Errata in geometry books
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
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