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Caput III: Problema Inuersum et Aequatio Cubica Laterum (§§ 19–24)¶

L. Eulerus — Solutio facilis problematum quorumdam geometricorum difficillimorum (Novi Commentarii, Tom. XI, MDCCLXVII, pp. 114–118)

[Folio § 19 | p. 114]

§ 19. Quantitates auxiliares duarum dimensionum P, Q, R¶

Quo nunc has formulas ad maiorem simplicitatem reuocemus, ponamus \(4pq - p^3 - 8r = 4s\) vt sit \(4\mathcal{A}\mathcal{A} = ps\) et \(4q = pp + \frac{8r}{p} + \frac{4s}{p}\); tum vero faciamus:

\[\frac{rr}{ps} = R, \qquad \frac{r}{p} = Q \qquad \text{et} \qquad pp = P\]

ita vt \(P, Q, R\) sint quantitates duas dimensiones inuoluentes. Quoniam igitur hinc est \(\frac{-s}{p} = \frac{QQ}{R}\) [sic; lege \(\frac{s}{p} = \frac{QQ}{R}\)] erit \(p = \sqrt{P}\); \(q = \frac{1}{4}P + 2Q + \frac{QQ}{R}\), et \(r = Q\sqrt{P}\), at-

[Folio § 19 (cont.) | p. 115]

que \(4\mathcal{A}\mathcal{A} = \frac{PQQ}{R}\) et interualla nostra ita exprimentur:

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

§ 20. Problema vnicum¶

Cum igitur horum quatuor punctorum terna nisi capiantur haec tria \(E, F\) et \(H\) iam contineant determinationem quarti, vnicum resultat problema, quod ita se habet.

Problema (Tab. II. Fig. 5.). Datis positione his quatuor punctis in quolibet triangulo assignabilibus 1°. Intersectione perpendicularium ex singulis angulis in latera opposita ductarum \(E\), 2°. Centro grauitatis \(F\), 3°. Centro circuli inscripti \(G\) et 4° centro circuli circumscripti \(H\); construere triangulum.

Quod problema ex hactenus erutis horum punctorum affectionibus satis concinne resoluere licebit.

§ 21. Solutio¶

Solutio. Cum positio horum quatuor punctorum per eorum distantias detur, vocemus:

[Folio § 21 (cont.) | p. 116]

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

nouimusque fore \(EF = 2g\) et \(EH = 3g\), itemque \(EG = \sqrt{6gg + 3hh - 2ff}\).

Nunc igitur statim habemus has tres aequationes

\[\begin{split}\text{I.} \quad ff &= \frac{1}{4}R - Q \\ \text{II.} \quad gg &= \frac{1}{4}R - \frac{1}{18}P + \frac{4}{9}Q + \frac{2QQ}{9R} \\ \text{III.} \quad hh &= \frac{1}{36}P - \frac{8}{9}Q + \frac{6QQ}{9R} \quad [\text{*sic*; lege } \frac{5QQ}{9R}]\end{split}\]

ex quarum resolutione colligimus:

\[\begin{split}R &= \frac{4f^4}{3gg + 6hh - 2ff}; \qquad Q = \frac{3ff(ff - gg - 2hh)}{3gg + 6hh - 2ff} \\ \text{et } P &= \frac{27f^4}{3gg + 6hh - 2ff} - 12ff - 15gg + 6hh\end{split}\]

vnde fit \(\frac{QQ}{R} = \frac{9(ff - gg - 2hh)^2}{4(3gg + 6hh - 2ff)}\).

§ 22. Aequatio cubica laterum¶

His valoribus inuentis inuestigentur tres sequentes expressiones:

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

indeque formetur haec aequatio cubica:

\[z^3 - pzz + qz - r = 0\]

cuius tres radices dabunt tria latera trianguli quaesiti, quo pacto eius constructio facillima habetur.

§ 23. Exemplum (a = 5, b = 6, c = 7)¶

Exemplum. Sumtis lateribus trianguli \(a = 5\), \(b = 6\) et \(c = 7\), vt sit area \(\mathcal{A} = 6\sqrt{6}\), inde colliguntur distantiae quaternorum punctorum:

[Folio § 23 (cont.) | p. 117]

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

vnde situs horum punctorum talis prodit vti in Tab. II. fig. 6. repraesentatur. Cum igitur habeamus:

\[ff = \frac{35}{32}; \qquad gg = \frac{155}{288} \qquad \text{et} \qquad hh = \frac{1}{9}\]

videamus num solutio inuenta ad triangulum assumtum perducat.

§ 24. Evolutio exempli¶

Hinc autem fit \(3gg + 6hh - 2ff = \frac{3}{32}\), tum vero \(ff - gg - 2hh = \frac{1}{3}\); \(4ff + 5gg - 2hh = \frac{219}{32}\); colligitur

\[R = \frac{1225}{24}; \qquad Q = \frac{35}{3}; \qquad P = 324 \qquad \text{et} \qquad \frac{QQ}{R} = \frac{24}{9} = \frac{8}{3}\]

vnde nanciscimur:

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

et aequatio cubica hinc oritur:

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

cuius tres radices manifesto sunt \(5, 6, 7\) quae sunt ipsa tria latera trianguli satisfacientis.