Re: [HM] "Dramatically Simple"

John Puddefoot (j.puddefoot@etoncollege.org.uk)
Fri, 23 Apr 1999 07:59:54 +0100

[APH]:
>> The following problem appeared in the American Math. Monthly 16
>> years ago:
>> Let A and B the midpoints of the sides DE and DF of an equilateral
>> triangle DEF. Extend AB to meet the circumcircle of DEF at C. Show
>> that B divides AC according to the golden section.
>> George Odom, AMM 90(1983) 482; Solution: 93(1986) 572

Antreas wrote:

* In Odom's figure:
Draw the altitude from D, intersecting AB, EF, and the circumference at
H, I, K, resp., and call: BC = x, IK = y. Also, let C' be the point that
AB intersects the arc ED.

D
/|\
/ | \
/ | \
C'-----A---H---B-----C
/ | \
/ | \
/ | \
E-------I-------F
|
|
K

Antreas' solution is clear, but the problem can be solved in one line with
no construction using what we call the 'Butterfly Theorem' that triangles
AEC'and ADC are similar. The proof is exactly as for David Fowler's two
square case:

BC = AC' = x

Let the sides of the triangle be 2

Then 1 = AD x AE = AC x AC' = x(1+x) QED

John Puddefoot,
Eton College,
UK