- Dear triangle geometers,

I'm sure I speak for many who thank Antreas for establishing

Hyacinthos@onelist.

Antreas mentioned that that name Hyacinthos honors E. Lemoine, of whose

full name Hyacinthe is a part. The Lemoine point is often called the

symmedian point. In Ross Honsberger's Episodes in Nineteenth and Twentieth

Century Euclidean Geometry (Mathematical Association of America, 1995), a

whole chapter is devoled to this point.

However, Honsberger doesn't mention (directly) a certain interesting

property of the Lemoine point. For any point P, let A'B'C' denote the

pedal triangle of P (i.e., A' is the point in which the line through P

perpendicular to line BC meets line BC). Let S(P) be the vector sum

PA'+PB'+PC'. Then S(P) is the zero vector if P is the Lemoine point.

I conjecture that the converse is true: that if P is a "point" (i.e.,

f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2

(barycentric coordinates of the Lemoine point).

By the way, many other vector sums involving triangle centers will be

included in ETC (Encyclopedia of Triangle Centers), which should appear

sometime before March 1, 2000.

Best holiday regards to all.

Clark Kimberling - Dear all,

Let me start in a similar way as Clark did, and thank Antreas for

coining the idea to start a discussion list and then finding a good way

to really start one. Thanks a lot, Antreas! I was a bit surprised by the

name Hyacinthos, which I first associated with bulbs in my garden, and

second with ``Hyacinth Bucket'' (say Bouquet) from the British comedy

series ``Keeping up appearances''... ;)

Attributing this list to E. Lemoine is excellent.

> From: Clark Kimberling <ck6@...>

Isn't this property exactly equivalent to the fact that the

>

> Dear triangle geometers,

>

> I'm sure I speak for many who thank Antreas for establishing

> Hyacinthos@onelist.

>

> Antreas mentioned that that name Hyacinthos honors E. Lemoine, of whose

> full name Hyacinthe is a part. The Lemoine point is often called the

> symmedian point. In Ross Honsberger's Episodes in Nineteenth and Twentieth

> Century Euclidean Geometry (Mathematical Association of America, 1995), a

> whole chapter is devoled to this point.

>

> However, Honsberger doesn't mention (directly) a certain interesting

> property of the Lemoine point. For any point P, let A'B'C' denote the

> pedal triangle of P (i.e., A' is the point in which the line through P

> perpendicular to line BC meets line BC). Let S(P) be the vector sum

> PA'+PB'+PC'. Then S(P) is the zero vector if P is the Lemoine point.

Lemoine/symmedian point K is the centroid of its pedal triangle? This

property of K is mentioned in O. Bottema's ``Hoofdstukken uit de

Elementaire Meetkunde''.

We can generalize this in the sence of P-trace (P-pedal) triangles: Let

X be a point, the inscribed triangle A'B'C' in ABC is called the P-trace

triangle of X if XA'//AP, XB'//BP and XC'//CP.

If P = f:g:h (barycentrics) then the P-version K_P of the Lemoine point

is f(g+h):g(f+h):h(f+g).

The centroid of the P-trace triangle of X = x:y:z is found as:

2x + f/(f+h) y + f/(f+g) z :

2y + g/(g+h) x + g/(g+f) z :

2z + h/(h+g) x + h/(h+f) y

It is easy to verify that in case that X=K_P then indeed this centroid

is equal to X.

> I conjecture that the converse is true: that if P is a "point" (i.e.,

This is not immediatly apparent to me.

> f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2

> (barycentric coordinates of the Lemoine point).

> By the way, many other vector sums involving triangle centers will be

Excellent news!

> included in ETC (Encyclopedia of Triangle Centers), which should appear

> sometime before March 1, 2000.

Another property of the Lemoine point: It is the center of the conic

inscribed in ABC, tangent to the sides of ABC in the orthic triangle.

(Problem 525 in Nieuw Archief voor Wiskunde, by O. Bottema).

Kind regards,

Floor. > Date: Thu, 23 Dec 1999 09:23:47 +0100

Of course that is an equivalent property.

> From: Floor van Lamoen <f.v.lamoen@...

>

> > From: Clark Kimberling <ck6@...>

> >

> > However, Honsberger doesn't mention (directly) a certain interesting

> > property of the Lemoine point. For any point P, let A'B'C' denote the

> > pedal triangle of P (i.e., A' is the point in which the line through P

> > perpendicular to line BC meets line BC). Let S(P) be the vector sum

> > PA'+PB'+PC'. Then S(P) is the zero vector if P is the Lemoine point.

>

> Isn't this property exactly equivalent to the fact that the

> Lemoine/symmedian point K is the centroid of its pedal triangle? This

> property of K is mentioned in O. Bottema's ``Hoofdstukken uit de

> Elementaire Meetkunde''.

>

[snip]

> > I conjecture that the converse is true: that if P is a "point" (i.e.,

It wasn't apparent to me either. However, the calculations are not

> > f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2

> > (barycentric coordinates of the Lemoine point).

> This is not immediatly apparent to me.

difficult, and this converse is true.

Write vec(X,Y) = vector from X to Y.

In homogenous barycentrics, P=(x:y:z).

In normalized barycentrics, if P=(x,y,z) with x+y+z=1, then

vec(O,P) = x vec(O,A) + y vec(O,B) + z vec(O,C)

for any choice of origin O.

Now the B-pedal point of P(x,y,z) is P_B = (x+y SC/bb , 0 , z+y SA/bb),

and this is normalized, since SA+SC=bb.

Then vec(P,P_B) = vec(O,P_B) - vec(O,P)

= y SC/bb vec(O,A) - y vec(O,B) + y SA/cc vec(O,C)

with similar formulas for vec(P,P_A) and vec(P,P_C)

So the equation S(P)=0 becomes

(-x + y SC/bb + z SB/cc) vec(O,A)

+ (x SC/aa - y + z SA/cc) vec(O,B)

+ (x SB/aa + y SA/bb - z) vec(O,C) = 0

Well, one solution is obviously (x,y,z)=const*(aa,bb,cc). Any others?

Choose the origin O to be the circumcenter. Then the three basis

vectors are dependent, and we have the dependency relation

aaSA vec(O,A) + bbSB vec(O,B) + ccSC vec(O,C) = 0

This is the only dependency relation between these three vectors.

So any other solution {x,y,z} would have to satisfy

(-x + y SC/bb + z SB/cc) = k aaSA

(x SC/aa - y + z SA/cc) = k bbSB

(x SB/aa + y SA/bb - z) = k ccSC

for some non-zero constant k. However, adding these 3 equations

shows k=0, so there are no other solutions

I admit the calculations weren't as easy as I expected,

but this converse is true.

--

Barry Wolk <wolkb@...>

Winnipeg Manitoba Canada