Witch Hat is Taller

Here's a problem ladened with shortcuts and assumptions. Present it and ask students what they need to know. Some of it will likely turn out to be superfluous, which might become apparent in the working.

Assumptions:

The cowboy's hat is hemispherical, with the annulus meeting it's great circle.

The witch's hat is conical.

Both hats have zero thickness.

Required information:

A variable for the radius of the head circumference, which I will call \(x\).

Solution

First off, the surface area of the inside and bottom of the hats are the same as the top of the hats, so focus purely on that. The annuli (rim part of the hats) are also the same, so we can now focus only on the hemisphere and cone.

The hemisphere has surface area \(\frac{1}{2} 4\pi x^2 = 2\pi x^2\)

If the cone has the same radius and height \(h\) then the surface area is \(\pi xl=\pi x \sqrt{h^2+x^2}\).

Equating these, we have \(2 \pi x^2 = \pi x \sqrt{h^2 +x^2}\\ \implies 2x=\sqrt{h^2 +x^2}\\ \implies 4x^2=h^2+x^2\\ \implies 3x^2=h^2\\ \implies \sqrt{3}x=h\).

The required ratio is therefore \(1:\sqrt{3}\)