000RM.dvi

9.3 Heron triangles with consecutive sides 307

Project: Triangles whose sides and one altitude are in arithmetic
progression

This is an extension of Problem 29 of Isaac Newton’sLectures on Alge-
bra([Whiteside, pp.234 – 237]).
(A) Newton considered a triangleABCwith an altitudeDC. Clearly,
DCis shorter thanACandBC. SettingAC =a,BC = x,DC=
2 x−a, andAB=2a−x, he obtained

16 x^4 − 80 ax^3 + 144a^2 x^2 − 10 a^3 x+25a^4 =0. (†)
“Divide this equation by 2 x−aand there will result 8 x^3 − 36 ax^2 +
54 a^2 x− 25 a^3 =0”. Newton did not solve this equation nor did he give
any numerical example. Actually, (†) can be rewritten as

(2x− 3 a)^3 +2a^3 =0,

so thatx= a 2 (3−^3

√

2), the other two roots being complex. By taking
a=2, we may assume the sides of the triangles to be

, , ,

and the altitude on the longest side to be.
The angles of the triangles are
, ,.

(B) Recalling the Heron triangle with sides 13, 14, 15 with altitude
12 on the side 14, we realize that these lengths can be in A.P. in some
other order. Note that the altitude in question is either the first or the
second terms of the A.P. (in increasing order). Assuming unit length for
this altitude, andx> 0 for the common difference, we have either

1.the three sides of the triangles are1+x,1+2x, and1+3x,or

2.the sides of the triangles are 1 −x,1+x, and1+2x, and the altitude

on the shortest side is 1.

In (1), the area of the triangle, by the Heron formula, is given by

2 =

3

16

(1 + 2x)^2 (1 + 4x).

On the other hand, =^12 · 1 ·(1 +kx)fork=1, 2 , 3. These lead to the
equations