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

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1 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 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,

2 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.

3 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.

4 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

5 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

6 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.

7 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

8 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

9 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 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.

Steiner s porism and Feuerbach s theorem

hapter 10 Steiner s porism and Feuerbach s theorem 10.1 Euler s formula Lemma 10.1. f the bisector of angle intersects the circumcircle at M, then M is the center of the circle through,,, and a. M a Proof.