AoPS Academy
if it is not parallel to any of the 
–axis, the 
–axis, or the line 
.
be a given integer. Determine all nonnegative integers 
such that there exist 
distinct lines in the plane satisfying both of the following:
and 
with 
, the point 
lies on at least one of the lines; and of the lines are sunny.
+1 w
denote the set of positive integers. A function 
is said to be bonza if![\[
f(a)~~\text{divides}~~b^a-f(b)^{f(a)}
\]](http://latex.artofproblemsolving.com/7/b/b/7bb6b2b4fcde900ee6f7ef90b906525d61312796.png)
for all positive integers and .
such that 
for all bonza functions 
and all positive integers .
be a circle and 
be circles internally tangent to at 
and 
. Assume that 
and 
are also externally tangent at point 
. Prove that the line through perpendicular to 
meets line 
on .
which is known to both players. On the th turn of the game (starting with 
) the following happens: is odd, Alice chooses a nonnegative real number 
such that![\[
x_1 + x_2 + \cdots + x_n \le \lambda n.
\]](http://latex.artofproblemsolving.com/f/5/e/f5e4f751946952d021736c6eafa0b7ec02d02055.png)
[*]If is even, Bazza chooses a nonnegative real number such that![\[
x_1^2 + x_2^2 + \cdots + x_n^2 \le n.
\]](http://latex.artofproblemsolving.com/f/f/4/ff40463336f4c5dc7381b14e71abe7d45b78a1c0.png)
[/list], the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players. for which Alice has a winning strategy and all those for which Bazza has a winning strategy.
be a cyclic quadrilateral such that 
. Let 
, and let 
be the incenters of triangles 
and 
, respectively. Let 
and 
be the midpoints of 
and 
, respectively. Prove that the segment 
bisects the segment 
.
have an altitude 
and a median 
. Points 
and 
lie inside 
such that 
and 
. Prove that the points 
lie on a circle.
with 
. The tangent line of 
at 
intersect the tangent lines of at 
and 
at 
and 
respectively. The line 
intersects at 
. It is known that 
. There also exists a point 
(not located on the same side of 
that contains ) such that 
and 
. Determine the value of 
.
denotes the circle that passes through points 
.
![\[
\cos \frac{\pi}{11} - \cos \frac{2\pi}{11} + \cos \frac{3\pi}{11} - \cos \frac{4\pi}{11} + \cos \frac{5\pi}{11}.
\]](http://latex.artofproblemsolving.com/7/5/5/7555e8d91752a703c7f866e2d080fe1c569ce486.png)
multiplied by 
and used telescoping, but I ended up with 
, which doesn't simplify neatly. Is there a cleaner approach?
.
,
,![$\angle BAC=\dfrac{4}{11}\pi \text{[rad]}$](http://latex.artofproblemsolving.com/f/d/3/fd3d005aca8bd78f5418c92d6dbf5a673d0c7e6e.png)
,![$\angle BAD_1=\dfrac{2}{11}\pi\text{[rad]}$](http://latex.artofproblemsolving.com/9/1/5/915cd6954323f3ba27edce08abbd4011abe4c66b.png)
,![$\angle CAD_3=\dfrac{3}{11}\pi\text{[rad]}$](http://latex.artofproblemsolving.com/e/e/3/ee390eff22145ec701ade0a9a59f7bc0ad9b2add.png)
,.
and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
be the set 
Let be the number of sets of two non-empty disjoint subsets of 
(Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when is divided by 
is valid while 
is not)
digit appear 
times: 
number
digit 
which 
appear 
times, 
appear times: 
number digit which appear 
times: 
number digit 
which appear times:
number
number
digit appear times: number digit which appear times, appear times: 
number digit which appear times: number digit which appear times: number
as there are three pairs of two digits each. This gives 
which is too big in my opinion. (There are other cases also but i think i am going wrong here itself)
in first step, in third step and in fifth step is difference with the way that you select example in first step, in third step and in fifth step. It is wrong because this way is same:select number 
, then change place this number (you count 
is number of way to change place couple this number)
if it satisfies conditions 
and 
. numbers 
there are two possible subcases i.e 
and 
numbers
numbers 
. numbers 
distinct digits.
digits, which is against our limitations.
: The number has distinct digits.
.
: The number has distinct digits. subcases:
: The number has 
of one digit and of another.
.
: The number has of each digit.
.
.
: The number has distinct digits.
.
.
number and total is