The ideal aspect ratio of powerpoint slides

The ideal aspect ratio of PowerPoint slides

Imagine you are in the hypothetical scenario of having to create a presentation but you do not know whether the projector uses the 16:9 or the 4:3 aspect ratio.

If you choose 4:3 for your slides it will waste space on the side of a 16:9 projector viewport like so:

You have those annoying black ("red") bars at either side.
On the other hand when you use 16:9 slides on a 4:3 projector:

You now have the opposite problem: Your slides are too wide for the screen, leaving you with black bars at the bottom and top. (highlighted in red)

Let's say you want to waste an equal portion of screen space in either case. What aspect ratio should your slides have?

Short Answer: $1.5396 : 1$

Long answer:

You want to waste equal Area in either case. In other words:

$\frac{A_{slides on 16:9 projector}}{A_{16:9 projector}} = \frac{A_{slides on 4:3 projector}}{A_{4:3 projector}}$

Without loss of generality we can assume that both projector screens are 9 units tall.
With that we can calculate their Area:

$A_{4:3 projector} = 12 \cdot 9 = 108$
$A_{16:9 projector} = 16 \cdot 9 = 144$

Let's call the perfect aspect ratio $α$ :

$α = \frac{w_{ideal}}{h_{ideal}}$

The size of the slides on the 16:9 monitor will have the same height as the projector screen:

$A_{slider on a 16:9 projector} = w \cdot h = \frac{w}{h} \cdot h^{2} = α \cdot h^{2} = α \cdot {(9)}^{2} = 81 α$

On the 4:3 projector the slides will use the full width:

$A_{slides on a 4:3 projector} = w \cdot h = w^{2} \cdot \frac{h}{w} = \frac{w^{2}}{α} = \frac{144}{α}$

Using our equation from above:

$\frac{A_{slides on 16:9 projector}}{A_{16:9 projector}} = \frac{A_{slides on 4:3 projector}}{A_{4:3 projector}}$

$\frac{81 \cdot α}{144} = \frac{\frac{144}{α}}{108}$

Rearranging:

$α = \sqrt{\frac{144}{108} \cdot \frac{144}{81}} = \frac{8}{9} \sqrt{3} = 1.5396$

There you have it.