On Wed, 05 Nov 2003 13:42:12 -0500, Jerry Avins <jya@ieee.org> wrote:> Rune Allnor wrote: >> Jerry Avins <jya@ieee.org> wrote in message news:<bo8mp4$3vi$1@bob.news.rcn.net>... >> > ... > >>>P.P.S. I know a simple way to trisect an angle with ruler, compass, and >>>pencil. I sometimes use it. It works well. I'm not (for this) a nut. >> >> >> Yeah, right. I suppose the margin of your post was just too small to >> describe your trick?[*] Anyway, if the occation ever arises where you can >> show me your method face to face, I'll by you a beer. >> >> Rune >> >> [*] Yep, you caught me. According to "authorities" (some maths book, I >> can't remember which one) it has been proven that trisecting the angle >> by means of the mentioned instruments is impossible. I can't give the >> proof myself, I can only refer to "authority". Somehow I suspect you >> knew I would respond like that... > > Rune, > > I wasn't trying to catch you. We needn't be face to face; I'll describe > the method so you can try it yourself. [All geometry problems that are > isomorphic to quadratic or linear equations can be solved with compass > and straightedge. In general, those isomorphic to cubic and higher > equations can not be so solved.]* Trisection is cubic*, so more powerful > tools are needed. Instead of a straightedge, I need a ruler, as I wrote. > A ruler has parallel sides and an end square to them. These extra- > Euclidian features play a part in the construction. I'll give you time > to play with the idea before I thrust it on you. I do not withhold it > now to tease. > > The method came to my attention in the late 1940s when it was published > on the front page of the second section of the then two-section New York > Times. It illustrated someone's clam to have solved the trisection > problem; The Times appeared to endorse that claim. Subsequent discussion > was interesting!A couple of us devised a technique in HS Geometry back in the late '70's, after the instructor told us it couldn't be done. After her disbelief subsided, she had us prove it was correct, and why it did NOT disprove that it couldn't be done using geometric construction. The same technique came up in Engineering Graphics my first year of college, though they extended it to n-section, and weren't especially rigorous in proving it correct.

# Minimum Phase Impulse Response

Started by ●October 29, 2003

Reply by ●November 5, 20032003-11-05

Reply by ●November 5, 20032003-11-05

Jerry Avins <jya@ieee.org> wrote in message news:<bobgaa$lvo$1@bob.news.rcn.net>...> Rune Allnor wrote: > > Jerry Avins <jya@ieee.org> wrote in message news:<bo8mp4$3vi$1@bob.news.rcn.net>... > > > ... > > >>P.P.S. I know a simple way to trisect an angle with ruler, compass, and > >>pencil. I sometimes use it. It works well. I'm not (for this) a nut. > > > > > > Yeah, right. I suppose the margin of your post was just too small to > > describe your trick?[*] Anyway, if the occation ever arises where you can > > show me your method face to face, I'll by you a beer. > > > > Rune > > > > [*] Yep, you caught me. According to "authorities" (some maths book, I > > can't remember which one) it has been proven that trisecting the angle > > by means of the mentioned instruments is impossible. I can't give the > > proof myself, I can only refer to "authority". Somehow I suspect you > > knew I would respond like that... > > Rune, > > I wasn't trying to catch you. We needn't be face to face; I'll describe > the method so you can try it yourself.OK. I'll have a beer anyway...> [All geometry problems that are > isomorphic to quadratic or linear equations can be solved with compass > and straightedge. In general, those isomorphic to cubic and higher > equations can not be so solved.]* Trisection is cubic*, so more powerful > tools are needed. Instead of a straightedge, I need a ruler, as I wrote. > A ruler has parallel sides and an end square to them. These extra- > Euclidian features play a part in the construction. I'll give you time > to play with the idea before I thrust it on you. I do not withhold it > now to tease.Here's my best shot. View with a fixed-width font to see the figure. There should be an <end line> immediately after each 'x'. C x / x E-+ - /- F x : /: x : / :- x : / : x D-: / :- H x :/ : x ----/-----+-------- B x /A G x Consider the lines AB and AC that intersect at an angle <BAC, to be trisected, in the point A. - Construct a normal to AB that intersects AB in the point A - Use the scale of the ruler and mark a point D on this normal at a distance L from AB. - Use the scale of the ruler and mark a point E on this normal at a distance 3L from AB. - Construct two parallels to AB, one that intersects the normal in the point D (not drawn in the figure), and another one that intersects the normal in point E. - The parallel through E intersects AC in the point F - Construct a normal from F to AB that intersects AB in the point G - This normal also intersects the parallel through D. Designate the intersection point H. - Draw the line AH (not shown in the figure). - Now we have <BAH = 1/3 <BAC I've been thinking in terms of compasses and straightedges, and a length reference L. It's easy to go through the same procedure if equiped with a ruler that is parallel and at right angles at the ends.> The method came to my attention in the late 1940s when it was published > on the front page of the second section of the then two-section New York > Times. It illustrated someone's clam to have solved the trisection > problem; The Times appeared to endorse that claim. Subsequent discussion > was interesting!I'm once again amazed by what you found in newspapers. You told a similar story about how you found that recipe for making a transistor out of two tubes, a piece of string and a bent nail, didn't you? Rune> Jerry > _______________________________ > * I haven't proved that, but accept it from Authority.

Reply by ●November 5, 20032003-11-05

Rune Allnor wrote:> > C x > / x > E-+ - /- F x > : /: x > : / :- x > : / : x > D-: / :- H x > :/ : x > ----/-----+-------- B x > /A G x > > > - Now we have <BAH = 1/3 <BAC >Unless I'm missing something, what you actually have is tan(<BAH) = 1/3 tan(<BAC) which is not the same thing. Paul

Reply by ●November 5, 20032003-11-05

Paul Russell wrote:> Rune Allnor wrote: > >> >> C x >> / x >> E-+ - /- F x >> : /: x >> : / :- x >> : / : x >> D-: / :- H x >> :/ : x >> ----/-----+-------- B x >> /A G x >> >> >> - Now we have <BAH = 1/3 <BAC >> > > Unless I'm missing something, what you actually have is > > tan(<BAH) = 1/3 tan(<BAC) > > which is not the same thing. > > PaulI don't have time now to give it the attention it deserves, but that's why I asked how it works for large angles. I can prove the construction I use. More tomorrow. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●November 6, 20032003-11-06

Paul Russell <prussell@sonic.net> wrote in message news:<nOfqb.4017$Wy2.46860@typhoon.sonic.net>...> Rune Allnor wrote: > > > > C x > > / x > > E-+ - /- F x > > : /: x > > : / :- x > > : / : x > > D-: / :- H x > > :/ : x > > ----/-----+-------- B x > > /A G x > > > > > > - Now we have <BAH = 1/3 <BAC > > > > Unless I'm missing something, what you actually have is > > tan(<BAH) = 1/3 tan(<BAC) > > which is not the same thing.You didn't miss anything, I did. You are absolutely right. Rune

Reply by ●November 6, 20032003-11-06

Rune Allnor wrote: ...> Here's my best shot. View with a fixed-width font to see the figure. > There should be an <end line> immediately after each 'x'. > > C x > / x > E-+ - /- F x > : /: x > : / :- x > : / : x > D-: / :- H x > :/ : x > ----/-----+-------- B x > /A G x > > Consider the lines AB and AC that intersect at an angle <BAC, to be > trisected, in the point A. > > - Construct a normal to AB that intersects AB in the point A > - Use the scale of the ruler and mark a point D on this normal > at a distance L from AB. > - Use the scale of the ruler and mark a point E on this normal > at a distance 3L from AB. > - Construct two parallels to AB, one that intersects the normal > in the point D (not drawn in the figure), and another one that > intersects the normal in point E. > - The parallel through E intersects AC in the point F > - Construct a normal from F to AB that intersects AB in the point G > - This normal also intersects the parallel through D. Designate the > intersection point H. > - Draw the line AH (not shown in the figure). > > - Now we have <BAH = 1/3 <BAC... Rune, You must have been Very tired! You indeed have tan(<BAH) = 1/3 tan(<BAC). Not only do you get the wrong angle, but you produce a right angle if you start with a one. Moreover, you don't need a marked ruler to trisect a line; compass and straightedge do nicely. Four tuts (two tut tuts)! Bisect the angle in the usual way. Draw two circles on the bisector whose diameters are the width of the ruler, one of them centered on the vertex. With the ruler covering both circles, draw lines with both edges, (the pair being obviously parallel to the bisector). Even though I didn't prescribe it that way, Euclidean construction could have been used to draw the figure so far. What follows can't be done with according to Euclid. Keeping the ruler centered over the circle about the vertex, slide it until one corner lies on one of the angle's lines and the other corner on the same end lies on the closer parallel to the bisector. The end of the ruler is now the cord of the trisected arc. Take it from there. An elegant tool to do this job is a ruler with a slit just wide enough to slide on a pin, and a mark in the middle of its unslit end. After bisecting the angle, put the pin at the vertex and set the ruler with the pin in the slit and the end mark on the bisector. Draw [at least one of] the parallel lines, and move the ruler to the position described above. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●November 6, 20032003-11-06

Jerry Avins <jya@ieee.org> wrote in message news:<bodjpo$kaq$1@bob.news.rcn.net>...> Rune Allnor wrote: > > ... > > > Here's my best shot. View with a fixed-width font to see the figure. > > There should be an <end line> immediately after each 'x'. > > > > C x > > / x > > E-+ - /- F x > > : /: x > > : / :- x > > : / : x > > D-: / :- H x > > :/ : x > > ----/-----+-------- B x > > /A G x > > > > Consider the lines AB and AC that intersect at an angle <BAC, to be > > trisected, in the point A. > > > > - Construct a normal to AB that intersects AB in the point A > > - Use the scale of the ruler and mark a point D on this normal > > at a distance L from AB. > > - Use the scale of the ruler and mark a point E on this normal > > at a distance 3L from AB. > > - Construct two parallels to AB, one that intersects the normal > > in the point D (not drawn in the figure), and another one that > > intersects the normal in point E. > > - The parallel through E intersects AC in the point F > > - Construct a normal from F to AB that intersects AB in the point G > > - This normal also intersects the parallel through D. Designate the > > intersection point H. > > - Draw the line AH (not shown in the figure). > > > > - Now we have <BAH = 1/3 <BAC > > ... > > Rune, > > You must have been Very tired!Nope, not tired. Comatose, judging from that attempt...> You indeed have tan(<BAH) = 1/3 tan(<BAC). > Not only do you get the wrong angle, but you produce a right angle if you > start with a one. Moreover, you don't need a marked ruler to trisect a > line; compass and straightedge do nicely. Four tuts (two tut tuts)! > > Bisect the angle in the usual way. Draw two circles on the bisector whose > diameters are the width of the ruler, one of them centered on the vertex. > With the ruler covering both circles, draw lines with both edges, (the > pair being obviously parallel to the bisector). > > Even though I didn't prescribe it that way, Euclidean construction could > have been used to draw the figure so far. What follows can't be done with > according to Euclid. > > Keeping the ruler centered over the circle about the vertex, slide it > until one corner lies on one of the angle's lines and the other corner on > the same end lies on the closer parallel to the bisector. The end of the > ruler is now the cord of the trisected arc. Take it from there.Neat! So you essentially use the width of the ruler to divide that arc of the angle in three equal parts... your soultion has all the trademarks of an ingeniousity (if that's a proper word in English): Once you see it, it's so obvious that you wonder why there was a problem in the first place. And I know there is just no way I could have come up with that myself.> An elegant tool to do this job is a ruler with a slit just wide enough to > slide on a pin, and a mark in the middle of its unslit end. After > bisecting the angle, put the pin at the vertex and set the ruler with the > pin in the slit and the end mark on the bisector. Draw [at least one of] > the parallel lines, and move the ruler to the position described above.After I realized the flaw of my first attempt, I devised a way to trisect the angle with a piece of string and a compass: - Use the compass to draw the arc of the angle - Put a piece of string on the paper along the arc and cut it to the same length as the arc - Fold the string in three equal lengths, like a "tight Z" - The length of the folded string is the arc length of the trisected angle Unfortunately, I could get there without the bent nail... Rune

Reply by ●November 7, 20032003-11-07

Jerry Avins wrote:> Andor wrote:...> > Archimedes knew how to do this already, with just a straight edge and > > (two I think) pencilmarks. Is that hint enough? > > That -- and Rune's way -- sounds simpler than my way. Do those ways work > for large angles, say, 90 degrees +/- a little?Yep. We did this in algebra class as a special topic - constructable numbers. That was quite fun. You can see Archimedes solution here: http://www.cut-the-knot.org/pythagoras/archi.shtml Regards, Andor

Reply by ●November 7, 20032003-11-07

Andor wrote:> Jerry Avins wrote: > >>Andor wrote: > > ... > >>>Archimedes knew how to do this already, with just a straight edge and >>>(two I think) pencilmarks. Is that hint enough? >> >>That -- and Rune's way -- sounds simpler than my way. Do those ways work >>for large angles, say, 90 degrees +/- a little? > > > Yep. We did this in algebra class as a special topic - constructable > numbers. That was quite fun. You can see Archimedes solution here: > > http://www.cut-the-knot.org/pythagoras/archi.shtml > > Regards, > AndorThat is rich with insights. I particularly like that the reverse construction -- tripling the angle -- is Euclidean. (Cubing is easier than taking cube roots.) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������