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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 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 = 3PQO or f =
1
𝜃
3
• Now construct a line OL || QB.
• BOL = f =
1
𝜃
3
• 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:
b
b
b
tan f 
tan  
and

tan 3f 
ax
a
a
• Use now the formula
Proof (Cont.)
• Substituting
•
•
•
•
and
into the above
equation leads to:
𝑥 3 − 3 𝑎2 + 𝑏 2 𝑥 − 2𝑎(𝑎2 + 𝑏 2 )=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 𝑥 3 − 3 7.24 𝑥 − 2 × 1.8 × 7.4 = 0
therefor 𝑥 3 − 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 𝑥 3 − 3𝑏2 𝑥 = 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 𝑥 2 + 𝑏2 = (2𝑏)2 𝑎𝑛𝑑 𝑥 2 = 3𝑏2 .
• In the case of b=1 𝑥 2 = 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-mysteriestrisecting-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
[email protected]