Re: [HM] "Dramatically Simple"

Antreas P. Hatzipolakis (xpolakis@otenet.gr)
Fri, 23 Apr 1999 00:58:51 +0300 (EET DST)

[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

[David Fowler]:
> Once you see the point of Odom's construction, you easily see this
> variation:
>
> Stick two squares together along one side AB. Draw the circumcircle
> around the resulting 2x1 rectangle. Extend AB to meet the circle in C.

This is algebraically much simpler than the Odom's.

* 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

Let us assume that DE = EF = FD = 1.

With respect to point I, we have:

EI * IF = IK * ID,
and since
EI = IF = 1/2, IK = y, and DI = sqrt(3)/2
we get:
1/2 * 1/2 = y * sqrt(3)/2

With respect to point H, we have:

HC * HC' = HD * HK ==> (HB + BC)(HA + AC') = HD(HI + IK),

and since:

AC' = BC = x, HB = HA = (EI)/2 = 1/4, and HD = HI = (DI)/2 = sqrt(3)/4,

we get:
(1/4 + x)^2 = sqrt(3)/4 * (sqrt(3)/4 + y)

Solving the system of the equations, we get:

y = sqr(3)/6, and x = (sqrt(5) - 1)/4,

and since AB = 1/2, then B divides AC in extreme and mean ratio.

* In David Fowler's figure:
Similarly, with respect to point A, we have:
(AB intersects the circle at C, C'; AB = AD = AE, etc = 1)

C
|
|
D--------A--------E
| | |
| | |
| | |
+--------B--------+
|
|
C'

AC * AC' = DA * AE ===> x * (x+1) = 1 * 1 ==> x = (sqrt(5)-1)/2

.... and with nested unit squares, we can discover phi but in Fibonaccis:

Three nested unit squares:

A-------+--------+--------+
| | | |
| | | |
| | | |
B-------1--------2--------3

angle(A1B) =: w_1, angle(A2B) =: w_2, angle(A3B) =: w_3

Theorem:
w_1 = w_2 + w_3

This theorem can be proven in ... almost infinitely many ways !

Charles Trigg published 54 proofs in the Journal of Recreational
Mathematics, vol. 4, April 1971, pp. 90-99.
See also: M. Gardner, Mathematical Circus. Knopf, 1979, pp. 131-132.

Let's now generalize the theorem using n nested unit squares:
We have:
w_1 = w_2 + w_3

w_3 = w_5 + w_8

w_8 = w_13 + w_21

. . . . . . . . .

w_f(2k) = w_f(2k+1) + w_f(2k+2)

where f(n) is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, ...

Antreas