AAMT 2015

Angle Trisection

Karim Noura

MED

Bayside P-12 College Melbourne

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 l1 and l2 parallel to the bottom edge lb such that l1 is equidistant to l2 and lb.

Step 3: Let P be the lower left vertex and let Q be the intersection of l2

and the left edge. Create the fold that places Q onto m (at Q') and P onto l1 (at P').

Trisecting the Angle (cont.)

Step 4: Leaving the paper folded, create the line l3 by folding the paper along the folded-over portion of l1.

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 l1 is the perpendicular

bisector of the edge between P and Q.

Then,

→Segment Q'P' is a reflection of segment

QP and l3 is the extension of the reflected

line l1. So l3 is the perpendicular bisector of

Q'P'.

→∆PQ'R = ∆PP'R (SAS congruence).

Proof of Angle Trisection (cont.) Let R` be the intersection of l1 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.) • By using the properties of isosceles triangles and exterior

angles we can prove that AOB = 3PQO or f = 1

3𝜃

• Now construct a line OL || QB.

• BOL = f = 1

3𝜃

• Use a compass to mark M so arc BL = arc LM and the

trisection is complete.

• But, how can we determine the position of point Q?

Proof (Cont.)

• Now refer the whole situation to the Cartesian system (X, Y) where S(0, 0) is the origin, A(a, b) and Q(-𝑥, 0), where a, b & 𝑥 are positive numbers .

• Using trigonometry: and

• Use now the formula

tanb

a xf

tan

b

a tan 3

b

af

Proof (Cont.) • Substituting and into the above

equation leads to: 𝑥3 − 3 𝑎2 + 𝑏2 𝑥 − 2𝑎(𝑎2 + 𝑏2)=0

• This equation can have up to three solutions, one of which must be a positive value.

• Thus for the given angle (Ɵ), 𝑟, 𝑎 & 𝑏 are arbitrary values.

• The value of 𝑥 can by found by solving the above equation.

• Therefore the position of Q is determined Q(- 𝑥, 0) and the trisection is completed.

Proof (Cont.) • As the whole diagram is referred to a Cartesian

therefore, we are allowed to measure the values of a & b for any given angle (using a graph paper is very useful).

• For example if A has the coordinates (1.8, 2) a=1.8 & b=2, This gives 𝑥3 − 3 7.24 𝑥 − 2 × 1.8 × 7.4 = 0 therefor 𝑥3 − 21.72𝑥 − 26.64 = 0 then the solutions are: 𝑥1 = 5.18 , 𝑥2 = −1.34 & 𝑥3 = −3.85 . The acceptable solution is 𝑥1 = 5.18 with Q being (-5.18, 0) which makes sense.

Right angle

• Consider the right angle to validate the above solution.

• If the angle <ASB is a right angle (𝜃 = 90˚) 𝑡ℎ𝑒𝑛 𝑎 =0. The equation 𝑥3 − 3𝑏2𝑥 = 0 has three solutions: 0 , 𝑏 3 and −𝑏 3, then Q is being (−𝑏 3, 0).

• To validate this result with classic geometry; we notice that if a=0 & b= r the point P will be in the middle of SQ and 𝑥 = 2𝑟; the triangle BSQ is a right-angled triangle therefore by applying the Pythagoras formula we get 𝑥2 + 𝑏2 = (2𝑏)2 𝑎𝑛𝑑 𝑥2 = 3𝑏2.

• In the case of b=1 𝑥2 = 3 𝑎𝑛𝑑 𝑥 = ∓ 3 𝑡ℎ𝑒 𝑥 = 3 is an acceptable solution and Q being (− 3 , 0).

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)

• http://www.mathopenref.com/constbisectangle.html

• http://plus.maths.org/content/mathematical-mysteries-trisecting-angle

• http://www2.washjeff.edu/users/mwoltermann/Dorrie/36.pdf

• https://www.sciencenews.org/article/trisecting-angle-origami

• http://plus.maths.org/content/power-origami-0

• http://www.cut-the-knot.org/pythagoras/archi.shtml • http://mathschallenge.net/library/constructions/trisecting_angle

Thanks

• For more information please email me: Karim Noura

kn4education@gmail.com