Orthic Triangle
the inscribed triangle of minimum circumference


Fagnano's problem:
In a given acute-angled triangle ABC, inscribe a triangle UVW whose perimeter is as small as possible.

In this proof (by Coxeter) consider any triangle UVW in triangle ABC and reflect UW and UV in AB and AC respectively.

Now: UV + VW + WU = U''V + VW + U'V the path from U' to U'', such a path is minimum when it is a straight line :

Hence for all the inscribed triangles with vertex U on BC the one with smallest perimeter occurs when V and W lie on the straight line U'U'' click here for a GSP animation.

By this technique we can find the triangle of minimum circumference for each choice of U. The problem that remains is to find the position of U for which the U'U'' is a minimum. Now AU' and AU'' are both equal to AU (by our construction) hence triangle AU'U'' is isosceles and its base will be a minimum when its equal sides are a minimum - ie when AU is a minimum - when AU is the shortest possible distance. Clearly this occurs when AU is the altitude from A to BC. Since we could easily have started with B or C rather than A, it follows that CW and BV are altitudes also.


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