Chapter 5. Menelaus theorem. 5.1 Menelaus theorem

Transcription

1 hapter 5 Menelaus theorem 5.1 Menelaus theorem Theorem 5.1 (Menelaus). Given a triangle with points,, on the side lines,, respectively, the points,, are collinear if and only if = 1. W Proof. (= ) LetW be the point on such that W//. Then, = W, and = W. It follows that = W W = W W = 1. ( =) Suppose the line joining and intersects at. From above, = 1 =. It follows that =. The points and divide the segment in the same ratio. These must be the same point, and,, are collinear.

2 202 Menelaus theorem Example 5.1. The external angle bisectors of a triangle intersect their opposite sides at three collinear points. c b a Proof. If the external bisectors are,, with,, on,, respectively, then = c b, = a c, = b a. It follows that = 1 and the points,, are collinear.

3 5.2 enters of similitude of two circles enters of similitude of two circles Given two circles O(R) and I(r), whose centers O and I are at a distance d apart, we animate a point on O(R) and construct a ray through I oppositely parallel to the ray O to intersect the circle I(r) at a point. The line joining and intersects the line OI of centers at a point T which satisfies OT : IT = O : I = R : r. This pointt is independent of the choice of. It is called the internal center of similitude, or simply the insimilicenter, of the two circles. T T O I If, on the other hand, we construct a ray through I directly parallel to the ray O to intersect the circlei(r) at, the line always intersectsoi at another pointt. 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 Desargue s theorem Given three circles with centers,, and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear.

4 204 Menelaus theorem 5.3 eva s theorem Theorem 5.2 (eva). Given a triangle with points,, on the side lines,, respectively, the lines,, are concurrent if and only if = +1. P Proof. (= ) Suppose the lines,, intersect at a point P. onsider the line P cutting the sides of triangle. y Menelaus theorem, P P = 1, or P P = +1. lso, consider the linep cutting the sides of triangle. y Menelaus theorem again, P P = 1, or P P = +1. Multiplying the two equations together, we have ( =) Exercise. = +1.

5 5.4 Some triangle centers Some triangle centers The centroid If D,E,F are the midpoints of the sides,, of triangle, then clearly F F D D E E = 1. The medians D, E, F are therefore concurrent. Their intersection is the centroid G of the triangle. onsider triangle D with transversalge. y Menelaus theorem, 1 = G GD D E E = G GD It follows that G : GD = 2 : 1. The centroid of a triangle divides each median in the ratio 2:1. F G E I D The incenter Let,, be points on,, such that,, bisect angles, and respectively. Then = b a, = c b, = a c. It follows that = b a c b a c = +1, and,, are concurrent. Their intersection is the incenter of the triangle. pplying Menelaus theorem to triangle with transversal I, we have 1 = I I = I I b b+c a b = I I = b+c a.

6 206 Menelaus theorem The Gergonne point Let the incircle of triangle be tangent to the sides at, at, and at respectively. Since = = s a, = = s b, and = = s c, we have = s b s c s c s a s a s b = 1. y eva s theorem, the lines,, are concurrent. The intersection is called the Gergonne point G e of the triangle. s a s a G e I s c s b s b s c Lemma 5.3. The Gergonne point G e divides the cevian in the ratio G e G e = a(s a) (s b)(s c). Proof. pplying Menelaus theorem to triangle with transversal G e, we have 1 = G e G e = G e G e (s c) s b a s a = G e G e = a(s a) (s b)(s c).

7 5.4 Some triangle centers The Nagel point If,, are the points of tangency of the excircles with the respective sidelines, the lines,, are concurrent by eva s theorem: = s c s b s c s a s a s b = 1. The point of concurrency is the Nagel point N a. I b I c s b s c s a Na s a s c s b I a Lemma 5.4. If the -excircle of triangle touches at, then the Nagel point divides the cevian in the ratio N a = a N a s a. Proof. pplying Menelaus theorem to triangle with transversal N a, we have 1 = N a N a = N a N a (s c) s a a s c = N a = a N a s a.

8 208 Menelaus theorem 5.5 Isotomic conjugates Given points on, on, and on, we consider their reflections in the midpoints of the respective sides. These are the points on, on and on satisfying =, = ; =, = ; =, =. learly,,, are concurrent if and only if,, are concurrent. P P Proof. ( )( ) ( )( = )( ) = 1. The points of concurrency of the two triads of lines are called isotomic conjugates.

9 5.5 Isotomic conjugates 209 Example 5.2. (The Gergonne and Nagel points) I b I c G e I N a I a Example 5.3. (The isotomic conjugate of the orthocenter) Let H denote the isotomic conjugate of the orthocenter H. Its traces are the pedals of the reflection of H in O. This latter point is the delongchamps point L o. L o O H H

10 210 Menelaus theorem Example 5.4. (ff-rocard points) onsider a point P = (u : v : w) with traces,, satisfying = = = µ. This means that w v +w a = u w +u b = v u+v c = µ. Elimination of u,v,w leads to 0 µ a µ 0 = b µ 0 µ µ c µ 0 = (a µ)(b µ)(c µ) µ3. Indeed, µ is the unique positive root of the cubic polynomial (a t)(b t)(c t) t 3. This gives the point P = ( (c µ b µ 3 : ( a µ c µ ( ) 3 b µ 3 :. a µ P P The isotomic conjugate ( (b µ P = has traces,, that satisfy c µ 3 : ( c µ a µ = = = µ. 3 : ( a µ b µ ) 3 These points are called the ff-rocard points. 1 They were briefly considered by. L. relle. 2 1 P. ff, n analogue of the rocard points, mer. Math. Monthly, 70 (1963) L. relle, 1815.