[alogo] Fagnano's problem

Next figure gives an intuitive proof of the following problem of Fagnano (1755):
The orthic triangle DEF is the one of least perimeter inscribed in an accute angled triangle ABC.

[0_0] [0_1] [0_2] [0_3] [0_4]
[1_0] [1_1] [1_2] [1_3] [1_4]

Reflect triangle ABC successively 5(6) times along its sides, each time on a different side. Take a side, BC say, and watch its position after each reflexion. After the sixth reflexion it returns to its initial (parallel to it) position. This because the angles by which it revolves sum up to 2pi. The same happens with a side, IK say, of an inscribed triangle IJK. Thus following consecutive sides on the reflected IJK's we obtain a broken line (aquamarine) II''' of length equal to twice the perimeter of IJK. In particular, because of the characteristic property of the (yellow) orthic triangle (the sides/altitudes of ABC are its bisectors), the broken line in this case becomes a straight one of length equal to 2*s, s being the perimeter of the orthic triangle. Line II''' joining the end points of the broken line is parallel and equal to FF'''. Thus, twice the perimeter of IJK is always greater or equal to twice the perimeter of the orthic (2*s).
Besides there is no other triangle with perimeter equal to that of the orthic. This follows from the fact that the middle of II''' is on A'B' only for the case of F' corresponding to the orthic triangle. We have though infinite many double reflected closed paths of length equal to 2*s and sides parallel to those of the orthic triangle. In fact, take any segment like II''' lying totally inside the area of the triangle and its reflected copies and being parallel and equal to FF'''. By taking the reflexions of this segment on the sides of ABC and its reflected copies we bend the segment into a closed broken line path, with sides parallel to those of the orthic and total length 2*s.
The arguments used above prove analogously the following property for polygons inscribed in triangles:
Given an integer multiple of 3, n=3*k > 0, and a polygon p = P0P1...Pn with Pn=P0 having each tripple of successive points on three different sides, we can successively reflect ABC on appropriate sides (each time on the side containing some repeated reflexion of Pi) 2*n-1 times and develop p to a broken line P0...Pn', such that: (i) the segment P0Pn' has length 2*k times the perimeter of the orthic and (ii) is parallel to a side of the orthic triangle.
The solution given here, due to H.A.Schwarz, has some implications for the billiard ball problem, see the references below. Another solution to Fagnano's problem is given in Fagnano2.html .

See Also



Hans Rademacher & Otto Toeplitz Von Zahlen und Figuren Berlin, Springer, 2001, pp. 19-27

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