Notes on Barycentric Homogeneous Coordinates

Transcription

1 Notes on arycentric Homogeneous oordinates Wong Yan Loi ontents 1 arycentric Homogeneous oordinates 2 2 Lines 3 3 rea 5 4 Distances 5 5 ircles I 8 6 ircles II 10 7 Worked Examples 14 8 Exercises 20 9 Hints Solutions References 42 1

2 1 arycentric Homogeneous oordinates 2 1 arycentric Homogeneous oordinates Let be a triangle on the plane For any point P, the ratio of the (signed) areas [P ] : [P ] : [P ] is called the barycentric coordinates or areal coordinates of P Here [P ] is the signed area of the triangle P It is positive, negative or zero according to both P and lie on the same side, opposite side, or on the line Generally, we use (x : y : z) to denote the barycentric coordinates of a point P The barycentric coordinates of a point are homogeneous That is (x : y : z) = (λx : λy : λz) for any nonzero real number λ If x+y +z = 1, then (x : y : z) is called the normalized barycentric coordinates of the point P For example, = (1 : 0 : 0), = (0 : 1 : 0), = (0 : 0 : 1) The triangle is called the reference triangle of the barycentric homogeneous coordinate system F y z x E P x P z D y Fig 1: arycentric coordinates Theorem 1 Let [P ] = x, [P ] = y, [P ] = z and [] = 1 so that x + y + z = 1 Let the extensions of the P, P, P meet the sides,, at D, E, F respectively Then (a) E : E = x : z, etc (b) P : P D = (y + z) : x (c) P = z + x (d) If a, b, c are the position vectors of the points,, respectively, then p = xa + yb + zc Here a = O is the position vector of with respect to a fixed origin O, etc (e) The normalized barycentric coordinates (x : y : z) of the point P is unique Proof (a) E : E = [P ] : [P ] = x : z (b) Let [P D] = α, [P D] = β Then P P D = y P α and P D = z β Thus P P D = y + z α + β = y + z x (c) follows from (b) (d) p = b + P = b + z + x = b + z(c b) + x(a b) = xa + yb + zc (e) follows from (d) 11 Ratio Formula Let P 1 = (x 1 : y 1 : z 1 ) and P 2 = (x 2 : y 2 : z 2 ) with x 1 + y 1 + z 1 = 1 and x 2 + y 2 + z 2 = 1 If P divides P 1 P 2 in the ratio P 1 P : P P 2 = β : α, then the point P has barycentric coordinates (αx 1 + βx 2 : αy 1 + βy 2 : αz 1 + βz 2 )

3 2 Lines 3 12 ommon points (1 : 0 : 0) centroid (1 : 1 : 1) incentre (a : b : c) symmedian point (a 2 : b 2 : c 2 ) - excentre ( a : b : c) orthocentre (tan : tan : tan ) = (S S : S S : S S ) circumcentre (sin 2 : sin 2 : sin 2) = (a 2 S : b 2 S : c 2 S ) Nagel point (s a : s b : s c) Gergonne point Isogonal conjugate of (x : y : z) Isotomic conjugate of (x : y : z) ((s b)(s c) : (s c)(s a) : (s a)(s b)) (a 2 /x : b 2 /y : c 2 /z) = (a 2 yz : b 2 zx : c 2 xy) (1/x : 1/y : 1/z) Here a =, b =, c =, s = 1 2 (a + b + c), and S = 1 2 (b2 + c 2 a 2 ) etc 2 Lines 21 Equation of a line If P = (x : y : z) is a point on the line joining P 1 = (x 1 : y 1 : z 1 ) and P 2 = (x 2 : y 2 : z 2 ), then (x : y : z) = (αx 1 + βx 2 : αy 1 + βy 2 : αz 1 + βz 2 ), for some α, β (The normalizing factors for P, P 1, P 2 can be absorbed into α, β) Thus This implies that ( 1 α ) β x y z x 1 y 1 z 1 x 2 y 2 z 2 x y z x 1 y 1 z 1 x 2 y 2 z 2 = ( ) = 0 Expanding the determinant about the first row, we have y 1 z 1 y 2 z 2 x x 1 z 1 x 2 z 2 y + x 1 y 1 x 2 y 2 z = 0 That is (x : y : z) satisfies the homogeneous linear equation (y 1 z 2 y 2 z 1 )x (x 1 z 2 x 2 z 1 )y + (x 1 y 2 x 2 y 1 )z = 0 onversely, any point P = (x : y : z) satisfies this equation can be shown to lie on the line P 1 P 2 We may compute the coefficients of this line by the following determinant: x [ ] 1 y 1 z 1 x 2 y 2 z 2 = y 1 z 1 y 2 z 2 : x 1 z 1 x 2 z 2 : x 1 y 1 x 2 y 2,

4 2 Lines 4 and represent the line using the 3 coefficients as homogeneous coordinates enclosed in a square bracket The 2 representations px + qy + rz = 0 and [p : q : r] are equivalent and will be used interchangeably For example, the line is x = 0 or [1 : 0 : 0] 22 Intersection of two lines Let l 1 : p 1 x + q 1 y + r 1 z = 0 and l 2 : p 2 x + q 2 y + r 2 z = 0 be 2 lines The intersection point is the simultaneous solution to the two equations, which is (q 1 r 2 q 2 r 1 : p 1 r 2 + p 2 r 1 : p 1 q 2 p 2 q 1 ) We may facilitate the computation by means of the following determinant: p ( 1 q 1 r 1 p 2 q 2 r 2 = q 1 r 1 q 2 r 2 : p 1 r 1 p 2 r 2 : p 1 q 1 p 2 q 2 ), 23 ollinearity Three points P 1 = (x 1 : y 1 : z 1 ), P 2 = (x 2 : y 2 : z 2 ) and P 3 = (x 3 : y 3 : z 3 ) are collinear if and only if x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 = 0 24 oncurrence Three lines l 1 = [p 1 : q 1 : r 1 ], l 2 = [p 2 : q 2 : r 2 ] and l 3 = [p 3 : q 3 : r 3 ] are concurrent if and only if p 1 q 1 r 1 p 2 q 2 r 2 p 3 q 3 r 3 = 0 25 ommon lines [1 : 0 : 0] internal bisector from [0 : c : b] external bisector from [0 : c : b] median from [0 : 1 : 1] perpendicular from [0 : S : S ] perpendicular bisector of [c 2 b 2 : a 2 : a 2 ] Euler line [S (S S ) : S (S S ) : S (S S )]

5 3 rea 5 3 rea If P, Q, R are points with normalized barycentric coordinates (x 1 : y 1 : z 1 ), (x 2 : y 2 : z 2 ), x 1 y 1 z 1 (x 3 : y 3 : z 3 ) respectively, then [P QR] = x 2 y 2 z 2 x 3 y 3 z 3 [] Proof Let δ be the determinant determinant is 1, we have x 1 y 1 1 δ = x 2 y 2 1 x 3 y 3 1 = x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 x 1 y 1 1 x 2 x 1 y 2 y 1 0 x 3 x 1 y 3 y 1 0 Similarly, δ = (y 2 y 1 )(z 3 z 1 ) (z 2 z 1 )(y 3 y 1 ), and δ = (z 2 z 1 )(x 3 x 1 ) (x 2 x 1 )(z 3 z 1 ) Since the sum of the entries in each row of the = (x 2 x 1 )(y 3 y 1 ) (y 2 y 1 )(x 3 x 1 ) Let n be the unit normal vector to the x y plane along the positive z direction Let O be the origin Then a b + b c + c a = ( a b sin O + b c sin O + c a sin O)n = 2([O] + [O] + [O])n = 2[]n Here O is the signed angle measured from the vector O to O Let x = x 1 a + y 1 b + z 1 c, y = x 2 a + y 2 b + z 2 c, z = x 3 a + y 3 b + z 3 c Then 2[XY Z]n = XY XZ sin Y XZ n = XY XZ = ((x 2 x 1 )a + (y 2 y 1 )b + (z 2 z 1 )c) ((x 3 x 1 )a + (y 3 y 1 )b + (z 3 z 1 )c) = ((x 2 x 1 )(y 3 y 1 ) (y 2 y 1 )(x 3 x 1 )) a b ((y 2 y 1 )(z 3 z 1 ) (z 2 z 1 )(y 3 y 1 )) b c ((z 2 z 1 )(x 3 x 1 ) (x 2 x 1 )(z 3 z 1 )) c a = δ(a b + b c + c a) = 2δ[]n onsequently, [XY Z] = δ[] 4 Distances 41 Displacement vectors Let P = (p 1 : p 2 : p 3 ) and Q = (q 1 : q 2 : q 3 ) be two points in normalized barycentric coordinates The displacement vector is the vector P Q = (q 1 p 1 : q 2 p 2 : q 3 p 3 ) Expressed in the terms of a, b, c, we have P Q = (q 1 p 1 )a + (q 2 p 2 )b + (q 3 p 3 )c Note that if P Q = (x : y : z) is a displacement vector, then x + y + z = 0

6 4 Distances 6 42 Inner product of two displacement vectors Let P Q = (x 1 : y 1 : z 1 ) and EF = (x 2 : y 2 : z 2 ) be two displacement vectors Then P Q EF = 1 2 ( a 2 (z 1 y 2 + y 1 z 2 ) + b 2 (x 1 z 2 + z 1 x 2 ) + c 2 (y 1 x 2 + x 1 y 2 ) ) Proof Take be the origin so that b = 0, a = c, c = a Let U and V be the points such that U = P Q and V = EF as free vectors respectively Note that P Q EF = U V Let U = (λ 1 : µ 1 : δ 1 ) and V = (λ 2 : µ 2 : δ 2 ) in normalized barycentric coordinates Then U = (λ 1 : µ 1 1 : δ 1 ) = λ 1 a + δ 1 c Since U = P Q, we have λ 1 = x 1, µ 1 1 = y 1, δ 1 = z 1 Thus U = (x 1 : y : z 1 ) Similarly, V = (x 2 : y : z 2 ) Therefore, U = x 1 a + z 1 c and V = x 2 a + z 2 c Then P Q EF = U V = (x 1 a+z 1 c) (x 2 a+z 2 c) = x 1 x 2 a 2 +z 1 z 2 c 2 +(x 1 z 2 +z 1 x 2 )a c = x 1 ( x 2 c 2 + z 1 z 2 a 2 + (x 1 z 2 + z 1 x 2 )a c = x 1 x 2 c 2 + z 1 z 2 a 2 + (x 1 z 2 + z 1 x 2 ) 1 2 (c2 + a 2 b 2 ) = a 2 (2z 1 z 2 + x 1 z 2 + z 1 x 2 ) b 2 (x 1 z 2 + z 1 x 2 ) + c 2 (2x 1 x 2 + x 1 z 2 + z 1 x 2 ) ) 1 2 Note that 2z 1 z 2 + x 1 z 2 + z 1 x 2 = 2z 1 z 2 + ( y 1 z 1 )z 2 + z 1 ( y 2 z 2 ) = (z 1 y 2 + y 1 z 2 ) 2x 1 x 2 + x 1 z 2 + z 1 x 2 = 2x 1 x 2 + x 1 ( x 2 y 2 ) + ( x 1 y 1 )x 2 = (y 1 x 2 + x 1 y 2 ) Therefore, P Q EF = 1 ( 2 a 2 (z 1 y 2 + y 1 z 2 ) + b 2 (x 1 z 2 + z 1 x 2 ) + c 2 (y 1 x 2 + x 1 y 2 ) ) 43 The length of a displacement vector If P Q = (x : y : z), where x + y + z = 0, is a displacement vector, then P Q 2 = (a 2 yz + b 2 zx + c 2 xy) 44 Perpendicular displacement vectors (a) P Q EF if and only if 0 = a 2 (z 1 y 2 + y 1 z 2 ) + b 2 (x 1 z 2 + z 1 x 2 ) + c 2 (y 1 x 2 + x 1 y 2 ) (b) Let (l : m : n) be a displacement vector displacement vector perpendicular to (l : m : n) is given by (a 2 (n m) + (c 2 b 2 )l : b 2 (l n) + (a 2 c 2 )m : c 2 (m l) + (b 2 a 2 )n) Example 41 displacement vector perpendicular to I = ( (b + c) : b : c) is (b c : b : c) Example 42 Let I and G be the incentre and the centroid of a triangle respectively Then IG is perpendicular to if and only if b = c or b + c = 3a Example 43 In a triangle, the incircle touches the side at E and at F The line through and the incentre I intersects the line EF at P Then P = ( a 2c : c a 2c : 1 2 ), and P is perpendicular to P

7 4 Distances 7 45 onway s Notation Denote S = 1 2 (b2 + c 2 a 2 ), S = 1 2 (c2 + a 2 b 2 ) and S = 1 2 (a2 + b 2 c 2 ) Let S be twice the area of the triangle Lemma 1 (i) S + S = c 2, S + S = a 2, S + S = b 2 (ii) S S = b 2 a 2, S S = c 2 b 2, S S = a 2 c 2 (iii) S S + S S + S S = S 2 (iv) a 2 S + b 2 S + c 2 S = 2S 2 (v) b 2 c 2 S 2 = S2 = S S + a 2 S Proof Let s prove (v) y (i) (ii) and (iii), b 2 c 2 S 2 = b2 (S + S ) S (b 2 S ) = b 2 S + S S = (S + S )S + S S = S 2 46 Examples Example 44 The perpendicular bisector of has the equation a 2 (z y) + (c 2 b 2 )x = 0, or [c 2 b 2 : a 2 : a 2 ] Solution The midpoint M of is (0 : 1 2 : 1 2 ) Let X = (x : y : z) be a point in normalized barycentric coordinates on the perpendicular bisector of Then XM = ( x : 1 2 y : 1 2 z) and = (0 : 1 : 1) y the condition on perpendicular displacement vectors, we have a 2 (z y) + (c 2 b 2 )x = 0 Example 45 The barycentric coordinates of the foot of perpendicular from a point P with normalized barycentric coordinates (x 0 : y 0 : z 0 ) onto is (0 : (a 2 + b 2 c 2 )x 0 + 2a 2 y 0 : (a 2 b 2 + c 2 )x 0 + 2a 2 z 0 ) = (0 : S a 2 x 0 + y 0, S a 2 x 0 + z 0 ) onto is ((b 2 c 2 + a 2 )y 0 + 2b 2 x 0 : 0 : (b 2 + c 2 a 2 )y 0 + 2b 2 z 0 ), onto is ((c 2 + a 2 b 2 )z 0 + 2c 2 x 0 : (c 2 a 2 + b 2 )z 0 + 2c 2 y 0 : 0) Example 46 The barycentric coordinates of the foot of perpendicular from onto is (0 : a 2 + b 2 c 2 : c 2 + a 2 b 2 ) = (0 : S : S ) Example 47 The reflection of P with normalized barycentric coordinates (x 0 : y 0 : z 0 ) across the line is the point ( a 2 x 0 : (a 2 + b 2 c 2 )x 0 + a 2 y 0 : (a 2 b 2 + c 2 )x 0 + a 2 z 0 ), or ( x 0 : 2S a 2 x 0 + y 0, 2S a 2 x 0 + z 0 ) Example 48 The reflection of the line [α, β, γ] across the line is the line [ 2 a 2 (S γ + S β) α, β, γ]

8 5 ircles I 8 5 ircles I 51 Parallel xis Theorem If O is the circumcentre of the reference triangle and P has normalized barycentric coordinates (x : y : z), then R 2 OP 2 = xp 2 + yp 2 + zp 2 = a 2 yz + b 2 zx + c 2 xy Proof Take the circumcentre O be the origin so that a = b = c = R We have p = xa+yb+zc with x + y + z = 1 s P = p a, we have P 2 = P 2 = p a 2 = p 2 + a 2 2 p a = OP 2 + R 2 2 p a Similarly, P 2 = OP 2 + R 2 2 p b and P 2 = OP 2 + R 2 2 p c Therefore, xp 2 + yp 2 + zp 2 = (x + y + z)op 2 + (x + y + z)r 2 2 p (xa + yb + zc) = OP 2 + R 2 2 p p = OP 2 + R 2 2 OP 2 = R 2 OP 2 Since P = p a = (x 1)a + yb + zc, we have P 2 = a 2 yz b 2 z(x 1) c 2 (x 1)y = (a 2 yz + b 2 zx + c 2 xy) + zb 2 + yc 2 Similarly, P 2 = (a 2 yz + b 2 zx + c 2 xy) + xc 2 + za 2 and P 2 = (a 2 yz + b 2 zx + c 2 xy) + ya 2 + xb 2 Therefore, xp 2 + yp 2 + zp 2 = (x + y + z)(a 2 yz + b 2 zx + c 2 xy) + x(zb 2 + yc 2 ) + y(xc 2 + za 2 ) + z(ya 2 + xb 2 ) = a 2 yz + b 2 zx + c 2 xy 52 Power of a point with respect to the circumcircle The power of P with respect to the circumcircle of is (a 2 yz + b 2 zx + c 2 xy), where P has normalized barycentric coordinates (x : y : z) 53 ircumcircle The equation of the circumcircle of the reference triangle is a 2 yz + b 2 zx + c 2 xy = 0 54 Power of a point with respect to a circle Let ω be a circle and P a point having normalized barycentric coordinates (x : y : z) Then the negative of the power of P with respect to ω can be expressed in the form a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz), where p, q, r are constants depending only on ω Proof Every circle ω is homothetic to the circumcircle of the reference triangle by a homothety, say h, with a center of similitude S = (u : v : w) (in normalized barycentric coordinates) and similitude ratio k s the point P has normalized barycentric coordinates (x : y : z), we have P h(p ) = kp + (1 k)s = k(x + tu(x + y + z) : y + tv(x + y + z) : z + tw(x + y + z)), where t = 1 k k, is the normalized barycentric coordinates of P Let R and R be the radii of the circumcircle of the reference triangle and ω respectively Then R = kr Let the circumcentre of the reference triangle be O and the center of ω be O Then O = h(o ) lso P O = kp O The power of P with respect to ω = P O 2 R 2 = k 2 (P O 2 R 2 ) = k 2 the power of P with respect to the circumcircle

9 5 ircles I 9 The negative of the power of P with respect to the circumcircle is obtained by substituting normalized barycentric coordinates of P into the equation of the circumcircle Therefore, the negative of the power of P with respect to ω is equal to a 2 (y + tv(x + y + z))(z + tw(x + y + z)) cyclic = a 2 (yz + t(wy + vz)(x + y + z) + t 2 vw(x + y + z) 2 ) cyclic = (a 2 yz + b 2 zx + c 2 xy) + t(x + y + z) a 2 (wy + vz)+ t 2 (x + y + z) 2 (a 2 vw + b 2 wu + c 2 uv) cyclic = (a 2 yz + b 2 zx + c 2 xy) + t(x + y + z) ( b 2 w + c 2 v + t(a 2 vw + b 2 wu + c 2 uv) ) x cyclic = a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz) 55 Equation of a circle The general equation of a circle is a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz) = 0 Note that the powers of the points,, with respect to the circle are p, q, r respectively For example, the powers of,, with respect to the incircle are (s a) 2, (s b) 2, (s c) 2 respectively Therefore, the equation of the incircle is a 2 yz + b 2 zx + c 2 xy (x + y + z) ( (s a) 2 x + (s b) 2 y + (s c) 2 z ) = 0 circumcircle a 2 yz + b 2 zx + c 2 xy = 0 incircle a 2 yz + b 2 zx + c 2 xy (x + y + z) ( (s a) 2 x + (s b) 2 y + (s c) 2 z ) = 0 -excircle a 2 yz + b 2 zx + c 2 xy (x + y + z) ( s 2 x + (s c) 2 y + (s b) 2 z ) = 0 nine-point circle a 2 yz + b 2 zx + c 2 xy 1 4 (x + y + z) ( ( a 2 + b 2 + c 2 )x + (a 2 b 2 + c 2 )y + (a 2 + b 2 c 2 )z ) = 0 circle centred at with radius ρ a 2 yz + b 2 zx + c 2 xy + (x + y + z) ( ρ 2 x + (ρ 2 c 2 )y + (ρ 2 b 2 )z ) = 0 circle through a 2 yz + b 2 zx + c 2 xy + (x + y + z)(bcx + cay + abz) = 0 the 3 excentres of mixtilinear incircle a 2 yz + b 2 zx + c 2 xy b2 c 2 s 2 (x + y + z)(x + (s/b 1) 2 x + (s/c 1) 2 z) = 0 opposite to b point of tangency with the circumcircle is ( a : s b : c 2 s c ) Example 51 The centroid lies on the incircle if and only if 5(a 2 + b 2 + c 2 ) = 6(ab + bc + ca) Example 52 The Nagel point lies on the incircle if and only if (a+b 3c)(b+c 3a)(c+a 3b) = 0 56 Radical axis The line px + qy + rz = 0 is the radical axis of ω and the circumcircle whenever ω is not concentric with the circumcircle For example, the radical axis of the circumcircle and the nine-point circle is the line (b 2 + c 2 a 2 )x + (c 2 + a 2 b 2 )y + (a 2 + b 2 c 2 )z = 0

10 6 ircles II The arc (a) The midpoint of the arc not containing has barycentric coordinates ( a 2 : b(b + c) : c(b + c)) (b) The midpoint of the arc containing has barycentric coordinates (a 2 : b(b c) : c(b c)) (c) The median through meets the circumcircle at the point with barycentric coordinates (a 2 : b 2 + c 2 : b 2 + c 2 ) 6 ircles II 61 The Feuerbach point onsider the difference of the coefficients of x in the equations of the incircle and nine-point circle, we have (s a) ( a2 + b 2 + c 2 ) = 1 2 (a b)(a c) Thus subtracting the equations of the incircle and the nine-point circle, we get which can be written as l : (a b)(a c)x + (b c)(b a)y + (c a)(c b) = 0, x b c + y c a + z a b = 0 Observe that there are two points P = ((b c) 2 : (c a) 2 : (a b) 2 ) and Q = (a(b c) 2 : b(c a) 2 : c(a b) 2 ) on l Using these two points, we can parameterize l, except P, as (x : y : z) = ( (a + t)(b c) 2 : (b + t)(c a) 2 : (c + t)(a b) 2) The centre N of the nine-point circle is the midpoint of OH, its barycentric coordinates is given by N = (a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 : b 2 (c 2 + a 2 ) (c 2 a 2 ) 2 : c 2 (a 2 + b 2 ) (a 2 b 2 ) 2 ) lso I = (a : b : c) Thus the line IN has parametric equation: (x : y : z) = ( a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 + ak : b 2 (c 2 + a 2 ) (c 2 a 2 ) 2 + bk : c 2 (a 2 + b 2 ) (a 2 b 2 ) 2 + ck ) If we take k = 2abc, then x = a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 2a 2 bc = a 2 (b c) 2 (b 2 c 2 ) 2 = (b c) 2 (a 2 (b + c) 2 ) = (a + b + c)(b c) 2 (a b c) = 4s(b c) 2 (s a) Similarly we get the expressions for y and z by cyclically permuting a, b, c Thus we have a point on the line IN ((b c) 2 (s a) : (c a) 2 (s b) : (a b) 2 (s c)) If we take t = s in l, we obtain the same point F Thus F is the intersection of the line IN and l F is the Feuerbach point of the triangle We can verify by direct substitution that the barycentric coordinates of the Feuerbach point satisfy the equation of the nine-point circle It follows that the nine-point circle and the incircle are tangent at the Feuerbach point

11 6 ircles II Excircles The equation of the common tangent l of the nine-point circle and the -excircle is x b c + y c + a z a + b = 0 There are two points P = ((b c) 2 : (c+a) 2 : (a+b) 2 ) and Q = ( a(b c) 2 : b(c+a) 2 : c(a+b) 2 ) on l We parameterize l as (x : y : z) = ( ( a + t)(b c) 2 : (b + t)(c + a) 2 : (c + t)(a + b) 2) Recall that I = ( a : b : c) Thus the line I N has parametric equation: (x : y : z) = ( a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 ak : b 2 (c 2 + a 2 ) (c 2 a 2 ) 2 + bk : c 2 (a 2 + b 2 ) (a 2 b 2 ) 2 + ck ) Taking k = 2abc, we get x = a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 2a 2 bc = 4s(b c) 2 (s a), y = b 2 (c 2 +a 2 ) (c 2 a 2 ) 2 +2ab 2 c = 4(c+a) 2 (s a)(s c), z = c 2 (a 2 +b 2 ) (a 2 b 2 ) 2 +2abc 2 = 4(a + b) 2 (s a)(s b) Thus we have the point F = ( s(b c) 2 : (s c)(c + a) 2 : (s b)(a + b) 2 ) on I N On the other hand, if we let t = a s in l, we get the same point Therefore F is the intersection point of l and I N We can verify by direct substitution that the barycentric coordinates of the point F satisfy the equation of the nine-point circle It follows that the nine-point circle and the -excircle are tangent at F Similarly, the nine-point circle is tangent to the other 2 excircles The points of tangency are F = ((s c)(b + c) 2 : s(c a) 2 : (s a)(a + b) 2 ), 63 Pedal triangle F = ((s b)(b + c) 2 : (s a)(c + a) 2 : s(a b) 2 ) Let P be a point with normalized barycentric coordinates (x 0 : y 0 : z 0 ) The determinant formed by the normalized barycentric coordinates of the foot of perpendiculars from P onto the sides of is 1 8a 2 b 2 c 2 0 (a 2 + b 2 c 2 )x 0 + 2a 2 y 0 (a 2 b 2 + c 2 )x 0 + 2a 2 z 0 (b 2 c 2 + a 2 )y 0 + 2b 2 x 0 0 (b 2 + c 2 a 2 )y 0 + 2b 2 z 0 (c 2 + a 2 b 2 )z 0 + 2c 2 x 0 (c 2 a 2 + b 2 )z 0 + 2c 2 y 0 0 = 1 4a 2 b 2 c 2 (b + c a)(c + a b)(a + b c)(a + b + c)(x 0 + y 0 + z 0 )(a 2 y 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 ) = 4 ( = 2[] abc s(s a)(s b)(s c)(a 2 y a 2 b 2 c 2 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 ) ) 2 (a 2 y 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 ) = 1 4R 2 (a 2 y 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 ) This proves Simson s theorem that P lies on the circumcircle if and only if the 3 feet of perpendiculars are collinear lso the area of the pedal triangle of P is [] 4R 2 (a 2 y 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 )

12 6 ircles II Equation of the circle with center (α : β : γ) and radius ρ Let the normalized barycentric coordinates of the centre O be (α : β : γ) Let P = (x : y : z), where x + y + z = 1, be a point on the circle Then OP = (x α : y β : z γ) Thus the equation of the circle is Expanding, we have a 2 (y β)(z γ) + b 2 (x α)(z γ) + c 2 (x α)(y β) = ρ 2 a 2 yz + b 2 zx + c 2 xy ((b 2 γ + c 2 β)x + (c 2 α + a 2 γ)y + (a 2 β + b 2 α)z) +a 2 βγ + b 2 γα + c 2 αβ + ρ 2 = 0 a 2 yz + b 2 zx + c 2 xy ((b 2 γ + c 2 β)x + (c 2 α + a 2 γ)y + (a 2 β + b 2 α)z) +(a 2 βγ + b 2 γα + c 2 αβ + ρ 2 )(x + y + z) = 0 The coefficient of x is (b 2 γ + c 2 β) + (a 2 βγ + b 2 γα + c 2 αβ + ρ 2 ) = b 2 γ(α + β + γ) c 2 β(α + β + γ) + (a 2 βγ + b 2 γα + c 2 αβ + ρ 2 ) = (b 2 γ 2 + 2S βγ + c 2 β 2 ρ 2 ) Thus the equation is a 2 yz + b 2 zx + c 2 xy (b 2 γ 2 + 2S βγ + c 2 β 2 ρ 2 )x = 0 cyclic Therefore, the general equation of the circle with centre (α : β : γ) and radius ρ in homogeneous form is a 2 yz + b 2 zx + c 2 xy (x + y + z) ( b 2 γ 2 + 2S βγ + c 2 β 2 ) (α + β + γ) 2 ρ 2 x = 0 cyclic On the other hand, the centre of the circle a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz) = 0 is (a 2 S S (r p)+s (p q) : b 2 S S (p q)+s (q r) : c 2 S S (q r)+s (r p)), and its radius is given by ρ 2 = 1 4S 2 ( (abc) 2 + 2(a 2 S p + b 2 S q + c 2 S r) + S (q r) 2 + S (r p) 2 + S (p q) 2) 65 oncentric circles The equation of the circle concentric with the circumcircle with radius kr is a 2 yz + b 2 zx + c 2 xy + (x + y + z)(γx + γy + γz) = 0, where γ = (k 2 1)a 2 b 2 c 2 (2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2 a 4 b 4 c 4 ) 1 Proof Let the normalized barycentric coordinates of the circumcentre be (α : β : γ) Thus the equation of the circle concentric with the circumcircle with radius kr is a 2 (y β)(z γ) + b 2 (x α)(z γ) + c 2 (x α)(y β) = (kr) 2,

13 6 ircles II 13 where (x : y : z) is the normalized barycentric coordinates of a point on the circle The left hand side is equal to a 2 yz + b 2 zx + c 2 xy R 2 Thus the above equation can be written as a 2 yz + b 2 zx + c 2 xy + (k 2 1)R 2 = 0 Note that R = abc 2S, where S is the twice the area of the triangle Thus R 2 = a 2 b 2 c 2 (2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2 a 4 b 4 c 4 ) 1 Using x + y + z = 1, we may write the equation as a 2 yz + b 2 zx + c 2 xy + (x + y + z)(γx + γy + γz) = 0 66 The equation of the tangent to a circle Let ω be the circle with equation a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz) = 0 Let P be a point on ω with homogeneous barycentric coordinates (x 0 : y 0 : z 0 ) The equation of the tangent to ω at P is given by a 2 (y 0 z + yz 0 ) + b 2 (z 0 x + zx 0 ) + c 2 (x 0 y + xy 0 ) +(x 0 + y 0 + z 0 )(px + qy + rz) + (x + y + z)(px 0 + qy 0 + rz 0 ) = 0 That is [2px 0 + (c 2 + p + q)y 0 + (b 2 + r + p)z 0 : (c 2 + p + q)x 0 + 2qy 0 + (a 2 + q + r)z 0 : (b 2 + r + p)x 0 + (a 2 + q + r)y 0 + 2rz 0 ] Proof Let the parametric equation of the tangent line to ω at P be x = x 0 + αt, y = y 0 + βt, z = z 0 + γt, where (α : β : γ) is a displacement vector along the direction of the tangent Note that α + β + γ = 0 Substituting (x 0 + αt : y 0 + βt : z 0 + γt) into the equation of the circle and using the fact that (x 0 : y 0 : z 0 ) satisfies the equation of the circle, we get a quadratic equation in t (a 2 βγ + b 2 γα + c 2 αβ)t 2 + [a 2 (y 0 γ + βz 0 ) + b 2 (z 0 α + γx 0 ) + c 2 (x 0 β + αy 0 ) +(x 0 + y 0 + z 0 )(pα + qβ + rγ)]t = 0 Since the tangent line intersects ω only at the point P, this quadratic equation has a double root t = 0, or equivalently, the coefficient of t is 0 Thus a 2 (y 0 γ + βz 0 ) + b 2 (z 0 α + γx 0 ) + c 2 (x 0 β + αy 0 ) + (x 0 + y 0 + z 0 )(pα + qβ + rγ) = 0 To find a linear equation satisfied by x, y, z, consider a 2 (y 0 z + yz 0 ) + b 2 (z 0 x + zx 0 ) + c 2 (x 0 y + xy 0 ) +(x 0 + y 0 + z 0 )(px + qy + rz) + (x + y + z)(px 0 + qy 0 + rz 0 ) = a 2 (y 0 (z 0 + γt) + (y 0 + βt)z 0 )) + b 2 (z 0 (x 0 + αt)+(z 0 + γt)x 0 )+c 2 (x 0 (y 0 + βt)+(x 0 + αt)y 0 ) +(x 0 + y 0 + z 0 )(p(x 0 + αt) + q(y 0 + βt) + r(z 0 + γt)) +(x 0 + y 0 + z 0 + (α + β + γ)t)(px 0 + qy 0 + rz 0 ) = 2a 2 y 0 z 0 + ta 2 (y 0 γ + βz 0 ) + 2b 2 z 0 x 0 + tb 2 (z 0 α + γx 0 ) + 2c 2 x 0 y 0 + tc 2 (x 0 β + αy 0 ) +(x 0 + y 0 + z 0 )(px 0 + qy 0 + rz 0 + t(pα + qβ + rγ)) +(x 0 + y 0 + z 0 )(px 0 + qy 0 + rz 0 ) = 2(a 2 y 0 z 0 + b 2 z 0 x 0 + c 2 x 0 y 0 ) + t(a 2 (y 0 γ + βz 0 ) + b 2 (z 0 α + γx 0 ) + c 2 (x 0 β + αy 0 ))

14 7 Worked Examples 14 = 0 +t(x 0 + y 0 + z 0 )(pα + qβ + rγ)) + 2(x 0 + y 0 + z 0 )(px 0 + qy 0 + rz 0 ) The last line is by the above relation and the fact that (x 0 : y 0 : z 0 ) satisfies the equation of the circle Example 61 The equation of the tangent to the circumcircle at is b 2 z + c 2 y = 0 67 The line at infinity and the circumcircle Recall that if (x : y : z) is a displacement vector which is a difference of two points and in normalized barycentric coordinates, then x + y + z = 0 If we identify all displacement vectors along using homogeneous coordinates so that (x : y : z) = (kx : ky : kz) for any k 0, then the displacement vector (x : y : z) represents either direction of the line displacement vector in homogeneous coordinates is also called the infinite point of the line See [1] The set of all infinite points constitutes a line called the line at infinity which has the equation x+y +z = 0 In the setting of the usual plane geometry, the isogonal conjugate of a point on the circumcircle is not defined More precisely, if P is a point on the circumcircle of the triangle, then the reflection of the line P about the bisector of, the reflection of the line P about the bisector of and the reflection of the line P about the bisector of are all parallel Thus they meet at an infinite point In other words, if P = (x : y : z) is a point on circumcircle of the triangle, then its isogonal conjugate (a 2 yz : b 2 zx : c 2 xy) lies on the line at infinity Thus a 2 yz + b 2 zx + c 2 xy = 0, which is the equation of the circumcircle 7 Worked Examples 1 [entromerican 2017] Let be a triangle and D be the foot of the altitude from Let l be the line that passes through the midpoints of and E is the reflection of D over l Prove that the circumcentre of the triangle lies on the line E Solution l E O F D It is known that D = (0 : S : S ), l = [1 : 1 : 1] displacement vector along l is ( 1 : 1 : 0) displacement vector perpendicular to l is (S : S : c 2 ) Thus a parametric equation of the line DE is (ts : S + ts : S tc 2 ) To find the intersection point F between l and DE, we substitute the parametric equation of DE into the equation of l Solving for t, we get t = S S S +S = S S = c2 b 2 +c 2 2c 2 2c 2

15 7 Worked Examples 15 Therefore, F = ( c2 b 2 S 2c 2 : S + c2 b 2 S 2c 2 : S c2 b 2 2 ) = ((c 2 b 2 )S : b 2 c 2 +c 2 S b 2 S : c 2 (2S c 2 + b 2 )) = ((c 2 b 2 )S : b 2 S + c 2 S : c 2 S + c 2 S ) Here we use the relations S + S = c 2 etc The normalized barycentric coordinates of F and D are F = 1 2a 2 c 2 ((c2 b 2 )S : b 2 S + c 2 S : c 2 S + c 2 S ), D = 1 a 2 (0 : S : S ) Thus we can compute E as 2F D That is E = 1 ((c 2 b 2 )S a 2 c 2 : b 2 S + c 2 S : c 2 S + c 2 S ) 1 (0 : c 2 S a 2 c 2 : c 2 S ) = ((c 2 b 2 )S : b 2 S : c 2 S ) Since = (1 : 0 : 0) and O = (a 2 S : b 2 S : c 2 S ) learly the determinant formed by these 3 points is zero onsequently,, O, E are collinear 2 Let O and G be the circumcentre and the centroid of a triangle Prove that G is perpendicular to OG if and only if b 2 + c 2 = 2a 2 Solution G O We have G = 1 3 ( 2 : 1 : 1) and OG = ( 1 3 a 2 S a 2 S +b 2 S +c 2 S, 1 3 b 2 S a 2 S +b 2 S +c 2 S, 1 3 c 2 S a 2 S +b 2 S +c 2 S ) = 1 ( 2a 2 S 6S 2 + b 2 S + c 2 S : a 2 S 2b 2 S + c 2 S : a 2 S + b 2 S 2c 2 S ) Then G OG = 1 36S 2 (a 2 (2a 2 S b 2 S c 2 S ) + b 2 ( 4a 2 S b 2 S + 5c 2 S ) + c 2 ( 4a 2 S + 5b 2 S c 2 S )) Therefore, 36S 2 G OG = 2a 2 (a 2 2b 2 2c 2 )S b 2 (a 2 + b 2 5c 2 )S c 2 (a 2 + c 2 5b 2 )S = 2a 2 (a 2 2b 2 2c 2 )S 2b 2 S S + 4b 2 c 2 S 2c 2 S S + 4b 2 c 2 S = 2a 2 (2b 2 + 2c 2 a 2 )S 2(b 2 + c 2 )S S + 4b 2 c 2 (S + S ) = 2a 2 (b 2 + c 2 a 2 )S 2a 2 (b 2 + c 2 )S 2(b 2 + c 2 )S S + 4a 2 b 2 c 2 = 4a 2 S 2 2(b2 + c 2 )(S S + a 2 S ) + 4a 2 b 2 c 2 = 4a 2 (b 2 c 2 S 2 ) 2(b2 + c 2 )(S S + a 2 S ) = 4a 2 S 2 2(b 2 + c 2 )S 2 = 2(2a 2 b 2 c 2 )S 2 That is G OG = 1 18 (2a2 b 2 c 2 ) onsequently, G OG = 0 if and only if 2a 2 b 2 c 2 = 0

16 7 Worked Examples 16 3 [Donova Mathematical Olympiad 2010] Given a triangle, let,, be the perpendicular feet dropped from the centroid G of the triangle onto the sides,, respectively Reflect,, through G to,, respectively Prove that the lines,, are concurrent Solution G Direct computation gives = (2a 2 : S : S ), = (S : 2b 2 : S ), = (S : S : 2c 2 ) Thus = [0 : S : S ], = [S : 0 : S ], = [ S : S : 0] The determinant formed by these 3 lines is clearly zero Thus,, are concurrent 4 [JMO Shortlist 2015] round the triangle the circle is circumscribed, and at the vertex tangent t to this circle is drawn The line p, which is parallel to this tangent intersects the lines and at the points D and E, respectively Prove that the points,, D, E belong to the same circle Solution E t p D Let D = (0 : 1 α : α) The tangent t to the circumcircle of is [b 2 : a 2 : 0] The points (0 : 0 : 1) and ( a 2 : b 2 : 0) lie on t Therefore a displacement vector along t is ( a2 b : 2 : 1) = (a 2 : b 2 : a 2 b 2 ) We can parametrize p by (0 : 1 α : b 2 a 2 b 2 a 2 α) + s(a 2 : b 2 : a 2 b 2 ) = (sa 2 : 1 α sb 2 : α + s(a 2 b 2 )) Substituting this into the line = [0 : 1 : 0], we have 1 α sb 2 = 0 so that s = 1 α Thus b 2 (1 α)a2 E = ( b 2 : 0 : α + 1 α b 2 (a 2 b 2 )) = ((1 α)a 2 : 0 : b 2 a 2 (1 α))

17 7 Worked Examples 17 y substituting the coordinates of,, D into the general equation of a circle, the equation of the circumcircle of D is found to be a 2 yz + b 2 zx + c 2 xy (x + y + z)a 2 (1 α)z = 0 heck that E satisfies this equation Thus,, D, E are concyclic Remark The result follows easily from alternate segment theorem 5 [Mongolia 2000] The bisectors of,, of a triangle intersect its sides at points 1, 1, 1 Prove that, 1, 1, 1 are concyclic if and only if Solution + = First 1 = (0 : b : c), 1 = (a : 0 : c), 1 = (a : b : 0) Substituting the coordinates of these 3 points into the general equation of a circle, the equation of the circumcircle of is a 2 yz + b 2 zx + c 2 xy + (x + y + z)(px + qy + rz) = 0, where p = bc 2 ( a b+c b c+a c a+b ), q = ca 2 ( b c+a c a+b a ab b+c ), and r = 2 ( c a+b a b+c b c+a ) Note that = (0 : 1 : 0) lies on this circle if and only if q = 0, which gives a b+c = b c+a c a+b 6 [enelux 2017] In the convex quadrilateral D we have = and D = 90 Suppose that = 2D Prove that the angle bisector of is perpendicular to D Solution D K E M

18 7 Worked Examples 18 Let M be the midpoint of and E the bisector of Since MD is an isosceles trapezium, D is the reflection of M across the perpendicular bisector of First note that = are obtuse angles If = < 90, then DM > 90 Since D = 90, this contradicts the given condition that D is a convex quadrilateral Take be the reference triangle Then M = 1 a 2 (1 : 1 : 0), E = ( a+b : b a+b : 0), and the perpendicular bisector of is [c 2 b 2 : a 2 : a 2 ] We can parametrize MD as (1 : 1 : 0) + t(0 : 1 : 1) = (1 : 1 t : t) Substituting this into the equation of the perpendicular bisector of, we have (c 2 b 2 ) a 2 (1 t) + a 2 t = 0 Solving for t, we get t = a2 +b 2 c 2 Thus the midpoint of DM is K = 1 2a 2 2 (1 : 1 a2 +b 2 c 2 : a2 +b 2 c 2 ) Thus 2a 2 2a 2 D = 2K M = ( 1 2 : 1 2 a2 +b 2 c 2 : a2 +b 2 c 2 ) 2a 2 2a 2 Then D = ( 1 2 : a2 +b 2 c 2 : 1 a2 +b 2 c 2 ) = ( a 2 : b 2 c 2 : c 2 +a 2 b 2 ) Similarly, 2a 2 2a 2 D = (a 2 : b 2 c 2 : a 2 b 2 + c 2 ) Since D = 90, we have 0 = D D = 2a 2 (ab + b 2 c 2 )(ab b 2 + c 2 ) Since is obtuse, we have b > c so that ab + b 2 c 2 0 Therefore, we must have ab b 2 + c 2 = 0 or b 2 c 2 = ab Thus D = (a 2 : ab : a 2 ab) = (a : b : a b) Note that E = ( a a+b : b a+b : 1) = (a : b : a + b) onsequently, D is parallel to E, or equivalently, E is perpendicular to D 7 [hina 2010] In a triangle, >, I is its incentre, M is the midpoint of and N is the midpoint of The line through parallel to IM meets at D, and the line through parallel to IN meets at E The line through I parallel to DE meets the line at P If Q is the foot of the perpendicular from P onto the line I, prove that Q lies on the circumcircle of Solution Ạ D E N I M P First IM = ( 1 2 : 0 : 1 2 ) ( a a+b+c : b a+b+c : c a+b+c ) = (b + c a : 2b : a + b c) We can parametrize D as (0 : 1 : 0) + t(b + c a : 2b : a + b c) Substituting this Q

19 7 Worked Examples 19 into the line = [0 : 1 : 0] and solving for t, we have t = 1 b+c a 2b Thus D = ( 2b : 0 : a+b c 2b ) Similarly, E = ( b+c a 2c : a b+c 2c : 0) Therefore, DE = ( b+c a 2c b+c a 2b : a b+c 2c : a+b c 2b ) = ((b c)(b + c a) : b(c + a b) : c(a + b c)) We can parametrize IP as (a : b : c) + t((b c)(b + c a) : b(c + a b) : c(a + b c)) To find P, we substitute a this into the line = [1 : 0 : 0] and solve for t We have t = (b c)(b+c a) Therefore, P = (0 : b ab(c+a b) (b c)(b+c a) : c + ac(a+b c) (b c)(b+c a) ) = (0 : b(c2 + a 2 b 2 ) : c(a 2 + b 2 c 2 )) = (0 : bs : cs ) displacement vector perpendicular to I is given by (b c : b : c) parametrize P Q as (0 : bs : cs ) + t(b c : b : c) Thus we can To find Q, we substitute this into the line I = [0 : c : b] and solve for t We have t = a2 2 Therefore, Q = ( a2 (b c) 2 : b(s a2 2 ) : c(s a2 2 )) = ( a2 (b c) 2 : b(c2 b 2 ) 2 : c(b 2 c 2 ) 2 ) = ( a 2 : b(b + c) : c(b + c)), which lies on the circumcircle In fact, Q is the midpoint of the arc not containing

20 8 Exercises 20 8 Exercises 1 Prove that in any triangle, the centroid G, the incentre I and the Nagel point N are collinear 2 [Newton s line] Let D be a quadrilateral Let H, I, G, J, E, F be the midpoints of,, D, D, D, respectively Let IJ intersect HG at M, intersect D at U, intersect D at V Let N be the midpoint of UV Prove that E, F, M, N are collinear 3 Prove that in any triangle the 3 lines each of which joins the midpoint of a side to the midpoint of the altitude to that side are concurrent 4 In a triangle, = 90, the bisector of meets the altitude D at the point E, and the bisector of D meets the side D at F The line through F perpendicular to intersects at G Prove that, E, G are collinear 5 In a triangle, M is the midpoint of and D is the point on such that D bisects The line through perpendicular to D intersects D at E and M at G Prove that GD is parallel to 6 In an acute-angled triangle, N is a point on the altitude M The line N, N meet and respectively at F and E Prove that EMN = F MN 7 [Pascal s theorem] Let, F,, D,, E be six points on a circle in this order Let F intersect D at P, F intersect E at Q and D intersect E at R Prove that P, Q, R are collinear 8 In a triangle, 90, M is the midpoint of and H is the orthocentre The feet of the perpendiculars from H onto the internal and external bisectors of are N and L respectively Prove that M, N, L are collinear 9 Let be an acute-angled triangle with orthocenter H The circle with diameter H intersects the circumcircle of the triangle at the point N distinct from Prove that the line NH bisects the segment 10 Let be an acute-angled triangle with incentre I The circle with diameter I intersects the circumcircle of the triangle at the point N distinct from Let the incircle of the triangle touch the side at D Prove that the line ND bisects the arc not containing 11 Let be a triangle with circumcentre O Points E, F lie on, respectively The line EF cuts the circumcircles of E and F again at M, N respectively Prove that OM = ON

21 8 Exercises [IMO 2017] Let R and S be different points on a circle Ω such that RS is not a diameter Let l be the tangent line to Ω at R Point T is such that S is the midpoint of the line segment RT Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects l at two distinct points Let be the common point of Γ and l that is closer to R Line J meets Ω again at K Prove that the line KT is tangent to Γ 13 [IMO 2016] Triangle F has a right angle at Let be the point on line F such that F = F and F lies between and Point D is chosen so that D = D and is the bisector of D Point E is chosen so that E = ED and D is the bisector of E Let M be the midpoint of F Let X be the point such that MXE is a parallelogram Prove that D, F X and M E are concurrent 14 [IMO 2014] Let P and Q be on segment of an acute triangle such that P = and Q = Let M and N be the points on P and Q, respectively, such that P is the midpoint of M and Q is the midpoint of N Prove that the intersection of M and N is on the circumference of triangle 15 [IMO 2012] Given triangle the point J is the centre of the excircle opposite the vertex This excircle is tangent to the side at M, and to the lines and at K and L, respectively The lines LM and J meet at F, and the lines KM and J meet at G Let S be the point of intersection of the lines F and, and let T be the point of intersection of the lines G and Prove that M is the midpoint of ST 16 [IMO 2010] Given a triangle, with I as its incentre and Γ as its circumcircle, I intersects Γ again at D Let E be a point on the arc D, and F a point on the segment, such that F = E < 1 If G is the midpoint of IF, prove that the meeting point 2 of the lines EI and DG lies on Γ 17 [PMO 2017] Let be a triangle with < Let D be the intersection point of the internal bisector of angle and the circumcircle of Let Z be the intersection point of the perpendicular bisector of with the external bisector of angle Prove that the midpoint of the segment lies on the circumcircle of triangle DZ 18 [IMO 2016 Shortlist] Let be a triangle with = and let I be its incentre The line I meets at D, and the line through D perpendicular to meets I at E Prove that the reflection of I in lies on the circumcircle of triangle DE 19 [Nordic 2017] Let M and N be the midpoints of the sides and, respectively, of an acute triangle, Let ω be the circle centered at M passing through, and let ω be the circle centered at N passing through Let the point D be such that D is an isosceles trapezoid with D parallel to ssume that ω and ω intersect in two distinct points P and Q Show that D lies on the line P Q

22 8 Exercises [hina 2017] In the non-isosceles triangle, D is the midpoint of side, E is the midpoint of side, F is the midpoint of side The line (different from line ) that is tangent to the inscribed circle of triangle and passing through point D intersect line EF at X Define Y, Z similarly Prove that X, Y, Z are collinear

23 9 Hints 23 9 Hints 1 heck that the determinant formed by the 3 points is 0 2 Take be the reference triangle and D = (u : v : 1 u v) Find the barycentric coordinates of E, F, M, N 3 Let 1 be the midpoint of the altitude from onto, and 2 the midpoint of Show that 1 2 = [tan tan : tan + tan : tan tan ] 4 Following the condition of the question, show that G = (a : 0 : c) 5 displacement vector perpendicular to D is ( b + c : b : c) Using this, show that E = (b c : b : c) and G = (b c : c : c) 6 Let l be the line through parallel to Let the extensions of MF and ME meet l at P and Q respectively Then l = [0 : 1 : 1] and M = (0 : S : S ) Let N = (a 2 (1 t) : S t : S t), for some t Find the barycentric coordinates of P, Q, and show that is the midpoint of P Q 7 Take be the reference triangle Let D = (d 1 : d 2 : d 3 ), E = (e 1 : e 2 : e 3 ), F = (f 1 : f 2 : f 3 ) Then show that P = (d 1 f 2 : d 2 f 2 : d 2 f 3 ), Q = (e 1 f 1 : e 2 f 1 : e 1 f 3 ), R = (d 1 e 3 : d 3 e 2 : d 3 e 3 ) heck that the determinant formed by the 3 points is 0 8 displacement vector for HL(= N ) is (b + c : b : c) Then parametrize HL as (S S : S S : S S ) + t(b + c : b : c) Using this, show that L = (2bcS S + c(b + c)s S + b(b + c)s S : bs (bs cs ) : cs (bs cs )) Similarly, show that N = (2bcS S + c(b c)s S b(b c)s S : bs (bs + cs ) : cs (bs + cs )) 9 Let M be the midpoint of Take N to be the point on the circumcircle of such that NM = 90 Try to show N, H, M are collinear Let N = (x : y : z) in normalized barycentric coordinates The displacement vectors N = (x 1 : y : z) and MN = (x : y 1 2 : z 1 2 ) are perpendicular Thus a2 (y(z 1 2 )+(y 1 2 )z)+b2 ((x 1)(z 1 2 )+zx)+c2 ((x 1)(y 1 2 )+xy) = 0 Using a2 yz +b 2 zx+c 2 xy = 0, this can be simplified to S y +S z = 0 Take y = S, z = S and substitute this into a 2 yz + b 2 zx + c 2 xy = 0 to get x = a2 S S (b 2 c 2 )S Therefore N = ( a2 S S (b 2 c 2 )S : S : S ) = (a 2 S S : (b 2 c 2 )S S : (b 2 c 2 )S S ) 10 ompute the barycentric coordinates of N Show that N = (a 2 (s b)(s c) : b(b c)(s a)(s c) : c(b c)(s a)(s b)) 11 Let E = (1 t : 0 : t) and F = (1 s : s : 0) Show that the equation of the circumcircle of E is a 2 yz + b 2 zx + c 2 xy (x + y + z)(1 t)b 2 z = 0 Parametrize the line EF as (1 t : 0 : t) + α(t s : s : t) Substituting this into the equation of the circumcircle of

24 9 Hints 24 E to get α = a2 st+b 2 t(t s)+c 2 s(1 t) a 2 st+b 2 t(t s) c 2 s(t s) Denote this value of α by α M, and the corresponding barycentric coordinates of M by (x M : y M : z M ) The power of M with respect to the circumcircle of is (a 2 y M z M + b 2 z M x M + c 2 x M y M ) = (x M + y M + z M )(1 t)b 2 z M = (1 t)b 2 z M To compute the power, we only need the explicit value of z M Then show that M and N have the same power with respect to the circumcircle of 12 Let RSJ be the reference triangle, where R = (1 : 0 : 0), S = (0 : 1 : 0), J = (0 : 0 : 1) Thus Ω : a 2 yz + b 2 zx + c 2 xy = 0 and Γ : a 2 yz + b 2 zx + c 2 xy 2c 2 x(x + y + z) = 0 The tangent l at R is b 2 z+c 2 y = 0 We may parametrize l by x = 1, y = b 2 t, z = c 2 t Thus = (1 : b 2 t : c 2 t), for some t Since lies on Γ, we have a 2 b 2 c 2 +b 2 c 2 t c 2 b 2 t 2c 2 (1 b 2 t+ c 2 t) = 0, or equivalently, a 2 b 2 t 2 +2(c 2 b 2 )t+2 = 0 Line J = [b 2 t : 1 : 0] The tangent to Γ at T is a 2 (2z)+b 2 ( z)+c 2 ( y+2x) 2c 2 x+(x+y+z)(2c 2 ) = 0; which can be simplified to 2c 2 x+c 2 y+(2a 2 b 2 +2c 2 )z = 0 That is [2c 2 : c 2 : (2a 2 b 2 +2c 2 )] ompute the barycentric coordinates of K and show that it lies on Ω using the relation a 2 b 2 t 2 + 2(c 2 b 2 )t + 2 = 0 13 Let F be the reference triangle, where F = (1 : 0 : 0), = (0 : 1 : 0), = (0 : 0 : 1) Note that a 2 = b 2 c 2 Since F divides in the ratio b : c, we have = (1 + c b : 0 : c b ) The midpoint of is N = ( b+c 2b : 0 : b c 2b ) Next obtain the displacement vector (c2 b 2 : b 2 : c 2 ) which is perpendicular to Then parametrize the perpendicular bisector of by x = b+c 2b + t(c2 b 2 ), y = tb 2, z = b c 2b tc2 This is the line DN, whose intersection with is D Then D = ( (2c b)(b+c) b 2bc : 2c : b 2c 2b ) Hence D = 2N D = ( b+c 2c : b 2c : 1 2 ) The midpoint of D is ( (b+c)(b+2c) 4bc : b 4c : b 2c 4b ) displacement vector perpendicular to D is ((b + c)(b 2c) : b 2 : bc + 2c 2 ) Since E lies on the perpendicular bisector of D, we may take E = ( (b+c)(b+2c) 4bc + t(b + c)(b 2c) : b 4c tb2 : b 2c 4b + t(bc + 2c 2 )), for some t s DE is parallel to, the second coordinate of the displacement vector DE must be b 0 Therefore, 4c tb2 + b 2c = 0 giving t = 1 b+2c b 4bc onsequently, E = ( 2c : 2c : 0) Lastly, X = ( bc 2c2 +b 2 b 2bc : 2c : b+2c 2b ) From this, we find that D = [c : 0 : (b + c)], F X = [0 : 2c 2 + bc : b 2 ] and ME = [b : b + 2c : b] 14 Note that the triangles, P, Q are all similar The point of intersection of the lines M and N is ( a 2 : 2b 2 : 2c 2 ) 15 Show that S = (0 : sa (s b)c a(s c) : c a ) and T = (0 : b a : as b(s c) a(s b ) 16 Let E intersect at F Then F is the isogonal conjugate of F Let F = (0 : 1 α : α) Then F = (0 : αb 2 : (1 α)c 2 ) Then show that E = (α(1 α)a 2 : αb 2 : (1 α)c 2 ), EI = [bc(αb (1 α)c) : ca(1 α)(c + αa) : abα((1 α)a + b)] and GD = [ (b + c)(αb (1 α)c) : a(c + αa) : a((1 α)a + b)] Then X = (a(c + αa)((1 α)a + b) : b(αb (1 α)c)((1 α)a + b) : c(c + αa)(αb (1 α)c); and show that it lies on Γ 17 Let M be the midpoint of Then the circumcircle of MD has the equation a 2 yz+b 2 zx+ c 2 xy c2 2 (x + y + z)(y b cz) = 0 Show that Z lies on this circle

25 9 Hints Let be the reference triangle with b = c a Direct calculation gives E = (a( a 2 ab+ 4b 2 ) : 2b 3 : 2b 3 ) and I = (a(a + b) : b 2 : 3b 2 a 2 ) lso the equation of the circumcircle of DE is a 2 yz + b 2 zx + c 2 xy + b 2 (a 2 b 2 )(a+2b) (x + y + z)(2b3 x a 2 (a + b)z) = 0 19 The reflection of about the perpendicular bisector of is D = (a 2 : c 2 b 2 : b 2 c 2 ) Let L = 2c 2 + 2a 2 b 2 and K = 2a 2 + 2b 2 c 2 Then the lengths of the medians from and are L/2 and K/2 respectively The circle ω centred at the midpoint of with radius L/2 passes through the points = (0 : 1 : 0), (b + L : 0 : b L) and (b L : 0 : b + L) Substituting these points into the general equation of a circle, we find that the equation of ω equal to a 2 yz + b 2 zx + c 2 xy + (x + y + z)(s x + S z) = 0 Similarly, the equation of ω equal to a 2 yz + b 2 zx + c 2 xy + (x + y + z)(s x + S y) = 0 Then show that D lies on the radical axis P Q of ω and ω 20 Let 1 be the point of tangency of the incircle with the side, and let 1 2 be a diameter of the incircle Let the other tangent from D to the incircle meet the incircle at 3 (That is 3 1 ) Thus = 90 The extension of 2 3 meets the side at a point 4 such that D is the midpoint of 1 4 This means that 4 is the point of tangency of the -excircle with the side That is 4 = (0 : s b : s c) lso it is well-known that, 2, 4 are collinear Thus, 1, 2, 4 are collinear Use this information to find 3 Show that 3 = ( (b c)2 (c a) (s a) : s b : s c) Similarly, 3 = (s a : 2 s b : s c) (a b) and 3 = (s a : s b : 2 s c ) From this, show that X = (b c : s c : b s), Y = (c s : c a : s a), Z = (s b : a s : a b)

26 10 Solutions Solutions 1 Prove that in any triangle, the centroid G, the incentre I and the Nagel point N are collinear Solution This is because a b c = 0 In fact G divides the segment IN in s a s b s c the ratio 1:2 2 [Newton s line] Let D be a quadrilateral Let H, I, G, J, E, F be the midpoints of,, D, D, D, respectively Let IJ intersect HG at M, intersect D at U, intersect D at V Let N be the midpoint of UV Prove that E, F, M, N are collinear Solution V N J D G M F E I U H Let = (1 : 0 : 0), = (0 : 1 : 0), = (0 : 0 : 1) and D = (u : v : 1 u v) Then H = (1 : 1 : 0), G = (u : v : 2 u v), I = (0 : 1 : 1), J = (1 + u : v : 1 u v) Direct computation gives F = (1 : 0 : 1), E = (u : 1 + v : 1 u v) and M = (u + 1 : v + 1 : 2 u v) The determinant u 1 + v 1 u v = 0, since the third row is the sum of the first u + 1 v u v two rows Therefore, E, F, M are collinear Similarly, we can show that N = (u u 2 : v + v 2 : w w 2 ) = (u(1 u) : v(1 + v) : (u + v)(1 u v)), and the determinant u 1 + v 1 u v u(1 u) v(1 + v) (u + v)(1 u v) = 0 Therefore, E, F, N are collinear 3 Prove that in any triangle the 3 lines each of which joins the midpoint of a side to the midpoint of the altitude to that side are concurrent Solution

27 10 Solutions Take = (1 : 0 : 0), = (0 : 1 : 0), = (0 : 0 : 1) Let 1, 1 and 1 be the midpoints of the altitudes from onto, from onto and from onto respectively If 2, 2 and 2 are the midpoints of the sides, and respectively, we find that 1 2 = [tan tan : tan + tan : tan tan ], 1 2 = [ tan tan : tan tan : tan + tan ], and 3 2 = [tan + tan : tan tan : tan tan ] tan tan tan + tan tan tan The determinant tan tan tan tan tan + tan = 0, since the tan + tan tan tan tan tan sum of the 3 rows is the zero row Therefore, 1 2, 1 2 and 1 2 are concurrent 4 In a triangle, = 90, the bisector of meets the altitude D at the point E, and the bisector of D meets the side D at F The line through F perpendicular to intersects at G Prove that, E, G are collinear Solution E G D F Let be the reference triangle Note that a 2 = b 2 + c 2 Using the angle bisector theorem, direct computation gives D = (0 : b 2 : c 2 ), E = (ca : b 2 : c 2 ), F = (0 : b 2 : c(a + c)) and G = (a : 0 : c) s ca b 2 c 2 a 0 c = ac2 ac 2 = 0, the points, E, G are collinear 5 In a triangle, M is the midpoint of and D is the point on such that D bisects The line through perpendicular to D intersects D at E and M at G Prove that GD is parallel to

28 10 Solutions 28 Solution E D G M Let be the reference triangle Then M = (0 : 1 : 1) and D = (0 : b : c) The b displacement vector D = ( 1 : b+c : c b+c ) = (b + c : b : c) displacement vector perpendicular to D is ( b + c : b : c) We may write E = (0 : 1 : 0) + t( b + c : b : c) = (( b + c)t : 1 + bt : ct) for some t The equation of the line D is cy + bz = 0 Substituting the coordinates of E into the equation of the line D and solving for t, we obtain t = 1 b c 1 2b Therefore, E = ( 2b : 2 : c 2b ) = (b c : b : c) From this we obtain: line E = [c : 0 : b + c] and line M = [0 : 1 : 1] Thus G = (b c : c : c) Hence, GD = ( b + c : b c : 0) = ( 1 : 1 : 0) = which means GD is parallel to 6 In an acute-angled triangle, N is a point on the altitude M The line N, N meet and respectively at F and E Prove that EMN = F MN Solution P Q F N E M Let l be the line through parallel to Let the extensions of MF and ME meet l at P and Q respectively We have l = [0 : 1 : 1] and M = (0 : S : S ) Let N = (a 2 (1 t) : S t : S t), for some t Then N = [ts : 0 : a 2 (1 t)], E = (a 2 (1 t) : 0 : ts ), ME = [ts S : a 2 (1 t)s : a 2 (1 t)s ] Thus Q = ( a 2 (1 t)(s + S ) : ts S : ts S ) = (2a 4 (1 t) : ts S : ts S ) Similarly, N = [ ts : a 2 (1 t) : 0], F = (a 2 (1 t) : ts : 0), MF = [ts S : a 2 (1 t)s : a 2 (1 t)s ], and P = (2a 4 (1 t) : ts S : ts S ) onsequently, the midpoint of P Q is It follows that EMN = F MN 7 [Pascal s theorem] Let, F,, D,, E be six points on a circle in this order Let F intersect D at P, F intersect E at Q and D intersect E at R Prove that P, Q, R are collinear

29 10 Solutions 29 Solution Let be the reference triangle and ω its circumcircle Let D = (d 1 : d 2 : d 3 ), E = (e 1 : e 2 : e 3 ), F = (f 1 : f 2 : f 3 ) Direct computations give P = (d 1 f 2 : d 2 f 2 : d 1 f 2 d 2 f 2 d 2 f 3 d 2 f 3 ), Q = (e 1 f 1 : e 2 f 1 : e 1 f 3 ), R = (d 1 e 3 : d 3 e 2 : d 3 e 3 ) Then e 1 f 1 e 2 f 1 e 1 f 3 d 1 e 3 d 3 e 2 d 3 e 3 = d 1 d 3 e 2 e 3 f 1 f 2 +d 1 d 2 e 1 e 3 f 2 f 3 +d 2 d 3 e 1 e 2 f 1 f 3 d 1 d 2 e 2 e 3 f 1 f 3 d 1 d 3 e 1 e 2 f 2 f 3 d 2 d 3 e 1 e 3 f 1 f 2 Using the relations a 2 d 2 d 3 + b 2 d 3 d 1 + c 2 d 1 d 2 = 0, a 2 e 2 e 3 + b 2 e 3 e 1 + c 2 e 1 e 2 = 0, a 2 f 2 f 3 + b 2 f 3 f 1 + c 2 f 1 f 2 = 0, we can eliminate the terms d 2 d 3, e 2 e 3, f 2 f 3 and show that the above determinant has value 0 Thus P, Q, R are collinear lternatively, we may rewrite the expression of the above determinant as which is equal to 1 a 2 a 2 d 2 d 3 d 3 d 1 d 1 d 2 a 2 f 2 f 3 f 3 f 1 f 1 f 2 a 2 e 2 e 3 e 3 e 1 e 1 e 2 = 1 a 2 b 2 d 3 d 1 c 2 d 1 d 2 d 3 d 1 d 1 d 2 b 2 f 3 f 1 c 2 f 1 f 2 f 3 f 1 f 1 f 2 b 2 e 3 e 1 c 2 e 1 e 2 e 3 e 1 e 1 e 2 as the first column is a linear combination of the last two columns d 2 d 3 d 3 d 1 d 1 d 2 f 2 f 3 f 3 f 1 f 1 f 2 e 2 e 3 e 3 e 1 e 1 e 2 = 0,, 8 In a triangle, 90, M is the midpoint of and H is the orthocentre The feet of the perpendiculars from H onto the internal and external bisectors of are N and L respectively Prove that M, N, L are collinear Solution L M N H displacement vector for HL(= N) is (b + c : b : c) We can parametrize HL as (S S : S S : S S ) + t(b + c : b : c) The external bisector of is L = [0 : c : b] Substituting the parametric equation of HL into the equation of L, we get t = S (bs +cs ) 2bc From this, we obtain L = (2bcS S + c(b + c)s S + b(b + c)s S : bs (bs cs ) : cs (bs cs )) Similarly, we get N = (2bcS S + c(b c)s S b(b c)s S : bs (bs + cs ) : cs (bs + cs )) We also know M = (0 : 1 : 1)