Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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. Click to enlarge Download ps file 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. Download ps file Type 2. Construct a triangle with the side joining larger angles ( e.g. a+ and b+) to have length 1. for example: Download ps file Type 3. Construct two lines that intersects the side opposite to b++ at angle b+, thus forming an isosceles triangle with base angle =b+. Download ps file Download ps file 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: Click to enlarge Download ps file 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. Click to enlarge Download ps file 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. Download ps file 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. Download ps file Extend the sides of the isosceles triangles so that they intersect as shown: Click to enlarge Download ps file Download ps file 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. Click to enlarge Download ps file Join B and C, note that BDC = 60+z+y (vertically opposite angles) Click to enlarge Download ps file 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. Click to enlarge Download ps file The last step is to prove the following figure: Click to enlarge ABC has angles , Download ps file 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. Download ps file Click to enlarge Download ps file 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 ). Download ps file 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. Click to enlarge Download ps file 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. Click to enlarge Download ps file