2. Equilateral Triangles

Transcription

1 2. Equilateral Triangles Recall the well-known theorem of van Schooten. Theorem 1 If ABC is an equilateral triangle and M is a point on the arc BC of C(ABC) then MA = MB + MC. Proof Use Ptolemy on the cyclic quadrilateral ABM C. (Figure 1) In fact, using the Ptolemy inequality for quadrilaterals, we get the following van Schooten inequality. Figure 1: Theorem 2 Let ABC be an equilateral triangle. Then if M is any point in the plane of ABC we have 1 Pompeiu Triangle MA MB + MC. We get the following well-known theorem of D. Pompeiu as an immediate consequence of the previous inequality. Theorem Let M be any point in the plane of an equilateral triangle ABC. Then the distances MA, MB and MC can be the sidelengths of a triangle. 1

2 Proof It follows immediately since MA MB + MC The triangle is degenerate if M lies on the arc BC of the circumcircle C(ABC). A triangle with side lengths MA, MB and MC is called a Pompeiu triangle. When M is in the interior of ABC, then the pompeiu triangle can be explicitly constructed. Locate N so that the triangle BNM is equilateral. Now consider the triangles We have and AMB and BNC, AB = BC, BM = BN M BA = 60 M BC = C BN. Thus the triangles are similar, in fact CBN is got by rotating triangle ABM through 60 anti-clockwise in the diagram. Thus AM = CN, Figure 2: and so the triangle N M C has side lengths equal to M A, M B and M C. Thus N M C is the Pompeiu triangle. The measure of the angles of the Pompeiu triangle in terms of the angles subtended at M by the vertices of ABC and the area of the Pompeiu triangle are given by the following result of the distinguished Romanian born Sabin Tabirca Theorem 4 (Tabirca) If ABC is an equilateral triangle, and M is an interior point of ABC, then the angles of the Pompeiu triangle and its area are as follows: (a) the angles are the angles B MC 60, C MA 60 and A MB 60 ; 2

3 (b) the area S (Pompeiu triangle) is 1 (area of ABC) 4 MO 2, where O is circumcentre of the triangle ABC. Proof (a) In the triangle NMC, C MN = C MB N MB = C MB 60, C NM Notation: S(ABC). = C NB M NB = C NM 60 = A MB 60, and, finally, MĈN = 180 {C MN + C NM} = 180 {C MB 60 + A MB 60 } = 00 {60 A MC} = {A MC 60 }. The area of a triangle ABC is denoted by Figure : (b) In the diagram, NMC is the Pompeiu triangle which we now denote by T p. Then: S(T p ) where a = BC = CA = AB. = 1 ( CM. MN ) sin(c MN) 2 = 1 2 CM. BM sin(c MB 60 ) = 1 2 CM. BM {sin(c MB). 1 cos(c MB). 2 2 } = 1 4 CM. BM sin(c MB) CM MB cos(c MB) 4 = 1 2 S(CMB) 8 { CM 2 + BM 2 a 2 },

4 Thus S(T p ) = S(CMB) Similarly we can show that 8 { CM 2 + BM 2 a 2 }. S(T p ) = 1 2 S(CMA) 8 { CM 2 + MA 2 a 2 }, and S(T p ) = 1 2 S(BMA) 8 { BM 2 + MA 2 a 2 }. Adding, we get S(T p ) = 1 2 S(ABC) 8 {2( MA 2 + MB 2 + MC 2 ) a 2 }. Recall the Leibniz formula which states that for any triangle ABC with centroid G and point M MA 2 + MB 2 + MC 2 = MG { AB 2 + BC 2 + CA 2 } In the case of an equilateral triangle, G = 0, the centre of the circumcircle and a 2 = AB 2 = BC 2 = CA 2 so MA 2 + MB 2 + MC 2 = MO 2 +a 2. Thus S(T p ) = 1 2 S(ABC) 8 {6 MO 2 + 2a 2 a 2 } = 1 2 S(ABC) 8 6 MO a2. Since ABC is equilateral with side length a, S(ABC) = 4 a2, so S(T p ) = 1 2 S(ABC) 4 MO S(ABC) = S(ABC) 4 MO 2. Thus S(T p ) = 1 S(ABC) 4 MO 2, as required. 4

5 2 Fermat-Toricelli Point Let ABC be any triangle and on each side construct externally three equilateral triangles ABC 1, BCA 1 and CAB 1. Then we have the following theorem. Theorem 5 (a) The three circumcircles of the equilateral triangles intersect in a point T, i.e. C(ABC 1 ) C(BCA 1 ) C(CAB 1 ) = {T }. (b) The lines AA 1, BB 1 and CC 1 are concurrent at T, i.e. AA 1 BB 1 CC 1 = {T }. (c) AA 1 = BB 1 = CC 1 = T A + T B + T C. (d) For all points M in the plane of ABC, MA + MB + MC AA 1 = T A + T B + T C. Figure 4: i.e. the point T minimises the expression MA + MB + MC. The point T is called the Toricelli-Fermat point. Proof (a) Let {A, T } be the intersection points of the circles C(ABC 1 ) and C(ACB 1 ). Then since Because C 1 BT A is cyclic, A T B = 180 AĈ1B = 120. B 1 AT C is cyclic, A T C = 180 AB 1 C = 120. Thus BT C = 120 also. Thus B T C + BÂ1C = 180. Figure 5: 5

6 and so BA 1 CT is cyclic, i.e. T C(A 1 BC) (b) We claim that C 1, T, C are collinear points. A T C = 120, A T C 1 = A BC 1 = 60 giving AT C + AT C 1 = 180, i.e. C, T and C 1 are collinear. Similarly A, T, A 1 and B, T, B 1 are collinear. (c) We claim that CC 1 = T A + T B + T C. Since T C(AC 1 B) and AC 1 B is equilateral, then by van Schooten s theorem T C 1 = T A + T B. Thus CC 1 = CT + T C 1 = T C + T A + T B, as required. Similarly for AA 1 and BB 1. (d) Now let M be any point in the plane of ABC. Then, since ABC 1 is equilateral: MC 1 MA + MB Thus MA + MB + MC MC + MC 1 CC 1 = T A + T B + T C So the point of a triangle which minimises the sum of the distances to the three vertices is the Toricelli-Fermat point. One could ask the question of weighted distances to the vertices and ask which point(s) minimise weighted sums. This is the question we now investigate. 6

7 Generalised Fermat-Toricelli Theorem Let x, y and z be the side length of a triangle αβγ with x the length of the side opposite vertex α, y the length of the side opposite β and z the length of the side opposite γ. On an arbitrary triangle ABC construct externally triangles similar to αβγ with vertices positioned as indicated in Figure 6. (a) Then their circumcircles intersect at a point T 1, i.e. C(ABC 1 ) C(BCA 1 ) C(CAB 1 ) = {T 1 } (b) The lines AA 1, BB 1 and CC 1 are concurrent, i.e. AA 1 BB 1 CC 1 = {T 1 } (c) x AA 1 = y BB 1 = z CC 1 = x AT 1 = y BT 1 = z CT 1 (d) For any point M in the plane of ABC, Figure 6: x MA + y MB + z MC x AA 1 = x AT 1 + y BT 1 + z CT 1 Thus the point T 1 minimises the weighted distances of a point to the vertices. Proof The construction of the proof is similar to the proofs in the special case when x = y = z. (a) Let C(ABC 1 ) C(ACB 1 ) = {A, T 1 }. 7 Figure 7:

8 Since and AT 1 BC 1 is cyclic, A T 1 B = 180 γ, AT 1 CB 1 is cyclic, so A T 1 C = 180 β. Thus B T 1 C = 60 {A T 1 B + A T 1 C} = 60 {180 γ β = 180 { γ + β} = α. Thus T 1 BA 1 C is cyclic, i.e. T 1 C(BA 1 C), as required. (b) We claim that A T 1 C 1 + A T 1 C = 180 and from this it follows that T 1 lies on CC 1. In a similar way, we get that T 1 also belongs to the line segments BB 1 and AA 1. To show that A T 1 C 1 + A T 1 C = 180, we have, since, AT 1 CB 1 is cyclic, AT 1 C = 180 β. Also, A T 1 C 1 + A BC 1 = 180, as required. (c) Since the triangles A 1 BC and αβγ are similar, then A 1 B z = A 1C y = BC x. Thus A 1 B = BC z x, and A 1 C = BC y x. Since T 1 C(BCA 1 ), then, by Ptolemy, Figure 8: 8

9 T 1 A 1. BC = BT 1 CA 1 + CT 1 BA 1 = BT 1. BC y x + CT 1 BC z x Dividing across by BC and multiplying by x, we get x T 1 A 1 = y BT 1 + z CT 1. Thus x T 1 A 1 + y BT 1 + z CT 1 = x T 1 A + x T 1 A 1 = x AA 1. Similarly, we can show that x T 1 A +y T 1 B +z T 1 C = y BB 1 = z CC 1 (d) Now take a point M C(BCA 1 ). Then, by the Ptolemy inequality, BM CA 1 + CM. BA 1 > MA 1 BC Proceeding as in (c) above, we get x MA + y MB + z MC > x AA 1 = x T 1 A + y T 1 B + z T 1 C as required. Remarks 1. Now suppose that the positive weights x, y and z are not the sides of a triangle, i.e. suppose x y + z What then is the point which minimises the quantity x MA + y MB + z MC? where M is any point in the plane of ABC. To decide this, consider 9

10 x MA + y MB + z MC y( MA + MB ) + z( MA + MC ), since x y + z y AB + z AC = x AA + y AB + z AC Thus the point A(vertex) minimises the quantity x MA + y MB + z MC 2. Suppose we take then the weighted expression x = sin(bâc) = sin Â, y = sin(a BC) = sin B, z = sin(bĉa) = sin Ĉ, sin(â) MA + sin( B) MB + sin(ĉ) MC is minimised when M = O the centre of the circumcircle of ABC. If we take x = sin(â), y = sin( B) and z = sin(ĉ), then sin(â) MA + sin( B) MB + sin(ĉ) MC is minimised when M = H, the orhthocentre. B 4. If x = sin(â ), y = sin( ) and z = sin(ĉ ) then sin(â 2 B ) MA + (sin ) MB + sin(ĉ 2 2 ) MC is minimised when M = I, the incentre. 10