Fagnano's Problem III

Samelson's solution

Fagnano's problem seeks to inscribe into a given acute triangle a triangle of the least perimeter. We have already discussed two classical solutions due H. A. Schwarz and L.Fejér. Following is a third solution communicated to me recently by Hans Samelson, Professor Emeritus of Mathematics, Stanford University.

Reflect ABC twice in its sides as if rolling it each time over a side. Let's call the three thus obtained triangles t1, t2, t3 in the order of creation starting with the leftmost. For the vertices I shall use "dot" notations: t1.A, t1.B, t1.C are the three vertices of the first triangles, and so on. So that, for example, t1.C = t2.C = t3.C. Similarly, the sides of, say, triangle t1 are denoted as t1.a, t1.b, t1.c with the usual convention that side a lies opposite to vertex A, and so on.

Extend sides t1.c and t3.c till they intersect. Note that the angle between them is 180^o - 2C, so that they never parallel. Draw the angle bisector m at the point of intersection that crosses the triangles. (It's easily seen that m passes through t1.C = t2.C = t3.C. Indeed the distances from this point to t1.c and t3.c both equal the altitude h_c in ABC.) Project points t1.B and t3.B on m and continue the lines till they intersect t1.c and t3.c in points B'.

t3.c is obtained from t1.c with two reflections: one in t2.b, the other in t2.a. Therefore one is obtained from the other with a distance preserving transformation. This transformation naturally extends to the lines of t1.c and t3.c. Then it is clear that the two points B' are images of each other under that transformation. Because of the distance preserving property the midpoint of B't1.B maps into the midpoint of B't3.B. Denote those points M. The line MM is perpendicular to m and points M (t1.M and t3.M) map into each other.

If triangles t3 and t2 are folded back into triangle t1, the three pieces of line MM - its parts inside triangles t1, t2, and t3 - form an inscribed triangle with mirror property. It therefore coincides with the orthic triangle of t1. As we already know, the orthic triangle solves Fagnano's problem. Which in particular means that MM is the shortest of all lines connecting pairs of points on t1.c and t3.c that are the images of each other under the transformation introduced above. If we can show this directly, we shall obtain an additional solution to Fagnano's problem.

Form triangles t4, t5, and t6 as in H. A. Schwarz's solution. A point t1.P corresponds to point t3.P = t4.P which, in turn, corresponds to point t6.P. Among all points P, the point M has the property that the broken line connecting its 3 locations is in fact straight. (This is easy to see by just repeating the construction of M for triangles t4, t5, t6. ) All others consist of two legs - the sides of an isosceles triangle with base PP. As all such bases are parallel and equal, we conclude that among all lines joining points P on t1.c with their image on t3.c that which connects points M is the shortest. Q.E.D.

Note

Point t3.M = t4.M is located half way between t1.M and t6.M that lie on equal and parallel segments t1.c and t6.c. Therefore, t3.M can be constructed at the intersection of t3.c and the line parallel to t1.c midway between t1.c and t6.c. This is because among all lines PP that fold into hexagons, the line MM folds into a triangle (the orthic triangle) covered twice. Therefore, the two legs of MM have equal lengths.

Fagnano's problem

• Schwarz's solution
• Fejér's solution
• Samelson's solution
• Izvolsky's solution
• Fagnano's Problem in Reverse

Copyright © 1996-2009 Alexander Bogomolny