hapter 6 asic triangle centers 6.1 The Euler line 6.1.1 The centroid Let E and F be the midpoints of and respectively, and G the intersection of the medians E and F. onstruct the parallel through to E, and extend G to intersect at D, and this parallel at H.. F G E D H y the converse of the midpoint theorem, G is the midpoint of H, and H =2 GE Join H. y the midpoint theorem, H//F. It follows that HG is a parallelogram. Therefore, D is the midpoint of (the diagonal), and D is also a median of triangle. We have shown that the three medians of triangle intersect at G, which we call the centroid of the triangle. Furthermore, G =GH =2GD, G =H =2GE, G =H =2GF. The centroid G divides each median in the ratio 2:1.
302 asic triangle centers Exercise The centroid and the circumcenter of a triangle coincide if and only if the triangle is equilateral. Theorem 6.1 (pollonius). Given triangle, let D be the midpoint of. The length of the median D is given by 2 + 2 =2(D 2 + D 2 ). D Proof. pplying the law of cosines to triangles D and D, and noting that cos D = cos D, wehave 2 = D 2 + D 2 2D D cos D; 2 = D 2 + D 2 2D D cos D, 2 = D 2 + D 2 +2D D cos D. The result follows by adding the first and the third lines. G F E G D The inferior triangle of is the triangle DEF whose vertices are the midpoints of the sides,,. The two triangles share the same centroid G, and are homothetic at G with ratio 1 :2. The superior triangle of is the triangle bounded by the parallels of the sides through the opposite vertices. The two triangles also share the same centroid G, and are homothetic at G with ratio 2: 1.
6.1 The Euler line 303 6.1.2 The orthocenter and the Euler line The three altitudes of a triangle are concurrent. This is because the line containing an altitude of triangle is the perpendicular bisector of a side of its superior triangle. The three lines therefore intersect at the circumcenter of the superior triangle. This is the orthocenter of the given triangle. H G O The circumcenter, centroid, and orthocenter of a triangle are collinear. This is because the orthocenter, being the circumcenter of the superior triangle, is the image of the circumcenter under the homothety h(g, 2). The line containing them is called the Euler line of the reference triangle (provided it is non-equilateral). The orthocenter of an acute (obtuse) triangle lies in the interior (exterior) of the triangle. The orthocenter of a right triangle is the right angle vertex. 6.1.3 The nine-point circle Theorem 6.2. The following nine points associated with a triangle are on a circle whose center is the midpoint between the circumcenter and the orthocenter: (i) the midpoints of the three sides, (ii) the pedals (orthogonal projections) of the three vertices on their opposite sides, (iii) the midpoints between the orthocenter and the three vertices. Proof. (1) Let N be the circumcenter of the inferior triangle DEF. Since DEF and are homothetic at G in the ratio 1:2, N, G, O are collinear, and NG : GO =1:2. Since HG : GO =2:1, the four are collinear, and HN : NG : GO =3:1:2, and N is the midpoint of OH. (2) Let X be the pedal of H on. Since N is the midpoint of OH, the pedal of N is the midpoint of DX. Therefore, N lies on the perpendicular bisector of DX, and
304 asic triangle centers D Y F Z H N G E O E X F D NX = ND. Similarly, NE = NY, and NF = NZ for the pedals of H on and respectively. This means that the circumcircle of DEF also contains X, Y, Z. (3) Let D, E, F be the midpoints of H, H, H respectively. The triangle D E F is homothetic to at H in the ratio 1:2. Denote by N its circumcenter. The points N, G, O are collinear, and N G : GO =1:2. It follows that N = N, and the circumcircle of DEF also contains D, E, F. This circle is called the nine-point circle of triangle. Its center N is called the nine-point center. Its radius is half of the circumradius of. Exercise 1. Let H be the orthocenter of triangle. Show that the Euler lines of triangles, H, H and H are concurrent. 1 2. For what triangles is the Euler line parallel (respectively perpendicular) to an angle bisector? 2 3. Let P be a point on the circumcircle. What is the locus of the midpoint of HP? Why? 4. If the midpoints of P, P, P are all on the nine-point circle, must P be the orthocenter of triangle? 6.2 The incircle and excircles 6.2.1 The incircle The internal angle bisectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. 1 Hint: find a point common to them all. 2 The Euler line is parallel (respectively perpendicular) to the bisector of angle if and only if α = 120 (respectively 60 ).
6.2 The incircle and excircles 305 Let the bisectors of angles and intersect at I. onsider the pedals of I on the three sides. Since I is on the bisector of angle, IX = IZ. Since I is also on the bisector of angle, IX = IY. It follows IX = IY = IZ, and the circle, center I, constructed through X, also passes through Y and Z, and is tangent to the three sides of the triangle. s a s a Y Y Z I Z I s c s b X s b X s c This is called the incircle of triangle, and I the incenter. Let s be the semiperimeter of triangle. The incircle of triangle touches its sides,, at X, Y, Z such that Y =Z = s a, Z =X = s b, X =Y = s c. The inradius of triangle is the radius of its incircle. It is given by 6.2.2 The excircles r = 2Δ a + b + c = Δ s. The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. n excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle. The exradii of a triangle with sides a, b, c are given by r a = Δ s a, r b = Δ s b, r c = Δ s c. The areas of the triangles I a, I a, and I a are 1 2 ar a, 1 2 br a, and 1 2 cr a respectively. Since Δ= ΔI a +ΔI a +ΔI a,
306 asic triangle centers I b I c X Y r a Z r a r a I a we have Δ= 1 2 r a( a + b + c) =r a (s a), from which r a = Δ s a. 6.2.3 Heron s formula for the area of a triangle onsider a triangle with area Δ. Denote by r the inradius, and r a the radius of the excircle on the side of triangle. It is convenient to introduce the semiperimeter s = 1 (a + b + c). 2 I a I r a r Y Y (1) From the similarity of triangles IY and I Y, r = s a. r a s
6.2 The incircle and excircles 307 (2) From the similarity of triangles IY and I Y, r r a =(s b)(s c). (3) From these, (s a)(s b)(s c) r =, s s(s b)(s c) r a =. s a Theorem 6.3 (Heron s formula). Proof. Δ=rs. Proposition 6.4. tan α 2 = (s b)(s c) s(s a) Δ= s(s a)(s b)(s c)., cos α 2 = s(s a) bc, sin α (s b)(s c) 2 =. bc 6.2.4 Euler s formula: distances between circumcenter and tritangent centers Lemma 6.5. If the bisector of angle intersects the circumcircle at M, then M is the center of the circle through, I,, and I a. I O M Proof. (1) Since M is the midpoint of the arc, M = M = M. Therefore, MI = M + I = M + I = MI, and M = MI. Similarly, M = MI. (2) On the other hand, since II a and II a are both right angles, the four points, I,, I a M are concyclic, with center at the midpoint of II. This is the point M. I a
308 asic triangle centers I Y O M r a Y I a Theorem 6.6 (Euler). (a) OI 2 = R 2 2Rr. (b) OI 2 a = R2 +2Rr a. Proof. (a) onsidering the power of I in the circumcircle, we have R 2 OI 2 = I IM = I M = r sin α 2R sin α 2 =2Rr. 2 (b) onsider the power of I a in the circumcircle. Note that I a =. lso, I a M = M =2Rsin α. 2 ra sin α 2 OIa 2 = R 2 + I a I a M = R 2 + r a sin α 2R sin α 2 2 = R 2 +2Rr a. Theorem 6.7 (ngle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If X and X respectively the internal and external bisectors of angle, then X : X = c : b and X : X = c : b. Z c Z b X X
6.2 The incircle and excircles 309 Proof. onstruct lines through parallel to the bisectors X and X to intersect the line at Z and Z. (1) Note that Z = X = X = Z. This means Z =. learly, X : X = : Z = : = c : b. (2) Similarly, Z =, and X : X = : Z = : = c : b. Proposition 6.8. (a) The lengths of the internal and external bisectors of angle are respectively t a = 2bc b + c cos α 2 and t a = 2bc b c sin α 2. c t a b t a X X Proof. Let X and X be the bisectors of angle. (1) onsider the area of triangle as the sum of those of triangles X and X. We have 1 2 t a(b + c)sin α 2 = 1 bc sin α. 2 From this, t a = bc b + c sin α sin α = 2bc b + c cos α 2. 2 (2) onsider the area of triangle as the difference between those of X and X. Remarks. (1) 2bc is the harmonic mean of b and c. It can be constructed as follows. If the b+c perpendicular to X at X intersects and at Y and Z, then Y = Z = 2bc. b+c Z c t a b X Y
310 asic triangle centers (2) pplying Stewart s Theorem with λ = c and μ = ±b, we also obtain the following expressions for the lengths of the angle bisectors: ( ( ) ) 2 a t 2 a = bc 1, b + c ( ( ) 2 t 2 a a = bc 1). b c
hapter 7 onstruction problems I We study the problem of construction of a triangle,, from three given associated points, including (i) the vertices, (ii) the triangle centers O, G, H, I, and (iii) the traces M a, M b, M c of the centroid, H a, H b, H c of the orthocenter, and T a, T b, T c of the incenter. 1 M c G M b O H c H H b T c I T b M a H a T a (o) Medians (p) ltitudes (q) isectors The project, formulated by W. Wernick, 2 consists of 139 problems. With a few exceptions, the status of these problems are known, and are indicated as follows. [R] Redundant. Given the location of two of the points of the triple, the location of the third point is determined. [L] Locus Restricted. Given the location of two points, the third must lie on a certain locus. [S] Solvable. Known ruler and compass solutions exist for these triples. [] Not solvable by ruler and compass, but solvable by intersection of conics. [U] Unsolvable with ruler and compass, nor conics. 1 The traces M a, M b, M c are also the pedals of the circumcenter O on the sides. 2 Triangle constructions with three located points, Math. Mag., 55 (1982) 227 230.
312 onstruction problems I 7.1 lassification of construction problems (10) Two vertices and one center or trace (6) One vertex and two centers O G H I M a H a T a M b H b T b, L S S S R L L S L S (O, G) (O, H) (O, I) (G, I) (G, H) (H, I) S S S S S S (24) One vertex, one center, and one trace (21) One vertex and two traces M a H a T a M b H b T b O S S S L S S G L S S S H S L S S L I S L S S S (, M a,m b ) (, H a,h b ) (, T a,t b ) S S (, M b,m c ) (, H b,h c ) (, T b,t c ) S S (, M a,h a ) (, M a,t a ) (, H a,t a ) L S L (, M b,h b ) (, M b,t b ) (, H b,t b ) L L L (, M a,h b ) (, M a,t b ) (, H a,t b ) S S (, M b,h a ) (, M b,t a ) (, H b,t a ) L S S (, M b,h c ) (, M b,t c ) (, H b,t c ) L S S
7.1 lassification of construction problems 313 (4) Three centers (18) Two centers and one trace (O, G, H) (O, G, I) (O, H, I) (G, H, I) R (O, G) (O, H) (O, I) (G, H) (G, I) (H, I) M a S S S S H a S S S T a (36) One center and two traces (20) Three traces: (3) Three traces of the same kind (12) Two traces of the same kind O G H I (M b,m c ) S S S (H b,h c ) S (T b,t c ) S (M a,h a ) L L L S (M a,t a ) L S S S (H a,t a ) S S L S (M b,h c ) S S S (M b,t c ) (H b,t c ) (M a,m b,m c ) (H a,h b,h c ) (T a,t b,t c ) S S (M a,m b,h c ) (M a,m b,h a ) (M a,m b,t c ) (M a,m b,t a ) S S (H a,h b,m c ) (H a,h b,m a ) (H a,h b,t c ) (H a,h b,t a ) L S (T a,t b,m c ) (T a,t b,m b ) (T a,t b,h c ) (T a,t b,h a ) (5) Three traces of different kinds (M a,h a,t a ) (M a,h a,t b ) (M a,h b,t a ) (M b,h a,t a ) (M a,h b,t c ) L S S
314 onstruction problems I 7.2 onstruction problems with easy geometric solutions 1. (,,G). (a) M a = midpoint of. (b) divides M a G in the ratio M a : G =3: 2. 2. (,,H). (a) = orthocenter of triangle H. 3. (,,I). (a) = reflection of in I. (b) = reflection of in I. (c) =. 4. (, O, I). (a) onstruct the circumcircle O() with. (b) onstruct the circle (I) with the same radius to intersect the circumcircle at X. (c) Join X, I to intersect the circumcircle at Y, and construct the midpoint M of IY. The circle I(M) is the incircle. (d) onstruct the tangents from to the incircle to intersect the circumcircle again at and. 5. (, M a,h b ). (a) = M a (H b ) H b. (b) divides M a in the ratio 2: 1. 6. (H, M a,h b ). (a) onstruct the circle M a (H b ). (b) onstruct the line HH b to intersect the circle at. (c) is the antipodal point of on the circle. (d) = H b () (H). 7. (M a,m b,m c ). (a) = triangle bounded by the parallel to M b M c through M a, M c M a through M b, and M a M b through M c. 8. (H a,h b,h c ). (a) H = incenter of H a H b H c if this triangle is acute-angled, or the excenter corresponding to the obtuse angle vertex.
7.3 onstruction problems with easy algebraic solutions 315 (b) = the triangle bounded by the perpendiculars to HH a at H a, HH b at H b, and HH c at H c. 9. (M a, H b, T a ). (a) onstruct the circle M a (H b ) to intersect the line T a M a at and. (b) onstruct the harmonic conjugate T a of T a with respect to. (c) onstruct the circle with diameter T a T a to intersect the line H b at. There are two solutions corresponding to the choices of. 10. (M a, H b, T b ). (a) onstruct the circle M a (H b ). (b) onstruct the line H b T b to intersect the circle again. (c) is the antipodal point of on the circle. (d) Reflect in T b to intersect the line H b T b at. 7.3 onstruction problems with easy algebraic solutions 1. (O, I, T a ). We put the origin at I, and let T a =(0, a), O =(h, k). It is enough to determine the vertex =(0,q). The altitude H is the reflection of O in the angle bisector T. Since O is the line (q k)x+hy hq =0, the altitude is the line (q k)x hy+hq = 0. The perpendicular from T a to this line, i.e., hx +(q k)y +(q k)a =0, is the line. Let R and r be the circumradius and inradius, and d the distance between O and I. Wehave R 2 = h 2 +(q k) 2, r 2 = a2 (q k) 2 h 2 +(q k), 2 d 2 = h 2 + k 2. Note that Rr = a(q k), and R 2 d 2 = q(q 2k). From Euler s formula, we have q(q 2k) =2a(q k). This gives q = a + k + a 2 + k 2. 7.3.1 (O, H a,t a ) Put the origin at T, H a =(a, 0), and O =(h, k). The vertex =(a, q) for some q. The reflection of O in T a is the point ( ) a 2 h +2akq hq 2, a2 k +2ahq + kq 2. a 2 + q 2 a 2 + q 2
316 onstruction problems I Since this point lies on the altitude through, a2 h+2akq hq 2 a 2 +q 2 = a and (a + h)q 2 2akq + a 2 (a h) =0. From this, q = a (k ± ) h a + h 2 + k 2 a 2. 2. (I,M a,h a ). Put the origin at I and let M a =(a, k), H a =(b, k). The vertices and have coordinates (a + p, k) and (a p, k) for some appropriate p. The incircle being x 2 + y 2 = k 2, the tangents from and are the lines From these we obtain ( = 2k(a + p)x +((a + p) 2 k 2 )y = k((a + p) 2 + k 2 ), 2k(a p)x +((a p) 2 k 2 )y = k((a p) 2 + k 2 ). 2ak 2 a 2 + k 2 p, (a2 k 2 p 2 )k 2 a 2 + k 2 p 2 Since lies on the perpendicular to y = k at H a =(b, k), wehave 2ak 2 ) a 2 + k 2 p 2 = b. ). From this, p 2 = a 2 + k 2 2a b k2. 3. (, T b,t c ). Put the origin at and let T b =( b, mb) and T c =(c, mc). If the incenter is the point I =(0,t), then = T c T b I and = T b T c I are the points ( ) ( ) bt = 2mb t, mbt ct and = 2mb t 2mc t, mct. 2mc t The line has equation m(b c)tx +((b + c)t 4mbc)y +2mbct =0. The square distance distance from (0,t) is From this we have ((b + c)t 2mbct) 2 m 2 (b c) 2 +((b + c)t 4mbc) 2 = t2 1+m 2. mt 2 +(1 m 2 )(b + c)t m(3 m 2 )bc =0.
7.3 onstruction problems with easy algebraic solutions 317 4. (, T a,t b ). Put the origin at and let T a =(a, 0), T b =(h, k). IfT b =(h, k) is the reflection of T b in T a, and I =(t, 0) is the incenter, then ( ) ht = T b T bi = 2h t, kt, 2h t ( ) aht = T b T a = (a +2h)t 2ah, akt. (a +2h)t 2ah The line has equation ktx +((a + h)t 2ah)y akt =0. Since the square distances from I to the lines T b and are equal, (akt kt 2 ) 2 k 2 t 2 +((a + h)t 2ah) 2 = (kt)2 h 2 + k 2. (a +2h)t 2 2(2ah + h 2 k 2 )t + a(3h 2 k 2 )=0. 5. (O, H a,t a ). Put the origin at T, H a =(a, 0), and O =(h, k). The vertex =(a, q) for some q. Since and are on the line x-axis, the midpoint of the arc is the intersection of T a and the vertical line through O. It is the point = ( ) h, qh a. Thus ( ) 2 qh a k =(h a) 2 +(k q) 2. This gives (a + h)q 2 2akq + a 2 (a h) =0. From this, q = a (k ± ) h a + h 2 + k 2 a 2. 6. (, G, I) and (, I, M a ). Put the origin at M a, I =(a, 0) and =(h, k). If =(p, q), then =( p, q). The lines, and have equations qx py = 0, (k + q)x (h + p)y (hq kp) = 0, (k q)x (h p)y +(hq kp) = 0. The square distances from (a, 0) to these lines are a 2 q 2 (a(k + q) (hq kp)2 (a(k q)+hq kp)2 = =. p 2 + q2 (k + q) 2 +(h + p) 2 (k q) 2 +(h p) 2
318 onstruction problems I omparing the first expression with the other two, we have (kp hq)(a 2 (2pq + hq + kp)+2a(p 2 + q 2 )(k + q)+(p 2 + q 2 )(kp hq)) = 0, (kp hq)(a 2 (2pq hq kp)+2a(p 2 + q 2 )(k q) (p 2 + q 2 )(kp hq)) = 0. learly, kp hq 0(otherwise,, M a and are collinear). It follows that a 2 (2pq + hq + kp)+2a(p 2 + q 2 )(k + q)+(p 2 + q 2 )(kp hq) = 0, a 2 (2pq hq kp)+2a(p 2 + q 2 )(k q) (p 2 + q 2 )(kp hq) = 0. ddition gives apq + k(p 2 + q 2 )=0. It follows that the slope of is q = k p a± a 2 4k 2 2. From this, the line containing,, and the incircle can be constructed. The two tangents from to the incircle intersect this line at and.