TITLE: Intuition: $\frac{1}{p}+\frac {1}{q} = 1$ equivalent to $(p-1)(q-1)=1$.
QUESTION [8 upvotes]: It is easy to show mathematically that if $p,q \ne 0$ then
$$\frac{1}{p}+\frac {1}{q} = 1 \qquad \Leftrightarrow \qquad (p-1)(q-1)=1 .$$
But intuitively, I would never have guessed this relationship.  Does anyone have an intuitive argument that makes this obvious?
Added later: There are two very nice answers that reduce it to
$$ x+y = 1 \qquad \Leftrightarrow \qquad xy = (1-x)(1-y) .$$
They only actually show $\Rightarrow$.
One of the answers does this by reflecting around the line $x+y=1$.  The other answer uses probability, but essentially boils down to saying that if $x+y=1$, then the unordered pair $\{x,y\}$ is the same as the unordered pair $\{1-x,1-y\}$, that is, a reflection which sends $(x,y)$ to $(1-y,1-x)$.  So in some sense the two answers are identical, but expressed differently.
Once you know $\Rightarrow$, you can get $\Leftarrow$ as follows.  Suppose $(1-x)(1-y) = xy$.  Let $z = 1-y$.  From $\Rightarrow$ we get $(1-z)(1-y) = zy$.  Hence
$$ \frac{1-x}x = \frac y{1-y} = \frac{1-z}z $$
Hence $x=z$ and so $x=1-y$.
I would like to accept both answers, but I cannot.  So following principles of distribution of wealth (points), I am accepting the one that comes from the person with less points.

REPLY [2 votes]: You can use trigonometry. 
Denote: 
$$\frac1p =\sin^2 x\iff p-1=\cot^2x\\
\frac1q=\cos^2x \iff q-1=\tan^2x$$