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such that for all positive real numbers 
and 
, the following equation holds:![\[
\frac{x + f(y)}{x f(y)} = f\left( \frac{1}{y} + f\left( \frac{1}{x} \right) \right)
\]](http://latex.artofproblemsolving.com/7/d/f/7df25755bc33bef970fffb392fb5b6b789390881.png)
such that for all 
and 
,
such that 
or 
, there exist integers![\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]](http://latex.artofproblemsolving.com/6/8/e/68e3d73edec43b2cc480832adf23bbbb5f4c7bee.png)
such that![\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]](http://latex.artofproblemsolving.com/2/9/e/29edf6ab8cee2a526e7753a140c5a291e987856a.png)
be distinct real numbers such that![\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]](http://latex.artofproblemsolving.com/5/1/e/51eef77358db2b6c98f0fb0fb4e30db837a03642.png)
Find the maximum possible value of 
.
. 
.
is tangent to 
.
and 
Thus, 
and we are done with this claim. 
is similitude center of 
and 
. with radius . Due to the first claim, we have 
. Therefore, 
and 
will swap the places. Clearly, 
and 
will also swap the places and 
will stay the same. Thus, 
and 
will swap the places. It is well-known that if there exists an inversion that swaps two circles, then center of the inversion is also one of the centers of similitude of those two circles and thus we are done. 
are collinear. 
be a triangle. 
and 
are points, such that 
,
= 
and 
and 
passes through the fix point.
