# Trisection of an Angle

• Sep 1st 2010, 09:49 PM
sameer
Trisection of an Angle
Attachment 18795

Given:

<XOY is a given angle which has to be trisected.

Construction:

Draw a circle "O" intersecting OX and OY at A and B respectively.
Joi A, B. Draw a semi circle on AB, AB diameter and P center.
Divide the semi circle at C, D such that arc AC = arc CD = arc DB = radius AP
Join O,C and O,D. OC and OD intersect arc AB at E and F.

Proof:

OA = OE = OF = OB (radii of circle O)

arc AB and semi circle are standing on the same straight line AB.
Further arc AC = arc CD = arc DB (construction)

Hence
Arc AE = arc EF = arc FB (experimental conclusion)

These arcs subtant <AOE, <EOF, <FOB which are equal as equal arcs subtant equal angles at the centre
Therefore <AOE = <EOF = <FOB

Hence the trisection

Solved by:
Mohd Shafiuddin Ahmed
Ganapavaram - village & post
Andhra Pradesh, India - PIN - 508206
• Sep 2nd 2010, 06:09 AM
skeeter
Quote:

Originally Posted by sameer
Attachment 18795

Given:

<XOY is a given angle which has to be trisected.

Construction:

Draw a circle "O" intersecting OX and OY at A and B respectively.
Joi A, B. Draw a semi circle on AB, AB diameter and P center.
Divide the semi circle at C, D such that arc AC = arc CD = arc DB = radius AP
Join O,C and O,D. OC and OD intersect arc AB at E and F.

Proof:

OA = OE = OF = OB (radii of circle O)

arc AB and semi circle are standing on the same straight line AB.
Further arc AC = arc CD = arc DB (construction)

Hence
Arc AE = arc EF = arc FB (experimental conclusion) I'm not familiar with "experimental conclusion" as a reason in a geometric proof ... ???

These arcs subtant <AOE, <EOF, <FOB which are equal as equal arcs subtant equal angles at the centre
Therefore <AOE = <EOF = <FOB

Hence the trisection

Solved by:
Mohd Shafiuddin Ahmed
Ganapavaram - village & post
Andhra Pradesh, India - PIN - 508206

...
• Sep 2nd 2010, 06:37 AM
Ackbeet
In addition, what does "standing on the same straight line AB" mean?

Creating equal arcs on the semicircle might be possible - you're essentially constructing 60 degree arcs there. But the fact that you have equal arcs on the semicircle does not, in my opinion, show that you have corresponding equal arcs on the circle centered at O.

In other words, this line: "Further arc AC = arc CD = arc DB (construction)" is not at all obvious. It needs more work.

You do realize that this construction has been proven to be impossible in general?
• Sep 2nd 2010, 07:54 AM
ebaines
Hello Sameer. Unfortunately your proof that the angle is trisected essentially amounts to: "If you trisect an angle, then it is trisected." It doesn't do much in terms of advancing your attempt to develop a method to trisect an angle.

The problem is in your contruction step that states: "Divide the semi circle at C, D such that arc AC = arc CD = arc DB = radius AP" First, the lengths of these arcs would each be of arc length AP * π/3, which is not equal to length AP. Second, there is no way to determine where to place points C and D exactly. Hence the construction method fails.
• Sep 2nd 2010, 10:01 AM
CaptainBlack
Quote:

Originally Posted by sameer
Attachment 18795

Given:

<XOY is a given angle which has to be trisected.

Construction:

Draw a circle "O" intersecting OX and OY at A and B respectively.
Joi A, B. Draw a semi circle on AB, AB diameter and P center.
Divide the semi circle at C, D such that arc AC = arc CD = arc DB = radius AP
Join O,C and O,D. OC and OD intersect arc AB at E and F.

Proof:

OA = OE = OF = OB (radii of circle O)

arc AB and semi circle are standing on the same straight line AB.
Further arc AC = arc CD = arc DB (construction)

Hence
Arc AE = arc EF = arc FB (experimental conclusion)

These arcs subtant <AOE, <EOF, <FOB which are equal as equal arcs subtant equal angles at the centre
Therefore <AOE = <EOF = <FOB

Hence the trisection

Solved by:
Mohd Shafiuddin Ahmed