Advanced Algebra and Geometry

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2010

Last updated: November 16, 2010. Contents

1 Circles 101 1.1 A construction problem of tangent circles ...... 101 1.2 The radical axis of two circles ...... 104 1.3 Orthogonal circles ...... 105 1.4 Coaxial circles ...... 106 1.5 Miscellaneous exercises ...... 107 2.2 Steiner chains theorem ...... 114 2.5 Feuerbach’s theorem ...... 117 2.6 The Apollonius problem for the excircles ...... 120

3 Parabolas 201 3.1 Deﬁnitions ...... 201 3.2 Chords and tangents ...... 202 3.3 Normals ...... 204 3.4 Evolute of a parabola ...... 206 3.5 Triangle bounded by three tangents ...... 207 3.6 Miscellaneous exercises ...... 209

4 Ellipses and Hyperbolas 211 4.1 Ellipses ...... 211 4.2 Tangents and normals of an ellipse ...... 213 4.2.1 Evolute of an ellipse ...... 214 4.3 Hyperbolas ...... 216 4.4 Tangents and normals of a hyperbola ...... 217 4.4.1 Evolute of a hyperbola ...... 218 4.5 Rectangular hyperbolas ...... 219 4.6 A theorem on the tangents from a point to a conic ...... 219 4.7 Miscellaneous exercises ...... 221

5 General conics 225 5.1 Classiﬁcation of conics ...... 225 5.2 Pole and polar ...... 226 5.3 Condition of tangency ...... 227 5.4 Parametrized conics ...... 228 5.5 Rectangular hyperbolas and the nine-point circle ...... 229 iv CONTENTS

6 Basic triangle centers 301 6.1 The Euler line ...... 301 6.1.1 The centroid ...... 301 6.1.2 The orthocenter and the Euler line ...... 303 6.1.3 The nine-point circle ...... 303 6.2 The incircle and excircles ...... 304 6.2.1 The incircle ...... 304 6.2.2 The excircles ...... 305 6.2.3 Heron’s formula for the area of a triangle ...... 306 6.2.4 Euler’s formula: distances between circumcenter and tritangent centers ...... 307

7 Construction problems I 311 7.1 Classiﬁcation of construction problems ...... 312 7.2 Construction problems with easy geometric solutions ...... 314 7.3 Construction problems with easy algebraic solutions ...... 315 7.3.1 (O, Ha,Ta) ...... 315

8 Construction problems II 319 8.1 Polynomial roots as intersections of conics ...... 319 8.2 Examples with cubic polynomials ...... 319 8.2.1 (G, Hb,Hc) ...... 320 8.2.2 (A, Ma,Tb) ...... 321 8.2.3 (Ma,Mb,Tc) ...... 322 8.2.4 (A, H, Tb) ...... 322 8.2.5 (O, H, Ta) ...... 323 8.3 Constructibility of roots of a quartic polynomial ...... 324 8.3.1 Factorization of a quartic into a product of two quadratics . . . . . 324 8.3.2 A quartic as a composite of two quadratics ...... 325 8.4 An example with quartic polynomials: (O, Hb,Hc) ...... 326

9 Cubic and quartic equations 329 9.1 Solution of cubic equations ...... 329 9.2 Quartic equations ...... 330

10 Construction problems III 333 10.1 Euler’s construction of a triangle from (O, H, I) ...... 333 10.2 Conic solution of the (O, H, I) problem ...... 335 10.2.1 Ruler and compass construction of circumcircle and incircle . . . . 335 10.2.2 The general case: intersections with a conic ...... 336 10.2.3 The case OI = IH ...... 338

11 Some triangle centers 401 11.1 The symmedian point and the Brocard axis ...... 401 11.2 The Euler reﬂection point and the Kiepert parabola ...... 401 CONTENTS v

11.3 The Apollonian circles and the isodynamic points ...... 402 11.4 The Fermat points and the Kiepert hyperbola ...... 402 11.5 The Steiner circum-ellipse and the Steiner point ...... 403 11.6 The Steiner inellipse ...... 403

12 Some construction in triangle geometry 405 12.1 Simson lines ...... 405 12.2 Line of reﬂections ...... 405 12.3 Isogonal conjugates ...... 405 12.4 Isotomic conjugates ...... 406 12.5 Inscribed conics ...... 407 12.6 The Steiner inellipse and the Kiepert hyperbola ...... 407 Chapter 1

Circles

1.1 A construction problem of tangent circles

Lemma 1.1. Given two points A and B, the point P on the line PQdividing P and Q in : = : = y·A+x·B the ratio AP BP x y is P x+y .

O1 O2

Given two circles O1(r1) and O2(r2), suppose there is a third circle K(ρ) tangent to A(a) internally at P and to B(b) externally at Q. Note that K divides O1P internally in the ratio O1K : KP = r1 − ρ : ρ, so that

ρ · O1 +(r1 − ρ)P K = . r1

Similarly, the same point K divides O2Q externally in the O2K : KQ = r2 + ρ : −ρ,so that −ρ · O2 +(r2 + ρ)Q K = . r2 Eliminating K from these two equations, and rearranging, we obtain

−r2(r1 − ρ)P + r1(r2 + ρ)Q r2 · O1 + r1 · O2 = . (r1 + r2)ρ r1 + r2 102 Circles

This equation shows that a point on the line PQ is the same as a point on the line O 1O2, which is the intersection of the lines PQ and O1O2. Note that the point on the line O1O2 depends only on the two circles O1(r1) and O2(r2). It is the point T+ which divides O1 and O2 in internally in the ratio O1T+ : T+O2 = r1 : r2. This is called the internal center of similitude of the two circles. It can be constructed as the intersection of the line O1O2 with the line joining the endpoints of a pair of oppositely parallel radii of the circles. From the equation

−r2(r1 − ρ)P + r1(r2 + ρ)Q = T+, (r1 + r2)ρ we note that each of P and Q determines the other, since the line PQ passes through T+. This leads to an easy construction of the point K as the intersection of the lines O1P and O2Q. From this, the circle K(ρ) can be constructed.

O1 O2 O1 O2 T+ T+

Exercise 1. Given three noncollinear points A, B, C, construct three mutually tangent circles with centers A, B, C.

2. Construct the two circles through a given point tangent to two parallel lines, assuming the point between the two lines.

3. Construct the two circles through a given point tangent to two concentric circles, assuming the point between the two circles.

4. Given two nonoverlapping circles O1(r1) and O2(r2), show how to construct a circle externally tangent to each of them.

5. Given two circles, one containing the other, and O2(r2) and a point between them, construct the two circles through the point tangent to the two circles. 1.1 A construction problem of tangent circles 103

Proposition 1.2. Let O1O2 = d. ( + )2 − 2 · = · r1 r2 d T+P T+Q r1r2 2 . (r1 + r2)

Proof. Applying the law of cosines to triangles PO1T+ and KO1O2,wehave 2 2 r1d 2 r1 + − T+P 2 2 2 r1+r2 (r1 − ρ) + d − (r2 + ρ) = . r1d 2r1 · 2(r1 − ρ)d r1+r2

From this, 2(( + )2 − 2) + 2 = r1 r1 r2 d · r2 ρ T+P 2 . (r1 + r2) r1 − ρ Similarly, 2(( + )2 − 2) − 2 = r2 r1 r2 d · r1 ρ T+Q 2 . (r1 + r2) r2 + ρ From these the result is clear.

Corollary 1.3. If two circles each tangent to both circles O1(r1) and O2(r2) are tangent to each other externally at a point T , then T lies on the circle, center T+ and radius the 2 2 (r1+r2) −d square root of r1r2 · 2 . (r1+r2)

P K1 T K T K H T

O1 O2 O1

T+ T+ O2

Construction of neighbors The neighbors of K(ρ) can be constructed as follows. Let T be an intersection of the circles (K) and (T+). (i) Construct the line TK, and mark the point H such that KH = r2 and TH = ρ + r2. (ii) Construct the perpendicular bisector of the segment O2H to intersect the line TK at K1.

The circle K(T ) is tangent to K(ρ), O1(r1) and O2(r2). 104 Circles

Exercise

1. In this exercise, we construct the two circles passing through a given point P tangent to the two given circles O1(r1) and O2(r2).

(a) Let a half-line through T+ intersect the two circles at A and B. Construct the circle through A, B, P , and let it intersect the line T+P at Q.

(b) Construct a circle through P and Q to intersect O1(r1), and let the line joining the intersections intersect T+P at T . (c) Construct a circle with center T and radius the square root of TP · TQ to intersect O1(r1) at A1 and A2. The two circles each passing through P , Q and one of A1, A2 are tangent to both O1(r1) and O2(r2).

1.2 The radical axis of two circles

A quadratic equation of the form

x2 + y2 +2gx +2fy + c =0 represents a circle, center (−g,−f) and radius the square root of g2 +f 2 −c. It is imaginary when this latter quantity is negative. Every circle equation can be normalized so that the coefﬁcients of x2 and y2 are both 1 (and there is no xy term). We call this the standard equation of a circle. The power of a point P with respect to a circle C(O, r) is the quantity OP 2 − r2.IfP is external to C(O, r), and T a point on the circle such that PT is tangent to the circle, then the power of P with respect to C is the squared length of PT.

Lemma 1.4. Let C be the circle with standard equation

f(x, y):=x2 + y2 +2gx +2fy + c =0.

For every point P =(u, v), f(u, v) is the power of P with respect to the circle.

The radical axis of two circles is the locus of points with equal powers with respect to the two circles. The radical axis of the two circles

2 2 x + y +2g1x +2f1y + c1 =0, 2 2 x + y +2g2x +2f2y + c2 =0 is clearly the straight line

2(g1 − g2)x +2(f1 − f2)y +(c1 − c2)=0. 1.3 Orthogonal circles 105

Construction If two circles intersect, their radical axis is clearly the line containing their intersections. In case of tangency, it is the common tangent at the intersection. Here is a construction of the radical axis in the general case.

Proposition 1.5. Given two circles C(O1,r1) and C(O2,r2), let P1, P2 be points on the respective circles and on the same side of the line O1O2, such that O1P1 and O2P2 are perpendicular to O1O2. If the perpendicular bisector of P1P2 intersects O1O2 at Q, and Q is the point on O1O2 such that O1Q = QO2 and O2Q = QO1, then the perpendicular to O1O2 at Q is the radical axis of the two circles.

O1 Q Q O2

Exercise 1. Prove that the radical axis of two nonconcentric circles is perpendicular to the line joining the centers of the circles.

1.3 Orthogonal circles

Two circles are orthogonal if they intersect at right angles (at each intersection). Proposition 1.6. The two circles with standard equations 2 2 x + y +2g1x +2f1y + c1 =0, 2 2 x + y +2g2x +2f2y + c2 =0 are orthogonal if and only if

2g1g2 +2f1f2 − c1 − c2 =0. Proof. The orthogonality condition is 2 2 2 2 0= (r1 + r2) − (g1 − g2) − (f1 − f2) 2 2 2 2 2 2 =(g1 + f1 − c1)+(g2 + f2 − c2) − (g1 − g2) − (f1 − f2) =2g1g2 +2f1f2 − c1 − c2. 106 Circles

Exercise 1. Construct a circle orthogonal to three given circles whose centers are not collinear.

2. Given three points A, B, C which form the vertices of an acute angled triangle, construct three mutually orthogonal circles with centers A, B, C.

Consider the possibility of a circle orthogonal to two given nonconcentric circles. We may assume the radical axis of the two given circles to be the y-axis. The standard equations of the circles are

2 2 x + y +2g1x + c0 =0, 2 2 x + y +2g2x + c0 =0.

If there is a circle with standard equation x2 + y2 +2gx +2fy + c =0orthogonal to both of these circles, then

2g1g − c0 − c =0,

2g2g − c0 − c =0.

Since g1 = g2, we must have g =0and c = −c0. 2 (1) If the circles intersect on the y-axis at (0, ±b), then c0 = −b , then the common orthogonal circle is x2 + y2 +2fy + b2 =0for some f satisfying f 2 ≥ b2. 2 (2) If the circles do not intersect, c0 must be positive, say, c0 = b . In this case, the common orthogonal circle is x2 + y2 +2fy − b2 =0. This always passes through the points (±b, 0).

1.4 Coaxial circles

A system of circles is coaxial if every pair of circles from the system have the same radical axis. Therefore, a coaxial system of circles is deﬁned by the radical axis and any one of the circles.

Example 1.1. Let c be a constant and λ a parameter. The equation x2 + y2 +2λx + c =0 deﬁnes a coaxial system of circles with centers on the x-axis. The common radical axis is the y-axis, the line x =0. (i) The circles cut the radical√ axis if c<0. In this case, the common points of the circles in the systems are (0, ± −c). 1.5 Miscellaneous exercises 107

=0 (ii) If c , the circles are all tangent to the y-axis. √ (iii) If c>0, the circles do not have common points. For λ = ± c, the two circles degenerate into two points, called the limiting points of the coaxial system.

Example 1.2. Let c =0be a ﬁxed number. The two systems of coaxial circles

x2 + y2 +2λy + c =0, x2 + y2 +2μx + c =0 are every circle in one system is orthogonal to every circle in the other system. These two systems of coaxial circles are said to be conjugate.

Exercise 1. Let c be a constant. Show that every circle from the coaxial system x2+y2+2λx+c = 0 is orthogonal to every circle in the coaxial system x2 + y2 +2μy − c =0.

2. A and B are two ﬁxed points, and P is a variable point such that AP = k · BP. (a) Show that the locus of P is a circle. (b) Show that for different values of k, the circles form a coaxial system with A and B as limiting points.

1.5 Miscellaneous exercises

1. Prove that, for all values of the constants p and q, the circle

(x − a)(x − a + p)+(y − b)(y − b + q)=r2

bisects the circumference of the circle

(x − a)2 +(y − b)2 = r2.

Make use of this to ﬁnd the equation of the circle which bisects the circumference of the circle x2 + y2 +2y − 3=0and touches the line x − y =0at the origin. 108 Circles

2. Find the equation of the circle which is orthogonal to all of the circles

x2 + y2 +3x +5y =0, x2 + y2 +2x +4y +3= 0, x2 + y2 +4x +5y − 1= 0.

3. Given a circle C with center O and a line , let A be the pedal of O on . For a variable point P on C , let the circle with diameter AP intersect the C (again) at X and (again) at Y . Show that the line XY passes through a ﬁxed point as P varies on C .

4. Let C1 and C2 be two circles on the same side of , none of them intersecting . Locate the point Q on for which the sum of the lengths of the tangents from Q to the circles is minimum.

5. Let P be a ﬁxed point outside the circle C (O, r). A line not containing P intersects the circle at two points X and Y such that PX· PY = OP 2 −r2. Show that passes through a ﬁxed point. 114 Circles

Replacement of p.114

2.2 Steiner chains theorem

Proposition 2.9. Given a coaxial system of non-intersecting circles with limiting points K 1 and K2, inversion in a circle with center at one of the limiting points transforms the circles into a system of concentric circles.

Proof. Let K and K be the limiting points of the coaxial system of circles. Consider two circles C1, C2 (in the conjugate coaxial system) passing through K and K . Their images are two lines 1 and 2 (through the inversive image Q of K ). Now, an arbitrary circle in the given system, being orthogonal to both C1 and C2, has its image orthogonal to both 1 and 2. This image must be a circle center center Q. This shows that the circles in the given system are transformed into concentric circles.

Remark. If the circle of inversion passes through K , then the images are concentric circles with center K.

K K

Theorem 2.10. (Steiner chains theorem). If two circles O1(r1) and O2(r2) are such that beginning with some point on (O1), a (steiner) chain of n circles can be constructed each tangent to its two immediate neighbors and the two given circles, then such a chain of n circles can be constructed beginning with any point on (O1).

Theorem 2.11. A Steiner chain of n circles can be constructed between two given circles O1(r1) and O2(r2) if and only if the distance d between their centers satisﬁes

2 2 2 π d =(r1 − r2) − 4r1r2 tan . n Proof. (1) Suppose the circles are concentric. A Steiner chain of n circles exists if and only if π r1 − r2 sin = . n r1 + r2 2.2 Steiner chains theorem 115

From this, 2 π 2 2 π sin (r1 − r2) (r1 − r2) tan2 = n = = . cos2 π ( + )2 − ( − )2 4 n n r1 r2 r1 r2 r1r2 (2) For the general case, we assume the given circles generate the coaxial system of circles x2 + y2 +2gx+ c2 =0. A typical circle in this system has center (−g,0) and radius r given by r2 = g2 − c2.

K 0 K −g −c −g + r c

Inversion in the circle K(K ) transforms the endpoint (−g + r, 0) of the circle into the − + 4c2 0 point c c−g+r , . The image circle has radius

4c2 2c(c + g − r) 2c(c + g) ρ = − 2c = = c − g + r c − g + r r as can be easily veriﬁed. (3) Suppose the given circles of radii r1 and r2 have centers (−g1, 0), (−g2, 0) respec- 2c(c+g1) 2c(c+g2) ρ1 = ρ2 = tively. The radii of their image circles are r1 and r2 . Here, ( − )2 (( + ) − ( + ) )2 ρ1 ρ2 = c g2 r1 c g1 r2 4ρ1ρ2 4r1r2(c + g1)(c + g2) 2( + )( + )( − 2 − ) = c g1 c g2 g1g2 c r1r2 4r1r2(c + g1)(c + g2) 2( − 2 − ) = g1g2 c r1r2 4r1r2 ( 2 + 2 − ( − )2 − 2 2 +( − )2 − ( 2 + 2) = g1 g2 g1 g2 c r1 r2 r1 r2 4r1r2 2 2 (r1 − r2) − (g1 − g2) = . 4r1r2 The two image circles are concentric at K. The result follows from (1), noting that the distance between the centers of the circles. 2.5 Feuerbach’s theorem 117

Appendix to Chapter 2: Feuerbach’s theorem and the Apollonius problem for the excircles

2.5 Feuerbach’s theorem

Theorem 2.7. The nine-point circle of a triangle is tangent internally to the incircle and externally to each of the excircles.

Ic F Cc Fb

Fc Bb I N

Ac B Aa C Ab Fa

Proof. Consider the incircle (I) and the A−excircle (Ia) of triangle ABC, tangent to BC at X and X respectively. These two points are symmetric with respect to the midpoint Ma of BC. Let T be the intersection of AIa and BC. It is well known that A and T divide I and Ia harmonically. Considering the projections of these points on the line BC, we conclude 2 that Ha and T divide X and X harmonically. By Proposition ?, MaHa · MaT = MaX .It follows that inversion in the circle C := Ma(X) interchanges Ha and T . Note that the circle of inversion C is orthogonal to each of the circles (I) and (Ia). These circles are therefore invariant under the inversion. Apart from the three sidelines, the circles (I) and (Ia) have a fourth common tangent L, which is the reﬂection of the line BC in AIa. The inversive image of L is a circle through Ma (the center of inversion) and Ha (the image of T ). We claim that this circle is actually the nine-point circle (N) through the midpoints of the three sides. To see that, note that L is parallel to the tangent to circumcircle at A. Since this latter line is orthogonal to OA, which is parallel to the line NMa (under a homothety at the centroid), L is orthogonal to the line MaN. 118 Circles

C B Ha X T Ma X

L Mc Mb O I N

T C B Ha Ma X

Therefore, the image of L is a circle orthogonal to the line MaN (which is invariant under the inversion). Since this circle passes through Ha and Ma, and has center on the line MaN, it must be the nine-point circle (N). From this we conclude that the nine-point circle is tangent to the incircle and the A- excircle.

The point of tangency with the incircle is on the line joining Ma to the reﬂection of X 2.5 Feuerbach’s theorem 119

Fe L

Mc Mb

T C B Ha X Ma X Fa

Ia in the angle bisector AI. This is called the Feuerbach point of the triangle. If the incircle touches CA at Y and AB at Z, the same Feuerbach point also lies on the lines joining Mb to the reflection of Y in BI, and Mc to the reflection of Z in CI. On the other hand, the line joining Ma to the reflection of X in AI intersects the A- excircle at the point of tangency with the nine-point circle. Similar constructions give the points of tangency of the nine-point circle with the other two excircles. 120 Circles

2.6 The Apollonius problem for the excircles

Proposition 2.8. The radical center of the three excircles is the Spieker center, the incenter of the inferior triangle MaMbMc.

Proof. The midpoint Ma of BC clearly has equal powers with respect to the B- and C- excircles. Since the line joining the excenters Ib and Ic is perpendicular to the bisector of angle A, the radical axis of the excircles is the parallel through Ma to the A-bisector. Equivalently, it is the Ma-bisector of the inferior triangle MaMbMc. Similarly, the radical axes of the C- and A- excircles, and that of the A- and B-excircles, are the bisectors of the Mb- and Mc bisectors of the same inferior triangle. The three radical axes are concurrent at the incenter of triangle MaMbMc.

Since the three excircles are mutually disjoint, the radical center Sp is a point exterior to each of them. There is a unique radical circle (Sp) orthogonal to each of the excircles.

Mc Mb

B Ma C

Ia Since the nine-point circle (N) is tangent externally to each of the excircles, inversion in the radical circle (Sp) gives a circle tangent internally to the excircle. If Fa is the point ( ) ( ) ( ) of tangency of N and Ia , the line SpFa intersects Ia at another point Fa. Similarly ( ) ( ) construct Fb and Fc on Ib and Ic , making use of the points of tangency Fb and Fc with ( ) N . The circle FaFbFc is tangent internally to the excircles. This is called the Apollonius circle of triangle ABC. 2.6 The Apollonius problem for the excircles 121

A Fa c F Fb

N Sp

B Fa C

Fb Each sideline is a common tangent of the three excircles. Inversion in the radical circle (Sp) yields three more circles through Sp, each tangent internally to two excircles and externally to the third. These are called the Bevan circles.

B C

Ia 122 Circles

Here are the eight circles each tangent to the three excircles.

Sp N

B C

Ia Chapter 3

Parabolas

3.1 Deﬁnitions

Given a point F (focus) and a line L (directrix), the locus of a point P which is equidistant from F and L is a parabola P. Let the distance between F and L be 2a. Set up a Cartesian coordinate system such that F =(a, 0) and L has equation x = −a. The origin O is clearly on the locus. Its distances from F and L are both a.

Q P (x, y)

directrix

− a O F =(a, 0)

Let P (x, y) be a point equidistant from F =(a, 0) and L. Then (x−a)2 +y2 =(x+a)2. Rearranging terms, we have y2 =4ax. 202 Parabolas

The parabola P can also be described parametrically: x = at2, y =2at.

We shall refer to the point P (t):=(at2, 2at) as the point on P with parameter t. It is the intersection of the line y =2at and the perpendicular bisector of FQ, Q =(−a, 2at) on the directrix.

Exercise 1. Find the equation of the perpendicular bisector of FP and show that it envelopes the parabola.

2. (Concyclic points on a parabola) Show that the points P (t1), P (t2), P (t3), P (t4) on 2 the parabola y =4ax are concyclic if and only if t1 + t2 + t3 + t4 =0. 0 x + y =1 3. Let a> be a constant. Show that the line t a−t envelopes a parabola, and locate the focus and the directrix.

3.2 Chords and tangents

The line joining the points P (t1) and P (t2) has equation

2x − (t1 + t2)y +2at1t2 =0.

Letting t1,t2 → t, we obtain the equation of the tangent at P (t):

x − ty + at2 =0.

( ) 1 (− 2 0) The tangent at P t has slope t and intersects the axis of the parabola at at , .It can be constructed as the perpendicular bisector of FQ, where Q is the pedal of P on the directrix.

Exercise

1. (Focal chord) The line joining two distinct points P (t1) and P (t2) on P passes through the focus F if and only if t1t2 +1=0.

2. (Intersection of two tangents) Show that the tangents at the points t1 and t2 on the 2 parabola y =4ax intersect at the point (at1t2,a(t1 + t2)). 3.2 Chords and tangents 203

3. Justify the construction suggested by the diagram below for the construction of the tangent at a point P on a parabola.

O F

4. (Tangents from a point to a parabola) Given a point P , construct the circle with diameter PF and let it intersect the tangent at the vertex at Q1 and Q2. Then the lines PQ1 and PQ2 are the tangents from P to the parabola.

O F

Q2 T2

5. Let P1P2 be a focal chord of a parabola P with midpoint M. The tangents at P1 and P2 intersect at Q. Show that (i) Q lies on the directrix, (ii) these tangents are perpendicular to each other, and (iii) QM is parallel to the axis of the parabola, and (iv) QM and intersects P at its own midpoint.

6. Justify the following construction of the chord of a parabola which has a given point M as its midpoint: Let Q be the pedal of M on the directrix. Construct the circle M(Q) to intersect the perpendicular from M to QF . These two intersections are on the parabola, and their midpoint is M. 204 Parabolas

L P1

Q M

O F

7. A circle on any focal chord of a parabola as diameter cuts the curve again at two points P and Q. Show that as the focal chord varies, the line PQ passes through a ﬁxed point.

8. (Chords orthogonal at the vertex) Let PQ be a chord of a parabola with vertex O such that angle POQis a right angle. Find the locus of the midpoint of PQ.

9. Find the locus of the point whose two tangents to the parabola y 2 =4ax make a given angle α.

10. PQ is a focal chord of a parabola with focus F . Construct the circles through F tangent to the parabola at P and Q. What is the locus of the second intersection of the circles (apart from F )?

3.3 Normals

The normal at P (t) of the parabola y2 =4ax is the perpendicular to the tangent at the same point. It has slope −t; its equation is

tx + y − (2at + at3)=0.

It intersects the parabola again at P (t), where 2 t = −t − . t 3.3 Normals 205

Q O F

t Exercise 1. (Reflection property) Prove the following reflection property of the parabola. Let P be a point on a parabola with focus F . The reflection of the focal chord FP in the normal at P is parallel to the axis of the parabola.

O F

2. (Orthogonal normals) Find the condition for the normals at the points t1 and t2 on the parabola y2 =4ax to be orthogonal. Hence, ﬁnd the locus of the intersection of the orthogonal normals.

3. (Normals intersecting on parabola) Find the condition on t1 and t2 so that the normals 2 to the parabola y =4ax at the points t1 and t2 intersect on the parabola. 4. Show that if the normals at A and B on the parabola y2 =4ax intersect at a point C on the parabola, then (a) the tangents at A and B intersect at a point on the line x =2a, (b) the centroid of the triangle ABC lies on the axis of the parabola.

5. (Three concurrent normals) Show that the normals at the points P (t1), P (t2), P (t3) 2 on the parabola y =4ax are concurrent if and only if t1 + t2 + t3 =0. 6. The normals at three points A, B, C on a parabola are concurrent. Show that the centroid of triangle ABC lies on the axis of the parabola. 206 Parabolas

3.4 Evolute of a parabola

2 The normals to the parabola P : y =4ax at the points t1 and t2 intersect at the point 2 2 (2a + a(t1 + t1t2 + t2, −at1t2(t1 + t2)). By putting t1 = t2 = t, we obtain the intersection of the normal at P and its “immediate neighbor”, namely,

Q(t)=(2a +3at2, −2at3).

The circle through P (t), with center Q(t), is the limiting position of a circle through P and two immediate neighbors. This is called the circle of curvature of the parabola at P .

N T O F K M

Suppose the tangent and normal at P intersect the axis at T and N respectively. Con- struct (1) the parallel through T to the normal, to intersect the ordinate through P (perpendicular to the axis) at M, (2) the parallel through M to the axis, to intersect the normal at K. The point K is the center of curvature of the parabola at P . The equation of the circle of curvature at t is

x2 + y2 − 2(2a +3at2)x +4at3y − 3a2t4 =0.

2 3 The radius of curvature is 2a(1 + t ) 2 . The locus of the center of curvature is called the evolute of the parabola. It is the semicubical parabola 27ay2 =(x − 2a)3.

P (t)

O 2 F Qa (t) 3.5 Triangle bounded by three tangents 207

Exercise 1. Find the minimum normal chord of the parabola y 2 =4ax.

2 2. Show that the normal to the parabola y =4x at the point t is tangent to its evolute − t ( ) and intersects it again at Q 2 (apart from Q t ).

P (t)

(− t ) Q 2 (− Ot ) F 2a P 2 Q(t)

3. (Pedal curve) Let A be a ﬁxed point. For a variable point P on the parabola y2 =4ax, let Q be the pedal of A on the tangent at P . (a) Find a parametrization of the locus of Q. (b) Describe this locus with reference to the evolute of the parabola.

3.5 Triangle bounded by three tangents

Theorem 3.1. The circumcircle of the triangle bounded by three tangents to a parabola passes through the focus of the parabola.

L3 P2

A2 F

P3 L2 208 Parabolas

=1 2 3 = − + 2 =0 Proof. For i , , , the tangent at ti is the line Li x tiy ati . We ﬁnd λ1, λ2, λ3 such that λ1L2 · L3 + λ2L3 · L1 + λ3L1 · L2 =0 represents a circle. This requires the coefﬁcients of x2 and y2 to be equal, and that of xy equal to 0:

(1 − t2t3)λ1 +(1− t3t1)λ2 +(1− t1t2)λ3 =0, (t2 + t3)λ1 +(t3 + t1)λ2 +(t1 + t2)λ3 =0. Solving these equations, we have 1 − t3t1 1 − t1t2 1 − t1t2 1 − t2t3 1 − t2t3 1 − t3t1 λ1 : λ2 : λ3 = : : t3 + t1 t1 + t2 t1 + t2 t2 + t3 t2 + t3 t3 + t1 2 2 2 =(t2 − t3)(1 + t1):(t3 − t1)(1 + t2):(t1 − t2)(1 + t3). 2 2 With these values of λ1, λ2, λ3, we compute the common coefﬁcient of x and y as 2 (t2 − t3)(1 + t1)=−(t2 − t3)(t3 − t1)(t1 − t2). Also, coefﬁcient of x 2 2 2 = (t2 − t3)(1 + t1) · a(t2 + t3)

= a(t2 − t3)(t3 − t1)(t1 − t2)(1 + t2t3 + t3t1 + t1t2), coefﬁcient of y 2 2 2 = − (t2 − t3)(1 + t1) · a(t2 + t3)

= a(t2 − t3)(t3 − t1)(t1 − t2)(t1 + t2 + t3 − t1t2t3), the constant term 2 2 2 2 = (t2 − t3)(1 + t1) · a t2t3 2 = − a (t2 − t3)(t3 − t1)(t1 − t2)(t2t3 + t3t1 + t1t2).

Cancelling a common factor −(t2 − t3)(t3 − t1)(t1 − t2), we obtain the equation of the circle as 2 2 2 x + y − a(1 + σ2)x − a(σ1 − σ3)y + a σ2 =0, where σ1, σ2, σ3 are the elementary symmetric functions of t1, t2, t3. This circle clearly passes through the focus F =(a, 0).

Exercise 1. Prove that the orthocenter of the triangle formed by three tangents to a parabola lies on the directrix. 2. Normals PX, PY, PZ are drawn to a parabola, focus F . Prove that the reﬂection of P in F lies on the circle circumscribing the triangle formed by the tangents at X, Y , Z. 3.6 Miscellaneous exercises 209

3.6 Miscellaneous exercises

1. (Parabolas with a common vertex and perpendicular axes) The two parabolas P : y2 =4ax and P : x2 =4by have a common vertex O and perpendicular axes. Let A be their common point other than O. The tangent at A to P intersects P at B and the tangent at A to P intersects P at C. Show that the line BC is a common tangent of the parabolas.

2. (Conformal focal parabolas) Two parabolas have a common focus F ; their directrices intersects at a point A. Show that the perpendicular bisector of AF is the common tangent of the two parabolas.

3. Tangents drawn to two confocal parabolas from a point on the common tangent intersect at the same angle as the axes of the parabolas.

4. (a) Find the condition for the two parabolas y2 =4ax and y2 =4b(x − c) to have a common tangent. (b) Construct the common tangent of the two parabolas.

5. Let A be a point on a circle (O), and t is the tangent at A. Triangle POQis such that P is on (O), Q is on t, and ∠POQ=90◦. Find the envelope of the perpendicular to AP through Q as triangle POQvaries.

6. From a point P normals are drawn to a parabola. Find the locus of P so that the intersections with the parabola form a right triangle.

7. TP and TQare tangents to a parabola. The circle TPQintersects the parabola again at P and Q. Prove that (i) the lines PQand P Q intersect on the directrix, (ii) if T is the pole of P Q, then TT is bisected at the focus. 210 Parabolas

Appendix: center of curvature More generally, consider the normal of C : x = f(t),y= g(t) at the point t, with equation given by f (t)(x − f(t)) + g(t)(y − g(t)) = 0. At a neighboring point t, the normal is given by the same equation with t replaced by t.If we write t = t+ ε for a small ε, and replace f(t) by f(t)+f (t)ε, f (t) by f (t)+f (t)ε, and similarly for g(t) and g(t), then the normal at this neighboring point is the line

f (t)(x − f(t)) + g(t)(y − g(t)) + ε(f (t)(x − f(t)) + g(t)(y − g(t)) =0.

Therefore, solving the system of equations

f (t)(x − f(t)) + g(t)(y − g(t)) = 0, f (t)(x − f (t)) + g(t)(y − g(t)) = 0, we obtain the intersection of two neighoring normals, namely, = ( ) − ( ) · f (t)2+g(t)2 x f t g t f (t)g(t)−f (t)g(t) , 2 2 (3.1) = ( )+ ( ) · f (t) +g (t) y g t f t f (t)g(t)−f (t)g(t) .

This is called the center of curvature at P (t). With this point as center, the circle through P is tangent to the curve C. The radius of this circle is 2 2 3 (f (t) + g (t) ) 2 ρ = . f (t)g(t) − f (t)g(t)

( ) := 1 The curvature at P t is κ ρ . Chapter 4

Ellipses and Hyperbolas

4.1 Ellipses

Ellipse with two given foci

Given two points F and F in a plane, the locus of point P for which the distances PF and PF have a constant sum is an ellipse with foci F and F . Assume FF =2c and the constant sum PF+PF =2a for a>c. Set up a coordinate system such that F =(c, 0) and F =(−c, 0). A point (x, y) is on the ellipse if and only if (x − c)2 + y2 + (x + c)2 + y2 =2a.

This can be reorganized as 1

x2 y2 + =1 with b2 := a2 − c2. (4.1) a2 b2

b P a

c a 2 2 O F − a A O F A a F c F c

1 2 2 2 2 From (x + c) + y =2a − (x − c) + y ,wehave 2 2 2 2 2 2 2 (x+ c) + y =4a +(x − c) + y − 4a (x − c) + y ; 4a (x − c)2 + y2 =4a2 − 4cx; a2((x − c)2 + y2)=(a2 − cx)2; (a2 − c2)x2 + a2y2 = a2(a2 − c2). Writing b2 := a2 − c2,wehaveb2x2 + a2y2 = a2b2. 212 Ellipses and Hyperbolas

The directrices and eccentricity of an ellipse

2 =( ) x2 + y =1 2 If P x, y is a point on the ellipse a2 b2 , its square distance from the focus F is 2 | | = c − a PF a x c .

2 2 − a A O F A a c F c

:= c = · − a Writing e a ,wehavePF e x e . The ellipse can also be regarded as the locus = a of point P whose distances from F (focus) and the line x e (directrix) bear a constant 2 | | = + a =(− 0) ratio e (the eccentricity). Similarly, PF e x c . The point F ae, and the = − a line x e form another pair of focus and directrix of the same ellipse.

The auxiliary circle and eccentric angle The circle with the major axis as diameter is called the auxiliary circle of the ellipse. Let P be a point on the ellipse. If the perpendicular from P to the major axis intersects the auxiliary circle at Q (on the same side of the major axis), we write Q =(a cos θ, a sin θ) and call θ the eccentric angle of P . In terms of θ, the coordinates of P are (a cos θ, b sin θ). This is a useful parametrization of the ellipse.

Q b

P Q

−a θ a O −c O c

−b

2 2 2 2 2 2 2 2 2 x b 2 2 2 c 2 2 PF =(x − c) + y =(x − c) + b 2 − 1 = 1+ 2 x − 2cx + c − b = 2 x − 2cx + a = a a a c 2 a x − a . 4.2 Tangents and normals of an ellipse 213

4.2 Tangents and normals of an ellipse

Construction of tangent at P

x1x y1y =( 1 1) + =1 If P x ,y , the tangent to the ellipse at P is the line a2 b2 . The corresponding 2 2 2 point Q on the auxiliary circle is Q =(x1,y2). The tangent to the circlex +y = a at Q 2 a2 1 + 2 = = 0 is the line x x y y a . Both tangents intersect the x-axis at T x1 , . Therefore, from the tangent of the circle (which is perperpendicular to the radius OQ), we have the point T . Then, PT is the tangent to the ellipse at P .

Q b

−a θ −c O c a T

−b

Remarks. (1) This construction can be reversed to locate the point of tangency on a line which is known to be tangent to an ellipse. (2) In terms of the eccentric angle of P , these tangents are

cos θ · x +sinθ · y =a, (circle) cos θ sin θ · x + · y =1, (ellipse). a b

Construction of tangents from an external point

P Q

F O F Q 214 Ellipses and Hyperbolas

Exercise 1. The circles with diameters PF and PF are tangent to the auxiliary circle at two points on the tangent to the ellipse at P .

b P

a F O F

2. Prove the following reﬂection property of the ellipse.

2 2 a A O F A a c F c

4.2.1 Evolute of an ellipse

2 x2 + y =1 Consider the ellipse a2 b2 with parametrization x = a cos θ, y = b sin θ, wherea>b(so that the foci are on the x-axis). The center of curvature at P = P (θ) is the a2−b2 · cos3 − a2−b2 · sin3 point a θ, b θ . It can be constructed as follows. (1) Let the tangent at P intersect the axes at X and Y . (2) Complete the rectangle OXQY . (3) Construct the perpendicular from O to the line PQ, to intersect the normal at K. The point K is the center of curvature at P . 4.2 Tangents and normals of an ellipse 215

Q Y

O X

The evolute is the parametric curve

a2 − b2 a2 − b2 x = · cos3 θ, y = − · sin3 θ. a b Eliminating t,wehave 2 2 4 (ax) 3 +(by) 3 = c 3 . This curve is called an astroid.

Q 216 Ellipses and Hyperbolas

4.3 Hyperbolas

Hyperbola with two given foci Given two points F and F in a plane, the locus of point P for which the distances PF and PF have a constant difference is a hyperbola with foci F and F .

c O a2 a F F c

Assume FF =2c and the constant difference |PF − PF| =2a for a

x2 y2 − =1 with b2 := c2 − a2. (4.2) a2 b2

The directrices and eccentricity of a hyperbola

2 x2 y If P =(x, y) is a point on the hyperbola 2 − 2 =1, its square distance from the focus a b 2 3 | | = c − a := c = · − a F is PF a x c . Writing e a ,wehavePF e x e . The hyperbola can = a also be regarded as the locus of point P whose distances from F (focus) and the line x e 2 1 | | = + a (directrix) bear a constant ratio e> (the eccentricity). Similarly, PF e x c . The =(− 0) = − a point F ae, and the line x e form another pair of focus and directrix of the same hyperbola. There is an easy parametrization of the hyperbola: P (θ)=(a sec θ, b tan θ). Given a point T on the auxiliary circle C (O, a), construct (i) the tangent to the circle at T to intersect the x-axis at A, (ii) the line x = b to intersect the line OT at B, (iii) the “vertical” line at A and the “horizontal” line at B to intersect at P . The point P has coordinates (a sec θ, b tan θ), and is a point on the hyperbola H . If Q is the pedal of T on the x-axis, then the line PQis tangent to the hyperbola. 2 2 2 3 2 2 2 2 2 x b 2 2 2 c 2 2 PF =(x − c) + y =(x − c) + b 1 − 2 = 1 − 2 x − 2cx + c + b = 2 x − 2cx + a = a a a c 2 a x − a . 4.4 Tangents and normals of a hyperbola 217

B P

θ −c O Q a b c A

4.4 Tangents and normals of a hyperbola

2 H : x2 − y =1 Let P be a point on the hyperbola a2 b2 . The circles with diameters PF1 and PF2 are tangent to the auxiliary circle C (O, a). The line joining the points of tangency is tangent to H at P .

a F O F

Exercise 1. If P and Q are on one branch of a hyperbola with foci F and G, show that F and G are on one branch of a hyperbola with foci P and Q.

2. Let P be a point on a hyperbola. The intersections of the tangent at P with the 218 Ellipses and Hyperbolas

asymptotes, and the intersection of the normal at P with the axes lie on a circle through the four points passes through the center of the hyperbola.

a F O F

a2+b2 sec a2+b2 tan The center of the circle is 2a θ, 2b θ . The locus is the hyperbola

4(a2x2 − b2y2)=(a2 + b2)2.

4.4.1 Evolute of a hyperbola

2 x2 − y =1 ( )=( sec tan ) For the hyperbola a2 b2 with parametrization x, y a t, b t , the evolute is the parametric curve a2 + b2 a2 + b2 x = · sec3 t, y = − · tan3 t. a b

O Q 4.5 Rectangular hyperbolas 219

4.5 Rectangular hyperbolas

2 Example 4.1. Four points on the rectangular hyperbola xy = c with parameters t1, t2, t3, t4 form an orthocentric system if and only if t1t2t3t4 = −1.

Proof. Consider three points on the hyperbola with parameters t1, t2, t3. The perpendiculars from each one of them to the line joining the other two are the lines

2 t1t2t3x − t1y + c(1 − t1t2t3)= 0, 2 t1t2t3x − t2y + c(1 − t1t2t3)= 0, 2 t1t2t3x − t3y + c(1 − t1t2t3)= 0. c 4 1 2 3 4 = −1 ( )= 4 If t is such that t t t t , then it is easy to see that x, y ct , t4 satisﬁes each of the above equations. Therefore, the orthocenter is the point with parameter t4 on the hyperbola.

2 Example 4.2. Four points on the rectangular hyperbola xy = c are with parameters t1, t2, t3, t4 are concyclic if and only if t1t2t3t4 =1.

Proof. If the four points are on a circle with equation

2 2 x + y +2gx +2fy + c0 =0 then t1, t2, t3, t4 are the roots of the equation

2 2 2 c 2cf c t + +2cgt + + c0 =0, t2 t or 2 4 3 2 2 c t +2cgt + c0t +2cft + c =0. = c2 =1 From this we conclude that t1t2t3t4 c2 . Example 4.3. PQis a diameter of a rectangular hyperbola. The circle, center P , passing through Q, intersects the hyperbola at three points which form the vertices of an equilateral triangle.

Proof. Let P and Q be the points with parameters τ and −τ. Since the points with parameters −τ, t1, t2, t3 are concyclic, −τ · t1t2t3 =1, and t1t2t3 · τ = −1. This means that P is the orthocenter of the triangle with vertices t1, t2, t3. This coincides with the circumcenter, and the triangle is equilateral.

4.6 A theorem on the tangents from a point to a conic

Theorem 4.1. Let P be the intersection of the tangents at T1 and T2 of an ellipse with foci F and F . The lines PF and PF make equal angles with the tangents PT1 and PT2. 220 Ellipses and Hyperbolas

F F

=( ) x1x + y1y =1 Proof. (1) Let T1 x1,y1 . The tangents at T1 and T2 are the lines a2 b2 and x2x + y2y =1 =( 0) a2 b2 . The distances from F c, to these tangents are

1 − cx1 a2 d1(F )= and 2 2 x1 + y1 a4 b4 1 − cx2 a2 d2(F )= . 2 2 x2 + y2 a4 b4 It follows that 2 2 2 x2 + y2 sin T1PF d1(F ) a − cx1 4 4 = = · a b . sin ( ) 2 − 2 2 FPT2 d2 F a cx2 x1 + y1 a4 b4 Similarly, for F =(−c, 0),wehave 2 2 2 x1 + y1 sin T2PF d2(F ) a + cx2 4 4 = = · a b . sin ( ) 2 + 2 2 F PT1 d1 F a cx1 x2 + y2 a4 b4

sin T1PF sin T2PF = (2) It is routine to verify that sin FPT2 sin F PT1 ; equivalently,

2 2 2 2 x1 y1 a − cx1 a + cx1 4 + 4 · = a b 2 2 2 2 . a − cx2 a + cx2 x2 + y2 a4 b4 This follows from 2 2 2 2 4 2 2 2 2 x y x 1 x a − c x (a − cx1)(a + cx1) 1 + 1 = 1 + 1 − 1 = 1 = , a4 b4 a4 b2 a2 a4b2 a4b2 4.7 Miscellaneous exercises 221

2 2 x2 + y2 and an analogous expression for a4 b4 . (3) For convenience, let ∠T1PT2 =Θ, ∠T1PF = θ and ∠T2PF = ϕ.Wehave sin θ sin ϕ = , sin(Θ − θ) sin(Θ − ϕ) sin(Θ − θ)sinϕ =sin(Θ− ϕ)sinθ, cos(Θ − θ − ϕ) − cos(Θ − θ + ϕ)= cos(Θ− θ − ϕ) − cos(Θ − ϕ + θ), cos(Θ − θ + ϕ)= cos(Θ− ϕ + θ).

Since Θ ± (θ − ϕ) <π, we conclude that Θ − θ + ϕ =Θ− ϕ + θ and θ = ϕ. Remark. Similar calculations show that the theorem is also true for hyperbolas and parabola. For a parabola, the line joining P to the “second” focus is parallel to the axis. Corollary 4.2. The foci of an inscribed conic of a triangle are isogonal conjugate points.

4.7 Miscellaneous exercises

2 ( ) E : x2 + y =1 1. (a) Find the equation of the normal at the point x0,y0 of the ellipse a2 b2 . x + y =1 E 2 2 + 2 2 = (b) Show that if the line p q is normal to the ellipse , then a p b q (a2 − b2)2. 2. An ellipse with semi-axes a and b slides on the positive x- and y-axes. Find the locus of its center.

2 E : x2 + y =1 3. (Director circle) The tangents to the ellipse a2 b2 with slope m are √ y = mx ± m2a2 + b2.

Deduce that the locus of points from which the two tangents to E are perpendicular is the circle x2 + y2 = a2 + b2.

2 = + E : x2 + y =1 4. (Pair of tangents) (a) Show that the line y mx c touches the ellipse a2 b2 if (and only if) c2 = m2a2 + b2. (b) Deduce that if (p, q) is outside the ellipse, and p2 = a2, then the two tangents from (p, q) to the ellipse are given by

(qx − py)2 = b2(x − p)2 + a2(y − q)2.

5. (Chords of ellipse subtending a right angle at the center) 2 x + y =1 E : x2 + y =1 The line p q intersects the ellipse a2 b2 at two points P and Q such ∠ = π that POQ 2 if and only if 1 1 1 1 + = + . p2 q2 a2 b2 222 Ellipses and Hyperbolas

Construct the circle, center O, tangent to a line joining two vertices of the ellipse (not on the same axis). A chord of the ellipse cut out by a tangent of this circle subtends a right angle at the center.

6. Let B and B be the endpoints of the minor axis of an ellipse. P and Q are points on the ellipse such that BP and BQ are perpendicular. Show that the lines BP and BQ intersect on a ﬁxed line.

√1 7. Given an ellipse of eccentricity > 2 , construct two circles with centers on the major axis, each passing through the center and tangent to the ellipse at two points.

8. Given an ellipse, construct an isosceles triangle with apex at a vertex on the minor axis and incircle concentric with the ellipse. 4.7 Miscellaneous exercises 223

9. In a right triangle of sides a and b, an ellipse of semi-axes h, k (parallel to the sides) is inscribed. Show that 2(a − h)(b − k)=ab.

C B 10. An ellipse is tangent to the four sides of a rectangle at their midpoints. Construct the circle tangent to the ellipse and two adjacent sides of the rectangle.

11. Find the locus of the vertices of a rectangular hyperbola with center O and tangent to the line x = a. Hint: Use polar coordinates.

12. If P and Q are on one branch of a hyperbola with foci F and G, show that F and G are on one branch of a hyperbola with foci P and Q.

13. Let P be a point on a hyperbola. The intersections of the tangent at P with the asymptotes, and the intersections of the normal at P with the axes lie on a circle through the center of the hyperbola.

a F O F Chapter 5

General conics

5.1 Classiﬁcation of conics

Given ﬁve generic points on the Cartesian plane, there is a unique second degree curve C : ax2 +2hxy + by2 +2gx +2fy + c =0 containing the points. We call this the conic determined by the 5 points. A straight line in the plane generically intersects the conic at two points, real or imaginary. When the two points coincide, the line is tangent to the conic. 2 Let Δ:=h − ab be the discriminant of the conic C. Δ =0 bg−hf −hg+af (1) We ﬁrst assume . By a translation of the origin to the point Δ , Δ , we may assume the conic given by

2 +2 + 2 − J =0 ax hxy by Δ , where ahg 2 2 2 J := abc +2fgh− af − bg − ch = hbf . gfc If J =0, the conic is a pair of straight lines, possibly imaginary. bg−hf −hg+af The point Δ , Δ is the center of the conic. It is clearly the midpoint of the chord cut out by a line through it. 2 2 We begin with a diagonalization of the quadratic form ax +2hxy+by . The symmetric ah matrix always has two real eigenvalues λ1 ≤ λ2 with orthogonal unit eigenvectors. hb The eigenvalues are the roots of the characteristic equation a − λh =0. hb− λ The unit eigenvectors (u, v) and (−v, u) are such that a − λ1 h u a − λ2 h −v =0 and =0. hb− λ1 v hb− λ2 u 226 General conics

These two equations can be combined into a single one: ah u −v u −v λ1 0 = , hb vu vu 0 λ2 and 2 2 ah x ax +2hxy + by = xy hb y u −v λ1 0 uv x = xy vu 0 λ2 −vu y 2 2 = λ1(ux + vy) + λ2(−vx + uy) . 2 Note that λ1λ2 = ab − h = −Δ. The equation of the conic becomes 2 2 λ1Δ(ux + vy) λ2Δ(−vx + uy) + =1. J J = −Δ Δ 0 Δ 0 λ1Δ Since λ1λ2 , this is a hyperbola if > , and an ellipse if < and J and λ2Δ (λ1+λ2)Δ = (a+b)Δ 0 ( + ) 0 J are both positive. This latter condition means J J > , i.e., a b J< . (2) If Δ=0, the above diagonalization still works, though one of the eigenvalues is zero. We write the equation of the conic in the form λ(ux + vy)2 +2gx +2fy + c =0, where λ =0and (u, v) is a unit vector. This can be further rewritten as λ(ux + vy)2 +2g(ux + vy)+2f (−vx + uy)+c =0. If f =0, the conic is a pair of parallel lines (possibly imaginary or coincident). If f =0, we further rewrite the conic equation in the form 2 g λ ux + vy + +2f (−vx + uy + s)=0, λ and recognize it as a parabola.

5.2 Pole and polar We shall identify a point P with coordinates (x, y) with the 1 × 3 matrix xy1 . The conic C can be represented by a symmetric 3 × 3 matrix ⎛ ⎞ ahg M = ⎝hbf⎠ gfc so that a point P lies on the conic if and only if PMPt =0. Two points P and Q are said to be conjugate with respect to C if PMQt =0. Clearly, given C and P , the points conjugate to P with respect to C form a line. This is the polar of P with respect to C. 5.3 Condition of tangency 227

Proposition 5.1. Let P and Q be conjugate with respect to C. If the line PQintersects the conic at two points, these points divide P and Q harmonically.

Proof. Given P and Q we seek a point λP + μQ with λ + μ =1lying on the conic, i.e.,

0=(λP + μQ)M(λP + μQ)t = λ2PMPt + μ2QMQt since PMQt = QMP t =0. If follows that if X = λP + μQ is one such point, then so is = λP −μQ Y λ−μ . These two points divide P and Q harmonically. Corollary 5.2. If the polar of P intersects C at X and Y , then PX and PY are tangent to C.

5.3 Condition of tangency

We ﬁnd the condition that a line px + qy + r =0is tangent to the conic C. Assuming q =0. Elimination of y from the equations of the line and the conic leads to the quadratic equation in x:

(aq2 − 2hpq + bp2)x2 +2(gq2 − hqr + brp − fpq)x +(br2 − 2fqr + cq2)=0.

This has a double root if and only if

(gq2 − hqr + brp − fpq)2 − (aq2 − 2hpq + bp2)(br2 − 2fqr + cq2)=0.

Equivalently,

(bc − f 2)p2 +(ac − g2)q2 +(ab − h2)r2 +2(fg − ch)pq +2(hf − bg)pr +2(gh − af)qr =0.

In matrix notation, we rewrite this condition as ⎛ ⎞ ⎛ ⎞ 2 bc − f fg − ch hf − bg p pqr⎝fg − ch ac − g2 gh − af⎠ ⎝q⎠ =0. hf − bg gh − af ab − h2 r

We denote⎛ ⎞ by adj(M) the 3 × 3 matrix in the above equation and call it the adjoint of p M.IfL = ⎝q⎠, the above tangency condition can be expressed as Ltadj(M)L =0. r With respect to the conic, (i) the polar of a point P is the line L = PM, (ii) the pole of a line L is the point P =adj(M)L (normalized so that the third component is 1). If P lies on the conic, its polar is tangent to the conic. If a line L is tangent to the conic, its point of tangency is P =adj(M)L. 228 General conics

5.4 Parametrized conics

Proposition 5.3. A parametrized curve of the form

2 2 a1t + b1t + c1 a2t + b2t + c2 x = ,y= at2 + bt + c at2 + bt + c is a conic.

Proof. Solving for t and t2,wehave

(ac2 − a2c)x +(a1c − ac1)y +(a2c1 − a1c2) t = , (a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1) (b2c − bc2)x +(bc1 − b1c)y +(b1c2 − b2c1) t2 = . (a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1)

It follows that

2 ((ac2 − a2c)x +(a1c − ac1)y +(a2c1 − a1c2))

=((b2c − bc2)x +(bc1 − b1c)y +(b1c2 − b2c1))

· ((a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1)).

Proposition 5.4. A family of lines

p(t)x + q(t)y + r(t)=0 in which p(t), q(t), r(t) are quadratic functions of t envelopes a conic.

Proof. The line intersects its “immediate neighbor” at the point

q(t)r(t) − r(t)q(t) r(t)p(t) − p(t)r(t) x = ,y= . p(t)q(t) − q(t)p(t) p(t)q(t) − q(t)p(t)

Here, both numerators and the common denominator are quadratic functions of t. The result follows from the preceding proposition. 5.5 Rectangular hyperbolas and the nine-point circle 229

5.5 Rectangular hyperbolas and the nine-point circle

Theorem 5.5. (a) A conic through the vertices and the orthocenter of a triangle is a rectangular hyperbola. (b) The locus of the centers of the rectangular hyperbolas through an orthocentric quadruple is the nine-point circle of the quadruple.

A(0,a)

N 0 − bc H , a

B(b, 0) O C(c, 0)

Proof. (a) Let A =(0,a), B =(b, 0), C =(c, 0). The orthocenter is the point H = 0 − bc , a . Consider a conic through these points, with equation

Ax2 +2hxy + By2 +2gx +2fy + C =0.

Since it contains B and C,

b2A +2bg + C =0, c2A +2cg + C =0.

From these, C (b + c)C A = ,g= − . bc 2bc Since it contains A and H,

a2B − 2af + C =0, b2c2B − 2abcf + a2C =0.

From these, C (a2 − bc)C B = − ,f= . bc 2abc 230 General conics

The equation of the conic is C C (b + c)C (a2 − bc)C · x2 +2hxy − · y2 − · x + · y + C =0. bc bc bc abc a Cx2 − 2bchxy + Cy2 + a(b + c)Cx − (a2 − bc)Cy − abcC =0. To determine the type of the conic, we diagonalize the quadratic form Cx2 − 2bchxy − Cy2. Since the eigenvalues of the symmetric matrix C −bch −bch −C √ are ± C2 − b2c2h2, this conic is a rectangular hyperbola. (b) Indeed, it is possible to ﬁnd the center of the rectangular hyperbola. By putting x = x + p and y = y + q, we transform the equation of the conic into

a(Cx2 − 2bchxy − Cy2) −a(2Cp +2bchq − (b + c)C)x +(−2abchp +2aCq − (a2 − bc)C)y +a Cp2 − 2bchpq − Cq2 + a(b + c)Cp − (a2 − bc)Cq − abcC =0. We choose p and q such that 2Cp +2bchq − (b + c)C =0, −2abchp +2aCq − (a2 − bc)C =0. The point (p, q) is the center of the rectangular hyperbola. Note that it is possible to eliminate C and h:

(2p − (b + c))C +2bch · q =0, (2aq − (a2 − bc))C − 2bch · ap =0. 2p − (b + c) q =0. 2aq − (a2 − bc) −ap

−2a(p2 + q2)+a(b + c)p +(a2 − bc)q =0. This means that the center (p, q) of the hyperbola lies on the circle + 2 − 2 + 2 − b c · + a bc · =0 x y 2 x 2 y . a b+c − a2−bc This is a circle with center 4 , 4a . It clearly passes through the points (0,0) and the midpoints of BC. = a − b − c =0 If we put y 2 , this becomes x 2 x 2 . Therefore, it passes through the b a c a midpoints 2 , 2 and 2 , 2 of AB and AC. This is the “nine-point circle” of the triangle. 5.5 Rectangular hyperbolas and the nine-point circle 231

Exercise 1. Each of the following conics is a pair of straight lines. Identify the lines and their intersection.

(a) x2 − 2xy − 3y2 − 2x − 6y − 3=0, (b) x2 − y2 − 4x +2y +3=0, (c) 3x2 +6xy +3y2 +2x +2y − 1=0.

2. Find the real intersections of the following pairs of conics.

(a) x2 +2xy − y2 +1=0and x2 +2xy + y2 − 2y =0, (b) x2 +2xy − y2 +1=0and x2 + y2 − 2x − 2y − 1=0, (c) x2 − 2xy − y2 − 1=0and x2 +2xy − y2 − 2y − 1=0, (d) x2 − y2 +1=0and x2 +2xy + y2 − 2x − 1=0, (e) x2 − y2 − 1=0and x2 +2xy + y2 +2x − 2y +1=0, (f) x2 +2xy − y2 +1=0and x2 + y2 − 2x − 1=0, (g) x2 − 2xy − y2 − 1=0and x2 − 2x − 2y =0.

3. Show that the angle ϕ between the pair of straight lines ax2 +2hxy + by2 =0is given by √ Δ tan ϕ = , a + b and that the pair of angle bisectors is given by

hx2 − (a − b)xy − hy2 =0.

x + y =1 4. Find the envelope of the line t a−t . 5. A variable line intersects the x- and y-axes at two points X and Y such that the segment XY has unit length. Is the envelope of the line a conic? x + y =1 6. Find the locus of the pole of the line a b with respect to the variable circles tangent to the coordinate axes and with center on the line x − y =0. Chapter 6

Basic triangle centers

6.1 The Euler line

6.1.1 The centroid Let E and F be the midpoints of AC and AB respectively, and G the intersection of the medians BE and CF. Construct the parallel through C to BE, and extend AG to intersect BC at D, and this parallel at H..

E F G

B D C

By the converse of the midpoint theorem, G is the midpoint of AH, and HC =2· GE Join BH. By the midpoint theorem, BH//CF . It follows that BHCG is a parallelogram. Therefore, D is the midpoint of (the diagonal) BC, and AD is also a median of triangle ABC. We have shown that the three medians of triangle ABC intersect at G, which we call the centroid of the triangle. Furthermore, AG =GH =2GD, BG =HC =2GE, CG =HB =2GF. The centroid G divides each median in the ratio 2:1. 302 Basic triangle centers

Exercise The centroid and the circumcenter of a triangle coincide if and only if the triangle is equilateral. Theorem 6.1 (Apollonius). Given triangle ABC, let D be the midpoint of BC. The length of the median AD is given by AB2 + AC2 =2(AD2 + BD2).

B D C Proof. Applying the law of cosines to triangles ABD and ACD, and noting that cos ADB = − cos ADC,wehave AB2 = AD2 + BD2 − 2AD · BD · cos ADB; AC2 = AD2 + CD2 − 2AD · CD · cos ADC, AC2 = AD2 + BD2 +2AD · BD · cos ADB. The result follows by adding the ﬁrst and the third lines.

C A B A

B C F E G

A B D C

The inferior triangle of ABC is the triangle DEF whose vertices are the midpoints of the sides BC, CA, AB. The two triangles share the same centroid G, and are homothetic at G with ratio −1:2. The superior triangle of ABC is the triangle ABC 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 ABC 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.

C A B

O G H

B C

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 ABC 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 BC. 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 Basic triangle centers

D Y F E G N O Z H E F B X D C

NX = ND. Similarly, NE = NY, and NF = NZ for the pedals of H on CA and AB respectively. This means that the circumcircle of DEF also contains X, Y , Z. (3) Let D, E, F be the midpoints of AH, BH, CH respectively. The triangle DEF is homothetic to ABC 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 ABC. Its center N is called the nine-point center. Its radius is half of the circumradius of ABC.

Exercise 1. Let H be the orthocenter of triangle ABC. Show that the Euler lines of triangles ABC, HBC, HCA and HAB 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 AP , BP, CP are all on the nine-point circle, must P be the orthocenter of triangle ABC?

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.

1Hint: ﬁnd a point common to them all. 2The Euler line is parallel (respectively perpendicular) to the bisector of angle A if and only if α = 120 ◦ (respectively 60◦). 6.2 The incircle and excircles 305

Let the bisectors of angles B and C intersect at I. Consider the pedals of I on the three sides. Since I is on the bisector of angle B, IX = IZ. Since I is also on the bisector of angle C, 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.

A A

s − a

Y Y

Z I Z I s − c

s − b

B X C B s − b X s − c C

This is called the incircle of triangle ABC, and I the incenter. Let s be the semiperimeter of triangle ABC. The incircle of triangle ABC touches its sides BC, CA, AB at X, Y , Z such that

AY =AZ = s − a, BZ =BX = s − b, CX =CY = s − c.

The inradius of triangle ABC is the radius of its incircle. It is given by 2Δ Δ r = = . a + b + c s

6.2.2 The excircles The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. An 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 Δ Δ Δ r = ,r= ,r= . a s − a b s − b c s − c 1 1 1 The areas of the triangles IaBC, IaCA, and IaAB are 2 ara, 2 bra, and 2 cra respectively. Since Δ=−ΔIaBC +ΔIaCA +ΔIaAB, 306 Basic triangle centers

X B C Y

ra Z ra

Ia we have 1 Δ= (− + + )= ( − ) 2ra a b c ra s a , = Δ from which ra s−a .

6.2.3 Heron’s formula for the area of a triangle

Consider a triangle ABC with area Δ. Denote by r the inradius, and ra the radius of the excircle on the side BC of triangle ABC. It is convenient to introduce the semiperimeter = 1 ( + + ) s 2 a b c .

ra r

A Y C Y

(1) From the similarity of triangles AIY and AI Y , r s − a = . ra s 6.2 The incircle and excircles 307

(2) From the similarity of triangles CIY and I CY ,

r · ra =(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). Δ= s(s − a)(s − b)(s − c).

Proof. Δ=rs. Proposition 6.4. α (s − b)(s − c) α s(s − a) α (s − b)(s − c) tan = , cos = , sin = . 2 s(s − a) 2 bc 2 bc

6.2.4 Euler’s formula: distances between circumcenter and tritangent centers Lemma 6.5. If the bisector of angle A intersects the circumcircle at M, then M is the center of the circle through B, I, C, and Ia.

O I

C B

Proof. (1) Since M is the midpoint of the arc BC, ∠MBC = ∠MCB = ∠MAB. There- fore, ∠MBI = ∠MBC + ∠CBI = ∠MAB + ∠IBA = ∠MIB, and MB = MI. Similarly, MC = MI. (2) On the other hand, since ∠IBIa and ICIa are both right angles, the four points B, I, C, IaM are concyclic, with center at the midpoint of IIA. This is the point M. 308 Basic triangle centers

Y O I

C B Y M ra

Theorem 6.6 (Euler). (a) OI2 = R2 − 2Rr. 2 = 2 +2 (b) OIa R Rra.

Proof. (a) Considering the power of I in the circumcircle, we have r α R2 − OI2 = AI · IM = AI · MB = · 2R · sin =2Rr. sin α 2 2

(b) Consider the power of Ia in the circumcircle. = ra = =2 sin α Note that IaA sin α . Also, IaM MB R 2 . 2

2 = 2 + · OIa R IaA IaM r α = R2 + a · 2R sin sin α 2 2 2 = R +2Rra.

Theorem 6.7 (Angle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If AX and AX respectively the internal and external bisectors of angle BAC, then BX : XC = c : b and BX : XC = c : −b.

Z c b

B X C X 6.2 The incircle and excircles 309

Proof. Construct lines through C parallel to the bisectors AX and AX to intersect the line AB at Z and Z. (1) Note that ∠AZC = ∠BAX = ∠XAC = ∠ACZ. This means AZ = AC. Clearly, BX : XC = BA : AZ = BA : AC = c : b. (2) Similarly, AZ = AC, and BX : XC = BA : AZ = BA : −AC = c : −b.

Proposition 6.8. (a) The lengths of the internal and external bisectors of angle A are respectively 2bc α 2bc α t = cos and t = sin . a b + c 2 a |b − c| 2

b c ta ta

B X C X

Proof. Let AX and AX be the bisectors of angle A. (1) Consider the area of triangle ABC as the sum of those of triangles AXC and ABX. We have 1 1 ( + )sinα = sin 2ta b c 2 2bc α. From this, bc sin α 2bc α t = · = · cos . a + sin α + 2 b c 2 b c (2) Consider the area of triangle as the difference between those of ABX and ACX.

2bc Remarks. (1) b+c is the harmonic mean of b and c. It can be constructed as follows. If the = = 2bc perpendicular to AX at X intersects AC and AB at Y and Z, then AY AZ b+c . A

ta c b Z

C B X

Y 310 Basic triangle centers

(2) Applying Stewart’s Theorem with λ = c and μ = ±b, we also obtain the following expressions for the lengths of the angle bisectors: 2 a t2 = bc 1 − , a b + c 2 a t 2 = bc − 1 . a b − c Chapter 7

Construction problems I

We study the problem of construction of a triangle A, B, C from three given associated points, including (i) the vertices, (ii) the triangle centers O, G, H, I, and (iii) the traces Ma, Mb, Mc of the centroid, Ha, Hb, Hc of the orthocenter, and Ta, Tb, Tc of the incenter. 1

A A A

Hb Tb Mc Mb Tc O G Hc H I

B Ma C B Ha C B Ta C

(o) Medians (p) Altitudes (q) Bisectors

The project, formulated by W. Wernick, 2 consists of 139 problems. With a few excep- tions, 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. [C] Not solvable by ruler and compass, but solvable by intersection of conics. [U] Unsolvable with ruler and compass, nor conics.

1 The traces Ma, Mb, Mc are also the pedals of the circumcenter O on the sides. 2Triangle constructions with three located points, Math. Mag., 55 (1982) 227–230. 312 Construction problems I

7.1 Classiﬁcation of construction problems

• (10) Two vertices and one center or trace

O G H I Ma Ha Ta Mb Hb Tb B,C L S S S R L L S L S

• (6) One vertex and two centers

(O, G) (O, H) (O, I) (G, I) (G, H) (H, I) A S S S S S S

• (24) One vertex, one center, and one trace

A Ma Ha Ta Mb Hb Tb O S S S L S S G R L S S S C H S L S S L C I S S L S S S

• (21) One vertex and two traces

(A, Ma,Mb) (A, Ha,Hb) (A, Ta,Tb) S S S

(A, Mb,Mc) (A, Hb,Hc) (A, Tb,Tc) S S S

(A, Ma,Ha) (A, Ma,Ta) (A, Ha,Ta) L S L

(A, Mb,Hb) (A, Mb,Tb) (A, Hb,Tb) L L L

(A, Ma,Hb) (A, Ma,Tb) (A, Ha,Tb) S C S

(A, Mb,Ha) (A, Mb,Ta) (A, Hb,Ta) L S S

(A, Mb,Hc) (A, Mb,Tc) (A, Hb,Tc) L S S 7.1 Classiﬁcation of construction problems 313

• (4) Three centers

(O, G, H) (O, G, I) (O, H, I) (G, H, I) R C C C

• (18) Two centers and one trace

(O, G) (O, H) (O, I) (G, H) (G, I) (H, I)

Ma S S S S Ha S S S Ta C C S C • (36) One center and two traces

O G H I

(Mb,Mc) S S S (Hb,Hc) C C S (Tb,Tc) S (Ma,Ha) L L L S (Ma,Ta) L S S S (Ha,Ta) S S L S

(Mb,Hc) S S S (Mb,Tc) (Hb,Tc) • (20) Three traces:

– (3) Three traces of the same kind

(Ma,Mb,Mc) (Ha,Hb,Hc) (Ta,Tb,Tc) S S C

– (12) Two traces of the same kind

(Ma,Mb,Hc) (Ma,Mb,Ha) (Ma,Mb,Tc) (Ma,Mb,Ta) S S C

(Ha,Hb,Mc) (Ha,Hb,Ma) (Ha,Hb,Tc) (Ha,Hb,Ta) L S

(Ta,Tb,Mc) (Ta,Tb,Mb) (Ta,Tb,Hc) (Ta,Tb,Ha)

– (5) Three traces of different kinds

(Ma,Ha,Ta) (Ma,Ha,Tb) (Ma,Hb,Ta) (Mb,Ha,Ta) (Ma,Hb,Tc) L S S 314 Construction problems I

7.2 Construction problems with easy geometric solutions

1. (B,C,G).

(a) Ma = midpoint of BC.

(b) A divides MaG in the ratio MaA : AG =3:−2. 2. (B,C,H).

(a) A = orthocenter of triangle BCH.

3. (B,C,I).

(a) B = reﬂection of B in BI. (b) C = reﬂection of C in CI. (c) A = BC ∩ CB.

4. (A, O, I).

(a) Construct the circumcircle O(A) with. (b) Construct 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) Construct the tangents from A to the incircle to intersect the circumcircle again at B and C.

5. (A, Ma,Hb).

(a) C = Ma(Hb) ∩ AHb.

(b) B divides CMa in the ratio 2:−1.

6. (H, Ma,Hb).

(a) Construct the circle Ma(Hb).

(b) Construct the line HHb to intersect the circle at B. (c) C is the antipodal point of B on the circle. ⊥ (d) A = CHb ∩ (BC) (H).

7. (Ma,Mb,Mc).

(a) ABC = triangle bounded by the parallel to MbMc through Ma, McMa through Mb, and MaMb through Mc.

8. (Ha,Hb,Hc).

(a) H = incenter of HaHbHc if this triangle is acute-angled, or the excenter corresponding to the obtuse angle vertex. 7.3 Construction problems with easy algebraic solutions 315

(b) ABC = the triangle bounded by the perpendiculars to HHa at Ha, HHb at Hb, and HHc at Hc.

9. (Ma, Hb, Ta).

(a) Construct the circle Ma(Hb) to intersect the line TaMa at B and C. (b) Construct the harmonic conjugate Ta of Ta with respect to BC. (c) Construct the circle with diameter TaTa to intersect the line CHb at A. There are two solutions corresponding to the choices of C.

10. (Ma, Hb, Tb).

(a) Construct the circle Ma(Hb).

(b) Construct the line HbTb to intersect the circle again C. (c) B is the antipodal point of C on the circle.

(d) Reﬂect BC in BTb to intersect the line HbTb at A.

7.3 Construction problems with easy algebraic solutions

1. (O, I, Ta).

We put the origin at I, and let Ta =(0, −a), O =(h, k). It is enough to determine the vertex A =(0,q). The altitude AH is the reﬂection of AO in the angle bisector AT . Since AO is the line (q−k)x+hy−hq =0, the altitude is the line (q−k)x−hy+hq = 0. The perpendicular from Ta to this line, i.e., hx +(q − k)y +(q − k)a =0, is the line BC. Let R and r be the circumradius and inradius, and d the distance between O and I.Wehave R2 = h2 +(q − k)2, a2(q − k)2 r2 = , h2 +(q − k)2 d2 = h2 + k2.

Note that Rr = a(q − k), and R2 − d2 = q(q − 2k). From Euler’s formula, we have q(q − 2k)=2a(q − k). This gives √ q = a + k + a2 + k2.

7.3.1 (O, Ha,Ta)

Put the origin at TA, Ha =(a, 0), and O =(h, k). The vertex A =(a, q) for some q. The reﬂection of O in ATa is the point a2h +2akq − hq2 −a2k +2ahq + kq2 , . a2 + q2 a2 + q2 316 Construction problems I

a2h+2akq−hq2 = Since this point lies on the altitude through A, a2+q2 a and

(a + h)q2 − 2akq + a2(a − h)=0.

From this, a √ q = k ± h2 + k2 − a2 . a + h

2. (I,Ma,Ha).

Put the origin at I and let Ma =(a, −k), Ha =(b, −k). The vertices B and C have coordinates (a + p, −k) and (a − p, −k) for some appropriate p. The incircle being x2 + y2 = k2, the tangents from B and C are the lines

2k(a + p)x +((a + p)2 − k2)y = k((a + p)2 + k2), 2k(a − p)x +((a − p)2 − k2)y = k((a − p)2 + k2).

From these we obtain 2ak2 (a2 − k2 − p2)k A = , . a2 + k2 − p2 a2 + k2 − p2

Since A lies on the perpendicular to y = −k at Ha =(b, −k),wehave 2ak2) = b. a2 + k2 − p2 From this, 2a p2 = a2 + k2 − · k2. b

3. (A, Tb,Tc).

Put the origin at A and let Tb =(−b, mb) and Tc =(c, mc). If the incenter is the point I =(0,t), then B = ATc ∩ TbI and C = ATb ∩ TcI are the points bt mbt −ct mct B = , and C = , . 2mb − t 2mb − t 2mc − t 2mc − t The line BC has equation

−m(b − c)tx +((b + c)t − 4mbc)y +2mbct =0.

The square distance distance from (0,t) is

((b + c)t − 2mbct)2 t2 = . m2(b − c)2 +((b + c)t − 4mbc)2 1+m2 From this we have

mt2 +(1− m2)(b + c)t − m(3 − m2)bc =0. 7.3 Construction problems with easy algebraic solutions 317

4. (A, Ta,Tb). =( 0) =( ) =( − ) Put the origin at A and let Ta a, , Tb h, k .IfTb h, k is the reﬂection of Tb in ATa, and I =(t, 0) is the incenter, then − = ∩ = ht kt B ATb TbI 2 − , 2 − , h t h t aht akt C = AT ∩ BT = , . b a (a +2h)t − 2ah (a +2h)t − 2ah The line BC has equation

ktx +((a + h)t − 2ah)y − akt =0.

Since the square distances from I to the lines ATb and BC are equal, (akt − kt2)2 (kt)2 = . k2t2 +((a + h)t − 2ah)2 h2 + k2

(a +2h)t2 − 2(2ah + h2 − k2)t + a(3h2 − k2)=0.

5. (O, Ha,Ta).

Put the origin at TA, Ha =(a, 0), and O =(h, k). The vertex A =(a, q) for some q. Since B and C are on the line x-axis, the midpoint of the arc BCis the intersection = qh of ATa and the vertical line through O. It is the point A h, a . Thus 2 qh − k =(h − a)2 +(k − q)2. a This gives (a + h)q2 − 2akq + a2(a − h)=0. From this, a √ q = k ± h2 + k2 − a2 . a + h

6. (A, G, I) and (A, I, Ma).

Put the origin at Ma, I =(a, 0) and A =(h, k). If B =(p, q), then C =(−p, −q). The lines BC, AB and AC 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 a2q2 (a(k + q) − (hq − kp)2 (a(k − q)+hq − kp)2 = = . p2 + q2 (k + q)2 +(h + p)2 (k − q)2 +(h − p)2 318 Construction problems I

Comparing the ﬁrst expression with the other two, we have

(kp − hq)(a2(2pq + hq + kp)+2a(p2 + q2)(k + q)+(p2 + q2)(kp − hq)) = 0, −(kp − hq)(a2(2pq − hq − kp)+2a(p2 + q2)(k − q) − (p2 + q2)(kp − hq)) = 0.

Clearly, kp − hq =0(otherwise, A, Ma and B are collinear). It follows that

a2(2pq + hq + kp)+2a(p2 + q2)(k + q)+(p2 + q2)(kp − hq)= 0, a2(2pq − hq − kp)+2a(p2 + q2)(k − q) − (p2 + q2)(kp − hq)= 0.

Addition gives apq + k(p2 + q2)=0. q = √k It follows that the slope of BC is p −a± a2−4k2 . From this, the line containing 2 B, C, and the incircle can be constructed. The two tangents from A to the incircle intersect this line at B and C. Chapter 8

Construction problems II

8.1 Polynomial roots as intersections of conics

Proposition 8.1. The roots of a polynomial of degree 3 or 4 can be constructed as the intersection of two conics.

3 2 Proof. (1) Assume a3 =0. The roots of f3(x):=a0x + a1x + a2x + a3 are the intersec- 2 tions of the parabola y = a0x + a1x + a2 and the rectangular hyperbola xy = a3. (2) Consider 4 3 2 f4(x):=a0x + a1x + a2x + a3x + a4

2 with a0 =0. Let y =2a0x + a1x.Wehave

2 4 3 2 2 2 2 4a0f4(x)= 4a0x +4a0a1x + a1x +(4a2 − a1)x +4a0a3x +4a0a4 2 2 2 = y +(4a2 − a1)x +4a0a3x +4a0a4.

2 It follows that the roots of f4(x) are the intersections of the parabola y =2a0x + a1x and the conic 2 2 2 (4a2 − a1)x + y +4a0a3x +4a0a4 =0.

2 2 2 Remark. The conic (4a2 − a1)x + y +4a0a3x +4a0a4 =0is an ellipse, a parabola, or a 2 hyperbola according as 4a2 is greater than, equal to, or less than a1.

Proposition 8.2. The roots of a cubic polynomial over Q are constructible with ruler and compass if and only if one of them is rational.

8.2 Examples with cubic polynomials 320 Construction problems II

8.2.1 (G, Hb,Hc)

1 Let Hb =(a, 0), Hc =(−a, 0), and G =(h, k).IfA =(p, q), then a(p2 − q2 − a2) 2a(p − a)q B = AH ∩ (AH )⊥(H )= , , c b b 2 + 2 − 2 2 + 2 − 2 p q a p q a −a(p2 − q2 − a2) −2a(p + a)q C = AH ∩ (AH )⊥(H )= , . b c c p2 + q2 − a2 p2 + q2 − a2

The centroid of triangle ABC is the point p (p2 + q2 − 5a2)q G = , . 3 3(p2 + q2 − a2)

This is the given point (h, k) if and only if p =3h and

q3 − 3kq2 +(9h2 − 5a2)q − 3k(9h2 − a2)=0.

The roots of this cubic equation is in general not constructible using ruler and compass. =1 ( )= 2 1 For example, if a and h, k 3 , , this is the cubic

q3 − 3q2 − q − 9=0.

Since this does not have rational roots, its three roots are not constructible with ruler and compass. However, the cubic can be solved by intersection of conics. Putting q = y +1,wehave 3 − 4 − 12 = 0 2 − 4= 12 ( ) y y ,ory y . If we set each equal to x, then x, y is the intersection of the parabola y2 = x +4and the rectangular hyperbola xy =12.

xy =12 A

Hc Hb y2 = x +4 C

1 The lines AHc and the perpendicular have equations −qx+(p+a)y = aq and (p−a)x+qy = a(p−a). 8.2 Examples with cubic polynomials 321

Exercise

1. Show that if a =3h, there are always two constructible solutions which are right triangles.

2. Let (a, h, k)=(3, 2, 2). Show that there is only one solution, and it is constructible. Compute the coordinates of the vertices.

3. Let (a, h, k)=(15, 2, 6). Show that there are three solutions, all constructible. Com- pute the coordinates of the vertices.

8.2.2 (A, Ma,Tb)

We put Tb at the origin, A =(−a, 0), Ma =(h, k). Let C =(p, 0). Then B =(2h−p, 2k). Since BA : BC = TbA : TbC,wehave

(2h − p + a)2 +4k2 :4(h − p)2 +4k2 = a2 : p2.

This gives

p3 − (3a +4h)p2 +4(h2 + k2 +2ah)p − 4a(h2 + k2)=0, after eliminating a linear factor p + a. Again, this is not constructible with ruler and compass, but can be obtained from the intersection of the conics

4ay = x2 − (3a +4h)x +4(h2 + k2 +2ah) and xy = h2 + k2.

Exercise

1. Give an explicit example of (a, h, k) with solutions not constructible with ruler and compass.

2. Let (a, h, k)=(2, 2, 1). Show that there are three real solutions none of which is constructible.

3. Let (a, h, k)=(2, 3, 1). Show that there are three constructible solutions. One of these has vertices with rational coordinates. Compute these coordinates.

4. Let (a, h, k)=(3, 1, 3). Show that there is one real solution, which is constructible. Compute the coordinates of the vertices.

5. Let (a, h, k)=(15, 4, 3). Show that there are three constructible solutions. Compute the coordinates of the vertices. 322 Construction problems II

8.2.3 (Ma,Mb,Tc)

Put Ma =(a, 0), Mb =(−a, 0), and Tc =(h, k). If we write C =(p, q), then A = (−2a − p, −q), B =(2a − p, −q). It follows that q = −k. By the angle bisector theorem, AC : BC = ATc : BTc gives 4(a + p)2 +4q2 :4(a − p)2 :4q2 =(2a + p + h)2 :(2a − p − h)2.

This simpliﬁes into

p3 − (2a2 + h2 − 2k2)p +2h(a2 + k2)=0.

The roots can be constructed by intersecting the rectangular hyperbola xy = a2 + k2 and the parabola 2hy =(2a2 + h2 − 2k2) − x2.

Exercise 1. Let (a, h, k)=(2, 1, 1). Show that there is one real solution, and that this is not constructible.

2. Let (a, h, k)=(3, 1, 1). Show that there are three real solutions, none of which is constructible. 3. Let (a, h, k)=(2, 1, 3). Show that there is one real solution, which is constructible. Compute the coordinates of the vertices.

8.2.4 (A, H, Tb)

Put the origin at A, and let Tb =(a, 0), H =(h, k).IfB =(h, q), then h2 + kq C = AT ∩ (AB)⊥(H)= , 0 . b h

The condition ATb : TbC = AB : BC reduces to (h2 + kq)(kq3 + h(h − 2a)q2 +(h2 − a2)kq + h2(h − a)2)=0.

If h2 + kq =0, then C =(0, 0) coincides with A, an impossibility. Therefore,

kq3 + h(h − 2a)q2 +(h2 − a2)kq + h2(h − a)2 =0.

If a =1, (h, k)=(2, 2), this gives q3 +3q +2=0, which has no rational roots.

Exercise 1. Let (a, h, k)=(1, 2, 2). Show that the roots are the intersections of the the rectangular hyperbola xy =1and the parabola y2 +2x +3=0. 2. Let (a, h, k)=(2, 1, 2). Show that there are three real solutions, but none of them is constructible. 8.2 Examples with cubic polynomials 323

3. Let (a, h, k)=(1, 2, 1). Show that there is one real solution, which is constructible. Compute the coordinates of the vertices. 4. Let (a, h, k)=(2, 1, 1). Show that there are three real solutions. One of these has vertices with rational coordinates. Compute the coordinates of the vertices.

8.2.5 (O, H, Ta) =(0− ) =( ) =( ) Put the origin at the circumcenter O, and let Ta , a , H h, k .IfA p, q , = h−p k−q Ma 2 , 2 . Since OMa and TaMa are perpendicular, Ma lies on the circle x2 + y2 + ay =0.

This means (p − h)2 +(q − k − a)2 = a2. (8.1) Therefore, A lies on the circle, center T =(h, k + a), radius a. This circle clearly passes through (h, k). Note that T is such that OTaHT is a parallelogram. Now, we rewrite the coordinates of A as (h + a cos θ, k + a + a sin θ). then the line joining A to Ta intersects the circle at p(p2 + a2 − a2) p2(q +2a)+q(q + a)2 A = − , − . p2 +(q + a)2 p2 +(q + a)2 Since A is the midpoint of the arc BC, OA and AH are parallel. This means p2(q +2a)+q(q + a)2 q − k = . p(p2 + q2 − a2) p − h This can be rearranged as ((k +2a)p − hq)(p2 + q2)=2a(hp2 − apq + hq2) − a2(kp + hq). (8.2) From this we see that the point (p, q) can be constructed by intersecting the circle (8.1) with a conic, since replacing p2 +q2 in the ﬁrst factor of (9.1) by (p−h)2 +(q−k−a)2 −a2, we need only adjust the right hand side, which remains a quadratic in p and q. For example, with (a, h, k)=(5, 6, 4), we have the circle (p − 6)2 +(q − 9)2 − 25 = 0 and the cubic 7p3 − 3p2q +7pq2 − 3q3 − 30p2 +25pq − 30q2 − 50p − 75q =0.

This can be rewritten as (7p − 3q)((p − 6)2 +(q − 9)2 − 25) = 54p2 + 115pq − 84q2 − 694p + 201q.

The circle and the hyperbola intersect, among other points, at the given orthocenter (6, 4) and A =(9, 13). The other two vertices are the points √ √ 3(−1 ± 91) −9 ∓ 91 2 , 2 . 324 Construction problems II

8.3 Constructibility of roots of a quartic polynomial

8.3.1 Factorization of a quartic into a product of two quadratics We seek a factorization of a general quartic

f(x)=x4 + ax3 + bx2 + cx + d in the form f(x)=(x2 + λx + μ)(x2 + λx + μ) Comparison of coefﬁcients gives

λ + λ =a, μ + μ =b − (λλ), λμ + λμ =c, μμ =d.

There is an easy matrix multiplication that leads to a simple elimination ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 110 1 λμ 2 λ + λ μ + μ ⎝λ λ 0⎠ ⎝1 λ μ⎠ = ⎝λ + λ 2λλ λμ + λμ⎠ . μ μ 0 00 0 μ + μ λμ + λμ 2μμ Passing to the determinants, we have 2 λ + λ μ + μ λ + λ 2λλ λμ + λ μ =0. μ + μ λμ + λμ 2μμ

This means 2 ab− λλ a 2λλ c =0. b − λλ c 2d

(λλ)3 − 2b(λλ)2 +(b2 + ac − 4d)(λλ)+(a2d − abc + c2)=0. By taking any (real) root of this cubic equation for λλ, we obtain the values of λ, λ, μ, μ. Example 8.1. x4 − 6x2 − 3x +4=0. 20−6 − w 2 02w −3 = −2(w +1)(w +11w +9)=0. −6 − w −38 With λλ = −1 and λ + λ =0, we take λ =1, λ = −1 and μ + μ =5, μ − μ = −3, μ =4, μ =1. Therefore, the quartic factors as

(x2 + x +4)(x2 − x +1). 8.3 Constructibility of roots of a quartic polynomial 325 √ 2 1 (−1 ± −15) The roots are ω, ω , and 2 . This condition is necessary but not sufﬁcient. For example, x4 − 2 has 20−w 2 02w 0 = −2w(w +8)=0. −w 0 −4

But with λλ =0,wehaveλ = λ =0. The quartic, does not factor as (x2 + a)(x2 + b) for rational a, b.

Exercise Factor the following quartics.

1. x4 +2x2 − x +2

2. x4 − 2x3 − x2 − 2x +1.

3. x4 +2x3 + x2 − 1

4. 2x4 + x2 − x +1

5. 2x4 + x3 + x2 + x +2

8.3.2 A quartic as a composite of two quadratics When is a quartic equation

4 3 2 f4(x):=a0x + a1x + a2x + a3x + a4 =0 solvable in two stages, each a quadratic equation? In other words, we seek y = px2 +qx+r such that the above equation becomes

Ay2 + By + C =0 for some A, B, C. It is enough to assume y = px2 + qx. Now, substitution gives

2 Ap = a0,

2Apq = a1, 2 Aq + Bp = a2,

Bq = a3.

From the ﬁrst two conditions, p :2q = a0 : a1. We may take p =2a0 and q = a1. Then, = 1 = a3 A 4a0 and B a1 . The third condition is equivalent to

3 a1 =4a0(a1a2 − 2a0a3). 326 Construction problems II

Exercise Find the roots of each of the following quartics by solving quadratic equations.

1. x4 − 4x3 +3x2 +2x +1.

2. x4 − 4x3 +3x2 +2x − 1.

3. x4 − 4x3 +3x2 +2x +2.

4. 2x4 − 4x3 +4x2 − 2x +2.

5. 2x4 − 4x3 +4x2 − 2x +1.

8.4 An example with quartic polynomials: (O, Hb,Hc)

Consider the problem of construction of triangle ABC from (O, Hb,Hc). Put the origin at the midpoint of HbHc so that Hb =(a, 0) and Hc =(−a, 0). Let the circumcenter be O =(h, k). Since the nine-point center is equidistant from Hb and Hc,it is on the y-axis. It follows that the orthocenter H has coordinates (−h, q) for some q.Now, a(q2 − h2 + a2) 2a(h − a)q B = HH ∩ (HH )⊥(H )= , , b c c 2 − 2 − 2 2 − 2 − 2 q h a q h a −a(q2 − h2 + a2) −2a(h + a)q C = HH ∩ (HH )⊥(H )= , . c b b q2 − h2 − a2 q2 − h2 − a2

These two points are equidistant from the circumcenter. This gives

F (q):=q4 +2kq3 +4a2q2 +2(h2 − a2)kq − (h2 − a2)2 =0. = h2−a2 Note that A h, q .

(a, h, k) F (q) (i)(2, 1, 2) (q − 1)(q3 +5q2 +21q +9) (ii)(2, 1, 5) (a2 +6q − 9)(q2 +4q +1) (iii)(3, 5, 2) q4 +4q3 +36q2 +64q − 256 In (ii), there are two constructible solutions, corresponding to the real roots of q 2 +6q − 9. In (iii), the roots can be expressed in terms of square roots: If z = q2 +2q, then the quartic becomes z2 +32z − 256. Therefore, the roots are √ q = −1 ± 4 −1 ± 2.

Exactly two of these are real. Therefore there are two real solutions which are constructible. 8.4 An example with quartic polynomials: (O, Hb,Hc) 327

Exercise 1. Let (a, h, k)=(2, 2, 5). There are two nondegenerate solutions with rational coordinates of the vertices. Compute these coordinates.

2. Let (a, h, k)=(2, 2, 4). Show that there is only one real solution, and it is constructible.

3. Let (a, h, k)=(2, 2, 1). Show that there is no real solution.

4. Let (a, h, k)=(2, 1, 6). Show that there are 4 real solutions. Only one of these is constructible.

5. Let (a, h, k)=(9, 1, 22). Show that there are 4 constructible solutions. Chapter 9

Cubic and quartic equations

9.1 Solution of cubic equations

A cubic equation x3 + ax2 + bx + c =0 (9.1) can be transformed into y3 + py + q =0 (9.2) = − a = + by putting x y 3 . To solve (9.2) we choose u and v appropriately so that y u v satisﬁes the equation. Substitution gives

u3 + v3 +(3uv + p)(u + v)+q =0.

Therefore, if we set 3uv + p =0, then u3 + v3 = −q. This means that u3 and v3 are the roots of the quadratic equation 3 2 + − p =0 t qt 27 . := p3 + q2 The discriminant of this quadratic in t is D 27 4 . √ √ 3 3 (i) If D>0, there are two real roots t1 and t2, giving a real root t1 + t2 of (9.2). If ω is an imaginary cube root of unity, then √ √ 3 3 2 u = t1 · ω, v = t2 · ω and √ √ 3 2 3 u = t1 · ω ,v= t2 · ω also satisfy 3uv + p =0. These gives two more roots of the equation. Thus, the three roots of (9.2) are √ √ 3 3 y1 = t1 + t2, √ √ 3 3 2 y2 = t1 · ω + t2 · ω , √ √ 3 2 3 y3 = t1 · ω + t2 · ω. 330 Cubic and quartic equations

(ii) If D =0, then we may take t1 = t2 = t. In this case, we obtain three real roots √ √ 3 3 y1 =2 t, y2 = y3 = − t. (iii) If D<0, the three roots are all real, and they can be conveniently expressed in terms of trisection of an angle. In fact, (9.2) can be modeled on the trigonometric identity cos 3θ =4cos3 θ − 3cosθ. Let y = r cos θ and rewrite (9.2) as r3 cos3 θ + pr cos θ + q =0. 3 : =4:−3 =2 − p If we choose r and θ so that r pr , i.e., r 3 , this becomes cos 3θ =4 cos3 θ − 3cosθ 4 = ( 3 cos3 + cos ) 3 r θ pr θ r −4q 3q 3 = = · − < 1, r3 2p p since D<0.Ifα is an angle satisfying this trigonometric equation, then the three real roots of (9.2) are 2 =2 −p cos + kπ =0 1 2 yk 3 α 3 ,k , , .

Exercise 1. y3 − 6y +9=0. 1 2. y3 +12y +12=0. 2 3. y3 − 3y +1=0. 3

9.2 Quartic equations

Every quartic equation can be reduced to the standard form x4 + ax2 + bx + c =0. (9.3) To solve the equation (9.3), we ﬁnd a number t so that when the quartic is written in the ( 2 + t )2 − ( − ) 2 + − ( t2 − ) ( − ) 2 − +(t2 − ) form x 2 t a x bx 4 c , the quadratic t a x bx 4 c is the square of a linear polynomial in x. This requires (−b)2 − (t − a)(t2 − 4c)=0,or t3 − at2 − 4ct − (b2 − 4ac)=0. If this auxiliary cubic has a rational root, then it leads to factorization of the quartic into the product of two quadratics.

1−3,√1+ω,√1 − ω√. √ √ √ 2−2 3 3+ 3 4, −2 3 3ω + 3 4ω2, −2 3 3ω2 + 3 4ω 32sin10◦, 2sin50◦, and −2sin70◦. 9.2 Quartic equations 331

Example 9.1. x4 −5x2 +2x+3 = 0. The auxiliary cubic equation is t3 +5t2 −12t−64 = 0. This has a root t = −4. 4 Therefore, the quartic can be rearranged in the form 0= (x2 − 2)2 − (x2 − 2x +1)=(x2 − 2)2 − (x − 1)2 =(x2 − x − 1)(x2 + x − 3). From this we have √ √ 1 ± 5 −1 ± 13 = x 2 , 2 . Example 9.2. For x4 −x2 +4x+5 = 0, the auxiliary cubic equation is (t+2)(t2 −t−18) = 0. With t = −2,wehave

x4 − x2 +4x +5=(x2 − 1)2 +(x +2)2 =((x2 − 1) + i(x + 2))(x2 − 1) − i(x +2)) =(x2 + ix − (1 − 2i))(x2 − ix − (1 + 2i)) =0.

This gives √ √ − ± 3 − 8 ± 3+8 = i i i i x 2 , 2 .

Exercise Solve the quartic equations.

1. x4 − 5x2 +2x +3=0.

2. x4 − 4x2 +5x − 4=0.

3. x4 − 3x2 +4x − 3=0.

4. x4 − x2 +2x +2=0.

5. x4 +3x − 2=0.

6. x4 + x3 +4x − 3=0.

4It factors as (t +4)(t2 + t − 16) = 0. Chapter 10

Construction problems III

10.1 Euler’s construction of a triangle from (O, H, I)

Given OI = d, HI = e, and OH = k, Euler sought the sidelengths BC = a, CA = b, and AB = c as a roots of a cubic polynomial. If p := a + b + c, q := ab + bc + ca, r := abc, then a, b, c are the roots of the cubic equation z3 − pz2 + qz − r =0.

H 2g

G e h g

I f O Denote by R and ρ the circumradius and the inradius respectively. It is more convenient to replace orthocenter H by the nine-point center N. By Feuerbach’s theorem, n := NI = R − := 2 = 2 − 2 2 ρ. Also, by Euler’s formula, d OI is given by d R Rρ. In terms of d and n, we have d2 d2 − 4n2 R = ,ρ= . 2n 4n We relate p and q to R and ρ by the following Proposition 10.1. (a) −p2 +4q =4ρ2 +16Rρ, (b) p2 − 2q =9R2 − 4m2. Proof. (a) Let Δ be the area of the triangle. By Heron’s formula, 16Δ2 =(a + b + c)(b + c − a)(c + a − b)(a + b − c) = p(p − 2a)(p − 2b)(p − 2c) = p(p3 − 2(a + b + c)p2 +4(ab + bc + ca)p − 8abc) = p(p3 − 2p3 +4pq − 8r) = p(−p3 +4pq − 8r). 334 Construction problems III

Now, r = abc =4ΔR =(2pρ)R. It follows that 4p2ρ2 = p(−p3 +4pq − 16pRρ). (a) follows by cancellation of a common factor p2. (b) Applying the law of cosines to triangle AOH, noting that AH =2R cos A and ∠OAH = A − 2∠BAH = A − 2(90◦ − B)=B − C,wehave

OH2 = R2 +4R2 cos2 A − 4R2 cos A cos(B − C) = R2 − 4R2 cos A(cos(B + C)+cos(B − C)) = R2 − 2R2(cos(A + B + C)+cos(B + C − A)+cos(A − B + C)+cos(A + B − C)) = R2 +2R2(1 + cos 2A +cos2B +cos2C) = R2 +2R2(4 − 2sin2 A − 2sin2 B − 2sin2 C) =9R2 − 4R2(sin2 A +sin2 B +sin2 C) =9R2 − (a2 + b2 + c2).

It follows that p2 − 2q = a2 + b2 + c2 =9R2 − 4m2. Solving the two equations we have

27d4 − 40d2n2 +16n4 − 32m2n2 p2 =2(9R2 +8Rρ +2ρ2 − 4m2)= 4n2 1 9d4 − 20d2n2 +8n4 − 8m2n2 q = (9R2 +16Rρ +4ρ2 − 4m2)= . 2 4n2 From r =2Rρ · p,wehave d2(d2 − 4n2) r = · p. 4n2 In terms of d, e, and k,wehave

11d4 +4e4 +3k4 − 8e2k2 +2d2k2 − 12d2e2 p2 = , 2d2 +2e2 − k2 d4 +2e4 + k4 − 3e2k2 +2d2k2 − 6d2e2 q = , 2d2 +2e2 − k2 d2(−d2 − 2e2 + k2) r = · p. 2d2 +2e2 − k2 √ √ For√ example, Euler set d = OI = 2, e = HI = 3 and k =√OH =3. (Equivalently,√ = 2 = 3 = 1 =2 = 1 = 71 =22 =2 71 d , m 2 , and n 2 ). From these, R , ρ 2 , and p , q , r . The sidelengths a, b, c are the roots of the cubic equation √ √ x3 − 71x2 +22x − 2 71 = 0. √ = y+ 71 If we put x 3 , this becomes √ y3 − 15y +2 71 = 0. 10.2 Conic solution of the (O, H, I) problem 335 √ By putting y =2 5cosθ, this equation is transformed into √ √ 5 5(4 cos3 θ − 3cosθ)=− 71. or 71 cos 3 = − θ 125. = arccos 71 ≈ 41◦530 With α 125 , Euler gave the sidelengths as √ √ √ √ 71 − 2 5cos α 71 + 2 5cos 60◦ ± α 3 3 3 , 3 ,

Exercise Find the sidelengths of the triangle in the following cases. √ √ 1. (d, e, k)=( 3, 2, 3). √ √ 2. (d, e, k)=( 5, 2 2, 5). √ √ 3. (d, e, k)=(2 2, 2, 4). √ √ 4. (d, e, k)=( 7, 7, 5). √ 5. (d, m, n)=(2 3, 3, 1).

6. (d, m, n)=(4, 4, 1).

10.2 Conic solution of the (O, H, I) problem

10.2.1 Ruler and compass construction of circumcircle and incircle Let G and N be the points which divide OH in the ratio

OG : GN : NH =2:1:3.

These are the centroid and the nine-point center of the required triangle (if it exists). A necessary and sufﬁcient condition for the existence of ABC is given by the following theorem. Theorem 10.2 (Guinand). Let D be the open circular disk with diameter HG. A triangle ABC exists with circumcenter O, orthocenter H, and incenter I if and only if I ∈ D\{N}. Let R and ρ denote respectively the circumradius and inradius of triangle ABC. From 2 = ( − 2 ) = R − Euler’s formula OI R R r and Feuerbach’s theorem NI 2 r,wehave 2R · NI = OI2. This suggests the following easy ruler-and-compass construction of the circumcircle, nine-point circle, and incircle. 336 Construction problems III

Construction Suppose O, H, I satisfy I ∈ D \{N}. (1) Extend OI to X such that OX =2· OI; construct the circle ONX, and extend NI to intersect the circle again at Y . The length of IY is twice the circumradius, and four times the radius of the nine-point circle. (2) Construct the circumcircle (O) and the nine-point circle (N). (3) Construct the intersection F of the circle (N) with the half line NI, and the circle, center I, passing through F . This is the incircle and F is the (Feuerbach) point of tangency with the nine-point circle.

X I F

O N H

Figure 10.1: Incircle from O, H, I

10.2.2 The general case: intersections with a conic Set up a cartesian coordinate system with origin at O. Assume H =(k, 0) and I =(p, q). Since ∠HAI = ∠OAI, and likewise for B and C, the vertices A, B, C are on the locus of the point P for which ∠HPI = ∠OPI.IfP =(x, y), a routine calculation shows that the locus of P is a curve K: K(x, y)=0, where

K(x, y):=2qx3 − (2p − k)x2y +2qxy2 − (2p − k)y3 −2(p + k)qx2 +2(p2 − q2)xy +2(p − k)qy2 +2kpqx − k(p2 − q2)y. Note that K(k, 0) = 0, i.e., K contains the point H. By computing the circumradius R, we easily obtain the equation of the circumcircle (O): G(x, y)=0, where (p2 + q2)2 G(x, y):=x2 + y2 − . (2p − k)2 +4q2 It is possible to ﬁnd a linear function L(x, y) such that Q(x, y):=K(x, y) − L(x, y)G(x, y) 10.2 Conic solution of the (O, H, I) problem 337 does not contain third degree terms. For example, by choosing L(x, y):=2qx − (2p − k)y − 2qk, we have

Q(x, y)=−2pqx2 +2(p2 − q2)xy +2pqy2 k2(k − 4p)(p2 − q2)+k(3p2 − 5q2)(p2 + q2)+2p(p2 + q2)2 − · x (2p − k)2 +4q2 2q(kp((2p − k)2 +4q2)+(p2 + q2)2) + · y (2p − k)2 +4q2 2k(p2 + q2)2q − . (2p − k)2 +4q2

The finite intersections of (O) with K are precisely the same with the conic C defined by Q(x, y)=0. Note that the coefficients of x and y in L are dictated by the elimination of the third degree terms in K−L·G. We have chosen the constant term such that L(k, 0) = 0, so that L(x, y)=0represents the line HP parallel to NI. It follows that Q(k, 0) = 0, and the conic C contains the vertices and the orthocenter of the required triangle ABC.It is necessarily a rectangular hyperbola. This fact also follows from the factorization of the quadratic part of Q, namely, −2(px + qy)(qx − py). This means that C is a rectangular q − p hyperbola whose asymptotes have slopes p and q . These are parallel and perpendicular to the segment OI. To construct the rectangular hyperbola C, we identify its center O. This is the point with coordinates (u, v) for which the quadratic polynomial Q(x − u, y − v) has no first degree terms in x and y. A routine calculation gives k((2p − k)2 − p2 +5q2)+2p(p2 + q2) (p2 + q2 − kp)q O = , 2((2 − )2 +4 2) (2 − )2 +4 2 p k q p k q k (2p − k)(p2 + q2)+2kq2 (p2 + q2)q − kpq = + , . (10.1) 2 2((2p − k)2 +4q2) (2p − k)2 +4q2 Since C is a rectangular hyperbola, its center O lies on the nine-point circle (N). The following observation leads to a very simple construction of the center. Proposition 10.3. The Feuerbach point F lies on the asymptote perpendicular to OI. Proof. The nine-point circle has equation 2 k (p2 + q2)2 x − + y2 − =0. (10.2) 2 4((2p − k)2 +4q2) This intersects the line NI at two points, the Feuerbach point and its antipode (on the nine-point circle). The Feuerbach point is the point k (2p − k)(p2 + q2) (p2 + q2)q F = + , . 2 2((2p − k)2 +4q2) (2p − k)2 +4q2 − p It is easy to see that the line O F has slope q , and is perpendicular to the line OI. 338 Construction problems III

Corollary 10.4. The center O of the rectangular hyperbola C is the second intersection of the nine-point circle (N) with the perpendicular from F to OI.

F I

O N H O F

Figure 10.2: Center O of rectangular hyperbola C

It is well known that if C is a rectangular hyperbola passing through the vertices of a triangle ABC, its fourth intersection with the circumcircle is the reﬂection of the orthocenter of the triangle in the center of the hyperbola. Therefore, one of the intersections of C with the circle (O) is the reﬂection of H in O. The other three are the vertices of the required triangle.

F I

O N H O F C

Figure 10.3: Triangle ABC with given O, H, I

10.2.3 The case OI = IH If OI = IH, consider the intersection A of the half line NI with the circumcircle. We complete a triangle ABC with (O) as circumcircle and (I) as incircle. The orthocenter of triangle ABC lies on the reﬂection of AO in the line AI. This is the line AH. Since angle = 1 · = 1 · = R AHN is a right angle, for the midpoint M of AH,wehaveNM 2 AH 2 AO 2 . This means that M is a point on the nine-point circle. It is the midpoint of the segment joining A to the orthocenter. It follows that the orthocenter must be H. 10.2 Conic solution of the (O, H, I) problem 339

X I N C O H

B Figure 10.4: Triangle from O, H, I with OI = IH Chapter 11

Some triangle centers

11.1 The symmedian point and the Brocard axis

Given a triangle ABC, construct the tangents tA, tB and tC to the circumcircle at the ∗ ∗ ∗ ∗ ∗ vertices. These tangents bound the tangential triangle A B C with A = tB ∩ tC , B = ∗ ∗ ∗ ∗ tC ∩ tA and C = tA ∩ tB. The lines AA , BB , CC are concurrent. The point of concurrence K is called the symmedian point of the triangle. The line joining the circumcenter and the symmedian point is called the Brocard axis.

Exercise 1. Construct the lines through K parallel to the sidelines. Let the parallel to BC intersect AC and AB at Ya and Za respectively, the parallel to CA intersect AB and BC at Zb and Xb, and the parallel to AB intersect BC and CA at Xc and Yc. The six points Xb, Xc, Yc, Ya, Za, Zb are concyclic. This is called the second Lemoine circle of the triangle.

2. Let HaHbHc be the orthic triangle. Construct the parallel through K to the line HbHc to intersect AC and AB at Ya and Za respectively. Similarly deﬁne Zb, Xb and Xc, Yc . The six points Xb, Xc, Yc , Ya, Za, Zb are all equidistant from K. The circle containing them is called the ﬁrst Lemonine circle of the triangle.

11.2 The Euler reﬂection point and the Kiepert parabola

Let O and H be the circumcenter and orthocenter of triangle ABC, with reflections OA, HA in the line BC. Similarly define the reflections OB, HB and OC, HC . The reflections HA, HB, HC lie on the circumcircle. The lines OAHA, OBHB, OC HC intersect at a point E on the circumcircle called the Euler reflection point of the triangle.

Exercise 1. The reﬂections of E in the three sidelines lie on the Euler line. 402 Some triangle centers

2. Find the inversive image of E in the circle with diameter OK. (This circle is called the Brocard circle).

The parabola with focus E and directrix the Euler line is called the Kiepert parabola. It is tangent to the sidelines of the triangle. If the points of tangency with BC, CA, AB are X, Y , Z respectively, then the lines AX, BY , CA are concurrent.

11.3 The Apollonian circles and the isodynamic points

Given triangle ABC, let the internal bisectors of angles A, B, C intersect the sides BC, CA, AB at Ta, Tb, Tc respectively. Likewise, let Ta, Tb, Tc be the intersections with the external angle bisectors. The three circles with diameters TaTa, TbTb and TcTc are called the Apollonian circles of triangle ABC. They have two common points J+ and J−, called the isodynamic points of the triangle.

Exercise 1. The radical axis of the Apollonian circles is the Brocard axis.

2. Let X, Y , Z be the pedals (orthogonal projections) of J+ (or J−) on the sidelines BC, CA, AB respectively. The triangle XY Z is equilateral.

11.4 The Fermat points and the Kiepert hyperbola

Given triangle ABC, construct equilateral triangles BCA+, CAB+ and ABC+ externally on the sides of the triangle.

Proposition 11.1. (a) The segments AA+, BB+, CC+ have equal lengths. (b) The lines AA+, BB+, CC+ are concurrent. The point of concurrence is called the (positive) Fermat point F+. ◦ (c) The lines AA+, BB+, CC+ make 60 angles with each other. (d) The circumcircles of the equilateral triangles BCA+, CAB+ and ABC+ all contain the point F+.

Let A− be the reﬂection of A+ in the sideline BC, and likewise B−, C− the reﬂections of B+, C+ in CA, AB respectively.

Proposition 11.2. (a) The segments AA−, BB−, CC− have equal lengths. (b) The lines AA−, BB−, CC− are concurrent. The point of concurrence is called the (negative) Fermat point F−. ◦ (c) The lines AA−, BB−, CC− make 60 angles with each other. (d) The circumcircles of the equilateral triangles BCA−, CAB− and ABC− all contain the point F−. 11.5 The Steiner circum-ellipse and the Steiner point 403

Exercise 1. Construct the circle through the circumcenter O and the two Fermat points. Where does this circle intersect the Euler line?

2. Let Na, Nb, Nc be the centroids of the equilateral triangles BCA+, CAB+, ABC+.

(a) What can you say about the triangle NaNbNc?

(b) The circumcircles of ANbNc, BNcNa and CNaNb have a common point on the circumcircle of ABC. What is this point?

The conic through the vertices and the Fermat points is called the Kiepert hyperbola. It intersects the Euler line at the centroid G and the orthocenter H, and is therefore a rectangular hyperbola. The center of the Kiepert hyperbola is the midpoint of the Fermat points. This is called theKiepert center Ki.

11.5 The Steiner circum-ellipse and the Steiner point

The Steiner circum-ellipse of triangle ABC is the ellipse with center G (the centroid) passing through the vertices. Apart from the vertices, it intersects the circumcircle at a fourth point called the Steiner point S.

Exercise

Let Oa, Ob, Oc be the circumcenters of the triangles AA+A−, BB+B−, CC+C− respectively. The lines AOa, BOb, COc are concurrent. The point of the concurrence is the Steiner point S.

11.6 The Steiner inellipse

Let Ma, Mb, Mc be the midpoints of the sides BC, CA, AB of triangle ABC. The Steiner inellipse is the unique conic tangent to the three sides at these points. The center of the ellipse is the centroid G. The ellipse can be constructed by marking three extra points Ma, 2:−1 Mb, Mc dividing MaG, MbG, McG in the ratio .

Exercise

The Steiner inellipse intersects the nine-point circle at Ma, Mb, Mc and a fourth point. What is this fourth intersection? Chapter 12

Some construction in triangle geometry

12.1 Simson lines

Let P be a point on the circumcircle of triangle ABC. The pedals (orthogonal projections) of P on the sidelines BC, CA, AB are collinear. The line containing these pedals is called the Simson line of P (with respect to triangle ABC). Theorem 12.1. If P and Q are antipodal points on the circumcircle, then their Simson lines are orthogonal and intersect on the nine-point circle.

12.2 Line of reﬂections

(1) Let P be a point on the circumcircle of triangle ABC. The reflections of P in the sidelines are collinear. The line (P ) containing them passes through the orthocenter of triangle ABC. (2) Let be a line through the orthocenter H. The reflections of in the sidelines BC, CA, AB are concurrent at a point P on the circumcircle. If is the Euler line, P is called the Euler reflection point E of the triangle.

Theorem 12.2. (P)= and P(P ) = P .

Exercise Construct the parabola with focus E tangent to the three sidelines. (This is called the Kiepert parabola of the triangle). If X, Y , Z are the points of tangency with BC, CA, AB, then the lines AX, BY , CZ are concurrent. This point of concurrency lies on the (Steiner) ellipse passing through the vertices and with center G (the centroid).

12.3 Isogonal conjugates

Let P be a point not on any of the sidelines of triangle ABC. Reﬂect the line AP in the bisector of angle A, BP in the bisector of angle B, and CP in the bisector of angle C. The three reﬂections are concurrent at a point P ∗ called the isogonal conjugate of P . 406 Some construction in triangle geometry

Exercise 1. What are the isogonal conjugates of O, I?

2. The isogonal conjugate of the centroid G is the symmedian point K.

3. The isogonal conjugates of the Fermat points are the isodynamic points.

4. What is the isogonal conjugate of a point on the circumcircle?

Let be a line passing through the circumcenter O. The locus of the isogonal conjugates of points on is a conic ∗ through the orthocenter H. This conic is therefore a rectangular hyperbola. The asymptotes of the rectangular hyperbola are the Simson lines of the (antipodal) points where intersects the circumcircle. The center of the hyperbola lies on the nine-point circle.

Exercise 1. If a rectangular circum-hyperbola (passing through the vertices and the orthocenter H) intersects the circumcircle at a point Q (other than the vertices), then the center of the conic is the midpoint of HQ.

2. The Kiepert hyperbola is the isogonal conjugate of the Brocard axis.

3. The isogonal conjugate of the Euler line is called the Jerabek hyperbola. Its center Je (the Jerabek center) can be located as follows. Construct the circle through the pedals of G (the centroid) on the sidelines. This circle intersects the nine-point circle at two points. One of them is the Kiepert center; the other is the Jerabek center.

4. The isogonal conjugate of the line OI is called the Feuerbach hyperbola. Its center is the Feuerbach point Fe, the point of tangency of the incircle with the nine-point circle.

5. Construct the Euler lines of the triangles AHbHc, BHcHa and CHaHb. The three Euler lines are concurrent. What is the point of concurrence?

12.4 Isotomic conjugates

Let P be a point not on any of the sidelines of triangle ABC. Suppose the lines AP , BP, CP intersect the lines BC, CA, AB respectively at X, Y , Z. Let X be the point on BC satisfying BX = X C, and likewise Y and Z on CA and AB respectively satisfying CY = Y A and AZ = ZB. The lines AX , BY , CZ are concurrent at a point P • called the isotomic conjugate of P . The perspector of the incircle is called the Gergonne point. Its isotomic conjugate is called the Nagel point. This is the point of concurrence of the three lines, each joining one vertex of the triangle to the point of tangency of the opposite side with the excircle on that side. 12.5 Inscribed conics 407

12.5 Inscribed conics

Given a point P (not on any of the sidelines of triangle ABC), there is a unique conic tangent to the sidelines. Here is a construction of this inscribed conic. Construct (1) the point Q dividing PGin the ratio PQ : QG =3:−1, (2) the isotomic conjugate Q• of Q, (3) the intersections X = AQ• ∩ BC, Y = BQ• ∩ CA and Z = CQ• ∩ AB, (4) the points X , Y , Z dividing XP, YP, ZP in the ratio 2:−1. The conic through the six points X, Y , Z, X , Y , Z is tangent to the sidelines at X, Y , Z. The point Q is called the perspector of the inscribed conic. Theorem 12.3. The foci of an inscribed conic are isogonal conjugates.

Exercise 1. A parabola tangent to the three sidelines BC, CA, AB has focus F on the circumcircle. The directrix of the parabola is the line of reﬂections of F . 2. Find the perspector of the Kiepert parabola (with focus E and directrix the Euler line). 3. Construct the inscribed conic with center N, the center of the nine-point circle. Verify that (a) the foci of the conic are the circumcenter and the orthocenter, (b) this conic is the envelope of the perpendicular bisector of the segment HQ, where Q varies on the circumcircle.

12.6 The Steiner inellipse and the Kiepert hyperbola

Let BCA+ be the equilateral triangle constructed externally on the side BC of triangle ABC, and A− the reﬂection of A+ in BC. Similarly deﬁne B+, B− and C+, C−.

Exercise

1. The bisectors of angles A+AA−, B+BB−, C+CC− are parallel, and are parallel to the bisector of angle KGS (where K is the symmedian point and S the Steiner point). 2. The axes of the Steiner inellipse are along the bisectors of angle KGS. 1 ( ± ) 3. The semiaxes of the Steiner inellipse are 6 AA+ AA− . 4. Construct the foci of the Steiner inellipse.

Theorem 12.4. The axes of the Steiner inellipse are parallel to the asymptotes of the Kiepert hyperbola.