AoPS Academy
is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let 
be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .
and 
, independently at random from the range 
to 
. He then selects an operation out of 
, 
, and 
and performs that operation on them. Find the expected value of the answer.
has a right angle at 
. 
and 
. Point 
lies on 
such that 
and point 
lies on 
such that 
= 
. Find the perimeter of 
.
. A circle O with center C has a radius of 2. Let P be a point on the circle O.
?
?
and be positive real numbers such that![\[
\log_x 64 + \log_{y^2} 16 = \frac{5}{3} \quad \text{and} \quad \log_y 64 + \log_{x^2} 16 = 1
\]](http://latex.artofproblemsolving.com/d/7/9/d79facd90d9c5b53cdc4df9db18923ad34f472df.png)
What is the value of 
? and be positive real numbers such that![\[\log_x64+\log_{y^2}16=\dfrac{5}{3}\text{ and }\log_y64+\log_{x^2}16=1 \]](http://latex.artofproblemsolving.com/6/6/6/666e220af0dd4e1f77283dc5109a59d796c81bda.png)
Find 
be positive integers. How many pairs of 
are there such that 
is rational?
be a function that satisfies
for all positive integer pairs 
Find all possible functions 
as the set of all positive integers.)
initial condition. Taking 
we have 
hence either 
and 
or 
and 
. and .
we get 
. Taking 
we get 
a contradiction. and .
for all positive integers 
. Clearly it is true for 
. Suppose that it is true for 
then taking 
we have 
as desired.
then setting 
implies 
Then but 
, impossible. So and upon iteration, 
; which works.
to get that 
we can know 
so 
gives the value of 
gives the value of 
thus all 
is uniquely determined. In fact, 
works since 
and also in this case thus the only solution in this case if 
we know 
We also know 
so plugging gives 
However, we now have 
contradicting the assumption.
be a function that satisfiesfor all positive integer pairs Find all possible functions as the set of all positive integers.) be the assertion 
.
, then:
, a contradiction..
, by simple induction now we get 
for all 
, which fits. 
and 
Prove that
satisfying
Prove
.
.
.
.
.
from base 
shares angles 
with 
through AA Postulate. 
,
and 
.
,
(as both are right angles), and 
,
through SAS postulate. 
,
.
.
be positive reals such that 
,
of 
.
so done by induction.