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 $C$.
$\mbox{The orthic of an acute triangle}$
So, here’s the picture and we’ll prove the assertion by construction. Given an arbitrary inscribed triangle, we wish to show that its perimeter will become smaller when we move one of its vertices towards the foot of the dropped altitude from the opposite’s vertice of $\bigtriangleup_{ABC}$. That is the same as saying that the perimeter of an orthic becomes larger when one of its vertices is moved. We thus start with the above sketch and keep $P$ and $Q$ fixed while we allow $R$ to move along $\overline{AC}$.
$\mbox{Arbitrary inscribed triangle}$
We then mirror $\overline{QP}$ along $\overline{BC}$ and $\overline{PR}$ along $\overline{CA}$ to get an »exploded« variant of the triangle, which is a polygon with length equal to the perimeter of $\bigtriangleup_{PQR}$.
$\mbox{»Exploded« inscribed triangle}$
The claim follows from comparing this polygon with the »exploded« variant of the orthic, which is a straight line.
$\mbox{»Exploded« orthic}$
Fine. Now, this is very closely related to Euclidea’s level 8.1.
$\mbox{Only $Q$ is given}$
The app prescribes an angle and one point $Q$ and then asks us to construct a triangle with the other two vertices lying on the respective legs in such a manner that its perimeter becomes minimal. In short, the task is to construct an orthic, which is easy. We draw a line connecting the given vertice $A$ and the point $Q$, then we draw its vertical and thereby identify $C$ and $B$. These in turn give $P$ and $R$.