Little explorations with HP calculators (no Prime)

[Gerson W. Barbosa]

A more useful formula might be one for the area of the inscribed triangle.

Given the triangles ABC and DEF, where D, E and F lie on AC, AB and BC, respectively, and defining

\(x = \frac{\overline{AB}}{\overline{AE}}\), \(y = \frac{\overline{BC}}{\overline{BF}}\) and \(z = \frac{\overline{AC}}{\overline{DC}}\)


Then the area of the inscribed triangle DEF will be

\(S_{\Delta DEF}=\frac{((x y-x-y) (z-1)+z)\sqrt{(a+b+c)(-a+b+c) (a-b+c) (a+b-c)}}{4xyz}\)



In this formula

\(\frac{\sqrt{(a+b+c)(-a+b+c) (a-b+c) (a+b-c)}}{4}\)

is the area of the triangle ABC.

Notice that the order in which x, y and z have been defined is not important, but the line segments in the denominators must not be adjacent to any other.

These have been derived using the law of cosines and Heron's formula, as shown elsewhere in this thread.

We can use this to compute the area of the triangle DEF inscribed in a right-trangle ABC with sides a = 4, b = 2, c = 2√3, CD = 1, AE = 2/3*√3 and BF = 1:

\(S_{\Delta DEF}=\frac{((2\times 3-2-3) (4-1)+4)\sqrt{(4+2+2\sqrt{3})(-4+2+2\sqrt{3}) (4-2+2\sqrt{3}) (4+2-2\sqrt{3})}}{4\times 2\times 3\times 4}=\frac{7\sqrt{3}}{12}\)

Here we have defined x = 2, y = 3 and z = 4, but x = 3/2, y = 4/3 and z = 2 might also have been chosen, among four other possible sets.

The area of the triangle ABC is implicit in the formula above, (√192)/4 = 2√3, but can be determined more trivially by just multiplying the length of side c by half the length of side b.
Incidentally, the ratio between the larger and the smaller areas is exactly 24/7.

Edited for grammar.