308 The area of a triangle
- fork=1: 48 x^3 +56x^2 +16x−1=0,
- fork=2: 48 x^3 +44x^2 +8x−1=0,
- fork=3: 48 x^3 +24x−1=0.
The casek=3has been dealt with in Newton’s solution.
Fork=2, the polynomial factors as
so that we havex=. This leads to the Heronian triangle
with sides 13, 14, 15, and altitude 12 on the side 14. The angles of the
triangles are
, ,.
Fork=1, it is easy to see, using elementary calculus, that the poly-
nomial 48 x^3 +56x^2 +16x− 1 has exactly one real root, which is positive.
This gives a similarity class of triangle with the three sides and the
altitude on the shortest side in A.P. More detailed calculation shows that
the angles of such triangles are
, ,.
Now we consider (2), when the altitude in question is the second term
of the A.P. Instead of constructing an equation inx, we seek one such
triangle with sides 15,17 + 2z,18 + 3z, and the altitude16 +zon the
shortest side. By considering the area of the triangle in two different
ways, we obtain the cubic equation
z^3 − 120 z+16=0. (∗)This can be solved by writingz=4
√
10 sinθfor an angleθ. Using the
trigonometric identitysin 3θ=3sinθ−4sin^3 θ, we reduce this to
sin 3θ=so that the positive roots of (∗) are the two numbers
z= ,.We obtaintwosimilarity classes of triangles, respectively with angles
, , ,
and
, ,.
There are altogetherfivesimilarity classes of triangles whose three
sides and one altitude, in some order, are in arithmetic progression.