- In message #6614, I gave a lemma for the Arnold

theorem. Now I have noted that it is not really

needed in the proof.

Here is the ARNOLD THEOREM (for H. E. Arnold, who

gave a special case of it as problem AMM 3258 in

American Mathematical Monthly):

Let ABC be a triangle with orthocenter H, and

let P be an arbitrary point. Call X, Y, Z

the traces of P on the sides of ABC (i. e.

X = AP /\ BC, Y = BP /\ CA, and Z = CP /\ AB).

The perpendicular to HX through A intersects

BC at X'; analogously, define Y' and Z'. Then,

- the points X', Y', Z' are collinear;

- the line X'Y'Z' is orthogonal to the

line HP.

More on this theorem I have written in a HTML

comment to a problem in the MathPro database. The

database is

http://problems.math.umr.edu/index.htm

Since it is not interlinked, to find my comment one

should search for the text "The perpendiculars

dropped from the vertices of a triangle upon the

lines joining the feet of the angle bisectors".

Now to the PROOF of the Arnold theorem. Let AA',

BB', CC' be the altitudes of triangle ABC. As we

know, AH * HA' = BH * HB' = CH * HC'. Let Q be a

point on the line HP, for which PH * HQ = AH * HA'

= BH * HB' = CH * HC' (with directed segments).

Since AX' is orthogonal to HX and AH is orthogonal

to XX', the point H is the orthocenter of triangle

AXX'. Thus, H lies on the perpendicular from X' to

AX. Call D the foot of this perpendicular. Then,

angle AHD = angle X'HA', and angle ADH = angle

X'A'H = 90°; therefore, triangles AHD and X'HA'

are similar, and AH : HD = X'H : HA'. Thus,

AH * HA' = X'H * HD. Together with PH * HQ = AH *

HA', this gives PH * HQ = X'H * HD, what is

equivalent to PH : HD = X'H : HQ. Together with

angle PHD = angle X'HQ, this yields that triangles

PHD and X'HQ are similar; thus, angle X'QH = angle

PDH = 90°. Therefore, X'H is perpendicular to HQ,

i. e. to HP. This shows that X' lies on the

perpendicular to HP through Q. Analogously, Y' and

Z' lies on this perpendicular. This proves the

Arnold Theorem.

EXERCISES:

1) Orthic products formula.

Prove that AH * HA' = BH * HB' = CH * HC' =

4 R² cos A cos B cos C, where R is the

circumradius of triangle ABC.

2) The First Panakis theorem.

The point Q is the intersection of circles AA'P,

BB'P and CC'P. (A theorem of independent

importance.)

3) Panakis transformation as inversion.

The point Q is the inverse of P with respect to

the polar circle of triangle ABC (if the polar

circle exists, i. e. if triangle ABC is obtuse).

4) Arnold theorem and polarity.

The line X'Y'Z' in the Arnold theorem is the

polar of P with respect to the polar circle of

triangle ABC (if the polar circle exists, i. e.

if triangle ABC is obtuse).

5) Alternative proof of Arnold theorem for

obtuse triangles.

Give a proof of the Arnold theorem based on

Exercise 4) and using the properties of polars.

(This proof will be valid for obtuse triangles

ABC.)

6) The orthic axis (Altshiller-Court theorem

If P is the centroid of triangle ABC, then the

line X'Y'Z' is the orthic axis, i. e. the axis

of perspective of triangle ABC and the orthic

triangle A'B'C'.

(A quite hard exercise. I know of an ugly proof.)

Darij Grinberg - Dear Barry Wolk,

In Hyacinthos message #6661, you wrote:

>> Darij Grinberg wrote:

Many thanks... the next step would be (perhaps) generalizing to

>>

>> > I don't know of which computational verification is Barry Wolk

>> > speaking, but I think he uses barycentric coordinates.

>>

>> Correct.

>>

>> > Perhaps a reason why

>> > I had not found a trivial coordinate proof of the Arnold

>> > theorem is that I am still too lazy to learn representing

>> > perpendicular lines in barycentrics.

>>

>> I call a calculation "trivial" if I can do it without needing a

>> computer to handle the polynomials. In many problems the

>> polynomials tend to get ridiculously large.

>>

>> It is very easy to handle perpendicular lines. Take two points

>> at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.

>> Then lines through P are perpendicular to lines through P' iff

>> x x' SA + y y' SB + z z' SC = 0

>>

>> So given P, we can find P' by letting x' = y SB - z SC, y' and

>> z' similarly.

>>

>> And the line through H and P' has coordinates

>> [x SA : y SB : z SC]

>> --

>> Barry Wolk

Lamoen's Pl-perpendicularity, but I think this is simply replacing

SA, SB, SC by some constants.

Sincerely,

Darij Grinberg