Unit IV: Collinear Centers, Isosceles Degeneracy & Casus Irreducibilis (§§ 25–34)

Leonhard Euler — An Easy Solution of Certain Very Difficult Geometric Problems (1765/1767, pp. 117–123)

[Folio § 25 | p. 117]

§ 25. Case Where All Four Points Lie in a Straight Line

Case in which the four points are situated in a straight line (Tab. II, Fig. 7). In this case therefore, since we have:

\[FH = g; \quad FG = h; \quad GH = f; \quad EF = 2g; \quad EH = 3g \quad \text{and} \quad EG = 2g - h,\]

[Folio § 25 (cont.) | p. 118]

we shall have \(g = f - h\), whence upon making this substitution we collect:

\[R = \frac{4f^4}{(3h - f)^2}; \qquad Q = \frac{3f^2 h(2f - 3h)}{(f - 3h)^2}; \qquad P = \frac{3h(4f - 3h)^3}{(f - 3h)^2},\]

and therefore \(\frac{Q^2}{R} = \frac{9h^2(2f - 3h)^2}{4(f - 3h)^2}\). From these we further elicit:

\[\begin{split}p &= \frac{(4f - 3h)\sqrt{3h(4f - 3h)}}{f - 3h}, \\ q &= \frac{3fh(4f - 3h)(5f - 6h)}{(f - 3h)^2}, \\ r &= \frac{3f^2 h(2f - 3h)(4f - 3h)\sqrt{3h(4f - 3h)}}{(f - 3h)^3}.\end{split}\]

§ 26. Double Root of the Cubic and Isosceles Necessity

Since now the roots of this cubic equation \(z^3 - pz^2 + qz - r = 0\) furnish the three sides \(a, b, c\) of the sought triangle, in order to render it neater let us set

\[z = \frac{y\sqrt{3h(4f - 3h)}}{f - 3h},\]

and this equation will come forth:

\[y^3 - (4f - 3h)y^2 + f(5f - 6h)y - f^2(2f - 3h) = 0,\]

whose roots are manifestly \(f, f\), and \(2f - 3h\). Wherefore the sides of the sought triangle, which becomes isosceles, will be:

\[a = b = \frac{f\sqrt{3h(4f - 3h)}}{f - 3h}; \qquad \text{and} \qquad c = \frac{(2f - 3h)\sqrt{3h(4f - 3h)}}{f - 3h}.\]

§ 27. Direct Verification on an Isosceles Triangle

Now this case is also easy to solve in itself, since that straight line on which the given points lie necessarily cuts the triangle

[Folio § 27 (cont.) | p. 119]

into two similar halves, and therefore the triangle becomes isosceles. Indeed, setting \(b = a\) right from the start, we obtain \(\mathcal{A} = \tfrac{1}{4}c\sqrt{4a^2 - c^2}\) and \(AP = AQ = AR = AS = \tfrac{1}{2}c\), and further:

\[PE = \frac{c^3}{4\mathcal{A}}; \qquad QF = \frac{2\mathcal{A}}{3c}; \qquad RG = \frac{2\mathcal{A}}{2a + c}; \qquad SH = \frac{c(2a^2 - c^2)}{8\mathcal{A}},\]

whence, because the feet \(P, Q, R, S\) coincide at the midpoint of the base—let it be \(O\)—the intervals among these points will be:

\[\begin{split}OF - OE &= \frac{2(a^2 - c^2)}{3\sqrt{4a^2 - c^2}}; \qquad OG - OE = \frac{c(a - c)}{\sqrt{4a^2 - c^2}}; \qquad OH - OE = \frac{a^2 - c^2}{\sqrt{4a^2 - c^2}}; \\ OF - OG &= \frac{(a - c)(2a - c)}{3\sqrt{4a^2 - c^2}}; \qquad OH - OF = \frac{a^2 - c^2}{3\sqrt{4a^2 - c^2}}; \qquad OH - OG = \frac{a(a - c)}{\sqrt{4a^2 - c^2}}.\end{split}\]

§ 28. Ordering of the Collinear Points When a > c vs. a < c

Here it is fitting to contemplate two cases according as \(a > c\) or \(a < c\), for if \(a = c\), i.e. an equilateral triangle, all four points coalesce into one:

• I. If \(a > c\): the points will be arranged as Tab. II, Fig. 8 shows, where \(HF = \tfrac{1}{3}EH\) or \(EF = \tfrac{2}{3}EH\) and \(EG < \tfrac{1}{2}EH\); and in this case the midpoint \(O\) of the base falls on the straight line \(HE\) produced beyond \(E\), so that \(OE = \frac{c^2}{2\sqrt{4a^2 - c^2}}\).

• II. If \(a < c\): the points will be arranged as Fig. 9 shows, where again \(HF = \tfrac{1}{3}EH\) or \(EF = \tfrac{2}{3}EH\), whereas \(EG > \tfrac{1}{2}EH\). In this case, however, the midpoint \(O\) of the base falls on the straight line \(EH\) produced beyond \(H\), so that

[Folio § 28 (cont.) | p. 120]

\[HO = \frac{2a^2 - c^2}{2\sqrt{4a^2 - c^2}},\]

whence if \(2a^2 < c^2\) [an obtuse isosceles triangle], the point \(O\) even falls between \(H\) and \(E\).

§ 29. Explicit Isosceles Formulas Relative to the Midpoint of EH

Given therefore on a straight line the three points \(E, G\), and \(H\) such that \(G\) is situated between the extremes \(E\) and \(H\), one must inspect whether \(EG < \tfrac{1}{2}EH\) or \(EG > \tfrac{1}{2}EH\):

• In the former case, where \(EG < \tfrac{1}{2}EH\) (Tab. II, Fig. 8), let \(EH = 2d\) and \(EG = d - e\); hence there is found:

  \[a = b = \frac{d + e}{2e}\sqrt{(d + 3e)(3d + e)}, \qquad c = \frac{d - e}{2e}\sqrt{(d + 3e)(3d + e)}, \qquad OE = \frac{(d - e)^2}{4e}.\]

• In the latter case, where \(EG > \tfrac{1}{2}EH\) (Fig. 9), let \(EH = 2d\) and \(EG = d + e\); hence there is collected:

  \[a = b = \frac{d - e}{2e}\sqrt{(d - 3e)(3d - e)}, \qquad c = \frac{d + e}{2e}\sqrt{(d - 3e)(3d - e)}, \qquad OE = \frac{(d + e)^2}{4e} \quad [\text{printed } 2e \text{ in folio}],\]

  whence it is clear that this case cannot take place if \(d\) lies between the limits \(3e\) and \(\tfrac{1}{3}e\). Indeed, since we must have \(2a > c\), it is necessary that \(d > 3e\) [so \(EG < \frac{2}{3}EH = EF\), meaning \(G\) always lies strictly between \(E\) and \(F\)!].

§ 30. Concise General Formulation Omitting the Centroid F

From this case we may gather that in general, too (Fig. 5), a neater solution will come forth if, omitting the point \(F\), the three points \(E, G\), and \(H\) are considered. Let us therefore set:

\[EG = e, \qquad GH = f, \qquad \text{and} \qquad EH = k;\]

then we shall have \(FH = g = \tfrac{1}{3}k\), \(EF = \tfrac{2}{3}k\), and \(FG = h = \sqrt{\tfrac{1}{3}e^2 + \tfrac{2}{3}f^2 - \tfrac{2}{9}k^2}\), and hence we obtain:

\[R = \frac{4f^4}{2f^2 + 2e^2 - k^2}, \qquad Q = \frac{f^2(k^2 - f^2 - 2e^2)}{2e^2 + 2f^2 - k^2},\]

[Folio § 30 (cont.) | p. 121]

\[P = \frac{27f^4}{2e^2 + 2f^2 - k^2} + 2e^2 - 8f^2 - 3k^2 = \frac{4e^4 + 11f^4 + 3k^4 - 12e^2f^2 + 2f^2k^2 - 8e^2k^2}{2e^2 + 2f^2 - k^2},\]

and furthermore \(\frac{Q^2}{R} = \frac{(k^2 - f^2 - 2e^2)^2}{4(2e^2 + 2f^2 - k^2)}\), whence we obtain:

\[p = \sqrt{P}, \qquad q = \frac{2e^4 + f^4 + k^4 - 6e^2f^2 - 3e^2k^2 + 2f^2k^2}{2e^2 + 2f^2 - k^2}, \qquad r = Q\sqrt{P},\]

and the roots of the equation \(z^3 - pz^2 + qz - r = 0\) give the sides of the sought triangle; which equation, upon setting \(z = y\sqrt{P}\), passes into this rational-coefficient form:

\[y^3 - y^2 + \frac{(2e^4 + f^4 + k^4 - 6e^2f^2 - 3e^2k^2 + 2f^2k^2)y - f^2(k^2 - 2e^2 - f^2)(2e^2 + 2f^2 - k^2)/( \dots )}{P} = 0\]

[written in the folio with common denominator \(4e^4 + 11f^4 + 3k^4 - 12e^2f^2 - 8e^2k^2 + 2f^2k^2\)].

§ 31. Diorismos: Positivity Inequalities on (e, f, k)

Here, however, I observe that these given quantities \(e, f, k\) must not only be so assumed that they form a triangle, but since the sides \(a, b, c\) of the sought triangle must be regarded as positive, \(P\) as well as \(Q\) and \(R\) must also receive positive values. Not only, therefore, must \(k^2 < 2e^2 + 2f^2\), but also \(k^2 > 2e^2 + f^2\); and further, so that \(P\) may become positive, it is necessary that

\[3k^2 > 4e^2 + f^2 + 2\sqrt{e^4 + 11e^2f^2 - 8f^4},\]

and when this condition is compared with the preceding ones [requiring \(3(2e^2 + 2f^2) > 4e^2 + f^2 + 2\sqrt{e^4 + 11e^2f^2 - 8f^4}\) and reality of the radical], there follows that \(f^2\) must exceed the lower bound of real configurations—otherwise the problem would admit no solution.

§ 32. Reducible Family e = f (60° Angle) and Worked Example

Example. Let \(e^2 = f^2\) [i.e. \(EG = GH\)]; we shall have:

\[R = \frac{4f^4}{4f^2 - k^2}; \qquad Q = \frac{f^2(k^2 - 3f^2)}{4f^2 - k^2}; \qquad P = \frac{3(k^2 - f^2)^2}{4f^2 - k^2}; \qquad \frac{Q^2}{R} = \frac{(k^2 - 3f^2)^2}{4(4f^2 - k^2)},\]

and therefore:

\[p = \frac{(k^2 - f^2)\sqrt{3}}{\sqrt{4f^2 - k^2}}; \qquad q = \frac{k^4 - f^2k^2 - 3f^4}{4f^2 - k^2}; \qquad r = \frac{f^2(k^2 - 3f^2)(k^2 - f^2)\sqrt{3}}{4(4f^2 - k^2)\sqrt{4f^2 - k^2}}.\]

[Folio § 32 (cont.) | p. 122]

Now let \(e^2 = f^2 = 2\) and \(k^2 = 7\); there will come forth \(p = 5\sqrt{3}\), \(q = 23\), and \(r = 10\sqrt{3}\), and the sides of the sought triangle will be the roots of this cubic equation:

\[z^3 - 5z^2\sqrt{3} + 23z - 10\sqrt{3} = 0,\]

which upon setting \(z = \frac{y}{\sqrt{3}}\) passes into \(y^3 - 15y^2 + 69y - 90 = 0\), one root of which is \(y = 6\), whence the remaining two are \(y = \frac{9 \pm \sqrt{21}}{2}\), and thus the sides of the sought triangle are:

\[a = \frac{3\sqrt{3} + \sqrt{7}}{2}; \qquad b = \frac{3\sqrt{3} - \sqrt{7}}{2}; \qquad c = 2\sqrt{3}.\]

§ 33. Universal Factorization Whenever EG = GH

Indeed, even more generally, while \(e^2 = f^2\) remains, however \(k\) be chosen, if we set \(z = \frac{y}{\sqrt{3(4f^2 - k^2)}}\), we have this equation to solve:

\[y^3 - 3(k^2 - f^2)y^2 + 3(k^4 - f^2k^2 - 3f^4)y - 9f^2(k^2 - 3f^2)(k^2 - f^2) = 0,\]

which is satisfied first by \(y = 3f^2\), and the remaining two roots come from the quadratic equation \(y^2 - 3(k^2 - 2f^2)y + 3(k^2 - 3f^2)(k^2 - f^2) = 0\), namely \(y = \frac{3(k^2 - 2f^2) \pm k\sqrt{3(4f^2 - k^2)}}{2}\); and thus the three sides of the sought triangle become:

\[a = \frac{(k^2 - 2f^2)\sqrt{3}}{2\sqrt{4f^2 - k^2}} + \frac{1}{2}k, \qquad b = \frac{(k^2 - 2f^2)\sqrt{3}}{2\sqrt{4f^2 - k^2}} - \frac{1}{2}k, \qquad c = \frac{f^2\sqrt{3}}{\sqrt{4f^2 - k^2}}.\]

[Folio § 34 | p. 123]

§ 34. The Irreducible General Case & Solution by Angle Trisection

In the remaining cases the business is not dispatched so easily, because the cubic equation does not admit rational factors [casus irreducibilis]. To show this by an example, let \(e^2 = 3\), \(f^2 = 2\), and \(k^2 = 9\); whence the sides of the triangle are the roots of this cubic equation:

\[z^3 - z^2\sqrt{71} + 22z - 2\sqrt{71} = 0;\]

to resolve which, let an acute angle \(\alpha\) be sought whose cosine is given by the trigonometric depressed-cubic formula [see Critical Notes on Euler’s printed \(+\sqrt{\frac{71}{75}}\) vs. \(-\sqrt{\frac{71}{125}}\)], and once this is found, the sides of the triangle will be expressed in the form:

\[a = \frac{1}{3}\sqrt{71} + \frac{2}{3}\sqrt{5}\cos\left(60^\circ - \frac{1}{3}\alpha\right), \quad b = \frac{1}{3}\sqrt{71} + \frac{2}{3}\sqrt{5}\cos\left(60^\circ + \frac{1}{3}\alpha\right), \quad c = \frac{1}{3}\sqrt{71} - \frac{2}{3}\sqrt{5}\cos\frac{1}{3}\alpha;\]

and thus by the trisection of an angle the problem will always be resolved expeditiously enough.