Download Morley`s Miracle - Teachers of India

Morley's Miracle
Morley's original proof stemmed from his results on algebraic curves
tangent to a given number of lines. As usual in mathematics, numerous
attempts have been made to find a simple, elementary proof that could
match the level of knowledge and proficiency required to grasp the
statement of the theorem. The simplest proofs proceed backwards
starting with an equilateral triangle. They differ in subsequent steps.
Most such proofs highlight some additional features of the configuration
but complicate matters unnecessarily as a few most trivial proofs
convincingly demonstrate.
Before giving several backward proofs, here is a direct one which, while
logically absolutely transparent, requires some high school trigonometry.
In all likelihood, this proof first appeared in A.Letac, Solution (Morley's
triangle), Problem No. 490, Sphinx, 9(1939) 46. I came across it in a
Russian book D.O.Shklyarsky, N.N.Chentsov, Y.M.Yaglom,Selected
Problems and Theorems of Elementary Mathematics, v. 2, problem 97,
Moscow, 1952 and also in The Art of Mathematics by B. Bollobás
(Cambridge University Press, 2006, p. 126-127.)
The idea of the proof is fairly straightforward.
1. In triangles ARB, BPC, CQA, we know the bases - AB, BC, and AC and the adjacent angles. TheLaw of Sines then yields the segments
AR, BR, BP, CP, CQ, and AQ.
2. Next we apply the Law of Cosines to triangles AQR, BPR, and CPQ to
determine (and compare) the segments QR, PR, and PQ. The fact
that they come out equal proves the theorem.
For simplicity, let (angles) A = 3α, B = 3β, and C = 3γ. This implies that α
+ β + γ = 60°. Also, assuming that the radius of the circle circumscribed
around ΔABC equals 1, we get AB = 2sin(3γ), BC = 2sin(3α),AC = 2sin(3β).
Consider now ΔBPC. By the Law of Sines,
BP/sin(γ) = BC/sin(180° - β - γ)
= 2sin(3α)/sin(β + γ)
= 2sin(3α)/sin(60° - α)
Therefore, BP = 2sin(3α)sin(γ)/sin(60° - α). To simplify the expression
note that
sin(3α) = 3sin(α) - 4sin3(α)
= 4sin(α)[(√3/2)² - sin²(α)]
= 4sin(α)[sin²(60°) - sin²(α)]
= 4sin(α)(sin(60°) + sin(α))(sin(60°) - sin(α))
= 4sin(α) 2sin[(60° + α)/2]cos[(60° - α)/2] 2sin[(60° - α)/2]cos[(60° +
α)/2]
= 4sin(α)sin(60° + α)sin(60° - α)
Reaping the fruits of this effort,
BP = 8sin(α)sin(γ)sin(60° + α)
Similarly,
BR = 8sin(γ)sin(α)sin(60° + γ)
Proceeding to the second step and the Law of Cosines,
PR² = BP² + BR² - 2 BP BR cos(β),
from where
PR² = 64sin²(α)sin²(γ)[sin²(60° + α) + sin²(60° + γ) - 2sin(60° +
α)sin(60° + γ)cos(β)].
Note, however, that (60° + α) + (60° + γ) + β = 180°. Thus, there exists a
triangle with angles (60° + α), (60° + γ), and β. Indeed, there is a whole
family of similar triangles with those angles. Out of this family, choose
the one with the circumscribed radius equal to 1 (then, as above, by
the Law of Sines, its sides have a very simple form.) In that triangle,
apply the Law of Cosines:
sin²(β) = sin²(60° + α) + sin²(60° + γ) - 2sin(60° + α)sin(60° +
γ)cos(β)
Which gives
PR = 8sin(α)sin(β)sin(γ),
an expression which is symmetric in α, β, and γ. QR and PQ are similarly
found to be equal to the same expression. Therefore, PR = PQ = QR.
The proof below is an indirect proof based on the work of Conway dating from around 1995.
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From the above figure we know that 3a+3b+3c = 180o,
which means that a+b+c= 60o.
Let x+ = x+60o for any angle x. since a+b+c = 60o. Let
0+=a+b+c=60o., This shows that it is possible to
construct an triangle with 3 different types of angle
combinations:
+
+
+
0 ,0 ,0
;
+ +
+
+
+ +
Type 2: a,b ,c ; a ,b,c ; a ,b ,c
++
++
++
Type 3: a ,b,c; a,b ,c; a,b,c
Type 1:
since these seven combinations of angles all have a
sum of 180o.
Instead of working forward, Conway worked backwards.
He showed that from an equilateral triangle one can
construct a triangle with any angles, i.e. with arbitrary
a,b and c (that sums to 60o). According to Conway, we
can make the following constructions:
Type 1. Construct an equilateral triangle with length 1.
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Type 2. Construct a triangle with the side joining larger
angles ( e.g. a+ and b+) to have length 1. for example:
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Type 3. Construct two lines that intersects the side
opposite to b++ at angle b+, thus forming an isosceles
triangle with base angle =b+.
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The significance of constructing the isosceles triangle is
to prove JHI and DFC are congruent.
1. JHI = DFC = b+
2. JIH = DCF = c
3. By construction, we get JH = DF=1
The above proves that JHI and DFC are congruent.
This result is important because it shows that DC = JI
are equal, when the two triangles are matched together,
they become the common edge, then point J = point D
and point I = point C.
Other triangles are constructed in similar way, and we
can get the following:
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Note that the angles at the vertex of the equilateral
triangle (e.g. D) sums up to 360o.
Check: (a+)+(c+)+(b++)+(0+)
= a+60o+c+60o+b+120o+60o
=(a+b+c)+300o=360o.
Since the angles at D sums up to 360o, and since the
sides match due to congruency, we see that the
triangles can be assembled perfectly together to form a
larger triangle with angles 3a, 3b and 3c, and hence the
conclusion.
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The proof below is another backwards proof based on the work of Dergiades dating from 1991.
According to Alexander Bogomolny, this proof is first published in Greek in the bulletin of Diastasi,
a Greek Mathematical Society.<ref 2>
The original reference is from "A simple geometric proof of Morley's Theorem", Diastasi 1991 issue
1-2 pp 37-38. Thessaloniki-Greece. <ref 2>
The proof uses the following lemma:
LEMMA Let BAC = a, point D is the incentre of ABC if
and only if (a) the line DA bisects a and (b) CDB = 90 +
a/2.
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PROOF Suppose that angles , and are given with
+ + = 180. Dergiades' approach is to construct a
triangle with these angles for which Morley's Theorem is
true.
According to Dergiades, suppose we let
x=60-( /3),
y=60-( /3),
z=60-( /3).
Then 0<x,y,z<60 and x+y+z=120.
Consider an equilateral triangle. Construct 3 isoceles
triangles with base angles x, y, z respectively ( with the
constraints x+y+z =120 and 0<x,y,z<60) on the sides on
the equilateral triangle.
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Extend the sides of the isosceles triangles so that they
intersect as shown:
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Since D is the vertex of the equilateral triangle, and I is
the vertex of the isosceles triangle with base angle x, it
follows that the line ID is the angle bisector.
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Join B and C, note that BDC = 60+z+y (vertically
opposite angles)
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BDC = 60+z+y = 60+x+y+z-x=180-x. Since BIC = 1802x. and D is bisects BIC, the Lemma shows that D is
the incentre of CIB.
It follows that the lines drawn at points A B and C are
angle trisectors.
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The last step is to prove
the following figure:
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ABC has angles ,
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we can see that
a+s+t+x+x= 180
s+x+z+60=180, i.e. s+x+z=120
t+x+y+60=180, i.e. t+x+y=120
x+y+z=120(given)
by substitution, we have
a+(s+x)+(t+x)=180
a+(120-z)+(120-y)=180
a+120+(120-z-y)=180
a+120+x=180
and . In
It follows that:
a=60-x= /3.
Hence
ABC has angles ,
and .
The proof below is a direct proof based on the work of Bankoff. This
proof uses less geometry than the previous two. However it requires
some knowledge in trigonometry.
According to Alexander Bogomolny, this proof is first published
in Mathematics Magazine, 35 (1962) 223-224.<ref 2>
Sine Law
a/sin(A) = b/sin(B) = c/sin(C) = 2r, if r=0.5, c= sin C;
b=sin B; a=sinA.
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PROOF Bankoff starts out with a large triangle ABC,
with angle trisectors drawn at points A, B and C. From
the above figure we know that A+B+C = 180o.
Note that ADC
= 180- A/3 - C/3
= 180 - (A+C)/3
= 180 - (180-B)/3
= 120+ B/3
= (360 +B)/3
sin ADC =sin((360+B)/3) =sin((180-B)/3)--- (i)
For simplification, let A = 3 , B= 3 , C= 3 . this
implies + + =60o. if we assume that the radius of the
circle circumscribed around ABC = 1 (i.e. r=1).
using sine law we get
AB = 2sinC, AB= 2sin(3 );
BC= 2sinA, BC= 2sin(3 );
AC= 2sinB, AC= 2sin(3 ).
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If we apply sine law to BFC, we have
BF/ sin = BC / sin (180- - ).
by substitution:
BC/ sin (180- - ) = 2sin(3 )/sin( + ) = 2sin(3 ) /
sin(60- ).
Therefore BF/ sin = 2sin(3 )/ sin(60- ),
i.e. BF = 2sin(3 ) sin c / sin (60- ).
3
Using the identity sin(3x) = 3sin(x) - 4sin (x), we can simplify the
identity.
sin(3x)
2
2
= 4sin(x) [ (¡Ô3/2) -sin (x)]
2
2
= 4sin(x) [ sin ( 60) - sin (x)]
= 4sin(x) (sin 60 + sin(x)) (sin 60 - sin (x))
= 4sin(x) 2 sin[(60+x)/2] cos [(60-x)/2] 2sin[(60-x)/2] cos [(60+x)/2]
= 4sin(x) sin (60+x)sin(60-x).
therefore
BF= 8 sin( ) sin (60+ )sin(60- )sin ( )/ sin(60- )
BF= 8 sin( ) sin (60+ ) sin ( ) ---(ii)
Applying the Sine Law,
AD/ sin( )= AC/ sin((180-B)/3) = 2r.
Recall from above that with the assumption that the
circumradius r = 0.5, we have AC= b= sin(B). Also
=C/3. Therefore, we have
AD*sin((180-B)/3 = 2r sin(B) sin(C/3)
If we do similar work as in (ii), using B=3 and C=3 , we
will get
AD = 8r sin( ) sin( ) sin(60+ ), and
AE = 8r sin( ) sin( ) sin((60+ ).
Note that the ratio
AE/AD = sin(60+ )/sin(60+ ).
But ADE + AED = 180 - A/3
= (540-A)/3
= (540-(180-B-C))/3
= (360+B+C)/3
= (180+B)/3 + (180+C)/3. From here,
ADE = (180+B)/3 and AED = (180 + C)/3,
and similarly for triangles BFE and DFC.
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It thus follows that the sum of angles around F,
excluding DFE is 300o, or DFE = 60o. The other two
angles are similarly shown to be 60o.
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