Angle Trisection

by Archimedes of Syracuse
(circa 287 - 212 B.C.)

Archimedes of Syracuse is popularily known for the law he discovered on occasion of taking his bath. "Eurika" he exclaimed and made it into the history. (Along with Newton and Gauss he is counted among the greatest mathematicians of all times. As an engineer he frustrated numerous attempts by the Romans to capture the city of Syracuse.)

The problem of constructing an angle equal to the one third of the given one has been pondered since the times of antiquity. Probably to make the notion of 'geometric construction' more exciting the Ancient Greeks have restricted the allowed operations to using a straightedge and a compass. It's thus specifically forbidden to use a ruler for the sake of measurement. Three famous construction problems lingered until early 19th century when it was shown that it's impossible to solve them with the help of only a straightedge and a compass. The three problems are: to trisect a given angle, to double a cube, and to square a circle. However, one illicit solution that has been found in the works of Archimedes is demonstrated below. (This is Proposition 8 from his Book of Lemmas.)

The Java applet has two control rings. One (B) is dragged to specify the angle (<AOB) to trisect. Another (C) that slides along the x-axis, is used to actually solve the problem. The point D is such that CD = OB. The problem is solved when D falls on the circle. Can you see why?

2 January 2016, Created with GeoGebra

(Hippocrates of Chios gave a different and an earlier neusis construction, i.e. construction that uses tools other than straightedge and compass. We also have a 21st century construction which serves to show that some problems in mathematics never lose their appeal.)


  1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1996
  2. H. Dorrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY, 1965.
  3. W. Dunham, The Mathematical Universe, John Wiley & Sons, NY, 1994.
  4. J. H. Conway and R. K. Guy, The Book of Numbers, Springer-Verlag, NY, 1996.

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∠BCA equals half the difference of arches the lines CB and CA cut on the circle. Assume D lies on the circle. Then ∠DOK measures the arch DK whereas ∠AOB measures the arch AB. Therefore,

∠DCO = (∠AOB - ∠DOK)/2.

However, since CD = DO, ∠DCO = ∠DOK. Combining the two identities yields

∠DCO = (∠AOB - ∠DCO)/2.

Solving which gives 3∠DCO = ∠AOB.

Here is another approach by Jorge Hygino Braga Sampaio Jr. from Brasil.

Hi Alex.

The solution for the angle trisection can be presented in a more straightforward way. Consider the figure shown in the your page. First, extend ray DO until it crosses the circle at, say, E. ΔDCO is isosceles. so that ∠DCO = ∠DOC equal to, say 1q. Therefore, ∠BDO = 2q. ΔOBD is also isosceles which makes ∠OBD = 2q. This yields to ∠BOE = 4q, and finally, ∠BOA = 3q. Evidently, the big difficulty is to find point D!

Rouben Rostamian offered a third approach.

The triangle CDO in your diagram is isosceles. Let the size of each of its base angles be q. By the Exterior Angle Theorem, the exterior angle at vertex D is the sum of the two base angles, therefore it is 2q.

The triangle DOB is isosceles therefore its base angles are equal. We have shown that the angle at its vertex D is 2q, therefore the angle at its vertex B is also 2q.

In the triangle COB, the interior angles at vertices C and B are q and 2q, as shown above. By the Exterior Angle Theorem, the exterior angle at vertex O is q + 2q which is 3q. QED.

Archimedes' Book of Lemmas

  1. Proposition 1: If two circles touch at A, and if CD, EF be parallel diameters in them, ADF is a straight line.

  2. Proposition 2: Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in T. If now DE be drawn perpendicular to AB, and if AT, DE meet in F, then DF = FE..

  3. Proposition 3: Let P be any point on a segment of a circle whose base is AB, and let PN be perpendicular to AB. Take D on AB so that AN = ND. If now PQ be an arc equal to the arc PA, and BQ be joined, then BQ, BD shall be equal.

  4. Proposition 4: If AB be the diameter of a semicircle and N any point on AB, and if semicircles be described within the first semicircle and having AN, BN as diameters respectively, the figure included between the circumferences of the three semicircles is "what Archimedes called arbelos"; and its area is equal to the circle on PN as diameter, where PN is perpendicular to AB and meets the original semicircle in P.

  5. Proposition 5: Let AB be the diameter of a semicircle, C any point on AB, and CD perpendicular to it, and let semicircles be described within the first semicircle and having AC, CB as diameters. Then if two circles be drawn touching CD on different sides and each touching two of the semicircles, the circles so drawn will be equal.

  6. Proposition 6: Let AB, the diameter of a semicircle, be divided at C so that AC = 3/2·CB [or in any ratio]. Describe semicircles within the first semicircle and on AC, CB as diameters, and suppose a circle drawn touching the all three semicircles. If GH be the diameter of this circle, to find relation between GH and AB.

  7. Proposition 7: If circles are circumscribed about and inscribed in a square, the circumscribed circle is double of the inscribed square..

  8. Proposition 8: If AB be any chord of a circle whose centre is O, and if AB be produced to C so that BC is equal to the radius; if further CO meets the circle in D and be produced to meet the circle the second time in E, the arc AE will be equal to three times the arc BD.

  9. Proposition 9: If in a circle two chords AB, CD which do not pass through the centre intersect at right angles, then (arc AD) + (arc CB) = (arc AC) + (arc DB).

  10. Proposition 10: Suppose that TA, TB are two tangents to a circle, while TC cuts it. Let BD be the chord through B parallel to TC, and let AD meet TC in E. Then, if EH be drawn perpendicular to BD, it will bisect it in H.

  11. Proposition 11: If two chords AB, CD in a circle intersect at right angles in a point O, not being the centre, then AO2 + BO2 + CO2 + DO2 = (diameter)2.

  12. Proposition 12: If AB be the diameter of a semicircle, and TP, TQ the tangents to it from any point T, and if AQ, BP be joined meeting in R, then TR is perpendicular to AB.

  13. Proposition 13: If a diameter AB of a circle meet any chord CD, not a diameter, in E, and if AM, BN be drawn perpendicular to CD, then CN = DM.

  14. Proposition 14: Let ACB be a semicircle on AB as diameter, and let AD, BE be equal lengths measured along AB from A, B respectively. On AD, BE as diameters describe semicircles on the side towards C, and on DE as diameter a semicircle on the opposite side. Let the perpendicular to AB through O, the centre of the first semicircle, meet the opposite semicircles in C, F respectively. Then shall the area of the figure bounded by the circumferences of all the semicircles be equal to the area of the circle on CF as diameter.

  15. Proposition 15: Let AB be the diameter of a circle., AC a side of an inscribed regular pentagon, D the middle point of the arc AC. Join CD and produce it to meet BA produced in E; join AC, DB meeting in F, and Draw FM perpendicular to AB. Then EM = (radius of circle).

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