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.
be an acute-angled triangle with 
, let 
be its circumcentre. The line through 
perpendicular to 
intersects circle 
again at 
. The tangents at 
and 
of intersect at 
. 
intersects at 
. 
intersects 
at 
. Let 
be the midpoint of 
.
are concyclic.
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence. grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.![\[f(xy)+f(xf(x^2y))=f(x^2)+f(y^2)+f(f(xy^2))+x \]](http://latex.artofproblemsolving.com/d/7/b/d7bf030b047d49477eba887e8bb78c6e5b126fe7.png)
holds for all 
.
be 
real numbers such that 
for all 
and such that 
Find the maximum possible value of the expression
sides are colored gold or silver. Prove that there are at least 
different isosceles triangles whose vertices have the same color.
such that 
. Prove that![\[\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq 255\cdot abcd.\]](http://latex.artofproblemsolving.com/d/e/6/de61611f628ea796c706d05941d7ee9b3fc22916.png)
[hide=P#]a1b2c3[/hide] [hide=S#]a1b2c3[/hide]
grid is made. 
unit squares are created by separating the side with length 
in portions and separating the side with length 
in portions. Find, with proof, a formula for how many rectangles with sides parallel to the grid can be created in terms of and . For example, when 
and 
, it can be found by casework that 
rectangles can be created.
such that 
.
Where 
.
such that for all 
:o
so 
and since 
is just 
therefore 
so 
,![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
to 
have the digit 
and are divisible by 
?
now, we have covered all the numbers in the 7 ones place. (27, if I counted right) now, lets find numbers with tens digit 7 that do NOT have 7 in its one place, since we already counted the _77 in the first case.
this totals up to 25 numbers, so the total is
so 
?
without a calculator.
for 
:
:
.
and 
where the rest has been omitted because of the floor function. Now 
and multiplying by 
yields 
grid, each unit square is created by dividing the sides of lengths 
and 
.
horizontal lines and 
vertical lines:
that can be formed, we multiply the number of ways to choose the horizontal lines and the ways to choose the vertical lines:
satisfy 
,
and 
are positive integers, what is the probability that 
is divisible by 
?
,
have?
not 
has the most factors?
?
.
,
represents a distinct digit from 
to 
.
is our answer. (By python, 
does not work and surprisingly, 
is the only permutation number that is divisible by 
8907