Chapter 8. Feuerbach s theorem. 8.1 Distance between the circumcenter and orthocenter

Similar documents
Steiner s porism and Feuerbach s theorem

The circumcircle and the incircle

Chapter 2. The laws of sines and cosines. 2.1 The law of sines. Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle ABC.

Chapter 3. The angle bisectors. 3.1 The angle bisector theorem

Menelaus and Ceva theorems

Chapter 6. Basic triangle centers. 6.1 The Euler line The centroid

Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

The Menelaus and Ceva Theorems

Geometry. Class Examples (July 3) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

XIII GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN The correspondence round. Solutions

Survey of Geometry. Paul Yiu. Department of Mathematics Florida Atlantic University. Spring 2007

Geometry. Class Examples (July 8) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Singapore International Mathematical Olympiad Training Problems

Geometry. Class Examples (July 10) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Advanced Euclidean Geometry

BMC Intermediate II: Triangle Centers

Chapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem

Heptagonal Triangles and Their Companions

22 SAMPLE PROBLEMS WITH SOLUTIONS FROM 555 GEOMETRY PROBLEMS

Chapter 5. Menelaus theorem. 5.1 Menelaus theorem

Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.

SOME NEW THEOREMS IN PLANE GEOMETRY. In this article we will represent some ideas and a lot of new theorems in plane geometry.

Distances Among the Feuerbach Points

Construction of Ajima Circles via Centers of Similitude

A Note on Reflections

arxiv: v1 [math.ho] 10 Feb 2018

XIV GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN The correspondence round. Solutions

Three Natural Homoteties of The Nine-Point Circle

Geometry JWR. Monday September 29, 2003

Triangles III. Stewart s Theorem (1746) Stewart s Theorem (1746) 9/26/2011. Stewart s Theorem, Orthocenter, Euler Line

XI Geometrical Olympiad in honour of I.F.Sharygin Final round. Grade 8. First day. Solutions Ratmino, 2015, July 30.

Geometry. Class Examples (July 29) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

1/19 Warm Up Fast answers!

GEOMETRY OF KIEPERT AND GRINBERG MYAKISHEV HYPERBOLAS

Homogeneous Barycentric Coordinates

Survey of Geometry. Supplementary Notes on Elementary Geometry. Paul Yiu. Department of Mathematics Florida Atlantic University.

2013 Sharygin Geometry Olympiad

SOME PROPERTIES OF INTERSECTION POINTS OF EULER LINE AND ORTHOTRIANGLE

On Emelyanov s Circle Theorem

Inversion. Contents. 1 General Properties. 1 General Properties Problems Solutions... 3

Determining a Triangle

On the Circumcenters of Cevasix Configurations

LAMC Intermediate I March 8, Oleg Gleizer. Theorem 1 Any two inscribed angles subtending the same arc on a circle are congruent.

ON SOME GEOMETRIC RELATIONS OF A TRIANGLE

Hagge circles revisited

Problems First day. 8 grade. Problems First day. 8 grade

Menelaus and Ceva theorems

Affine Transformations

Recreational Mathematics

XII Geometrical Olympiad in honour of I.F.Sharygin Final round. Solutions. First day. 8 grade

Revised Edition: 2016 ISBN All rights reserved.

XIV Geometrical Olympiad in honour of I.F.Sharygin Final round. Solutions. First day. 8 grade

IX Geometrical Olympiad in honour of I.F.Sharygin Final round. Ratmino, 2013, August 1 = 90 ECD = 90 DBC = ABF,

A Note on the Barycentric Square Roots of Kiepert Perspectors

Isotomic Inscribed Triangles and Their Residuals

Isogonal Conjugates. Navneel Singhal October 9, Abstract

Construction of a Triangle from the Feet of Its Angle Bisectors

Triangle Centers. Maria Nogin. (based on joint work with Larry Cusick)

Chapter 4. Feuerbach s Theorem

Some Collinearities in the Heptagonal Triangle

Nagel, Speiker, Napoleon, Torricelli. Centroid. Circumcenter 10/6/2011. MA 341 Topics in Geometry Lecture 17

Plane geometry Circles: Problems with some Solutions

The Apollonian Circles and Isodynamic Points

MAHESH TUTORIALS. Time : 1 hr. 15 min. Q.1. Solve the following : 3

The Apollonius Circle and Related Triangle Centers

Classical Theorems in Plane Geometry 1

Collinearity/Concurrence

Integrated Math II. IM2.1.2 Interpret given situations as functions in graphs, formulas, and words.

Conic Construction of a Triangle from the Feet of Its Angle Bisectors

Complex Numbers in Geometry

On a Porism Associated with the Euler and Droz-Farny Lines

arxiv:math/ v2 [math.mg] 19 Dec 2001

The Inversion Transformation

Triangles III. Stewart s Theorem, Orthocenter, Euler Line. 23-Sept-2011 MA

Objective Mathematics

Chapter 4. The angle bisectors. 4.1 The angle bisector theorem

Power Round: Geometry Revisited

CLASS IX GEOMETRY MOCK TEST PAPER

NEW YORK CITY INTERSCHOLASTIC MATHEMATICS LEAGUE Senior A Division CONTEST NUMBER 1

7.1 Projections and Components

Examples: Identify three pairs of parallel segments in the diagram. 1. AB 2. BC 3. AC. Write an equation to model this theorem based on the figure.

Problem Solving and Recreational Mathematics

Cultivating and Raising Families of Triangles. Adam Carr, Julia Fisher, Andrew Roberts, David Xu, and advisor Stephen Kennedy

GEOMETRY AND COMPLEX NUMBERS (January 23, 2004) 5

A Distance Property of the Feuerbach Point and Its Extension

INVERSION IN THE PLANE BERKELEY MATH CIRCLE

SOME NEW THEOREMS IN PLANE GEOMETRY II

The Arbelos and Nine-Point Circles

Another Variation on the Steiner-Lehmus Theme

Harmonic Division and its Applications

Geometry in the Complex Plane


Review exercise 2. 1 The equation of the line is: = 5 a The gradient of l1 is 3. y y x x. So the gradient of l2 is. The equation of line l2 is: y =

Vectors - Applications to Problem Solving

Homework Assignments Math /02 Fall 2017

Chapter-wise questions

The Kiepert Pencil of Kiepert Hyperbolas

Calgary Math Circles: Triangles, Concurrency and Quadrilaterals 1

Trigonometrical identities and inequalities

Transcription:

hapter 8 Feuerbach s theorem 8.1 Distance between the circumcenter and orthocenter Y F E Z H N X D Proposition 8.1. H = R 1 8 cosαcos β cosγ). Proof. n triangle H, = R, H = R cosα, and H = β γ. y the law of cosines, H = R 1 + 4 cos α 4 cosα cosβ γ)) = R 1 4 cosαcosβ + γ) + cosβ γ)) = R 1 8 cosα cosβ cosγ). Proposition 8.. r = 4R sin α sin β sin γ, s = 4R cos α cos β cos γ,

30 Feuerbach s theorem a a a Proof. 1) n triangle, = R sin γ, = β, and = 180 α+β. pplying the law of sines, we have a = From this, sin β sin α+β = R sin γ sin β cos γ = 4R sin γ cos γ sin β cos γ r = sin α = 4R sin α sin β sin γ. = 4R sin β sin γ. ) Similarly, in triangle a, a = 90 + β, we have a = 4R cos β cos γ. t follows that s = a cos α = 4R cos α cos β cos γ. 8. Distance between circumcenter and tritangent centers Lemma 8.1. f the bisector of angle intersects the circumcircle at M, then M is the center of the circle through,,, and a. Proof. 1) Since M is the midpoint of the arc, M = M = M. Therefore, M = M + = M + = M, and M = M. Similarly, M = M. ) n the other hand, since a and a are both right angles, the four points,,, a M are concyclic, with center at the midpoint of. This is the point M.

8. Distance between circumcenter and tritangent centers 31 M a Y M Y r a a Theorem 8. Euler). a) = R Rr. b) a = R + Rr a. Proof. a) onsidering the power of in the circumcircle, we have R = M = M = r sin α b) onsider the power of a in the circumcircle. Note that a = ra. lso, sin α a M = M = R sin α. R sin α = Rr. a = R + a a M = R + r a sin α R sin α = R + Rr a.

3 Feuerbach s theorem 8.3 Distance between orthocenter and tritangent centers Proposition 8.3. H = r 4R cosαcosβ cosγ, H a = r a 4R cosαcosβ cosγ. H X Proof. n triangle H, we have H = R cos α, = 4R sin β sin γ β γ. y the law of cosines, and H = H = H + H cos β γ = 4R cos α + 4 sin β γ sin 4 cosαsin β sin γ cos β γ ) = 4R cos α + 4 sin β γ sin 4 cosαsin β sin γ cos β cos γ 4 cosα β γ sin sin = 4R cos α + 4 sin β sin γ cosα sinβ sinγ 4 1 sin α = 4R cosαcos α sin β sin γ) + 8 sin α β γ ) sin sin = 4R cosα cosβ cosγ + 8 sin α β γ ) sin sin = r 4R cosαcosβ cosγ. ) sin β sin γ ) ) ) n triangle H a, a = 4R cos β cos γ. y the law of cosines, we have

8.4 Proof of Feuerbach s theorem 33 H a a a a Ha = H + a a H cos β γ = 4R cos α + 4 cos β γ cos 4 cosα cos β cos γ cos β γ ) = 4R cos α + 4 cos β γ cos 4 cosα β γ cos cos 4 cosα cos β cos γ sin β sin γ ) = 4R cos α + 4 cos β γ cos 4 1 sin α ) cos β γ ) cos cosα sinβ sinγ = 4R cosαcosα sin β sin γ) + 8 sin α β γ ) cos cos = 4R cosα cosβ cosγ + 8 sin α β γ ) cos cos = r a 4R cosαcosβ cosγ. 8.4 Proof of Feuerbach s theorem Theorem 8.3 Feuerbach). The nine-point circle is tangent internally to the incircle and externally to each of the excircles. H N a

34 Feuerbach s theorem Proof. 1) Since N is the midpoint of H, N is a median of triangle H. y pollonius theorem, N = 1 H + ) 1 4 H = 1 4 R Rr + r = ) R r. Therefore, N is the difference between the radii of the nine-point circle and the incircle. This shows that the two circles are tangent to each other internally. ) Similarly, in triangle a H, N a =1 H a + a ) 1 4 H = 1 4 R + Rr a + r a = ) R + r a. This shows that the distance between the centers of the nine-point and an excircle is the sum of their radii. The two circles are tangent externally.

8.4 Proof of Feuerbach s theorem 35 b b c c c F F b F c b N c F a a b a Exercise a 1. Suppose there is a circle, center, tangent externally to all three excircles. Show that triangle is equilateral.. Find the dimensions of an isosceles but non-equilateral) triangle for which there is a circle, center, tangent to all three excircles. a

36 Feuerbach s theorem c b a Excursus: Steiner s porism onstruct the circumcircle ) and the incircle ) of triangle. nimate a point on the circumcircle, and construct the tangents from to the incircle ). Extend these tangents to intersect the circumcircle again at and. The lines is always tangent to the incircle. This is the famous theorem on Steiner porism: if two given circles are the circumcircle and incircle of one triangle, then they are the circumcircle and incircle of a continuous family of poristic triangles. Y Z X

8.4 Proof of Feuerbach s theorem 37 Exercise 1. r 1 R. When does equality hold?. Suppose = d. Show that there is a right-angled triangle whose sides are d, r and R r. Which one of these is the hypotenuse? 3. Given a point inside a circle R), construct a circle r) so that R) and r) are the circumcircle and incircle of a family of poristic) triangles). 4. Given the circumcenter, incenter, and one vertex of a triangle, construct the triangle. 5. onstruct an animation picture of a triangle whose circumcenter lies on the incircle. 1 6. What is the locus of the centroids of the poristic triangles with the same circumcircle and incircle of triangle? How about the orthocenter? 7. Let be a poristic triangle with the same circumcircle and incircle of triangle, and let the sides of,, touch the incircle at X, Y, Z. i) What is the locus of the centroid of XY Z? ii) What is the locus of the orthocenter of XY Z? iii) What can you say about the Euler line of the triangle XY Z? 1 Hint: = r.