Chapter 3 Introduction to Triangle Geometry

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Chapter 3 Introduction to Triangle Geometry Chapter 3 Introduction to Triangle Geometry 3.1 Preliminaries 3.1.1 Coordinatization of points on a line Let B and C be two fixed points on a line L. Every point X on L can be coordi- natized in one of several ways: BX (1) the ratio of division t = XC , (2) the absolute barycentric coordinates: an expression of X as a convex com- bination of B and C: X = (1 t)B + tC, − which expresses for an arbitrary point P outside the line L, the vector PX as a combination of the vectors PB and PC. (3) the homogeneous barycentric coordinates: the proportion XC : BX, which are masses at B and C so that the resulting system (of two particles) has balance point at X. P B X C 26 Introduction to Triangle Geometry 3.1.2 Centers of similitude of two circles Consider two circles O(R) and I(r), whose centers O and I are at a distance d apart. Animate a point X on O(R) and construct a ray through I oppositely parallel to the ray OX to intersect the circle I(r) at a point Y . You will find that the line XY always intersects the line OI at the same point T . This we call the internal center of similitude, or simply the insimilicenter, of the two circles. It divides the segment OI in the ratio OT : T I = R : r. The absolute barycentric coordinates of P with respect to OI are R I + r O T = · · . R + r Y T T ′ O I Y ′ X If, on the other hand, we construct a ray through I directly parallel to the ray OX to intersect the circle I(r) at Y ′, the line XY ′ always intersects OI at another point T ′. This is the external center of similitude, or simply the exsimilicenter, of the two circles. It divides the segment OI in the ratio OT ′ : T ′I = R : r, and has absolute barycentric coordinates − ′ R I r O T = · − · . R r − 3.1.3 Tangent circles If two circles are tangent to each other, the line joining their centers passes through the point of tangency, which is a center of similitude of the circles. 3.1 Preliminaries 27 T ′ O T I O I 3.1.4 Harmonic division Two points X and Y are said to divide two other points B and C harmonically if BX BY = . XC −YC They are harmonic conjugates of each other with respect to the segment BC. Examples 1. For two given circles, the two centers of similitude divide the centers har- monically. 2. Given triangle ABC, let the internal bisector of angle A intersect BC at X. The harmonic conjugate of X in BC is the intersection of BC with the external bisector of angle A. 3. Let A and B be distinct points. If M is the midpoint of the segment AB, it is not possible to find a finite point N on the line AB so that M, N AN AM divide A, B harmonically. This is because NB = MB = 1, requiring AN = NB = BN, and AB = BN AN =− 0, a contradiction.− We shall agree− to say that if M and N divide−A, B harmonically, then N is the infinite point of the line AB. Exercise 1. If X, Y divide B, C harmonically, then B, C divide X, Y harmonically. 28 Introduction to Triangle Geometry A B X C X′ 2. Given a point X on the line BC, construct its harmonic associate with re- spect to the segment BC. Distinguish between two cases when X divides BC internally and externally. 1 3. The centers A and B of two circles A(a) and B(b) are at a distance d apart. The line AB intersect the circles at A′ and B′ respectively, so that A, B are between A′, B′. A′ B′ A B 4. Given two fixed points B and C and a positive constant k =1, the locus of the points P for which BP : CP = k is a circle. 6 | | | | 1Make use of the notion of centers of similitude of two circles. 3.1 Preliminaries 29 3.1.5 Homothety Given a point T and a nonzero constant k, the similarity transformation h(T,k) which carries a point X to the point X′ on the line TX satisfying TX′ : TX = k :1 is called the homothety with center T and ratio k. Explicitly, h(T,k)(P )=(1 k)T + kP. − Any two circles are homothetic. Let P and Q be the internal and external h r centers of similitude of two circles O(R) and I(r). Both the homotheties (Q, R ) and h(P, r ) transform the circle O(R) into I(r). − R 3.1.6 The power of a point with respect to a circle The power of a point P with respect to a circle C = O(R) is the quantity C(P ) := OP 2 R2. − This is positive, zero, or negative according as P is outside, on, or inside the circle C. If it is positive, it is the square of the length of a tangent from P to the circle. Theorem (Intersecting chords) If a line L through P intersects a circle C at two points X and Y , the product PX PY (of signed lengths) is equal to the power of P with respect to the circle. · T O X Y P T ′ 30 Introduction to Triangle Geometry 3.2 Menelaus and Ceva theorems 3.2.1 Menelaus and Ceva Theorems Consider a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively. Theorem 3.1 (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ = 1. XC · Y A · ZB − A Y Z X B C Theorem 3.2 (Ceva). The lines AX, BY , CZ are concurrent if and only if BX CY AZ = +1. XC · Y A · ZB 3.2.2 Desargues Theorem As a simple illustration of the use of the Menelaus and Ceva theorems, we prove the following rmDesargues Theorem: Given three circles, the exsimilicenters of the three pairs of circles are collinear. Likewise, the three lines each joining the insimilicenter of a pair of circles to the center of the remaining circle are concur- rent. We prove the second statement only. Given three circles A(r1), B(r2) and C(r3), the insimilicenters X of (B) and (C), Y of (C), (A), and Z of (A), (B) are the points which divide BC, CA, AB in the ratios BX r CY r AZ r = 2 , = 3 , = 1 . XC r3 Y A r1 ZB r2 3.2 Menelaus and Ceva theorems 31 A Z Y P B X C X′ A Y C P Y ′ Z X B Z′ It is clear that the product of these three ratio is +1, and it follows from the Ceva theorem that AX, BY , CZ are concurrent. 32 Introduction to Triangle Geometry 3.2.3 The incircle and the Gergonne point The incircle is tangent to each of the three sides BC, CA, AB (without extension). Its center, the incenter I, is the intersection of the bisectors of the three angles. The inradius r is related to the area ∆ by S =(a + b + c)r. A Y Z G e I B X C If the incircle is tangent to the sides BC at X, CA at Y , and AB at Z, then b + c a c + a b a + b c AY = AZ = − , BZ = BX = − , CX = CY = − . 2 2 2 These expressions are usually simplified by introducing the semiperimeter s = 1 2 (a + b + c): AY = AZ = s a, BZ = BX = s b, CX = CY = s c. − − − ∆ Also, r = s . It follows easily from the Ceva theorem that AX, BY , CZ are concurrent. The point of concurrency Ge is called the Gergonne point of triangle ABC. Triangle XYZ is called the intouch triangle of ABC. Clearly, B + C C + A A + B X = , Y = , Z = . 2 2 2 3.2 Menelaus and Ceva theorems 33 It is always acute angled, and A B C YZ =2r cos , ZX =2r cos , XY =2r cos . 2 2 2 Exercise 1. Given three points A, B, C not on the same line, construct three circles, with centers at A, B, C, mutually tangent to each other externally. B Y X C Z A 2. Construct the three circles each passing through the Gergonne point and tangent to two sides of triangle ABC. The 6 points of tangency lie on a circle. 2 3. Two circles are orthogonal to each other if their tangents at an intersection are perpendicular to each other. Given three points A, B, C not on a line, construct three circles with these as centers and orthogonal to each other. (1) Construct the tangents from A′ to the circle B(b), and the circle tangent to these two lines and to A(a) internally. (2) Construct the tangents from B′ to the circle A(a), and the circle tangent to these two lines and to B(b) internally. (3) The two circles in (1) and (2) are congruent. R r 2 s2 2 √(4 + ) + This is called the Adams circle. It is concentricwith the incircle, and has radius 4R+r r. · 34 Introduction to Triangle Geometry A Ge I B C 4. Given a point Z on a line segment AB, construct a right-angled triangle ABC whose incircle touches the hypotenuse AB at Z. 3 5.
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