﻿ Trisecting the Angle

Trisecting the Angle

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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Trisecting the angle is one of the three classic unsolved problems of antiquity. They are all known to be unsolvable under the rules used by the Greeks.

Marked Ruler Constructions

If you mark your ruler to make measurements, you can trisect the angle and duplicate the cube, but marking the ruler was not allowed under ancient Greek rules. The ruler was to be used only for drawing straight lines.

 This is perhaps the simplest trisection using a marked straightedge. It was discovered by Archimedes. Given the angle AOX, draw a circle of arbitrary radius centered at O. Extend one side of the angle through the opposite side of the circle at D (top).Mark off interval BC on the straightedge. BC = OX = radius of the circle. Slide the straightedge so that B lies on line DOX, C lies on the circle and the straightedge passes through A. Angle CBD is one-third of AOX. Note the element of trial and error inherent in positioning the straightedge.

Proof:

• Since BC=OC, angles CBD and COD are equal and angle BCO = 180-2CBD.
• And since OC = AO, angles OCA and OAC are equal.
• OCA + COD = 180 so OCA = 2CBD.
• The angles in triangle ABO sum to 180, so we have CBD + DOA + OAC = 180 = CBD + DOA + 2CBD.
• Rearranging, we get DOA = 180 - 3CBD, and since DOA + AOX = 180, AOX = 3CBD.

Below is another construction, possibly neater since it doesn't involve a circle.

 Top: Given angle AOB, construct a line from A parallel to OB, and drop AC perpendicular to OB. Mark off intervals DF and EF on a ruler, with DF = EF = OA. Middle: Slide the ruler so it passes through O, D rests on AC and E rests on the upper line. Angle BOE trisects angle AOB. Bottom: Proof: Triangles EFG, AFG, AFH and FDH are all similar. Thus angle AFD = 2EOB. Also AF = EF = AO, so triangle OAF is isoceles. Therefore angle AOF = 2EOB. Thus angle AOB = 3EOB and EOB trisects AOB.

Approximations: An Oldie But Goodie

The approximation below was devised by that Renaissance Renaissance Man, Albrecht Durer, in 1525.

 Construct the angle AOB to be trisected. Draw AB. With A as center, draw arc AB. Trisect AB and locate point C. Construct CD perpendicular to AB. With A as center, draw arc DE.  Trisect EC and locate point F. With A as center, draw arc FG. Angle AOG approximately trisects angle AOB

This approximation is obviously a series of trisections. Trisect chord AB, use CD for a first approximation at trisecting the arc, trisect the chord difference, and get a second approximation at trisecting the arc.

The error increases quite rapidly with angle AOB, but is so tiny for small angles that even a huge increase is of no consequence. (Compound interest on a penny, even at a high rate, takes a long time to add up!) At 30 degrees the error is .007 seconds of arc, increasing to one whole second at 60 degrees, about 15 seconds at 90 degrees, 9 minutes of arc at 120 degrees and a whopping 31 minutes of arc (half a degree) at 180 degrees.

Considering how simple this approximation is and how phenomenally accurate it is for angles less than 60 degrees (and not bad to beyond 90 degrees), it seems superfluous to show any others.

Created 6 September 2001, Last Update 30 August 2011

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