AoPS Academy
, 
and 
are the midpoints of sides 
and 
respectively. Point 
lies inside the triangle. Let 
. Prove that 
.
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
, let 
be the circumcentre, 
be the orthocentre, and 
be the centroid.
satisfy the following condition:
is constant.
. Find 
.
such that 
are two finite groups with 
then show that 
be real numbers in the open interval 
Show that there exist distinct indices 
such that 
are the side lengths of an acute triangle.
of class 
for which there exists an 
such that 
be a function differentiable at 0 with 
. Find
be a positive integer, let 
, and let 
be a complex root of 
. Prove that there exist polynomials 
and 
with integer coefficients such that 
.
let 
be the largest integer 
such that![\[\binom{m}{n-1}>\binom{m-1}{n}.\]](http://latex.artofproblemsolving.com/b/9/3/b935e59b8b46d8c5616312e5d51c0aa8d6b02f0e.png)
Evaluate![\[\lim_{n\to\infty}\frac{M(n)}{n}.\]](http://latex.artofproblemsolving.com/d/a/d/dadea3ddb4d22c0cdcba05fc8c246f90892fac6b.png)
students. Let, for 
, the 
student have 
many friends. For 
let us define 
to be the number of students who have strictly more than 
friends. Show that 
for all 
, but 
(and all other 
), so the equation becomes 
. Perhaps means the number of students with strictly more than friends. is counted once for all 
, namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
friends is counted in 
.i.e times . so in the RHS too it is counted times . so these are two different ways of counting the sum 
and hence it is proved
chaotic_iak already gave that "simple solution" of yours. Why bother repeating it?
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
-gon with vertices 
(
) as persons.We colour the segment 
red if and 
are friends,otherwise we colour it as blue.We shall now show that both the sides number twice the number of red sides..It is counted once when is counted and once when 
is counted.Thus LHS part is proved.
.While we are counting 
,we are only counting 
redlines emanating from 
.This proves our claim for RHS.(Note that thus summing over the 's gives the sum of exact number of redlines originating from each vertex).
means the number of students with strictly more than friends. , then what is 
? 0 ?
be a simple graph with 
vertex. Suppose 
And we connect 
if they are friend 
. Suppose 
.And 
denote number of vertices having degree more than 
.
.
.
.
.
.If it is isolated then 
,
count number of vertex having degree more than in the new graph with 
vertex. we are done. is not isolated. . Suppose it joins with 
,
. Then 
.
,
.
will as same as .
.One must see that number of edges in new graph is 
.
new edges
.
..
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
is NOT equal to , it is indeed 
. Observe the question carefully.. means the number of students having MORE than (i.e. 
) friends.. which according to your assumption that everyone hates everyone is indeed . For the same reason, the iteration of stops at 
and not because no one can possibly have 
friends.. hope I could make myself clear..
students. Let, for , the student have many friends. For let us define to be the number of students who have strictly more than friends. Show that 
means the number of students with strictly more than friends. , then what is ? 0 ?
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
![\[
\int_{-\infty}^{x}e^{-t^2/2}dt=\sqrt{2\pi}\left( \frac{1}{2}+\frac{1}{2}\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right)
\]](http://latex.artofproblemsolving.com/f/8/1/f817a6ea9980ffb03c35901f75afa19a23881985.png)
This is unexpected. And sad. Because I just proved a wrong problem. 
array with the columns named as 
and the rows named as 
.
.. Suddenly the problem becomes so much more interesting, isn't it ? 
.
,
boxes, i.e. all the boxes below 30. So to the count we add 30, and to the boxes, we're adding 30 balls.
.
number of students who have greater than or equal to 
)
if 
or,
if 
