This posting contains large animations.

We’re going to encode a video to HEVC, and we basically have two choices what hardware to use for that task: CPU or GPU.

Povray in action.

We’re going to 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. In the following sketch, two circles with same radius $\overline{AC}$ intersect in such a manner that their respective centers are located on the other circle’s boundary.

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.
Of course, the triangle needs to be acute, as for obtuse triangles, the orthocenter lies outside $\bigtriangleup_{ABC}$, and for a right triangle, the orthocenter coincides with either $A$, $B$, or $C$.

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.
Suppose that we’re given a straight line and an external point $P$. The exercise is to construct a second line through $P$ which cuts the given one at $\delta = \displaystyle\frac{\pi}{3} = 60^°$.

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? The answer is $$ \frac{l}{r} = {\verb+1.15872847301812+} $$ and the value can only be achieved by numerical means. Everything needed to derive this result is depicted and named in the following sketch.

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. We set

There’s a problem with Mail::GnuPG.pm for which I have filed a bug-request: