Unit III: The Inverse Reconstruction Problem & Side-Length Cubic (§§ 19–24)

Leonhard Euler — An Easy Solution of Certain Very Difficult Geometric Problems (1765/1767, pp. 114–118)

[Folio § 19 | p. 114]

§ 19. Dimension-2 Auxiliary Invariants P, Q, R

So that we may now recall these formulas to greater simplicity, let us set \(4pq - p^3 - 8r = 4s\), so that \(4\mathcal{A}^2 = ps\) and \(4q = p^2 + \frac{8r}{p} + \frac{4s}{p}\); next let us make:

\[\frac{r^2}{ps} = R, \qquad \frac{r}{p} = Q, \qquad \text{and} \qquad p^2 = P,\]

so that \(P, Q, R\) are quantities involving two dimensions [of length]. Since therefore from this we have \(\frac{s}{p} = \frac{Q^2}{R}\) [printed \(\frac{-s}{p}\) in the folio], we shall obtain:

\[p = \sqrt{P}, \qquad q = \frac{1}{4}P + 2Q + \frac{Q^2}{R}, \qquad \text{and} \qquad r = Q\sqrt{P},\]

[Folio § 19 (cont.) | p. 115]

and \(4\mathcal{A}^2 = \frac{PQ^2}{R}\), and our six intervals will be so expressed:

\[\begin{split}\text{I.} \quad EF^2 &= R - \frac{2}{9}P + \frac{16}{9}Q + \frac{8Q^2}{9R} \\ \text{II.} \quad EG^2 &= R - \frac{1}{4}P + 2Q + \frac{3Q^2}{R} \\ \text{III.} \quad EH^2 &= \frac{9}{4}R - \frac{1}{2}P + 4Q + \frac{2Q^2}{R} \\ \text{IV.} \quad FG^2 &= +\frac{1}{36}P - \frac{8}{9}Q + \frac{5Q^2}{9R} \\ \text{V.} \quad FH^2 &= \frac{1}{4}R - \frac{1}{18}P + \frac{4}{9}Q + \frac{2Q^2}{9R} \\ \text{VI.} \quad GH^2 &= \frac{1}{4}R - Q.\end{split}\]

§ 20. Statement of the Unified Inverse Problem

Since therefore any three of these four points—provided one does not take the collinear triad \(E, F\), and \(H\)—already contain the determination of the fourth, a single problem results, which runs as follows.

Problem (Tab. II, Fig. 5). Given in position these four points assignable in any triangle—1° the intersection of the perpendiculars drawn from the several angles upon the opposite sides \(E\), 2° the center of gravity \(F\), 3° the center of the inscribed circle \(G\), and 4° the center of the circumscribed circle \(H\)—to construct the triangle.

Which problem we shall be able to resolve concisely enough from the properties of these points elicited thus far.

§ 21. Rational Resolution for R, Q, and P

Solution. Since the position of these four points is given through their mutual distances, let us call:

[Folio § 21 (cont.) | p. 116]

\[GH = f, \qquad FH = g, \qquad \text{and} \qquad FG = h,\]

and we know that \(EF = 2g\) and \(EH = 3g\), and likewise \(EG = \sqrt{6g^2 + 3h^2 - 2f^2}\). Now therefore we immediately have these three equations:

\[\begin{split}\text{I.} \quad f^2 &= \frac{1}{4}R - Q \\ \text{II.} \quad g^2 &= \frac{1}{4}R - \frac{1}{18}P + \frac{4}{9}Q + \frac{2Q^2}{9R} \\ \text{III.} \quad h^2 &= \frac{1}{36}P - \frac{8}{9}Q + \frac{5Q^2}{9R} \quad [\text{printed } \frac{6Q^2}{9R} \text{ in folio}],\end{split}\]

from whose resolution we collect:

\[\begin{split}R &= \frac{4f^4}{3g^2 + 6h^2 - 2f^2}, \qquad Q = \frac{3f^2(f^2 - g^2 - 2h^2)}{3g^2 + 6h^2 - 2f^2}, \\ P &= \frac{27f^4}{3g^2 + 6h^2 - 2f^2} - 12f^2 - 15g^2 + 6h^2,\end{split}\]

whence we obtain:

\[\frac{Q^2}{R} = \frac{9(f^2 - g^2 - 2h^2)^2}{4(3g^2 + 6h^2 - 2f^2)}.\]

§ 22. Formation of the Side-Length Cubic Equation

Having found these values, let the three following expressions be computed:

\[p = \sqrt{P}, \qquad q = \frac{1}{4}P + 2Q + \frac{Q^2}{R}, \qquad \text{and} \qquad r = Q\sqrt{P},\]

and therefrom let this cubic equation be formed:

\[z^3 - pz^2 + qz - r = 0,\]

whose three roots will give the three sides of the sought triangle, whereby its construction is most easily obtained.

§ 23. Worked Numerical Example: The (5, 6, 7) Triangle

Example. Taking the sides of the triangle \(a = 5\), \(b = 6\), and \(c = 7\), so that the area is \(\mathcal{A} = 6\sqrt{6}\), therefrom are collected the squared distances of the four points:

[Folio § 23 (cont.) | p. 117]

\[EF^2 = \frac{155}{72}; \quad EG^2 = \frac{11}{8}; \quad EH^2 = \frac{155}{32}; \quad FG^2 = \frac{1}{9}; \quad FH^2 = \frac{155}{288}; \quad GH^2 = \frac{35}{32},\]

whence the position of these points comes forth such as is represented in Tab. II, Fig. 6. Since therefore we have:

\[f^2 = \frac{35}{32}, \qquad g^2 = \frac{155}{288}, \qquad \text{and} \qquad h^2 = \frac{1}{9},\]

let us see whether the solution found leads back to the assumed triangle.

§ 24. Verification of the Reconstructed Cubic

Now from these we obtain \(3g^2 + 6h^2 - 2f^2 = \frac{3}{32}\), and further \(f^2 - g^2 - 2h^2 = \frac{1}{3}\) and \(4f^2 + 5g^2 - 2h^2 = \frac{219}{32}\); hence there is collected:

\[R = \frac{1225}{24}, \qquad Q = \frac{35}{3}, \qquad P = 324, \qquad \text{and} \qquad \frac{Q^2}{R} = \frac{24}{9} = \frac{8}{3},\]

whence we obtain:

\[p = \sqrt{P} = 18, \qquad q = 107, \qquad \text{and} \qquad r = \frac{35}{3} \cdot 18 = 5 \cdot 6 \cdot 7 = 210,\]

and hence this cubic equation arises:

\[z^3 - 18z^2 + 107z - 210 = 0,\]

whose three roots are manifestly \(5, 6, 7\), which are the very three sides of the satisfying triangle.