<s>[INST] When are the curves $a =xy$ and $b = (1-x)(1-y)$ tangential? I've seen that the curves $xy = 1/4$ and $(1-x)(1-y) =1/4$ are tangential at the point $(x,y) = (1/2,1/2).$
Similarly I can see that the curves $xy = 1/16$ and $(1-x)(1-y) = 9/16$ are tangential at $(x,y) = (1/4,1/4).$
I therefore have the general conjecture that $a = xy$ and $b = (1-x)(1-y)$ are tangential when $a=z^2$ and $b = (1-z)^2,$ for some $z \in (0,1),$ with the tangential being $(x,y)=(z,z).$
However I try to prove this by solving the system of equations $y = \frac{a}{x}$ and $y=1-\frac{b}{1-x},$ but I am stuck in getting to an expression in terms of a parameter $z,$ as above.
 [/INST] Your conjecture works if there is a specific relationship between values of $a$ and $b$.
First curve is $y = \frac{a}{x}$
Second curve is $y = 1 - \frac{b}{1-x}$
For your general conjecture to be true we should have only one point where both curves meet,
So, $\frac{a}{x} = 1 - \frac{b}{1-x}$ should have exactly one solution for $x$ if they both meet and also they do not intersect.
$x^2 + (b-a-1)x + a = 0$ will have only one solution if its discriminant is zero.
i.e $(b-a-1)^2 = 4a \implies b = a + 1 \pm 2 \sqrt a$ ...(i)
(you have to specifically check whether $+$ works or $-$ works).
This relationship between $a$ and $b$ also means $x = y$ at the point of tangent.
I used your working which assumes $x = y = z$.
So, $z^2 = a \implies z = \pm \sqrt a, (1 - z)^2 = b \implies b = a + 1 \pm 2\sqrt a \,$ which is same as (i).
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