Forum GeometricorumVolume 13 (2013) 227–231. b b

b

b

FORUM GEOM

ISSN 1534-1178

Some Simple Results on Cevian Quotients

Francisco Javier Garcıa Capitan

Abstract. We find the loci of the cevian quotientsP/Q andQ/P when one ofthe points is fixed and the other moves along a given line. We also show that,for a given pointP , the locus ofQ for which the line joiningP/Q andQ/P isparallel toPQ is a conic throughP andG/P , and give two simple constructionsof the conic.

The term cevian quotient was due to John Conway [1]. Given two pointsP =(u : v : w) andQ = (x : y : z) in homogeneous barycentric coordinates withreference to a triangleABC, the cevian quotientP/Q is the perspector of thecevian triangle ofP and the anticevian triangle ofQ. It is the point

P/Q =(

x(

−x

u+

y

v+

z

w

)

: y(x

u−

y

v+

z

w

)

: z(x

u+

y

v−

z

w

))

.

A most basic property of cevian quotient is the following theorem.

Theorem 1 ([2, §2.12], [5,§8.3]). P/Q = Q′ if and only ifP/Q′ = Q.

This is equivalent toP/(P/Q) = Q. It can be proved by direct verificationwith coordinates. We offer an indirect proof, with the advantage of an explicitconstruction, for givenQ andQ′, of a pointP with P/Q = Q′ andP/Q′ = Q.

ForQ = (x : y : z) andQ′ = (x′ : y′ : z′) with anticevian trianglesXY Z andX ′Y ′Z ′, it is easy to check that the linesQX ′ andQ′X intersect on the sideline

BC, at the point(0 : xy′ + x′y : zx′ + z′x) =(

0 : 1zx′+z′x

: 1xy′+x′y

)

(see Figure

1). Similarly, the linesQY ′ andQ′Y intersect onCA at(

1yz′+y′z

: 0 : 1xy′+x′y

)

,

and the linesQZ ′ andQ′Z intersect onAB at(

1yz′+y′z

: 1zx′+z′x

: 0)

. These form

the cevian triangle of the pointP =(

1yz′+y′z

: 1zx′+z′x

: 1xy′+x′y

)

. It is clear that

P/Q = Q′ andP/Q′ = Q.

Remark.The pointP =(

1yz′+y′z

: 1zx′+z′x

: 1xy′+x′y

)

is called the cevian product

Q ∗Q′ of Q andQ′. Clearly,Q ∗Q′ = Q′ ∗Q.

Proposition 2. LetP be a fixed point. IfQ moves along a lineL , then the quotientP/Q traverses the bicevian conic ofP and the trilinear pole ofL .

Publication Date: December 3, 2013. Communicating Editor: Paul Yiu.The author thanks Paul Yiu for his help in improving this paper.

228 F. J. Garcıa Capitan

A

B

C

X′

Y ′

Z′

X

Y

Z

Q′Q

P

Figure 1.

Proof. Let P = (u : v : w) andQ move along the lineL with line coordinates[p : q : r]. If Q′ = P/Q = (x : y : z), thenQ = P/Q′ is on the lineL , and

px(

−x

u+

y

v+

z

w

)

+ qy(x

u−

y

v+

z

w

)

+ rz(x

u+

y

v−

z

w

)

= 0.

Clearing denominators and simplifying, we obtain

pvwx2+ qwuy2+ ruvz2−u(qv+ rw)yz− v(rw+ pu)zx−w(pu+ qv)xy = 0.

If x = 0, this becomesu(qy − rz)(wy − uz) = 0. The conic intersects the lineBC at (0 : v : w) and(0 : r : q). Similarly, it intersectsCA at (u : 0 : w) and(r : 0 : p), andAB at (u : v : 0) and(q : p : 0). This is the bicevian conic through

the traces ofP and(

1p: 1q: 1r

)

, the trilinear pole ofL . �

Corollary 3 ([4]). Let P be a fixed point. The locus ofQ for which the cevianquotientP/Q lies on the tripolar ofP is the inscribed conic with perspectorP .

Proposition 4. Let P be a fixed point. IfQ moves along a lineL , then the ce-vian quotientQ/P traverses the circumconic of the anticevian triangle ofP withperspectorPL /P , wherePL is the trilinear pole ofL .

Proof. Let P = (u : v : w) andQ move along the lineL with line coordinates

[p : q : r]. If Q′′ = Q/P = (x : y : z), thenQ =(

1wy+vz

: 1uz+wx

: 1vx+uy

)

is on

the lineL , andp

wy + vz+

q

uz + wx+

r

vx+ uy= 0.

Clearing denominators and simplifying, we obtain

pvwx2 + qwuy2 + ruvz2 + (pu+ qv + rw)(uyz + vzx+ wxy) = 0.

Some simple results on cevian quotients 229

It is easy to verify that this conic passes throughA′ = (−u : v : w),B′ = (u : −v :w), C ′ = (u : v : −w). It is a circumconic of the anticevian triangle ofP . Thetangents to the conic atA′, B′, C ′ are the linesLa : (qv+ rw)x+ quy+ ruz = 0,Lb : pvx + (pu + rw)y + rvz = 0, Lc : pwx + qwy + (pu + qv)z = 0 whichintersectsL onBC, CA, AB respectively. This is the conic tangent to the linesA′X ′, B′Y ′, C ′Z ′ at A′, B′, C ′ respectively. These lines bound a triangle withvertices

(

qv + rw

u: q : r

)

,

(

p :rw + pu

v: r

)

,

(

p : q :pu+ qv

w

)

.

These form a triangle perspective with the anticevian triangle ofP at

(u(−pu+ qv + rw) : v(pu− qv + rw) : w(pu+ qv − rw)),

the cevian quotient of(

1p: 1

q: 1

r

)

(the trilinear pole ofL ) by P . �

Let L /P be the conic in Proposition 4. This conic is a circle if and only if it isthe circumcircle of the anticevian triangle ofP . The lineL is the one containingthe intercepts of the tangents to this circle atA′, B′, C ′ on the respective sidelinesof triangleABC (see Figure 2). This has line coordinates

p : q : r = (u(v + w − u)(c2v2 − (b2 + c2 − a2)vw + b2w2)

v(w + u− v)(a2w2 − (c2 + a2 − b2)wu+ c2u2)

w(u+ v − w)(b2u2 − (a2 + b2 − c2)uv + a2v2).

A

B C

A′

B′

C′

XY

Z

P

Q

Q/P

Figure 2.

230 F. J. Garcıa Capitan

Here are some simple examples in whichL /P is the circumcircle of the antice-vian triangle ofP :

P L

centroid a2x+ b2y + c2z = 0incenter line at infinitysymmedian point

∑

cyclic a2(b2 + c2 − a2)x = 0

Proposition 5. LetP be a fixed point. The locus ofQ for which the line joining(P/Q) to (Q/P ) is parallel toPQ is the union of the cevian linesAP , BP , CPand a conicΓ(P )(1) homothetic to the circumconic with perspectorP ,(2) passing throughP and the cevian quotientG/P , and has(3) the midpoint ofP andG/P as center.

Proof. If P = (u : v : w) andQ = (x : y : z), the line joiningP/Q andQ/Pcontains the infinite point ofPQ if and only if

∣

∣

∣

∣

∣

∣

∣

x(

−xu+ y

v+ z

w

)

y(

xu− y

v+ z

w

)

z(

xu+ y

v− z

w

)

u(

−ux+ v

y+ w

z

)

v(

ux− v

y+ w

z

)

w(

ux+ v

y− w

z

)

(v + w)x− u(y + z) (w + u)y − v(z + x) (u+ v)z − w(x+ y)

∣

∣

∣

∣

∣

∣

∣

= 0.

Clearing denominators and simplifying, we obtain

2(wy−vz)(uz−wx)(vx−uy)(vwx2+wuy2+uvz2−u2yz−v2zx−w2xy) = 0.

ThereforeQ lies on one of the linesAP , BP , CP or the conicΓ(P ) defined by

vwx2 + wuy2 + uvz2 − u2yz − v2zx− w2xy = 0.

Rewriting this as

Γ(P ) : (u+ v+w)(uyz+ vzx+wxy)− (x+ y+ z)(vwx+wuy+ uvz) = 0,

it is clear thatΓ(P ) is homothetic to the circumconic with perspectorP , and it isroutine to verify that it containsP and the cevian quotientG/P = (u(−u+v+w) :v(u − v + w) : (u + v − w)w). The center of the conic is the midpoint ofP andG/P , namely,(

u(u2 − uv − uw − 2vw) : v(v2 − uv − 2uw − vw) : w(w2 − 2uv − uw − vw))

.

�

Remarks.(1) If P is the symmedian point, thenΓ(P ) is the Brocard circle withdiameterOK.

(2) If the linePQ containsA, then both cevian quotientsP/Q andQ/P are onthe same line.

It is easy to note that the conicΓ(P ) contains the points

A1 = (−u+ v + w : v : w), B1 = (u : u− v + w : w), (u : v : u+ v − w)

We present two simple constructions of these points, one by Peter Moses [3],and another by Paul Yiu [6].

Some simple results on cevian quotients 231

Construction (Moses). Intersect the ceviansAP , BP , CP with the parallelsthrough the centroidG to BC, CA, AB, at X, Y , Z respectively.A1, B1, C1

are the harmonic conjugates ofP in AX, BY , CZ respectively.

A

B C

C′

A′

Y ′

Z′

P

G/P

P ′

GA1

B1

C1

X′

X′

Figure 3.

Construction (Yiu). Let P ′ be the superior ofP , i.e., the point dividingPG inthe ratioPP ′ : P ′G = 3 : −2. Construct the parallels of the linePG through theverticesA,B,C, to intersect the sidelinesBC,CA,AB atX ′, Y ′,Z ′ respectively.ThenA1 = AP ∩X ′P ′, B1 = BP ∩ Y ′P ′, andC1 = CP ∩ Z ′P ′. See Figure 3.

References

[1] J. H. Conway, Hyacinthos message 1018, June 14, 2000.[2] C. Kimberling, Triangle centers and central triangles,Congressus Numerantium, 129 (1998)

1–285.[3] P. Moses, Private communication, October 19, 2013.[4] P. Yiu, Hyacinthos message 1030, June 16, 2000.[5] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,

2001; with corrections, 2013, available athttp://math.fau.edu/Yiu/Geometry.html

[6] P. Yiu, ADGEOM message 728, October 19, 2013.

Francisco Javier Garcıa Capitan: Departamento de Matematicas, I.E.S. Alvarez Cubero, Avda.Presidente Alcala-Zamora, s/n, 14800 Priego de Cordoba, Cordoba, Spain

E-mail address: garciacapitan@gmail.com