Povray in action.

Construct a line tangential to two given circles.

Given a square and an arbitrary length $l$, we’re going to construct a rectangle with one of its edges being of length $l$ such that the areas of the square and the rectangle coincide.

For a triangle $\bigtriangleup_{ABC}$ its *orthic* is defined to be the triangle created by the foot of the altitudes dropped from the vertices of $\bigtriangleup_{ABC}$. We will see that under all inscribed triangles, the orthic has minimum perimeter.

The app *Euclidea*, which I have mentioned in the last post, has an interesting level 4.2. I say interesting, because I was close to beaking my head while trying to solve it for the first time, although its solution is surprisingly easy.

Let there be given a pasture with circular shape and diameter $r$ and let a sheep be attached to a point on the perimeter by a rope of length $l$. Then, how long does $l$ need to be so that the sheep can reach exactly half the pasture?

Let’s do some very simple maths. Assume that we are given a *closed* curve $\gamma \subset \mathbb{R}^2$, parametrised via its arc-length $r \in [0,s]$, and two functions $f$ and $g$ defined there. We make the additional assumption that

$$\int_0^s f( r ) dr = 0$$

and we thereby express that the *mean value* of $f$ vanishes.