Intuition: $\frac{1}{p}+\frac {1}{q} = 1$ equivalent to $(p-1)(q-1)=1$. 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.
 A: I think you can get some pictorial intuition if you rewrite $\frac{1}{p}+\frac{1}{q}=1$ as $q+p=pq$. Pretending $p$ and $q$ are positive numbers greater than $1$ for now, you can draw a rectangle with side lengths $q-1$ and $p-1$ with area $A$, whatever it is. Then extend both legs by $1$ "unit", and make a bigger rectangle. You end up getting two "thinner" rectangles of area $p-1$ and $q-1$ on the sides of your original rectangle, plus a little unit square in the corner. I drew a pretty low quality picture below. 
Then geometrically, if $(q-1)(p-1)=1$, i.e., $A=1$, adding the area of four pieces shows the big outer rectangle has area $q+p$, so $pq=q+p$. 
Conversely, if $pq=q+p$, so that the big outer rectangle has area $q+p$, the thin upper rectangle and unit square give you $p$ units of area, the side thin rectangle gives you $q-1$ units of area, so $A$ must equal $1$ to give you the last unit of area. That is $(p-1)(q-1)=1$.

A: It's intuitive if you think of $1/p$ and $1/q$, which we'll denote by $P$ and $Q$. 


*

*The first equation tells you that $P+Q=1$, i.e. you can think of them as probabilities of complementary events, say $P=\operatorname{Prob}(A)$ and $Q=\operatorname{Prob}(\text{not }A)$. 

*The second equation, in terms of $P$ and $Q$ is
$$\frac{1-P}{P}\cdot\frac{1-Q}{Q}=
\frac{\operatorname{Prob}(\text{not }A)}{\operatorname{Prob}(A)}
\cdot
\frac{\operatorname{Prob}(A)}{\operatorname{Prob}(\text{not }A)}=1$$
A: You can use trigonometry. 
Denote: 
$$\frac1p =\sin^2 x\iff p-1=\cot^2x\\
\frac1q=\cos^2x \iff q-1=\tan^2x$$
A: I f we have a line $x+y=1$, ....(1),  then from symmetry the area of the rectangle made by $(0,0);(x,0);(0,y);(x,y)$ should be equal to the area of the rectangle made by $(1,1); (x,y) ;(x,1) ;(1,y)$. 
Because it's just the transformation of the previous rectangle by taking a reflection of the previous image about $x+y=1$ line.
So, $(1-x)(1-y)=xy$.....(2)
Now, taking 
$\frac{1}{p}=x, \frac{1}{q}=y$ we get 
$(p-1)(q-1)=pq$ ......(1)'
As, we started with $x+y=1$ that means $\frac{1}{p}+\frac{1}{q}=1$....(2)'
As (1) and (2) are equivalent so is (1)' and (2)'. 
This way we can establish an intuitive relation between
$\frac{1}{p}+\frac{1}{q}=1$ and $(p-1)(q-1)=pq$.
A: For the first equation to hold, if $1/p$ is close to 1, then $1/q$ is really tiny. Likewise, in the second equation, if $(p-1)$ is close to zero, then $(q-1)$ is really large. 
A: Construct three curves as follows:


*

*$y = \frac{1}{x}$

*$y = -\frac{1}{x}$ and then slide it along the $x$-axis so that at the point $x=p$, the vertical distance from this curve to curve (1) is 1.

*$x = a$ where $a$ is the vertical asymptote of chart (2).
By construction, at $x=p$ the distance between curve (1) and curve (2) is $\tfrac{1}{p} + \tfrac{1}{q} = 1$. And this occurs if and only if the rectangle with corners $(a,0)$ and $(p-1,p-1)$ has area $(p-1)(q-1) = 1$.
Here is the construction for $p=4$.
$\tfrac{1}{p} + \tfrac{1}{q} = 1$ if and only if $(p-1)(q-1) = 1$ for $p = 4$">
Remark ("Funky step") Turn your head 90 degrees and look at curve (2).  You'll see a chart for the function $y = \frac{1}{x}$ in the 'local' coordinates where $(a,0)$ is the origin. To help things, curve (3) is provided as the $x$-axis of this 'local' coordinate system.
Why the shaded rectangle has area $(p-1)(q-1) = 1$
Any rectangle that has one corner at $(a,0)$ and its opposite corner on curve (2) will have area 1.  We construct two such rectangles:


*

*Rectangle $R_p$ has top-left corner at $(a,0)$ and bottom-right corner at $(p-1,p-1)$.  This is the shaded rectangle

*Rectangle $R_q$ has top-left corner at $(a,0)$ and bottom-right corner at the point where the line $x=p$ intersects curve (2). This is the other rectangle drawn with solid lines in the picture 
Then: 


*

*$R_p$ has height $p-1$. Should be visually evident

*$R_p$ has width $q$. Why: $R_q$ has height $\frac{1}{q}$ from the original construction of curve (2).  It follows then that $R_q$ has width $q$.
and therefore


*$R_q$ has width $q-1$. Why: Let $S$ be the rectangle with bottom-left corner at the intersection of $R_p$ and $R_q$, and top-left corner at $(p,\tfrac{1}{p})$. Then $S$ is a square of side length 1.


Therefore $R_p$ is a rectangle of height $p-1$, width $q-1$, and area 1.
