ALTITUDES AND
ORTHOCENTERS
Assignment 8
By Gloria L. Jones
The objective in this assignment is to find the triangle of minimal perimeter that can be inscribed in a given triangle. Using GSP, one can translate vertices of the inscribed triangle and simultaneously record the perimeter. In doing this exploration, the following results were found for the given acute triangle:
Looking at the picture, one can conclude the three vertices of the inscribed triangle appear to fall at the foot of the three altitudes. Construction of the three altitudes nearly confirms this conjecture:
Should this conjecture prove to be true, one could conclude the orthic triangle (formed by connecting the feet of the altitudes) is the triangle of minimal perimeter that can be inscribed in a given acute triangle.
Given any acute triangle ABC and inscribed triangle TRP, reflect point P over segment AB and segment BC to get Pı and P² respectively:
Since ΔPıTU is congruent to ΔPTU, we know segment PıT is congruent to segment PT. By the same reasoning, we have segment P²R congruent to segment PR so it follows that the distance from PıTRP² is the same as the perimeter of inscribed ΔTRP. In order to minimize this distance between these four points, we would need them to be collinear as a straight line is the shortest distance between points.
It is important to find the replacement of point P along segment AC so as to insure PıP² is a minimum. ΔPıBT is congruent to ΔPBT and ΔP²BR is congruent to ΔPBR thus angle PıBP² =2 · angle ABC from the original triangle:
Since angle PıBP² will always equal 2· angle ABC and since angle ABC remains the same since it is part of the original triangle, then all possibilities for ΔPıBP² will be similar to each other by SAS Similarity. Since segment PıB is congruent to segment PB which is congruent to segment P²B, we want segment PB to be as small as possible to minimize segment PıP² and thus know that it must be the perpendicular segment from B to P (an altitude). By a similar procedure, we could show that points R and T of the inscribed triangle must also be along the altitudes. Thus the orthic triangle is the triangle of minimal perimeter that can be inscribed in a given acute triangle.
One must question whether the same is true for obtuse and right triangles. Given an obtuse ΔABC, we notice the orthic triangle in red is no longer considered an inscribe triangle for ΔABC:
Through manipulation of the sketch, we can conclude that the inscribed ΔPTR of minimal perimeter for obtuse ΔABC would be the triangle whose sides closely approach altitude BT, the only altitude that goes through the interior of the original ΔABC (labeled ΔPıTRı on sketch):
Similarly, it has been discovered that the same holds true for right triangles.
Thus the orthic triangle is the triangle of minimal perimeter that can be inscribed in a given triangle ONLY if that original triangle is acute.