# Fagnano's Problem

I.F.Fagnano posed in 1775 the following problem

To inscribe in a given acute-angle triangle the triangle of a minimum perimeter.

Fagnano's original solution used Calculus. But, once the answer became known, several purely geometric solutions were also discovered. Two solutions are cited most frequently. I'll add the solutions shortly. Till then, play with the applets and see if you can surmise what those things are about.

### H.A.Schwarz's solution

Reflect ΔABC repeatedly in its edges 5 times. Note that the final position of the side AB is parallel to its original position (the direction is also preserved.)

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Recollect that the orthic triangle has the property that ∠AHbHc = ∠CHbHa, and similarly for the angles at feet Ha and Hc. Therefore, the reflection of HaHb in the side AC is the continuation of HbHc, and vice versa, similarly for the other two sides.

Reflect now ΔABC first in AC, then in (the new position of) BC, and so on, as shown above. The six congruent triangles form a chain-like shape that starts with the side AB and ends with a line segment, which the position of the side AB after 5 reflections. This is also denoted AB. The two segments AB are parallel. (The first reflection rotates AB by trough twice ∠A counterclockwise; the second rotates AB counterclockwise through twice ∠B, then AB remains stationary, rotates clockwise through 2∠A and, lastly, clockwise through 2∠B.)

Let's have a look at what's happening with the inscribed triangles due to the 5 reflections. Because of the mirror property of the orthic triangle, the sides of the orthic triangle stretch into a straight line HcHc which is parallel to the two lines AA and BB. All three lines thus measure two perimeters of the orthic triangle. Sides of any other triangle RQP will form a broken line between two R points. The broken line RR is longer than the straight line RR which is the same length as HcHc. This shows that the orthic triangle solves Fagnano's problem.

The solution is unique because any triangle with the minimum perimeter should possess the mirror property. As we know, the mirror property is a characteristic feature of the orthic triangle.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Reflect point P in AC and AB to obtain P' and P''. The perimeter of ΔPQR equals the sum

P'Q + QR + RP''

Therefore, among all inscribed triangles with fixed vertex P, the one whose vertices Q and R lie on the line P'P'', has the minimum perimeter. Obviously, this triangle has the mirror property at vertices Q and R. Thus, for any position of the point P on BC, there is a unique triangle of minimal perimeter. Solution to Fagnano's problem must lie among such triangles. Its perimeter is given by P'P'' for some position of point P.

Note that ∠P'AP'' equals twice ∠A regardless of the position of point P. Therefore, all triangles P'AP'' are similar. The one with the shortest base P'P'' has also the shortest side AP'. But AP' = AP. Therefore, solution to Fagnano's problem exists and is such that AP is perpendicular to BC. By similarity, BQ must be perpendicular to AC and CR to AB.

### Remark The orthic triangle possesses the mirror property (albeit somewhat different) even in obtuse triangles. In obtuse or right triangles, however, no inscribed proper (with distinct vertices) triangle attains the minimum possible perimeter. There exist triangles whose perimeter is arbitarily close to twice the length of the shortest altitude but no proper triangle attains this value.

### Remark

A third solution provides additional information about the configuration of six triangles that appeared in Schwarz's solution. It's also interesting to observe what becomes of lines parallel to RR when six triangles are folded into the original one. Another solution employs a property of antiparallels.

### References

1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, NY, 1996.
2. H. S. M.Coxeter, S. L. Greitzer, Geometry Revisited, MAA 1967
3. H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
4. H. Dorrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY, 1965.
5. P. J. Nahin, When Least Is Best, Princeton University Press, 2007 (Fifth printing).
6. H. Rademacher and O. Toeplitz, The Enjoyment of Mathematics, Dover Publications, 1990 • 