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. 1) SinceM is the midpoint of the arc, M = M = M. Therefore, M = M + = M + = M, andm = 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. Theorem 10. Euler). a) = R Rr. b) a = R +Rr a.
34 Steiner s porism and Feuerbach s theorem Y M Y r 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 = Rsin α. a R sin α = Rr. a = R + a a M = R + r a sin α Rsin α = R +Rr a. 10. 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. Given a triangle with incircle ) and circumcircle ), let be an arbitrary point on circumcircle. Join to, to intersect the circumcircle again atm, and let Y, Z be the tangents to the incircle. onstruct a circle, center M, through to intersect the circumcircle at and. Denote by θ between and each tangent. t is known that M = Rr power of incenter in )). Since = r, we sinθ have M = Rsinθ. Therefore, M = M = Rsinθ, and by the law of sines, M = M = θ. t follows that and are tangent to the incircle ).
10. Steiner s porism 35 Z Y M Since M is the midpoint of the arc, the circle M ) passes through the incenter of triangle. This means that is the incenter, and ) the incircle of triangle. Theorem 10.3 Steiner). f two 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
36 Steiner s porism and Feuerbach s theorem 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 atx, Y,Z. i) What is the locus of the centroid ofxyz? ii) What is the locus of the orthocenter of XYZ? iii) What can you say about the Euler line of the triangle XYZ? 1 Hint: = r.
10.3 Distance between the circumcenter and orthocenter 37 10.3 Distance between the circumcenter and orthocenter Y F E Z H N X D Proposition 10.1. H = R 1 8cosαcosβcosγ). Proof. n triangleh, = R,H = Rcosα, and H = β γ. y the law of cosines, H = R 1+4cos α 4cosαcosβ γ)) = R 1 4cosαcosβ +γ)+cosβ γ)) = R 1 8cosαcosβcosγ). Proposition 10.. r = 4Rsin α sin β sin γ, s = 4Rcos α cos β cos γ, Proof. 1) n triangle, = Rsinγ, = β, and = 180 α+β. pplying the law of sines, we have = From this, sin β sin α+β = Rsinγ sin β cos γ = 4Rsin γ cos γ sin β cos γ r = sin α = 4Rsin α sin β sin γ. = 4Rsin β sin γ. ) Similarly, in triangle a, a = 90 + β, we have a = 4Rcos β cos γ.
38 Steiner s porism and Feuerbach s theorem a a a a t follows that s = a cos α = 4Rcos α cos β cos γ.
10.4 Distance between orthocenter and tritangent centers 39 10.4 Distance between orthocenter and tritangent centers Proposition 10.3. H = r 4R cosαcosβcosγ, H a = r a 4R cosαcosβcosγ. H X Proof. n triangle H, we have H = Rcosα, = 4Rsin β sin γ β γ. y the law of cosines, and H = H = H + H cos β γ = 4R cos α+4sin β γ sin 4cosαsin β sin γ ) β γ cos = 4R cos α+4sin β γ sin 4cosαsin β sin γ cos β cos γ β γ 4cosαsin sin = 4R cos α+4sin β sin γ cosαsinβsinγ 4 1 sin α = 4R cosαcosα sinβsinγ)+8sin α β γ ) sin sin = 4R cosαcosβcosγ +8sin α β γ ) sin sin = r 4R cosαcosβcosγ. ) sin β sin γ ) ) ) n triangle H a, a = 4Rcos β cos γ. y the law of cosines, we have
40 Steiner s porism and Feuerbach s theorem H a a a a Ha = H +a a H cos β γ = 4R cos α+4cos β γ cos 4cosαcos β cos γ ) β γ cos = 4R cos α+4cos β γ β γ cos 4cosαcos cos 4cosαcos β cos γ sin β sin γ ) = 4R cos α+4cos β γ cos 4 1 sin α )cos β γ ) cos cosαsinβsinγ = 4R cosαcosα sinβsinγ)+8sin α β γ ) cos cos = 4R cosαcosβcosγ +8sin α β γ ) cos cos = r a 4R cosαcosβcosγ.
10.5 Feuerbach s theorem 41 10.5 Feuerbach s theorem Theorem 10.4 Feuerbach). The nine-point circle is tangent internally to the incircle and externally to each of the excircles. H N a 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.
4 Steiner s porism and Feuerbach s theorem b b c c c F F b F c b N c F a a b a a a
10.5 Feuerbach s theorem 43 Exercise 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. c b a