HYACINTHOS 19

> Date: Thu, 23 Dec 1999 09:23:47 +0100
> From: Floor van Lamoen <f.v.lamoen@xxx.xxx
>
> > From: Clark Kimberling <ck6@xxxxxxxxxx.xxx>
> >
> > 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@cc.umanitoba.ca>
Winnipeg Manitoba Canada