AoPS Academy
, 
, 
cut three common chords concurrent at 
. Consider two distinct circles 
, 
which are internally tangent to all 
. Determine, with proof, which of the following two statements is true. is the insimilicenter of and is the exsimilicenter of and .
+2 w
be a parallelogram of center 
. Prove that for any point 
, there exist unique points 
and 
such that is the center of mass of 
.
+1 w with 
be given . On the extension of the segment 
beyond 
, a point 
is chosen arbitrarily. Let the line through and 
intersect the line 
at the point 
. Let 
be the intersection of the lines 
and 
. Prove that the line 
is tangent to the circumcircle of the triangle 
.
such that for all positive real numbers 
implies 
or 
,
,
for all real 
does not exist since 
is only defined for positive reals.
which is given neither.
be the assertion of the equation.
and 
,
so that 
,
such that 
.
and 
to get 
.
,which gives that 
.
is constant!
but since 
is 
we have 
hence we get a solution 
(never seen this before)
denote the assertion.
.
and 
we obtain 
,
.
denote the set of all 
such that 
for all 
.
,
we obtain 
,
,
be arbitrary and compare 
,
.
For fixed 
and 
,
the above fraction spans the interval 
.
to be arbitrarily small it follows that all of 
is in 
since 
,
.
which is impossible over 
for which 
.
for 
with 
.
and 
we get that![\[f(rx + f(1)) = f(x + f(1)) = f(rx + rf(1)) \implies f(x + f(1)) = f(x + rf(1))\]](http://latex.artofproblemsolving.com/b/8/9/b899b0ac342e1edfcf0bbe1355dfdef96e4c04bc.png)
which means 
.
with large enough 
for periodicity to get![\[f(xy + yT) = f(xy).\]](http://latex.artofproblemsolving.com/2/f/a/2fa573094e6a91fe979d327174ae84799ecb834d.png)
Setting 
here would yield 
.
term will be constant as well. Simplifying the equation, we would get 
.
for all 
and 
.
such 
for all 
it follows that ![\[ f(1 + f(x)) = f(1) \]](http://latex.artofproblemsolving.com/9/b/f/9bf23817bc729cf5d8c2b1a26cf0f0ef49cb7e8f.png)
so 
.
.
it follows that ![\[ f(sx + f(y)) = f(x + f(y)) \]](http://latex.artofproblemsolving.com/3/b/e/3bea5701eb96f89451594dc3695c6c73dba1ce56.png)
so![\[ \frac{sx + f(y)}{x + f(y)} \in S \]](http://latex.artofproblemsolving.com/1/2/c/12c751efda2b7b4533f4ca7bf9d2b2be8f11a36c.png)
which has the image of interval 
as 
gives the result. 
for 
gives that 
.
.
to prove that it works for 
too.
and 
.
Switch 
of the equation 
we have 
Let 
then 
then 
.
we have 
.
,
And so 
.
then 
.
then there exist 
such that 
and so 
.
.
.![\[ f(ax) = f(bx) \implies f(x) = f(rx), r=\frac{b}{a} > 1. \]](http://latex.artofproblemsolving.com/5/6/2/562850d6ae42843a2e6845ca99a386dd66ef38f9.png)
.
.
for all 
.
and we can increase indefinitely to cover the whole 
.
is a constant, contradiction.
.
,
,
,
.
such that 
.
and 
gives us that 
.
,
.
.
,![$[1,r]\in S$](http://latex.artofproblemsolving.com/0/9/b/09bde435a713191d6c4f0dcb542316023b419b0c.png)
.
.
we have 
.
be a positive real. Then, we can set 
so that 
.
for any positive integer 
and hence ![$[1,r^n]\in S$](http://latex.artofproblemsolving.com/f/5/0/f500639f2a0a387c09c82e6919c80aca77855e95.png)
for any 
.
with 
,
which means 
.
,
and 
such that 
.![\[f(x+f(y)) = f(rx+f(y)).\]](http://latex.artofproblemsolving.com/7/f/d/7fd1925816a584b69dc6ff99014a266fab43c9c4.png)
.
yields![\[f(xy) = f(xy+yp),\]](http://latex.artofproblemsolving.com/0/f/8/0f8f965bf5b9a29b9fd59afdfe95c139079997ad.png)
![\[f(f(1)+1) = f(1) \implies f(1) = 0,\]](http://latex.artofproblemsolving.com/b/1/8/b18f1c3ba29ac79972004a9bc000a0e12d5837b7.png)
give 
.
.
.
.
for fixed 
.
for 
.
for 
give 
so for any 
we have 
.
then 
for 
large enough so that 
.
for some constant 
.
.
,
.
gives![\[ f(x + f(y)) = f(cx + f(y))\]](http://latex.artofproblemsolving.com/e/2/2/e221d1d899be80d1d996ba65abb3c06ea1289a4c.png)
Thus, 
,
.
,
.
so that for all 
,
(possible by above).
,
gives that 
.
,
.
and integers 
,
.
,
.
for any positive integer 
,
gives that 
,
,
,
.
![\[P(1, y)\]](http://latex.artofproblemsolving.com/9/7/6/9766cd2065738435858437b00850c537a4b0bb10.png)
![\[f(1+f(y))=f(1)\]](http://latex.artofproblemsolving.com/d/6/8/d685c18464262c12297d43ee65eea4a11e7865dc.png)
![\[P(x, 1+f(y))\]](http://latex.artofproblemsolving.com/d/9/1/d9150134b3736b924e65a1d4b21668010bc0080c.png)
![\[f(1)f(x+f(1))=f(1)f(x(1+f(y)))\]](http://latex.artofproblemsolving.com/7/d/0/7d0ea0aeb3a60fea534225e0ca9b756f5e5c79e7.png)
![\[P(x, 1)\]](http://latex.artofproblemsolving.com/1/1/b/11b17463c0a09adb906c249c4b957ce3e26f3d63.png)
![\[f(1)f(x+f(1))=f(1)f(x)\]](http://latex.artofproblemsolving.com/4/8/1/481f2d4d03cb2772ae9b0f9cfa7bbb54b4008ced.png)
Thus 
when 
![\[P(kx, y)\]](http://latex.artofproblemsolving.com/a/4/b/a4bcda851b16f81c96ba5691b6ed2225bf0382ac.png)
![\[f(y)f(kx+f(y))=f(y)f((xy))\]](http://latex.artofproblemsolving.com/4/f/8/4f86290a05f581fec6dcc1ce3653a98f9b31e0b1.png)
Thus 
thus let 
so 
.
for any ![\[P(x+c, y)\]](http://latex.artofproblemsolving.com/9/4/2/942efc9a5bb7c7e66cf27e4267392107e42d04f1.png)
![\[f(y)f(x+f(y))=f(y)f(xy+yc)\]](http://latex.artofproblemsolving.com/4/3/e/43e06fd3945c69627534924206e5aaa08012a9d8.png)
Thus 
for any 
and this tells us that 
Taking 
we get that 
Let 
It is known 1 isn't the only element in 
which gives 
Hence, 
and some fixed 
Taking 
and 
we get 
which implies 
This implies 
for all 
.
gives that 
because if 
)
and 
where 
gives 
.![\[\frac{saf(1)+f(1)}{af(1)+f(1)}=\frac{sa+1}{a+1}=s-\frac{s-1}{a+1}\in S.\]](http://latex.artofproblemsolving.com/a/3/7/a37574d153255301c70c77bb34e785d952b2222d.png)
This achieves everything between 
and