Geometry lass Examples (July 10) Paul iu Department of Mathematics Florida tlantic University c b a Summer 2014
1 Menelaus theorem Theorem (Menelaus). Given a triangle with points,, on the side lines,, respectively, the points,, are collinear if and only if = 1. W Proof. (= ) Let W 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 Example F1: The external bisectors. 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 Example F1: Menelaus Theorem. Given triangle and points on, on, and on the extension of, such that,, are collinear. If = x, = z, and = y, and two of these lengths are given, calculate the remaining one. x a b y z c (a, b, c) x y z (3, 4, 5) 1 2 4 (3, 5, 6) 1 4 (3, 5, 7) 1 6 (4, 5, 6) 3 4 (4, 5, 7) 2 4 (5, 6, 7) 1 6 (6, 7, 8) 4 6
4 Example F2: (2, 3, 4) triangle. is a triangle with a =2, b =3, c =4. transversal intersects the sidelines at,, such that = = = t. alculate t. 2 3 4
5 Example F3: (7, 12, 18) triangle. is a triangle with a =7, b =12, c =18. transversal intersects the sidelines at,, such that = = = t. alculate t. 7 12 18
6 Example F4: (9, 10, 12) triangle. is a triangle with a =9, b =10, c =12. transversal intersects the sidelines at,, such that = = = t. alculate t. 9 10 12
7 Example F5. Given three circles with centers,, and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear.
8 Example F6: Line with equal intercepts on sidelines of a given triangle. Given triangle, construct a line intersecting at externally, at and at internally so that = =. x x x
Solution to Example F6: Line with equal intercepts on sidelines of a given triangle. Given triangle, construct a line intersecting at externally, at and at internally so that = =. 9 x r r I Solution. If = x, by Menelaus theorem, we require From this, Note that a + x x x b x x = x c x = 1. bc a + b + c. bc sin x = 2s sin = Δ s 1 sin = r sin. This means that I is parallel to, and suggests the following simple construction of the line. (1) onstruct the incenter I of triangle. (2) onstruct a line through I parallel to, to intersect at. (3) onstruct a circle with center, radius, to intersect externally at and internally at. Then,, are collinear with = = = r sin.
10 Example F7: Line with equal intercepts on sidelines of a given triangle. Given triangle, construct a line intersecting at externally, at and at internally so that = =. y y y
Solution to Example F7: Line with equal intercepts on sidelines of a given triangle. Given triangle, construct a line intersecting at externally, at and at internally so that = =. 11 c I c c y y y c Solution. If = = = y, by Menelaus theorem, we require a + y y y b y c y y = 1. From this, ca y = a + b c = r c sin, where r c is the radius of the excircle on the side.
12 eva s theorem Theorem (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 line P 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.
13 Example F8: eva s theorem. Given triangle and points on, on, on such that the cevians,, are concurrent. If = x, = z, and = y, and two of these lengths are given, calculate the remaining one. a x b y z c (a, b, c) x y z (3, 4, 6) 1 2 4 (3, 5, 6) 1 4 (3, 5, 7) 1 3 (4, 5, 6) 3 4 (4, 5, 7) 1 4 (5, 6, 7) 3 4 (6, 7, 9) 5 4
14 Example F9: (3, 4, 6) triangle. is a triangle with a =3, b =4, c =6.,, are points on,, respectively such that = = = t. If the cevians,, are concurrent, calculate t. 3 4 6
15 Example F10. is a right triangle. Show that the lines,, and Q are concurrent. P Q
16 Solution to Example F10. is a right triangle. Show that the lines,, and Q are concurrent. 0 0 0 Q Solution. Let intersect at 0, intersect at 0, and Q intersect at 0. 0 0 0 0 0 0 = 0 0 2 = 2 = 1. y eva s theorem, the lines 0, 0, 0 are concurrent.
Example The centroid..ifd, 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. 17 F G E D onsider the line GE intersecting the sides of triangle D. y the Menelaus theorem, 1 = G GD D E E = G GD 1 2 1 1. It follows that G : GD =2:1. The centroid of a triangle divides each median in the ratio 2:1.
18 Example The incenter. Let,, be points on,, such that,, bisect angles, and respectively. Then = b a, = c b, = a c. I It follows that = b a c b a c =+1, and,, are concurrent. Their intersection is the incenter of the triangle.
Example F:. Given a point P, let the lines P, P, P intersect,, respectively at,,. onstruct the circle through,,, to intersect the lines,, again at,,. Then the lines,, are concurrent. 19
20 Example F11. Suppose two cevians, each through a vertex of a triangle, trisect each other. Show that these are medians of the triangle. P
21 Solution to Example F11. Suppose two cevians, each through a vertex of a triangle, trisect each other. Show that these are medians of the triangle. P Given: Triangle with cevians and intersecting at P such that P trisects and. To prove: and are medians, i.e., and are midpoints of and respectively. Proof. (1) Since P is a trisection of and Q, P = 3 p for p =1or 2, P = q 3 for q =1or 2. (2) pplying Menelaus theorem to triangle P with transversal, we have... (3) Similarly, by applying Menelaus theorem to triangle P with transversal, we conclude that is the midpoint of, and is also a median.
22 Example F. Let,, be cevians of intersecting at a point P. (i) Show that if bisects angle, and =, then is isosceles. (ii) Show if if,, are bisectors, and P P = P, then is a right triangle.
27 Example F: Kiepert perspectors. If similar isosceles triangles, and (of base angle θ) are constructed on the sides of triangle, either all externally or all internally, the lines,, and are concurrent. θ α 1 α 2 θ θ P θ β 2 β 1 θ θ γ 1 γ2 Proof. pplying the law of sines to triangles and,wehave Likewise, sin β 1 sin β 2 sin α 1 = sin α 1 sin(β + θ) sin(γ + θ) sin α 2 sin(β + θ) sin(γ + θ) sin α 2 = sin(β + θ) sin(γ + θ) sin(β + θ) = sin(γ + θ). = sin(γ+θ) sin(α+θ) and sin γ 1 sin γ 2 = sin(α+θ) sin(β+θ). From these sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 =+1. and the lines,, are concurrent. The point of intersection is called the Kiepert perspector K(θ). In particular, it is called (1) a Fermat point if θ = ±60, (2) a Napoleon point if θ = ±30, (3) a Vecten point if θ = ±45.
28 Trigonmetric version of the eva Theorem Let be a point on the side of triangle such that the directed angles = α 1 and = α 2. y the sine formula, = / / = sin α 1/ sin β sin α 2 / sin γ = sin γ sin β sin α 1 = c sin α 2 b sin α 1. sin α 2 α 1 α 2 β 2 β 1 Likewise, if and be points on the lines, respectively, with = β 1, = β 2 and = γ 1, = γ 2,wehave = a c sin β 1 sin β 2, γ 2 γ 1 = b a sin γ 1 sin γ 2. These lead to the following trigonometric version of the eva theorem. Theorem (eva). The lines,, are concurrent if and only if Proof. sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 =+1. = sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2.