Show us how to trisect an angle of any size using geometry?

Actually....... The joke is *you*. With your attitude ("don't want to hear that it's an intractable problem and cannot be solved", etc.) you'll go a whole lot further in management than you will in a technical field (where your co-workers will just laugh at you)

But..... You never know. Maybe Aaron and I could put our degrees together and come up with a way to write the exact value of π for you as a quotient of two integers <g>

ROTFLMSFAO!!!!!!!!!!!!!!!!!

Doug

I am sorry to say that what you asked is impossible. It is possible to trisect many angles, e.g. a 90 degree angle, but it has been proven that a 60 degree angle is impossible to trisect. The proof as I know it involves Galois Theory which can also be used to prove that given a circle, straight edge and compass, it is impossible to construct a square with the same area as the circle. It also shows that there is no formula which will give you the roots to a general polynomial of degree 5, like the quadratic formula does for polynomials of degree 2. You may already know all of that, so I'll give you a link to a site that has a lot of strategies and ideas that come close to trisecting an arbitrary angle: http://www.jimloy.com/geometry/trisect.htm

I must admit, I haven't read through it. I've been spending my time trying to find a set whose size is between that of the natural numbers and the real numbers. If you come up with one, let me know.

A general angle cannot be trisected using only straihtedge and compass. The basic reason is that to do so requires solving a cubic equation, but circles and lines only allow the repeated solution of linear and quadratic equations. You can't solve a cubic in this way. In fact, it turns out to be impossible to trisect the 60 degree angle using only straightedge and compass. The standard proof uses Galois theory, which is a careful study of how polynomials can be solved.

HOWEVER, if you allow a *marked* straightedge, then it is possible. Here's how. Draw a circle with center at the corner of your angle with radius equal to the marked length on the straightedge. Let A and B be the intersections of this circle with the lines of your angle (so angle AOB is the angle you want to trisect). Continue the line AO through the circle and extend it to the other side. Now, take your straigthedge so that it goes through the point B and so that the marked points are on the circle and the extended line from A through O. Let the points on the line and on the circle be labeled C and D respectively. Then the angle DCO will trisect the angle BOA.

Proof: Draw the line from D to O. Then The lengths CD, DO, and OB are all the same (the marked length on the straightedge). Thus, the angles DCO and DOC are equal. But the angle ODB is the sum of these two angles, so is twice the value of DCO. Again, the angle OBD is equal to the angle ODB since OB and DO are equal lengths. Thus, the angle DOB is equal to a straight angle minus 4 times the angle DCO. But now, angle AOB plus angle BOD plus angle DOC is striaght and angle DOC is equal to angle DCO. This shows that angle AOB is 3 times angle DCO, as required.

Is this a joke? A protractor measures angles... so it's pretty easy to trisect an angle. Just divide by three and find that mark on the protractor.

I think maybe you were thinking of "straight-edge and *compass* constructions". If that's what you meant then it's provably impossible to accomplish. Doesn't matter how clever you are...

I could sketch the proof of the impossibility if you'd like...

many human beings might say no by way of fact it geometrically impossible to trisect an arbitrary attitude. in spite of the incontrovertible fact that some angles might nicely be trisected. All angles might nicely be trisected to any required degree of accuracy much less then suited. this has similarities to numerical techniques used to sparkling up issues like DE's or polynomials of better degree. Successive approximation. did you comprehend you could degree an arbitrary attitude with in basic terms a compass? For irrational angles, in basic terms an successive approximation must be used.

trisect means dividing in 3 equal parts. Generally an angle is never greater than 360 degrees. select the degree in question say 90 degrees. So 3 angle of 30 degree will be formed. you may come across an odd figure where the decimal numbers may be many, here you may have to compromise with the 100% accuracy. like 80 degrees which will be

80/3 = 26.66666666666

26.66 +26.66+26.66 which comes to 79.98

or

26.67 +26.67+26.66 which comes to 80 degrees

Oh it's quite possible, providing you make marks on your straitedge - just look at Archimedes' method in the link the previous poster gave. This is one of those instances where what you're allowed to do in the mathematical problem, and what you can actually accomplish given physical versions of the tools, are widely divergent.

Sorry, but it has been proved that this is impossible to do.