Morley’s Miracle

At the end of the 19^th century, mathematician Frank Morley discovered that the trisectors of each angle of any triangle meet in three new points that form a new triangle that is always an equilateral triangle.  The fact that this property holds for all triangles, including scalene triangles, was so unexpected that it became to be known as Morley’s Miracle.  (An interactive applet for Morley’s Miracle can be found at here.)

The proof is rather lengthy, but only requires a knowledge of high school trigonometry (including the law of sines, the law of cosines, Pythagorean identity, and sine and cosine addition and subtraction formulas).  The basic outline of the proof is to first use the law of sines on ∆ACE to express c₁ as a function of ∠A, ∠C, and b; then use the law of sines on ∆ABF to express b₃ as a function of ∠A, ∠B, and c; and finally use the law of cosines on ∆AEF to show that the value of a₂ is symmetrical when expressed in terms of the sides, angles, and area of ∆ABC.  The same calculations used to find a₂ can then be used to find b₂ and c₂, which by their symmetrical equations make them all equal to each other, showing that ∆DEF is an equilateral triangle.  The full proof can be found here.

Another surprising aspect of Morley’s Miracle is that the theorem cannot be extended to polygons with more sides than triangles.  Trisecting the four angles of a quadrilateral does not always produce a rhombus, and neither does quadrisecting the four angles.  It also does not appear that Morley’s Miracle can be extended to higher dimensions.  Trisecting the four angles of a tetrahedron does not always produce a regular tetrahedron.

                                                                                                                                                     

Quadrilateral with Trisected Angles

The fact that the trisectors of each angle of any triangle meet in three new points that can always form an equilateral triangle is pleasantly surprising.  The fact that this theorem cannot be extended to polygons with more sides or to shapes of higher dimension is also unexpected.  Both of these facts, and both of these surprises, make Morley’s Theorem truly miraculous.

Morley’s Miracle (Proof)

Morley’s Miracle states that the trisectors of each angle of any triangle meet in three new points that form a new triangle that is always an equilateral triangle. 

Proof:

sin(⅓C) / c1 = sin(∠AEC) / b	(law of sines on ∆AEC)
∠AEC + ⅓A + ⅓C = 180°	(angle sum of ∆AEC)
∠AEC = 180° – (⅓A + ⅓C)	(subtract)
sin(⅓C) / c1 = sin(180° – (⅓A + ⅓C)) / b	(substitute)
sin(⅓C) / c1 = sin(⅓A + ⅓C) / b	(sin(180° – x) = sin x)
c1 = b sin(⅓C) / sin(⅓A + ⅓C)	(rearrange)
A + B + C = 180°	(angle sum of ∆ABC)
⅓A + ⅓B + ⅓C = 60°	(divide by 3)
⅓A + ⅓C = 60° – ⅓B	(subtract)
c1 = b sin(⅓C) / sin(60° – ⅓B)	(substitute)
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / (4 sin(⅓B) sin(60° – ⅓B) sin(60° + ⅓B))	(multiply num and denom by 4 sin (⅓B) sin(60° + ⅓B))
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / d	(let d be the denon)
d = 4 sin(⅓B) sin(60° – ⅓B) sin(60° + ⅓B)	(d value)
d = 4 sin(⅓B) (sin(60°)cos(⅓B) – cos(60°)sin(⅓B)) (sin(60°)cos(⅓B) + cos (60°)sin(⅓B))	(sine addition and subtraction)
d = 4 sin(⅓B) (√3/2cos(⅓B) – ½sin(⅓B)) (√3/2cos(⅓B) – ½sin(⅓B))	(sine and cosine values)
d = 4 sin(⅓B) (¾cos2(⅓B) – ¼sin2(⅓B))	(difference of squares)
d = 3 sin(⅓B) cos2(⅓B) – sin3(⅓B)	(distribute 4 sin(⅓B))
d = sin(⅓B) cos2(⅓B) – sin3(⅓B) + 2 sin(⅓B) cos2(⅓B)	(rearrange)
d = sin(⅓B)(cos2(⅓B) – sin2(⅓B)) + 2sin(⅓B)cos(⅓B)cos(⅓B)	(factor out sin(⅓B))
d = sin(⅓B)(cos(⅓B)cos(⅓B) – sin(⅓B)sin(⅓B)) + (sin(⅓B)cos(⅓B) + sin(⅓B)cos(⅓B))cos(⅓B))	(rearrange)
d = sin(⅓B)cos(⅓B + ⅓B) + sin(⅓B + ⅓B)cos(⅓B))	(sine addition)
d = sin(⅓B)cos(2(⅓B)) + sin(2(⅓B))cos(⅓B))	(addition)
d = sin(⅓B  + 2(⅓B))	(sine addition)
d = sin(3(⅓B))	(addition)
d = sin(B)	(multiply)
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / sin(B)	(substitute)
c1 = (4b/sin(B)) sin(⅓B) sin(⅓C) sin(60° + ⅓B)	(rearrange)
K = ½ ac sin B	(area of ∆ABC)
4bK = 2 abc sin B	(multiply by 4b)
4b/sin B = 2abc/K	(divide by K sin B)
c1 = (2abc/K) sin(⅓B) sin(⅓C) sin(60° + ⅓B)	(substitute)
b3 = (2acb/K) sin(⅓C) sin(⅓B) sin(60° + ⅓C).	(similar argument to c1)
Let P = (2abc/K) sin(⅓B) sin(⅓C)	(P value assignment)
Let Q = 60° + ⅓B	(Q value assignment)
Let R = 60° + ⅓C	(R value assignment)
c1 = P sin Q	(substitution)
b3 = P sin R	(substitution)
a22 = b32 + c12 – 2b3c1cos(⅓A)	(law of cosines on ∆AEF)
a22 = (P sin R)2 + (P sin Q)2 – 2(P sin R)(P sin Q)cos(⅓A)	(substitution)
a22 = P2 sin2R + P2 sin2Q – 2 P2 sin R sin Q cos (⅓A)	(square)
a22 = P2(sin2R + sin2Q – 2 sin R sin Q cos (⅓A))	(factor P2)
A + B + C = 180°	(angle sum of ∆ABC)
⅓A + ⅓B + ⅓C = 60°	(divide by 3)
⅓A + ⅓B + ⅓C + 60° + 60° = 60° + 60° + 60°	(add 60° and 60°)
⅓A + (60° + ⅓B) + (60° + ⅓C) = 180°	(rearrange)
⅓A + Q + R = 180°	(substitution)
⅓A = 180° – (Q + R)	(subtract (Q + R))
a22 = P2(sin2R + sin2Q – 2 sin R sin Q cos (180° – (Q + R)))	(substitution)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q cos (Q + R))	(cos(180° – x) = -cos x)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q (cos Q cos R – sin Q sin R))	(cosine addition)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q cos Q cos R – 2 sin2R sin2Q)	(distribute)
a22 = P2(sin2R – sin2R sin2Q + sin2Q – sin2R sin2Q + 2 sin R sin Q cos Q cos R)	(rearrange)
a22 = P2(sin2R(1 – sin2Q) + sin2Q(1 – sin2R) + 2 sin R sin Q cos Q cos R)	(factor)
a22 = P2(sin2R cos2Q + sin2Q cos2R + 2 sin R sin Q cos Q cos R)	(1 – sin2Q = cos2Q)
a22 = P2(sin2R cos2Q + 2 sin R sin Q cos Q cos R + sin2Q cos2R)	(rearrange)
a22 = P2(sin R cos Q + sin Q cos R)2	(factor)
a22 = P2(sin(R + Q))2	(sine addition)
a22 = P2(sin(180° – (R + Q)))2	(sin(180° – x) = sin x)
a22 = P2(sin(⅓A))2	(substitution)
a2 = P sin(⅓A)	(square root)
a2 = ((2abc/K) sin(⅓B) sin(⅓C)) sin(⅓A)	(substitution)
a2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
b2 = (2bca/K) sin(⅓B) sin(⅓C) sin(⅓A)	(similar argument to a2)
c2 = (2cab/K) sin(⅓C) sin(⅓A) sin(⅓B)	(similar argument to a2)
b2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
c2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
a2 = b2 = c2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(substitution)
∆DEF is equilateral	(a2, b2, and c2 are congruent)

Q.E.D.