AoPS Academy
, 
, 
, 
and 
according to the diagram. The edges from points 
and 
, respectively are pairwise perpendicular. Prove that ![\[[ABCD]^2+[ABFE]^2+[ADHE]^2=[BCGF]^2+[CDHG]^2+[EFGH]^2,\]](http://latex.artofproblemsolving.com/c/f/b/cfb6dc6726af4711bcfe7827b6531471a75270fb.png)
where ![$[XYZW]$](http://latex.artofproblemsolving.com/9/c/2/9c25ef1b5cfb129136334ed811fb699a84f48a13.png)
denotes the area of quadrilateral 
.
![\[\alpha =1-\cfrac{1}{2a_1-\cfrac{1}{2a_2-\cfrac{1}{2a_3-\cdots}}}\]](http://latex.artofproblemsolving.com/3/0/9/309e0411f481b6d77ccdfe8f18af9852074375aa.png)
where coefficients 
are positive integers, infinitely many of which are greater than 
. Prove that for every positive integer 
at least half of the numbers 
are even.
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
, the maximum number of good cells over all possible starting configurations.
be a given positive integer. Luca, the lazy flea sits on one of the vertices of a regular 
-gon. For each jump, Luca picks an axis of symmetry of the polygon, and reflects herself on the chosen axis of symmetry. Let 
denote the number of different ways Luca can make jumps such that she returns to her original position in the end, and does not pick the same axis twice. (It is possible that Luca's jump does not change her position, however, it still counts as a jump.) if is odd. is even, then![\[P(n)=(n-1)!\cdot n!\cdot \sum_{d\mid n}\left(\varphi\left(\frac{n}d\right)\binom{2d}{d}\right).\]](http://latex.artofproblemsolving.com/3/0/9/30992e5982bce4a032d5e96e0fb5fce8ab6ff21a.png)
, but your problem clearly stated 
.
)
![\[ f(f(x)^2 + f(y)) = x f(x) + y \]](http://latex.artofproblemsolving.com/d/0/2/d022cecec0757daff00434c6a686225ccf384eef.png)
.
such that
.
to get 
for 
.
for 
and 
.
to yield
We then have
If 
and 
for nonzero 
and 
,
This makes 
or 
,
to find 
.
is not constant, we have 
and 
.
denote the assertion, and observe that there are no constant solutions. By 
,
,
,
.
,
is nonconstant, we must have 
.
,
,
,
,
,
,
for some real 
is the only solution.
![\[ f(0) = 1 \]](http://latex.artofproblemsolving.com/8/7/8/8784a89d399b9480a41862bb77e7645094282249.png)
![\[ f^k(x) = f(x\sqrt{k}) \quad \forall k \in \mathbb{N}, \forall x \geq 0. \]](http://latex.artofproblemsolving.com/7/3/1/731d31076d43a297e6a9b249cfffd88bf5871f18.png)
,
and 
if 
gives injective and varying 
.
is a constant value, and since 
and therefore 
.
.
gives that 
,
implies that 
,
.
for some constant 
,
.
from 
to reals where 
and $f^k(x)=f(x\sqrt k).
and 
and 
,
Since 
and 
,
This makes 
.
.
with 
in the given functional equation, we get![\[
g(x)^k = g(kx).
\]](http://latex.artofproblemsolving.com/e/c/2/ec229f3427eadce49a2b42957cd6cf6ee4210516.png)
also we can prove that it's positive always by contradiction, assume 
a 
,
> 0 take a sequence 
as n tends to 
from this we can contradict continuity therefore 
![\[
h(x) = \log g(x).
\]](http://latex.artofproblemsolving.com/7/e/f/7ef39ee05c2bf39ad71ee6a16459195daa8cf2b6.png)
and
,
,
that forms an Arithmetic Progression
Now, set 
Subtracting the two equations, we get that:-
for some 
value and 
be the set of real numbers. Determine all functions ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
such that 
and![\[
\Bigl(f(ab)\Bigr)^2 = \Bigl(af(b)+f(a)\Bigr)\Bigl(bf(a)+f(b)\Bigr) \quad \forall a,b \in (0,1).
\]](http://latex.artofproblemsolving.com/0/f/2/0f2d0a333bdd73ff468072cfd716f94865385710.png)
of pretty well known FEs 
we see 
.
and since 
for arbitrary 
we see 
.
and since 
is the identity.
works for all real 
.
twice to get 
which is equivalent to 
gives 
for nonnegative real 
gives 
(Cauchy FE) for nonnegative real 
so it still works for all nonnegative real 
for all nonnegative rational 
so 
be a positive rational
so 
which is equivalent to 
so 
.
such that![\[ f(x^2 f(y)^2) = f(x)^2 f(y) \]](http://latex.artofproblemsolving.com/d/9/3/d93711733d0b34679f88e7a4d5fa00470dd539d4.png)
for all 
,
denote the assertion.
,
,
.
;
.
,
.
.
is the square of a rational number. Suppose 
-
and 
are 
-
for every 
.