has to ask is 
chance of being asked before all selected people, hence 
,
,...,
} where 
, mark the point (,,...,) in -dimensional space.), (1,2,...,-1,), ... so you can take the average of each coordinate.
,
.![\[ F(n,r)={n+1\over r+1}. \]](http://latex.artofproblemsolving.com/e/9/d/e9da0a074ee712e67c3d999193a933bded5f98f5.png)
,
by virtue of the following argument.
which contains 
1s. For some value of 
t
ways to arrange the bits coming before the second 1, and 
ways to arrange the bits after the second 1. Our identity follows, and the problem is solved.
will be undefined because in 
:
The LHS is sum of all 
.
is where we count the sum of 
is the sum of 
.
is the 
and so 
so it's enough to prove that 
or 
Let 
and let 
be a real number such that 
.
we obtain
,
.
,
from the set 
.
whose subset is 
.
)
will be 
.
is 
,
.
as desired.
