MAV 2014 Angle Trisection Karim Noura MED Bayside P-12 College

Solving the Impossible Maths Problem Angle trisection is one of many classic problems in the history of mathematics. It is about constructing of an angle equal to one third of a given arbitrary angle. It is also about providing a clear presentation and straight forward demonstration for the solution.

A bit of history This problem was stated as the impossible problem (Pierre Wantzel 1837). “From Wikipedia, the free encyclopaedia” Mathematicians have been trying to solve this problem by using various methods and strategies. Folding papers (Origami) and classic geometry. This problem is not “Impossible “ anymore. It took me about two years of study, research and trials till I get, what I believe, a clear solution.

Bisecting an Angle Bisecting an arbitrary angle can be done by folding if the angle is drawn on a paper or on a soft board. Bisecting an angle can be done by construction using compass and ruler.

Trisecting an angle Similar strategies can be used to trisect a Straight angle (180˚) or a Right angle (90˚) utilising compass and ruler.

Trisecting angles with Origami Trisecting an arbitrary angle can be done, with Origami, by folding if the angle is drawn on a paper or on a soft board. Trisecting a straight angle Trisecting a right angle Trisecting an arbitrary angle

Trisecting straight & Right angles with Origami Step 1

Step 2

Step 3

Step 4

Step 5

Trisecting an arbitrary angle with Origami Step 1: Create (fold) a line m that passes through the bottom right corner of your sheet of paper. Let <A be the given angle. Step 2: Create the lines l 1 and l 2 parallel to the bottom edge l b such that l 1 is equidistant to l 2 and l b. Step 3: Let P be the lower left vertex and let Q be the intersection of l 2 and the left edge. Create the fold that places Q onto m (at Q') and P onto l 1 (at P').

Trisecting the Angle (cont.) Step 4: Leaving the paper folded, create the line l 3 by folding the paper along the folded-over portion of l 1. Step 5: Create the line that passes through P an P'. The angle trisection is now complete.

Proof of Angle Trisection We need to show that the triangles ∆PQ'R, ∆PP'R and ∆PP'S are congruent. Recall that l 1 is the perpendicular bisector of the edge between P and Q. Then, → Segment Q'P' is a reflection of segment QP and l 3 is the extension of the reflected line l 1. So l 3 is the perpendicular bisector of Q'P'. → ∆PQ'R = ∆PP'R (SAS congruence).

Proof of Angle Trisection (cont.) Let R` be the intersection of l 1 and the left edge. From our construction we see that RP`P is the reflection of R`PP` across the fold created in Step 3. → <RP'P = <R'PP' and ∆P'PR' = ∆PP'S (SSS congruence). → ∆PP'S = ∆PP'R (SAS congruence). → ∆PP'S = ∆PP'R = ∆PQ'R

Classic and co-ordinate geometry (Proof) Archimedes of Syracuse presented a geometrical situation that helps to trisect any arbitrary angle. For any angle such  AOB = Ɵ, construct a circle with centre O and radius r (value of r is your choice). Let Q be a point on BO extended so that AQ cuts the circle at P. Move Q till we get PQ = PO = r.

Proof (Cont.)

Right angle

Another scenario By using co-ordinate (analytic) geometry, trigonometry and algebra for this scenario we are able to trisect an arbitrary angle <AOC

Reference Nelson Maths Yr. 10 text-book for CSF II 2001 (page 160) Maths Quest 9 – 2001 (Page 381) trisecting-angle trisecting-angle

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