The Triangle of Minimal
Perimeter

By:

Alex Moore

Our goal in
this exploration is to prove the following: Given an acute triangle ABC, the Orthic triangle is the
inscribed triangle of minimal perimeter.

*Proof*: Let triangle ABC be any acute triangle, and let triangle DEF
be any inscribed triangle in ABC.
Let us first consider the placement of the point D on the segment
AB. To help us ÒunfoldÓ this
problem, let us first reflect D across the line BC and across the line AC. Let DÕ and DÓ denote the reflections of
the point D.

First
notice that triangle DÕED is isosceles by definition of reflection and so the
length of DÕE is the same as the length of ED. For the same reason the length of DF is also equal to the
length of DÓF. Therefore, DÕC =
DÓC. Since the perimeter of DEF is
equal to DF+FE+ED, we now have the perimeter of DEF is equal to
DÓF+FE+EDÕ.

Using
the old fact that the shortest distance between two points is a straight line,
the perimeter of DEF will be minimize, for fixed position of D, when both E and
F lie on DÕDÓ, so we move E and F to lie on DÕDÓ.

With
E and F in place we now shift our attention back to D. Our next task is to position D such
that DEF has minimal perimeter.
Clearly moving D with alter the lengths of DÕE and DÓF, however, the
angle DÕCDÓ with remain at a constant value of twice the angle measure of angle
BCA by properties of reflection.
As D translates along AB the length of DÕDÓ will change as well. Since the altitude from C to AB is the
shortest distance from C to AB, moving D to this point yields the shortest
length of CD, and therefore the shortest length of DÕC and CDÓ as well due to
reflection. Since triangle DÕCDÓ
is isosceles by reflection, this also gives the shortest length of DÕDÓ. Therefore, Moving D to the intersection
of AB with the altitude minimizes the perimeter of DEF since the perimeter has
length DÕDÓ. By the symmetry of
the problem we can apply the same construction to E and F to see these points
should also be at the foot of the altitudes.

Therefore,
the triangle inscribed in ABC of minimal perimeter is the Orthic triangle!

Now that we
have proven the result for acute triangles, does this work for obtuse
triangles? No! Why is that? If the triangle is obtuse then at least one of the altitudes
is outside of the triangle and therefore at least one of the vertices is
outside of the triangle. Hence the
Othic triangle is not an inscribed triangle for obtuse triangles.

Does the
result hold for right triangles?
No! For right triangles the
Orthic triangle is degenerate.
Therefore, if triangle ABC is acute, then the Orthic triangle is the
inscribed triangle of minimal perimeter.