Explanation of a flaw in
Roger A. Johnson, Advanced Euclidean
Geometry, 1929 / 1960 / 2007,
Page 169, §264, Proof of the last Theorem.
The problem lies in the degenerate cases.
What does "It is evident intuitionally
that a minimum exists" mean? And what does "inscribed
triangle" mean?
It is indeed easy to see (by analysis) that, among all triangles
P1P2P3 such that the points
P1, P2, P3 lie on the
straight lines A2A3,
A3A1, A1A2, there
exists one with minimum perimeter. It is also readily shown that
the same holds for triangles
P1P2P3 such that the points
P1, P2, P3 lie on the
closed segments A2A3,
A3A1, A1A2
(a closed segment is a segment with its two
endpoints). But it is absolutely not clear why there is a
triangle with minimum perimeter if the points
P1, P2, P3 are to lie on the
open segments A2A3,
A3A1, A1A2 (an
open segment is a segment without its two
endpoints).
As a consequence of this, whether we want to or not, we have to
take into account the case when some of the points
P1, P2, P3 coincide with
vertices of triangle A1A2A3.
And the proof is not guaranteed to work in this case anymore.
For instance, consider the case when
P1 = A3 and P2 = A3,
while P3 is the foot of the perpendicular from
A3 to A1A2. The construction
used in the proof, even if modified to make sense (it is hard
to speak about the lines P1P2 and
P1P3 making equal angles with
A2A3, since the line
P1P2 is not defined at all, but one can
still reasonably define Q1), fails to yield an
inscribed triangle with greater perimeter than
P1P2P3 in this case.
With some more work, one could characterize such "evil" cases with the result that the triangle P1P2P3 with minimum perimeter is either the pedal triangle of the orthocenter of A1A2A3, or the pedal triangle of the vertex A1, or that of the vertex A2, or that of A3. What remains to be shown is that it actually is the pedal triangle of the orthocenter of A1A2A3, and not one of the three other options. This requires a separate demonstration! (Besides, as we know, this holds only for acute- or right-angled triangles A1A2A3.)
One can fill in these missing steps to obtain a complete proof, but it will not be particularly short anymore. And it uses analysis (or intuition) at the step where the existence of the minimal-perimeter triangle P1P2P3 is postulated. All in all, it is a bad proof.
Errata in geometry books / footnote 1
Darij Grinberg