Fagnano’s Problem: To inscribe in a given acute-angled triangle the triangle of minimum perimeter.
Why does the geometric proof through reflection only work for acute triangles?
This website provides some reasoning, but it is really confusing. Here is a quote of the two parts that are confusing.
[If ∠C = 90∘, then HK intersects ΔABC at C; if ∠C > 90∘,then ΔCHK exists, but it is outside ΔABC : This is where the hypothesis that ΔABC is acute is needed.]
[If ∠C ≥ 90∘, then there is no minimum for P,but P > 2CZ and P can be made arbitrarily close to 2CZ. ]
http://www2.washjeff.edu/users/mwoltermann/Dorrie/90.pdf
A picture of the proof I refer to: Fagnano's Problem Geometric Proof w/Reflection
In an obtuse triangle, the orthic triangle cannot be inscribed, since two of its vertices lie outside the given triangle. Hence the triangle with minimum perimeter inscribable within an obtuse triangle is not the orthic triangle. We'll need an argument and solution different from that of the geometric proof in question.
[edit]:
To see that in fact there is no triangle of minimal perimeter inscribable in an obtuse triangle, let triangle $ABC$ have obtuse angle at $C$. Taking $X$, $Y$, $Z$ at random, let $K$ and $H$ be the reflections of $Z$ about $CA$, $CB$ respectively. Let $YX$ in triangle $XYZ$ intersect $CZ$ at $M$, and join $MK$, $MH$.
Then keeping $Z$ fixed while moving $Y$ and $X$ in any manner down toward $A$ and $B$, it's clear that $YZ$ approaches equality with $AZ$, $XZ$ with $BZ$, and $YX$ with $AB$. Hence the perimeter of $\triangle XYZ$ can be made arbitrarily close to $2AB$.
Alternatively, moving $X$ and $Y$ upward toward $C$, then since$$KY+YM\ge KM$$and$$HX+XM\ge HM$$then the perimeter of $\triangle XYZ$$$KY+YX+XH\ge KM+MH$$But by Euclid, *Elements* I, 21$$KM+MH>KC+CH=2CZ$$Therefore, the perimeter $P$ of $\triangle XYZ$ approaches $2CZ$ as a minimum, and$$2AB>P>2CZ$$
And we reach the same conclusion when $\triangle ABC$ is a right triangle.