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Pappus of Alexandria, Mathematical Collection IV §§49-51, pp. 290.24-298.2
Prop. 48. To construct an isosceles triangle with each of the angles at the base in a given ratio to the angle at the vertex.
Prop. 49, Part 1: Corollary to Prop. 48, to find an equiangular and equilateral polygon.
Prop. 49, Part 2: To find a circumference equal to a given line.
Prop. 50. Given a ratio and a line, to find a circular-arc on the line in the given ratio to the line.
Prop. 51. To find incommensurable angles.
Prop. 48. (pp. 288.15-290.23)
To construct an isosceles triangle having each of the angles at the base with a given ratio to the other angle. (analysis and synthesis)
(diagram 1) Let it have happened, and let ABG be constructed, (diagram 2) and let circle ADG be drawn about center A and through point A, G, and let AB be extended to D, and let DG be joined. (diagram 3) And so, since the ratio of angle GAB to angle ABG is given, and the angle at D is half angle ABG, the ratio of angle GAD to angle ADG is therefore given, so that so is the ratio of circular-arc DG to circular-arc AG. (diagram 4) And so, since circular-arc AGD of the semicircle is cut in a given ratio, G is given, and triangle ABG is given in species.
It will be synthesized as follows. (diagram 5) For let the given ratio which each of the the angles at the base must have to the other angle be the ratio of EZ to ZH,(diagram 6) and let ZH be bisected at Q, and let a circle ADG be displayed with center B and diameter AD, (diagram 7) and let circular-arc AGD be cut at G, so that as circular-arc DG is to circular-arc GA, so is EZ to ZQ (for this was previously inscribed (i.e., proved), i.e., how generally a given circular-arc may be cut in a given ratio), (diagram 8) and let BG, GA, GD be joined. (diagram 9) And so, since circular-arc DG is to circular-arc GA, i.e., as angle DAG to ADG, so is EZ to ZQ, (diagram 10 = general diagram) as well as to twice the following terms, therefore, as angle GAB is to angle ABG (= 2 angle ADG), so is EZ to ZH. Therefore a triangle, ABG, is constructed having each of the angles at the base having a given ratio to the remaining angle.
Prop. 49. (p. 290.24-6): Part 1 (Corollary to Prop. 48). When this is given it is obvious that it is possible to inscribe an equiangular and equilateral polygon in a circle having as many sides as someone requests.
Part 2 of Prop 49 (p. 292.1-11)
It is easy to see how a circle is found whose circular-arc is equal to the given line. (analysis)
For let the circular-arc of circle
A equal to straight-line G be found, and let any circle B be displayed,
and let the straight-line D equal to its circular-arc be found
through the squaring-line (quadratrix).
Therefore, as G is to D, so is the radius of circle A to the radius
of circle B. Therefore, the ratio of the radii to one another
is also given. And the radius is
B is given. Therefore the radius of A is also given, so that so
is A. The synthesis is obvious.
(diagram 1) Given in position and magnitude straight line AB, to inscribe though A, B a circular-arc of a circle having a given ratio to straight-line AB.
(analysis) (diagram 2) Let AGB be inscribed and (diagram 3) let there be displayed a fourth-part of a circle given in position, ZHE, and let there be inscribed squaring-line (quadratrix) ZQK, and on circular-arc ZE let angle EHL be constructed equal to the angle standing on circular-arc AG, and (diagram 4, end of 3) let perpendiculars LM, QN be drawn. (diagram 5) And so, because of the property of the line, as circular-arc ELZ is to straight-line ZH, i.e. as LH to HK, (diagram 6) so is circular-arc LE to straight-line QN. (diagram 7) But also as QH is to HL, QN is to LM. (diagram 8) And therefore (ex aequali), (diagram 9) as QH is to HK, so is circular-arc EL to straight-line LM. (diagram 10) In fact, let the center X of circular-arc AGB be taken, and perpendicular XRG to AB. Therefore, angle GXA is equal to angle EHL, and X, H are the centers. (diagram 11) Therefore, as circular-arc AG is to straight-line AR, i.e., QH to HK, so is circular-arc AGB to straight-line AB. (diagram 12) And the ratio of AGB (ms. ABG) to AB is given. Therefore so is the ratio of QH to HK. And HK is given. Therefore, so is HQ given. (diagram 13) Therefore, Q is on a circular-arc. But also it is on line ZQK. Therefore, Q is given. HQL is in position; therefore, angle EHL is given. And it is equal to angle GXA. Therefore, angle EHL is given. And it is equal to angle GXA, and GX is given in position, and A is given. Therefore AX is given in position, so that so is circular-arc AGB.
(synthesis) And the synthesis is obvious. (diagram 14) For it is required to make the ratio DH to HK the same as the given ratio and (diagram 15) to inscribe a circular-arc about center H and through D, (diagram 16) and to get Q where it cuts the quadratrix, (diagram 17) and to join QH (diagram 18) and bisecting AB (diagram 19) and erecting a right angle RX to draw AX enclosing with XR an angle equal to angle KHQ (diagram 20) and about center X to inscribe a circular-arc of a circle AGB through A having the same ratio to base AB as the given.
Note: not stated in the theorem is the requirement that ZE : ZH >= the given ratio > 1 : 1. Hence the proof will not work if the angle of the segment is to be greater than 180 degrees.
The synthesis is stated tersely, as is
reasonable given the elaborate analysis. The point is to show
that angle LHE is the required angle (these were proved in the
analysis, cf. diagram
1. QN : LM = QH : HL (by similar triangles)
2. arc ZLE : ZH = HL : HK (HL = ZH and by the quadratrix, Pappus, Collection iv, Prop. 31-32)
3. arc LE : arc ZLE = QN : ZH (by the quadratrix, Pappus, Collection iv, Prop. 30)
4. arc LE : QN = arc ZLE : ZH (4, by alternando)
5. By 2 and 4, arc LE : QN = HL : HK
6. By 1 and 5, arc LE : LM = QH : HK (ex aequali, here: a : b = c : d and b : e = f : c => a : e = f : d)
Hence, if one constructs, angle AXR = angle LHE, arc LE : LM = arc AG : AR.
Prop. 51. (pp. 296.9-298.2)
It is not incredible that one can find incommensurable angles. For through this and through the same circle incommensurable circular-arcs will be taken, even if we suppose as a rational one angle or circular arc, the remainder will become irrational.
(diagram 1) Let there be fourth-part ABG, and on it a quadratrix AEDZ, and let BE be drawn and EH parallel to BG, and let BQ, incommensurable in length with BH, be taken away, and let a parallel DQ be drawn, and let DB be joined. I say that angle EBZ is incommensurable with angle DBZ.
(diagram 2 = general diagram) Let perpendicular DN be drawn. Therefore, because of the line, as EK is to DN, so is angle EBZ to angle DBZ. But EK is incommensurable with DN (since also HB is with BQ). Therefore, one angle is also incommensurable with the other angle, even if we suppose angle EBZ is a rational [even if we suppose it half a right angle-Hultsch rightly deletes this], angle DBZ will be irrational.