Wikipedia:Reference desk/Archives/Mathematics/2017 March 16

= March 16 =

Trisecting an Angle - Original research
On the topic of one of the three classical Greek "straightedge and compass constructions" that are deemed "impossible", I would like to submit my own solution for trisecting any given angle with as much accuracy as a drawing would allow. The first thing that should be understood about compass and straightedge constructions is that you are not given numerical data or units to measure with, so talking about any given angle as a numeric in "degrees", while solving, is an error in itself. Also to be noted, this solution will not work for a straight line. 180° is easily trisected with 60° angles, but also, using a parallel line and constructing an equilateral triangle with the base on the parallel line and the third point located equidistant from the other two on the original (180°) line. If further explanation is required for 180°, I can provide it. ((A Plea: Please do not discount my method because 180° & 360° are not solvable; division by zero is unsolvable, but division is still an accepted mathematical function)) This works for both Acute and Obtuse angles. As Previously stated, a straight line requires a different method, the reason being the initial line between the radii cannot be differentiated from the angle (line) itself. For all reflex angles, this method is used on the smaller angle, and the trisecting line is continued through the origin, in order to trisect the reflex angle. 360º is also not trisectable using this method, for the same reason as 180º, and also has a known, numeric trisection (120º)
 * 1) Use the compass to draw an arc between the two radii of the angle. You now have two points that are equidistant from the angle's origin; one on each radii.
 * 2) Use the straight-edge to draw a line between these two points. You now have an isosceles triangle, size is irrelevant.
 * 3) Use this newly constructed line [that terminates at the angle's radii] to construct a 4x4 grid, with the [above] line as its base.
 * 4) Use the straight-edge to draw a line on this constructed grid from 0,4 through 2,1 and continue this line back to the the base of the grid; this line meets the base at the 2⅔ mark.
 * 5) Do the same on the opposite side of the grid, construct a line from 4,4 through 2,1 and continue this line back to the base of the grid; this line meets the base at the 1⅓ mark.
 * 6) You have now marked the points where the trisections exist at.
 * 7) Use the straight-edge to draw a line from the angle's origin through the 1⅓ mark and another line from the origin through the 2⅔ mark. The angle has been trisected.


 * This is original research and I do not expect it to be published on the official Wikipedia page for Angle_trisection, I am looking for someone to refute my claim, as I do not see the error, and this seems too easy for it to have been deemed "impossible"
 * I tried to upload a crude .png I drew of this, but my wiki account is too new Wcichello (talk) 20:45, 16 March 2017 (UTC)
 * Your construction may be good at constructing an approximation of the angle trisection, but it does not address the classical angle-trisection task which is to construct a theoretically exact trisection. You reveal this in the introduction to your post where you state that you have a "solution for trisecting any given angle with as much accuracy as a drawing would allow." That may be a very practical thing to do but it is not the required task&mdash;which is unsurprising given that exactness with straightedge and compass is not attainable. Newyorkbrad (talk) 20:50, 16 March 2017 (UTC)
 * Your grid trisects a line but not an angle. The middle angle becomes larger than the others, and there is not as much accuracy as a drawing would allow. Try it with a large angle like 135° and see what happens. PrimeHunter (talk) 21:18, 16 March 2017 (UTC)
 * Why is trisecting a line which is a chord, not the same as trisecting an angle? It seems obvious enough that is you trisect a chord, you should also be trisecting the angle. 148.182.26.69 (talk) 22:09, 16 March 2017 (UTC)
 * It might "seem obvious", but it's false. The middle segment is closer to the vertex than the other two segments.  Therefore, being the same length as the other two, it subtends a larger angle than the other two. --Trovatore (talk) 22:42, 16 March 2017 (UTC)
 * Indeed. As an example, draw an equilateral triangle ABC with sides length 1 unit. Divide one side BC into thirds BDEC and connect D and E to the opposite vertex A. Consider the middle triangle ADE. It is an isosceles triangle with height $$\sqrt{3}/2$$ and side DE with length 1/3. Angle DAE is therefore $$2\tan^{-1}(1/3\sqrt{3})$$ which is approximately 21.787 degrees. But if angle DAE trisected angle BAC it would be exactly 20 degrees.. Gandalf61 (talk) 22:47, 16 March 2017 (UTC)


 * Methinks the whole thing is based on the (incorrect) premise that tangent is a linear function (i.e. tan(t/3)=1/3*tan(t)). This is of course not true (and has nothing to do with "infinite precision" red herrings). Tigraan Click here to contact me 16:31, 17 March 2017 (UTC)

Thank you for your quick responses. I can see now that the center will have a larger "third" of the angle. I have not constructed it with a compass yet, but I can mentally see that the one third point on the line will not match up to a 1/3 point on my initial arc. One thing I would like to state about Newyorkbrad's conjecture- I believe that your idea of the "required task" is clouded by a sort of mathematic dogma. The Greeks weren't looking for an answer of "I give up" or "it's impossible"; This was a drawing excercise, and a sort of semi-trick question, meant to inspire creativity, understanding (of math) and inspire conversations such as this one. IP 148.182.26.69 and I have now learned how inaccurate my method would be. But even if you're using a hypothetical compass, and a hypothetical straight-edge, you are still supposed to come up with a way to hypothetically draw the specified solution. Then the teacher(s) tell you why you are incorrect! Thank you again Wcichello (talk) 16:06, 17 March 2017 (UTC)
 * For any future endeavours, I would recommend giving GeoGebra a try. It's an easy-to-use program that would have shown you that your construction doesn't trisect the angle. —  Wasell ( T ) 10:26, 18 March 2017 (UTC)