Three Geometrical Problems

Excerpts

• Three Geometrical Problems

  Among the more interesting geometrical problems of antiquity are three questions which attracted the special attention of the early Greek mathematicians.

• Three Geometrical Problems

  To duplicate a cube the length of whose side is $a$, we have to find a line of length $x$, such that $x^3 = 2a^3$.

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  He did not give a geometrical construction, but he reduced the question to that of finding two means between one straight line ($a$), and another twice as long ($2a$). If these means are $x$ and $y$, we have $a: x = x: y = y: 2a$, from which it follows that $x^3 = 2a^3$. It is in this form that the problem is always presented now.

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  The insertion of Plato’s name is an obvious anachronism.

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  It is probable that the Greeks were aware that the latter ratio is incommensurable, in other words, that no two integers can be found whose ratio is the same as that of $\sqrt[3]{2}: 1$, but it did not therefore follow that they could not find the ratio by geometry: in fact, the side and diagonal of a square are instances of lines whose numerical measures are incommensurable.

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  It is however a mere accident that $\pi$ is defined usually in that way, and it really represents a certain number which would enter into analysis from whatever side the subject was approached.

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  In reality the fact that the ratio of the length of the circumference of a circle to its diameter is the number denoted by $\pi$ does not afford the best analytical definition of $\pi$, and is only one of its properties.

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  The use of a single symbol to denote this number $3.14159\ldots$ seems to have been introduced about the beginning of the eighteenth century.

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  We may say that the $\pi$-calculators who used the first method regarded $\pi$ as equivalent to a geometrical ratio, but those who adopted the modern method treated it as the symbol for a certain number which enters into numerous branches of mathematical analysis.

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  With a polygon of $n$ sides this process gives a value of $\pi$ correct to at least the integral part of $(2\log n - 1.19)$ places of decimals.

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  The reason is that Archimedes, having calculated the lengths of the sides of inscribed and circumscribed regular polygons of $n$ sides, assumed that the length of $1/n$th of the perimeter of the circle was intermediate between them; whereas Snell constructed from the sides of these polygons two other lines which gave closer limits for the corresponding arc.

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  If the experiment is repeated many hundreds of times, the ratio of the number of favourable cases to the whole number of experiments will be very nearly equal to this fraction: hence the value of $\pi$ can be found.

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  Inscribe in the given circle a square, and to three times the diameter of the circle add a fifth of a side of the square, the result will differ from the circumference of the circle by less than one-seventeen-thousandth part of it.

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  “Only prove to me that it is impossible,” said one of them, “and I will set about it immediately”; and doubtless the statement that the problem is insoluble has attracted much attention to it.

Equations

• Three Geometrical Problems

  a : x = x : y = y : b

  Descartes's intersection of the curves gives x and y as the two mean proportionals between a and b.

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  x^3 = 2a^3

  The side x of the required cube, whose volume is double that of a cube of side a, satisfies x cubed equals twice a cubed.

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  4x^3=3x-a

  If x is the sine of an angle one-third of a given angle whose sine is a, then x satisfies this cubic equation.

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  x^2 + y^2 + ax + by + c = 0

  The general form of the equation of a circle in Cartesian coordinates.

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  \alpha x + \beta y +\gamma = 0

  The general form of the equation of a straight line in Cartesian coordinates.

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  x: a = \sqrt[3]{2}: 1

  The ratio of the side of the required cube to the side of the given cube is the cube root of two.

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  a: x = x: y = y: 2a

  Two mean proportionals x and y lie between a and 2a, so that the problem of the duplicated cube is that of finding two means between a line and one twice as long.

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  r = 2a \sin\theta

  In polar coordinates, the equation of the surface traced by Archytas's semicircle.

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  r \sin\theta = 2a \cos\phi

  The equation of the cylinder used in Archytas's construction, in polar coordinates.

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  \sin\theta \cos\phi = \frac{1}{2}

  The equation of the right cone used in Archytas's construction.

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  \sin^3\theta = \frac{1}{2}

  At the point where the three surfaces meet, sin cubed theta equals one half.

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  (r\sin\theta)^3=2a^3

  The cube on side r sin theta has twice the volume of the cube on side a.

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  PC : PB = PB : PA = PA : PD

  In Plato's figure, the segments PC, PB, PA and PD are in continued proportion.

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  y^2 = 2ax

  Equation of the first parabola in Menaechmus's first solution, with latus rectum double that of the second.

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  x^2=ay

  Equation of the second parabola in Menaechmus's first solution.

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  x^3 = 2l^3

  The abscissa of the intersection of Menaechmus's parabola and rectangular hyperbola satisfies x cubed equals twice l cubed.

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  y^3 = 4l^3

  The ordinate of the intersection of Menaechmus's curves satisfies y cubed equals four l cubed.

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  x^2 = ly

  Equation of the parabola of latus rectum l in Menaechmus's second solution.

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  xy = 2l^2

  Equation of the rectangular hyperbola in Menaechmus's second solution.

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  l : x = x : y = y : 2l

  The abscissa and ordinate of the intersection are the mean proportionals between l and 2l.

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  OA : Bb = Bb : Aa = Aa : OB

  Apollonius's construction gives Bb and Aa as the two mean proportionals between OA and OB.

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  x^2 + y^2 = ay + bx

  Equation of the circle used in Descartes's and Gregory's constructions.

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  xy = ab

  Equation of the hyperbola with the two sides of the rectangle as asymptotes, in Gregory's construction.

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  BC : OD = OD : CE = CE : OA

  In Newton's construction, OD and CE are two mean proportionals between BC and OA.

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  AB : GC = GC : GA = GA : CD

  In Vieta's construction, GC and GA are the two mean proportionals between AB and CD.

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  QR = 2\dotm OP

  The condition on the construction for trisecting an angle by Pappus's first method: QR is twice OP.

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  AOR=\frac{1}{3}AOB

  If the construction can be made, the angle AOR is one third of the given angle AOB.

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  \tan^{-1} (b/x)=\frac{1}{3}\tan^{-1}(b/a)

  The angle with tangent b/x is one third of the angle with tangent b/a.

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  (x-a)^2 + (y-b)^2 = 4(a^2 + b^2)

  Equation of the circle in the analytic form of Pappus's first trisection.

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  PR = x-a

  The length PR equals the abscissa x of the chosen intersection point minus a.

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  AOE = \frac{1}{3} AOB

  If the construction can be made, the angle AOE is one third of the given angle AOB.

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  SOP = \frac{1}{3}SOA

  Pappus's hyperbola construction trisects the angle SOA, giving the angle SOP as one third of it.

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  y^2 = \frac{1}{4}x

  Equation of the parabola intersected with a circle in Descartes's trisection.

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  x^2 + y^2 - \frac{13}{4}x + 4ay = 0

  Equation of the circle intersected with a parabola in Descartes's trisection.

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  4y^3 = 3y - a

  The ordinates of the intersection points satisfy this cubic; the smaller positive root is the sine of one third of the angle whose sine is a.

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  AH = 2 \dotm HL

  In Clairaut's construction, AH is twice HL.

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  AP : PM = AH : HL = 2 : 1

  By the focus and directrix property, the ratio AP to PM equals 2 to 1.

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  AP = 2 \dotm PM = PQ

  AP is twice PM, and PM is half of PQ.

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  AP = PQ = QR

  By symmetry the three chords AP, PQ and QR are equal, so the angles AOP, POQ and QOR are equal.

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  6336/2017\frac14 <\pi<14688/4673\frac12

  Archimedes' bounds from the 96-sided polygons: pi lies between 6336/2017 1/4 and 14688/4673 1/2.

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  \sin\theta < \theta < \tan\theta

  The proposition Archimedes' polygon method is equivalent to: sine of an angle is less than the angle, which is less than its tangent.

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  \theta= \pi/96

  In Archimedes' method the angle θ is set equal to π/96.

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  \pi = 3^{\circ} 8' 30''

  Ptolemy asserted that pi equals 3 degrees 8 minutes 30 seconds.

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  \pi = 3 + \frac8{60} + \frac{30}{3600} =\allowbreak 3\frac{17}{120} =\allowbreak 3.141\dot6

  Ptolemy's sexagesimal value written as a decimal fraction, 3 17/120 = 3.1416 approximately.

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  b^2=\frac{1}{2}-\frac{1}{2}(1-a^2)^{\frac{1}{2}}

  Aryabhata's relation: the squared side b of the inscribed 2n-gon in a circle of unit diameter follows from the side a of the inscribed n-gon.

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  2 \sin^2\frac{1}{2}\theta = 1-\cos \theta

  Half-angle identity used by Vieta to halve polygon sides repeatedly.

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  \frac{2}{\pi} = \frac{\surd 2}{2} \frac{\surd (2+\surd 2)}{2} \frac{\surd \{2+\surd (2+\surd 2)\}}{2} \dotsm\;

  Vieta's infinite product giving 2/π as a product of nested square roots.

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  1-\cos A = 2 \sin^2\frac{1}{2}A

  Identity van Ceulen used to double polygon sides when computing perimeters.

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  3 \sin\theta /(2 + \cos\theta) < \theta < (2 \sin\frac{1}{3}\theta + \tan\frac{1}{3}\theta)

  Snell's bounds on an angle (and so on an arc), by which a polygon of n sides gives pi to more places than Archimedes' method.

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  \frac{\pi}{2}=\frac{2\dotm 2\dotm 4\dotm 4\dotm 6\dotm 6\dotsm} {1\dotm 3\dotm 3\dotm 5\dotm 5\dotm 7\dotm 7\dotsm}

  Wallis's infinite product for pi/2.

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  \frac{\pi}{4}=1+\frac{1^2}{2} \genfrac{}{}{0pt}{}{}{+} \frac{3^2}{2} \genfrac{}{}{0pt}{}{}{+} \frac{5^2}{2} \genfrac{}{}{0pt}{}{}{+}\ldots\;

  Brouncker's continued-fraction-type series for pi/4, as quoted by Wallis.

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  \theta = \tan\theta - \frac{1}{3}\tan^3\theta + \frac{1}{5}\tan^5\theta - \dotsb

  Gregory's arctangent series, true only for θ between minus π/4 and π/4.

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  \tfrac{1}{4}\pi = 4 \tan^{-1}\tfrac{1}{5} -\tan^{-1}\tfrac{1}{239}

  Machin's arctangent formula for pi/4, used to compute pi to 100 places.

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  \frac{1}{4}\pi =\allowbreak \tan^{-1}\frac{1}{2} +\allowbreak \tan^{-1}\frac{1}{3}

  Hutton's arctangent formula for pi/4.

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  \frac{1}{4}\pi = 5 \tan^{-1}\frac{1}{7} + 2\tan^{-1}\frac{3}{79}

  Euler's arctangent formula for pi/4.

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  \frac{1}{4}\pi=4\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99}

  Rutherford's arctangent formula for pi/4, used for his 1841 calculation.

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  \frac{1}{4}\pi= \tan^{-1}\frac{1}{2} +\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}

  Dase's arctangent formula for pi/4, used for his 1844 calculation.

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  \frac{1}{4}\pi = 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}

  Clausen's arctangent formula for pi/4, used for his 1847 calculation.

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  \frac{\pi}{6} = \frac{1}{2} + \frac{1}{2}\dotm \frac{1}{3\dotm 2^3} + \frac{1\dotm 3}{2\dotm 4} \dotm \frac{1}{5\dotm 2^5} + \dotsb\;

  A rapidly converging series for pi/6 quoted in the chapter.

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  \frac{\pi}{4} = 2 + 22\tan^{-1}\frac{1}{28} + \tan^{-1}\frac{1}{443} - 5\tan^{-1}\frac{1}{1393} - 10\tan^{-1}\frac{1}{11018}\;

  Escott's arctangent formula for pi/4 as printed in the book; a rough hand estimate did not reproduce pi/4 from these coefficients, so it needs a numerical check (possible misprint in the source).

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  \tan^{-1}\tfrac{1}{5}

  Placeholder check, not an equation: omitted.

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  \frac{3}{4}(\sqrt{3} + \sqrt{6})

  Cusa's believed value of pi, which the chapter reports is about 3.1423 and which Regiomontanus is said to have shown wrong.

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  2l/\pi a

  Probability that a stick of length l dropped on a plane ruled with lines a apart lies across a line; used to estimate pi experimentally.

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  6/\pi^2 = 154/250

  Relating the probability 6/π² that two random numbers are prime to each other to the observed frequency 154/250 gives π about 3.12.

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  \pi=3.12

  The value of pi that the chapter derives from the coprime-pair experiment.

Problems

No exercises in this chapter.