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.
games in a row, you score 
points instead of 
points, until you hit getting 
points per win, where you just get 8 points per win consistently. A tie equals 
point, and a loss is zero. A tie or a loss will break your win streak.
+1 w
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
be a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let 
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be an acute triangle with 
, and let be the projection from 
onto . Let be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at . Prove that 
.
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
satisfy that: 
and 
. Find the maximum: 
be a convex quadrilateral. Suppose that the circles with diameters and 
intersect at points 
and . Let 
and 
. Prove that the points , , and are concyclic. and are not the diagnols)
heads and 
tails, his score will be: 
. We also know 
. Let's list out possible values of and and see which pairs give to be a perfect square(we will be using complementary counting). If 
, this is impossible. If 
, this is also impossible. If 
, this will give us that 
, which is a perfect square. If 
, this will give us that 
, which is a perfect square. If 
, this is impossible. If 
, this is impossible. So, Mike will only get a perfect square if he gets either heads and 
tails or heads and tails. If Mike gets heads and tails, there are 
possible ways to do this(
factorial ways to arrange the flips, then divide by the factorial ways to arrange the heads and the factorial ways to arrange the tails). Similarly, there are possible ways to do the second case. So we can have a total of 
total possible occurrences for which Mike WILL get a perfect square. The probability is simply 
. Taking the complementary of this, we get our final probability of 
.
\end{table}
such that 
.
:
:
,
.
,
for this case
to get 
