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.
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 just ![$[\sum_{i=1}^{n}\sigma_0(i)]-n=\sum_{i=1}^{n}[\sigma_0(i)-1]$](http://latex.artofproblemsolving.com/4/5/5/4555f8d23420fe2d9db1d8b5c288ef33ade9f4c9.png)
. By CRT, there must exist consecutive integers 
that are divisible by 
respectively, where 
is the th prime. Consequently, each 
is divisible by , and each 
is 
mod . Therefore, the 
, reduced mod , span the set {
}, so (exactly) one 
is divisible by .
, so in fact there are infinitely many satisfying your condition.
is just . By CRT, there must exist consecutive integers that are divisible by respectively, where is the th prime. Consequently, each is divisible by , and each is mod . Therefore, the , reduced mod , span the set {}, so (exactly) one is divisible by ., so in fact there are infinitely many satisfying your condition. should not only be a multiple of 
, but actually take the form of 
, where 
. This is something to fix - by specifying the remainder w.r.t 
. But CRT cannot apply directly. I'm still working on this. What do you think?
such that 