## Re: Re: Three Concurrent Lines in a Circle.

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• ... First Paragraph: The late Leon Bankoff (he died in 1997) was a Beverly Hills, California, dentist who also was a world expert on plane geometry....In 1979
Message 1 of 6 , Dec 31 4:00 PM
Julio Gonzalez Cabillon wrote:

>At 04:49 PM 28/12/1999 -0700, Richard Guy wrote:
>>
>> somewhere in this year's Coll. Math. J. R.
>>
>
>See:
>
>Gardner, Martin:
>"The Asymmetric Propeller", _College Math Journal_, vol 30 (1999),
>no 1, pp 18-22.

First Paragraph:
The late Leon Bankoff (he died in 1997) was a Beverly Hills, California,
dentist who also was a world expert on plane geometry....In 1979 he told me
the asymmetric propeller theorem. He intended to discuss them in an article,
but never got around to it. This is a summary of what he told me.

Also:
Bankoff, Leon - Erdos, Paul - Klamkin, Murray: The Asymmetric Propeller.
Mathematics Magazine 46 (1973) 270-272.

Antreas
• ... The Asymmetric Propeller Martin Gardner A theorem, seventy years old at least and of unknown origin, says that if three congruent equilateral triangles are
Message 2 of 6 , Dec 31 4:00 PM
Richard Guy wrote:

>somewhere in this year's Coll. Math. J. R.

The Asymmetric Propeller
Martin Gardner

A theorem, seventy years old at least and of unknown origin,
says that if three congruent equilateral triangles are have
corners meeting, the midpoints of the lines joining the other
two vertices of the triangles are vertices of an equilateral
triangle. The late Leon Bankoff discovered that the triangles
don't have be congruent and don't have to meet at a point.
Martin Gardner describes the results, and conjectures that the
triangles don't have to be triangular--squares seem to work
as well.
The College Mathematics Journal, January 1999.

Antreas
• ... H. S. M. Coxeter: Given six consecutive points A, B, C, D, E and F on a circle, prove that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE and CF are
Message 3 of 6 , Dec 31 4:00 PM
Den Roussel wrote:

>When Cevians are drawn from the vertices of a triangle ABC, they
>intersect the opposite sides at A'B'C' and the circumcircle at
>A''B''C''.
>
>Ceva's theorem shows that the product of the ratios of the segments of
>the sides is equal to 1.
>
>(AB'/B'C) (CA'/A'B) (BC'/C'A) = 1
>
>In like manner, the product of the ratios of the consecutive sides of
>the inscribed hexagon is equal to 1.
>
>(AB''/B''C) (CA''/A''B) (BC''/C''A) = 1
>
>In other words, when three concurrent lines are drawn in a circle, then
>the hexagon they form has the property that the product of three
>non-adjacent sides is equal to the product of the other three sides.
>
>I'm not sure if this is known or not.

H. S. M. Coxeter:

Given six consecutive points A, B, C, D, E and F on a circle, prove
that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE and CF are concurrent.
_The Mathematics Student Journal_ 27:5 (1980) 3.

Antreas
• - ... See: Gardner, Martin: The Asymmetric Propeller , _College Math Journal_, vol 30 (1999), no 1, pp 18-22. Regards, JGC -
Message 4 of 6 , Dec 28 3:02 PM
-

At 04:49 PM 28/12/1999 -0700, Richard Guy wrote:
>
> somewhere in this year's Coll. Math. J. R.
>

See:

Gardner, Martin:
"The Asymmetric Propeller", _College Math Journal_, vol 30 (1999),
no 1, pp 18-22.

Regards, JGC

-
• ... Which according to ZfM, the paper seems to provide two proofs of the theorem: If OAB, OCD and OEF are equilateral triangles of the same orientation then
Message 5 of 6 , Dec 28 3:37 PM
Antreas quoted:

> Bankoff, Leon - Erdos, Paul - Klamkin, Murray: The Asymmetric Propeller.
> Mathematics Magazine 46 (1973) 270-272.
>

Which according to ZfM, the paper seems to provide two proofs of the
theorem:

If OAB, OCD and OEF are equilateral triangles of the same
orientation then the midpoints of BC, DE and FA are the
vertices of an equilateral triangle.

Regards, JGC
• See also Martin Gardner s article on the Asymmetric Propellor, somewhere in this year s Coll. Math. J. R.
Message 6 of 6 , Dec 28 3:49 PM