# Automedian triangle

An automedian triangle (black) with side lengths in the proportion 13:17:7, its three medians (brown), and a triangle similar to the original one whose sides are translated copies of the medians

In plane geometry, an automedian triangle is a triangle in which the lengths of the three medians (the line segments connecting each vertex to the midpoint of the opposite side) are proportional to the lengths of the three sides, in a different order. The three medians of an automedian triangle may be translated to form the sides of a second triangle that is similar to the first one.

## Characterization

The side lengths of an automedian triangle satisfy the formula 2a2 = b2 + c2, analogous to the Pythagorean theorem characterizing right triangles as the triangles satisfying the formula a2 = b2 + c2. That is, in order for the three numbers a, b, and c to be the sides of an automedian triangle, the sequence of three squared side lengths b2, a2, and c2 should form an arithmetic progression.[1]

## Construction from right triangles

If x, y, and z are the three sides of a right triangle, sorted in increasing order by size, and if 2x < z, then z, x + y, and y − x are the three sides of an automedian triangle. For instance, the right triangle with side lengths 5, 12, and 13 can be used to form in this way an automedian triangle with side lengths 13, 17, and 7.[2]

The condition that 2x < z is necessary: if it were not met, then the three numbers a = z, b = x + y, and c = x − y would still satisfy the equation 2a2 = b2c2 characterizing automedian triangles, but they would not satisfy the triangle inequality and could not be used to form the sides of a triangle.

Consequently, using Euler's formula that generates primitive Pythagorean triangles it is possible to generate primitive integer automedian triangles as

${\displaystyle |m^{2}-2mn-n^{2}|}$
${\displaystyle m^{2}+2mn-n^{2}}$
${\displaystyle m^{2}+n^{2}}$

with ${\displaystyle m}$ and ${\displaystyle n}$ coprime, ${\displaystyle m+n}$ odd and to satisfy the triangle inequality ${\displaystyle n  or  ${\displaystyle m>(2+{\sqrt {3}})n}$.

There is a primitive automedian triangle that is not generated from a right triangle and this is an equilateral triangle of unit length.

The Euler line of an automedian triangle is perpendicular to one of the medians.[2]

If the medians of an automedian triangle are extended to the circumcircle of the triangle, then the three points LMN where the extended medians meet the circumcircle form an isosceles triangle. The triangles for which this second triangle LMN is isosceles are exactly the triangles that are themselves either isosceles or automedian. This property of automedian triangles stands in contrast to the Steiner–Lehmus theorem, according to which the only triangles two of whose angle bisectors have equal length are the isosceles triangles.[2]

Additionally, suppose that ABC is an automedian triangle, in which vertex A stands opposite the side a. Let G be the point where the three medians of ABC intersect, and let AL be one of the extended medians of ABC, with L lying on the circumcircle of ABC. Then BGCL is a parallelogram, the two triangles BGL and CLG into which it may be subdivided are both similar to ABC, G is the midpoint of AL, and the Euler line of the triangle is the perpendicular bisector of AL.[2]

As with any reference triangle, the triangle whose sides equal the medians of an automedian reference triangle has 3/4 the area of the reference triangle.[3]

## History

The study of integer squares in arithmetic progression has a long history stretching back to Diophantus and Fibonacci; it is closely connected with congruent numbers, which are the numbers that can be the differences of the squares in such a progression.[1] However, the connection between this problem and automedian triangles is much more recent. The problem of characterizing automedian triangles was posed in the late 19th century in the Educational Times (in French) by Joseph Jean Baptiste Neuberg, and solved there with the formula 2a2 = b2 + c2 by W. J. Greenstreet.[4]

## Special cases

Apart from the trivial cases of equilateral triangles, the triangle with side lengths 17, 13, and 7 is the smallest (by area or perimeter) automedian triangle with integer side lengths.[2]

There is only one automedian right triangle, the triangle with side lengths proportional to 1, √2, and √3.[2] This triangle is the second triangle in the spiral of Theodorus. It is the only right triangle in which two of the medians are perpendicular to each other.[2]