rom point A draw a vertical line and another line angling off to the right by angle β which you wish to trisect. At an arbitrary point B on the vertical line construct a right angle and draw the extended horizontal line BC. The distance h is the length of the hypoteneuse of the triangle BAC, namely AC.
Draw a large number of rays from A, marking their upper endpoints D such that the distance above the extended horizontal line BC is in each case 2h. The loci of the upper endpoints describe a curve known as the Conchoid of Nicomedes ( circa 200 BC).
In polar coordinates, the Conchoid of Nicomedes has r = a + b*secθ. See below :-
Forgive the inaccuracy of my hand-sketched (=not constructed) conchoid, please.
ow we have the conchoid, we can proceed with the actual trisection. I refer you to my sketch on the left.
From the end of the hypoteneuse at C raise a right angle vertical line from the horizontal side BC. Label the point where it intersects the conchoid D. Actually, this is the only tiny area where we need to construct a segment of the conchoid, we don't need the rest of it :-)
Draw a line DA connecting D on the conchoid with the origin at A. It intersects the horizontal BC at point E. DE=2h, remember.
Now angle BAE is one third of the angle BAC which we have successfully trisected :-)
The "impossible" trick? Note that I subtly deviated from Euclid's compass and straightedge only requirement. When I wrote, I implicitly changed the straightedge into a ruler, sneaking my change past most people :(
Although my construct shown above does in fact trisect the angle β (I won't bore you with the proof, suffice that it exists), it does remain impossible to do the trisection while complying with Euclid's compass and straightedge only requirement.
Comments (3) :
Siglinde (Cheb) complains "Obviously it only works for acute angles, not for obtuse or even right angles, so it is at best a partial solution :-(" Mea Culpa.
Ivan (Moscow) says "You also would have to interpolate the conchoid between constructed points, which is cheating on Euclid too!". Yup, you caught that one too. Barbara (15, Florida) noted gleefully " Even my math teacher didn't know that! I practised at home with 60° for an hour then showed her today and got a result measured with a protractor of 20° ! Great Show-and-Tell! Thankyou, Stu :-)"
Now go visit my blog please, or look at other interesting maths stuff :-)