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## Lemoine point

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• 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.
Message 1 of 3 , Dec 22, 1999
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
Message 2 of 3 , Dec 23, 1999
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@...>
>
> 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.

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''.

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.,
> 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.

> 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.

Excellent news!

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.
• ... Of course that is an equivalent property. ... [snip] ... It wasn t apparent to me either. However, the calculations are not difficult, and this converse is
Message 3 of 3 , Dec 27, 1999
> Date: Thu, 23 Dec 1999 09:23:47 +0100
> 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''.

Of course that is an equivalent property.

>
[snip]
> > 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).

> This is not immediatly apparent to me.

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