AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, 
is the angle bisector of 
. The circumcircle of 
intersects 
in 
respectively and the line through 
parallel to 
cuts 
in 
respectively. Prove that point 
lie on a circle.
kids in a class. Is it true that for all positive integers 
it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
, the maximum number of good cells over all possible starting configurations.
lamps 
in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp 
and its neighbours (only one neighbour for 
or 
, two neighbours for other 
) are in the same state, then is switched off; – otherwise, is switched on.
Prove that there are infinitely many integers 
for which all the lamps will eventually be off.
Prove that there are infinitely many integers for which the lamps will never be all off.
integers 
, each less than or equal to 
, prove that at least one of them is divisible by some other member of the set.
be positive integers such that no one of them is divisible by any other member of the sequence. Then![\[
a_1 \geq 2^k,
\]](http://latex.artofproblemsolving.com/1/f/e/1fecb6d6811bfc4aff7ccefa20412f0a5b860066.png)
where 
is defined by the inequalities![\[
3^k < 2n < 3^{k+1}.
\]](http://latex.artofproblemsolving.com/2/8/f/28f12a76029727a828167f1cb8e118b84d710d62.png)
This estimate for 
is the best possible.
be a sequence of positive integers. Then![\[
\max \left( (a_i, a_j) \right) > \frac{38n}{147} - c,
\]](http://latex.artofproblemsolving.com/6/5/d/65df2fa2c65d90d0548cb3c11dd36ea42c4060e6.png)
where 
is a constant independent of 
, and 
denotes the greatest common divisor of 
and 
. This is the best possible estimate.
, there exists only degree polynomial 
satisfying 
and ![$ (x + 1)[f(x)]^2 - 1$](http://latex.artofproblemsolving.com/9/c/1/9c1e80d035c923dd13711bff624d8a1a3e6a3b15.png)
is an odd function.
be the inscribed circle of a triangle 
. Let 
and 
be the tangency points of and the sides 
and 
, respectively, and let 
and 
intersect at 
and 
, respectively, such that 
and are all distinct. The tangent line of at 
intersects 
at 
, the tangent line of at 
intersects 
at 
, and the tangent line of at M intersects 
at 
. Prove that 
and are collinear.
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
, your right-neighbor’s is 
, and the next person over is 
, then for every trio in the circle they see
be an acute triangle with 
. Denote by 
and 
points on the segment 
such that 
. 
is a point on segment . 
intersects 
and 
at 
and 
, respectively. The angle bisectors of 
and 
intersect at . If 
and 
, prove that 
is cyclic.
area of the face opose of vertex 
,
midpoint of edge ![$[XY]\in\{[BC]; [AD];[AC];[BD]\}$](http://latex.artofproblemsolving.com/e/1/5/e15c5aff097ff2fdc4fd8a868ca203c7e635063f.png)
, then:
be the midpoints of 
Since 
is a parallelogram, then 
lie on a same plane 
It's known that 
divides the tetrahedron into two pentahedra with equal volumen, hence:![$\frac{_1}{^2}[ABCD]=[ODCMK]+[ODCLN]+[OCML]+[OKDN] \pm [OMLNK]$](http://latex.artofproblemsolving.com/4/b/6/4b6b4247d21838d27b2ba2af9f85e3e8e75599b6.png)
![$[ODCMK]=\frac{_3}{^4}[OCDB],$](http://latex.artofproblemsolving.com/d/2/f/d2fa75cd74722ba5739a485387b6b7fba76f762c.png)
![$ [ODCLN]=\frac{_3}{^4}[OCDA],$](http://latex.artofproblemsolving.com/f/3/a/f3a598d79481dd88c1536a10e09059f877570769.png)
![$ [OCML]=\frac{_1}{^4}[OABC]$](http://latex.artofproblemsolving.com/0/e/e/0eeba1519359c7006f093796ef48db28c128a02a.png)
and ![$[OKDN]=\frac{_1}{^4}[OABD].$](http://latex.artofproblemsolving.com/7/9/8/7980d72d13a00a3cee4a4fca6950f1c6a2ce35b8.png)
Thus, substituting the volumen ![$[ABCD]$](http://latex.artofproblemsolving.com/c/d/3/cd3361fb9586094387d885d23e9efce948e8b291.png)
(in the first equation) as the sum of the volumes ![$[OCDB],$](http://latex.artofproblemsolving.com/f/1/3/f13c6e7973e49394f400b8f6062e0e0317f398a6.png)
![$[OCDA],$](http://latex.artofproblemsolving.com/8/c/5/8c5f2d385db6bc69cc4cb76f66790a576c35cd37.png)
![$[OABC],$](http://latex.artofproblemsolving.com/f/5/d/f5d64cd0feead81c6f8efa8c011795395f82a3a6.png)
![$[OABD],$](http://latex.artofproblemsolving.com/7/4/c/74cdeda1aac8b5dd96d28bc2cd4300525500f0c5.png)
yields![$[OABC]+[OABD]-[OCDA]-[OCDB]= \pm \ 4 \cdot [OMLNK] \ (\star)$](http://latex.artofproblemsolving.com/5/c/2/5c2e4523503c6e14e4953873cdcaf6616a75dd0a.png)
have equal O-altitudes (inradius of ABCD), then the given condition is equivalent to ![$[OABC]+[OABD]=[OCDA]+[OCDB].$](http://latex.artofproblemsolving.com/4/7/3/4738ea15fb3fdac1d1cc38999270fc351d066a23.png)
Together with 
we get ![$[OMLNK]=0$](http://latex.artofproblemsolving.com/1/5/e/15e03150943b83a68c5ee02009a30c9ff047c417.png)
is a prime and 
,![\[
n! \sum_{pi+j=n} \frac{1}{p^i i! j!} \equiv 0 \pmod{p}.
\]](http://latex.artofproblemsolving.com/1/6/6/166f07aa9e69c0eb35833971acb9b04932fea087.png)
be the center of insphere of a tetrahedron 
equals the sum of areas of faces 
and 
.
and 