AoPS Academy
, 
is the Gergon point and the incircle 
touch 
, 
, 
at 
, 
, 
respectively.
, 
, 
be 
, 
, 
respectively. in 
and then we get the circle 
in 
and then the circle 
in 
and then the circle 
, , are coaxial.
be non-negative real numbers such that 
Prove that
Prove that 
Where 
(this implies 
) and 
.
s.t. 
whenever 
for some 
, where 
is the sequence of primes, and then extend it to the rest of 
so that 
holds. Then it's clear that will automatically hold.
, this implies 
, hence we must deifne the function on all primes, let 
denote the 
prime number we define 
as:
. This implies that is injective. 
. Let 
, so 
. Then 
implies 
. A function on primes like Amir's works.
and 
is injective.
, we know that 
is surjective, therefore 
exists for all positive rationals 
.
as 
in 
as and as into the original equation and we obtain:
as and as into the original equation and we obtain:.
you infer 
.
.
Plugging in x = 1 we get f(f(y)) = f(1)/y and hence f(y1) = f(y2)
, this implies , hence we must deifne the function on all primes, let denote the prime number we define as:
, this implies , hence we must deifne the function on all primes, let denote the prime number we define as:
for all primes.
. So you have to manipulate the primes in some other method, with a group of elements acting as a medium. (In other words, f(f(p)) maps A 
B C, where you know that {p} = A, B is unknown, and {
} = C.
}, {
}, {
} and {
}. Playing around with directed arrows that map elements between the sets gives you the function.
. It's pretty trivial to see that this function is necessarily bijective, as 
and surjectivity is direct from 
. Again, 
by injectivity. ). around the problem to get 
. Replace 
with 
to get 
.
as 
to get 
.
.
is our domain aka integer prime factorizations- this problem is calling for a number-theoretic way to attack the problem.
in our original problem- simplifying a little and spamming the multiplicative property, we get 
. Hmm... so how do we get the primes to randomly appear in the denominator- oh wait this is trivial, just make a
cycle of sorts- let the 
prime, 
, map to 
if, say, the index is odd, and let it map to 
if even. Now all we have to do is put it back into the equation and check, and this construction indeed seems to work.
and 
.
and 
, 
, and 
, where 
is the 
th prime for all positive integers 
. It's easy to see we can extend this to all positive rationals.. Call a number 1-over-prime if it can be written as for some prime 
. Clearly 
for all primes and 1-over-primes 
. Now we have 
can be written as 
where the 
are some primes or 1-over-primes which multiply up to . 
, as desired.
be the set of positive rational numbers. Construct a function 
such that![\[ f(xf(y)) = \frac {f(x)}{y}
\]](http://latex.artofproblemsolving.com/5/9/1/59172e31cf6b7d885295f6e1daf6891a71739247.png)
for all 
, 
in .
be the assertion.
.
.
Inputting this to get,
.
; (b) solve the equation. mapping through primes, satisfying (a) as,![$f(p^{n_1}_1p^{n_2}_1 \dots p^{n_k}_k) = [f(p_1)]^{n_1} [f(p_2]^{n_2} \dots [f(p_k)]^{n_k},$](http://latex.artofproblemsolving.com/f/4/5/f4575fd98dff6d58d74a68c5cfd7fa9ea79f4a8f.png)
where is the jth prime and 
. Such a function wil satisfy (b) for each prime. A possible solution is: 
Clearly, 
. Thus satisfies.
and 
works, as 
be the 
th prime. is odd, 
is even, 
and are relatively prime positive integers, 
defined such that 
for all 
, where we define
and
where 
are the primes. Note that is uniquely determined due to prime factorisation, and is multiplicative and 
, therefore we're done. 
is surjective. Let be such that 
then 
Thus 
Then 
so is completely multiplicative.
One such solution is pairing primes arbitrarily, say you pair 
and then maps 
for fixed 
We can then construct for other numbers that are not prime powers. This clearly works as is multiplicative and indeed 
such that 
(since 
)