Solution: Balanced division of a sandwich

We publish the solution to the Balanced division of a sandwich divertimento.

The Fun

A sandwich is in the shape of an isosceles right triangle. It is to be cut into two pieces by a single straight cut, so that the two resulting figures have the same area. How should it be cut so that the cut is as short as possible?

The Solution

Consider an isosceles right triangle whose legs measure \(a\). The surface of the sandwich is
$$
S=\frac{a^2}{2}.
$$
There are two different ways to cut the sandwich: from one leg to the hypotenuse, or from one leg to the other.

Let’s suppose, first of all, that we cut the triangle from one leg to the hypotenuse. We obtain a triangle with sides \(x\), \(y\), \(z\). If the sandwich has been divided into two parts of equal area,
$$
\frac{S}{2}= \frac{1}{2} x y \sin 45 = \frac{\sqrt{2}}{4} x y,
$$
from which it follows
$$
xy= \sqrt{2}S = \frac{a^2}{\sqrt{2}}.
$$
Therefore, the product \(xy\) is constant when the two halves have the same area. On the other hand, by the cosine theorem one has that
$$
z^2 = x^2 + y^2 – 2xy \cos 45 =
x^2 + y^2 – \sqrt{2}xy =
(x-y)^2 + (2- \sqrt{2}) xy.
$$
If the product \(xy\) is kept constant, the minimum value of this expression is obtained for \(x=y\), that is,
$$
x=y= \frac{a}{\sqrt[4]{2}}, \qquad z^2=(2-\sqrt{2})\frac{a^2}{\sqrt{2}}=(\sqrt{2}-1)a^2.
$$

Let’s suppose now that we cut the triangle from one leg to the other. We obtain a triangle with sides \(x\), \(y\), \(z\). As in the previous case, if we divide the triangle into two parts of equal area,
$$
\frac{S}{2}= \frac{xy}{2}, \qquad \frac{a^2}{4}=\frac{xy}{2},
$$
from which \(2xy=a^2\). In this case, the product \(xy\) is also constant when the two halves have the same area. From the equality
$$
z^2 = x^2 + y^2 =
(x-y)^2 + 2 xy,
$$
it follows that if the product \(xy\) remains constant, the minimum value of this expression is obtained for \(x=y\), that is,
$$
x=y=\frac{ a}{\sqrt{2}}, \qquad z^2=2xy=a^2.
$$

Finally, since \((\sqrt{2}-1)a^2 <a^2\), we have that the smallest possible cut is the one that goes from one leg to the hypotenuse.