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Old 2005-03-14, 15:31   #1
mfgoode
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Cool Problems of Antiquity

I have been out of touch with Mersenneforum for about two months as I was on a business cum holiday trip to Ratlam (India). where there are few cyber cafes.
Amongst others I took with me two ‘companions’ viz:

1) The ‘ Mathematical Experience’ by Davis and Hersh

2) A ‘Concise history of Mathematics’ by Dirk J. Struik.

I read them both for the second time around..
I would recommend these books to anyone who is interested in Maths.
.In them I found four challenging problems known to the ancient Egyptians, Greeks and Orientals before Christ which can even confound math scholars today.

1) Given the sum (x + y) and product x y. Find x and y

2) The square number 16 and the rectangular no. 18 are alone of plane numbers, which have their perimeters equal to the areas enclosed by them. Prove it.

3) Find the volume of the frustrum of a square pyramid where ‘a’ and ‘b’ are the lengths of the sides of the squares and ‘h’ is its height., Assume vol. of square pyramid is known to be one third the Area of the base multiplied by its height..

4) Prove Herons formula for the Area (A) of a triangle is
A = Sqr. Root of [ s (s-a) (s-b) (s-c) ] where a, b, c, are its sides and ‘s’ = semi- perimeter.

I spent a few pleasant evenings deriving/solving them without referring to text books or the Net.

Try your hand at them too and prove yourself equal to the ancients!

Mally.
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Old 2005-03-14, 16:09   #2
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1. is easy. x+y=a, xy=b, x=(a +- sqrt(a^2 - 4*b))/2

Alex
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Old 2005-03-14, 16:38   #3
mfgoode
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Quote:
Originally Posted by akruppa
1. is easy. x+y=a, xy=b, x=(a +- sqrt(a^2 - 4*b))/2

Alex
Excellent Alex ! x=y = a/2 +- sqr.rt.[ (a/2)/2)^2 - b ] when simplified further

Mally.
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Old 2005-03-14, 16:58   #4
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Don't forget If x==0 or y==0 <=> b==0, the system has two solutions.
Either x=0 and y=a or y=0 and x=a.


3 looks nice and linear.
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Old 2005-03-14, 17:41   #5
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I remember proving Heron's Formula in high school. It took nearly a page. Basically, it can be done using the cosine rule and lots of brute force algebra.
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Old 2005-03-14, 22:55   #6
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2a) square number 16.

For each square number n^2, the area is n^2 and the perimeter is 4n. n^2 = 4n => n=4.

But it could be that n^2 = 16 falls out as a special case of rectangular number, which proof I haven't completed.

Last fiddled with by cheesehead on 2005-03-14 at 23:06
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Old 2005-03-15, 03:33   #7
mfgoode
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Quote:
Originally Posted by mfgoode
Excellent Alex ! x=y = a/2 +- sqr.rt.[ (a/2)/2)^2 - b ] when simplified further

Mally.
Correction: x=a/2 + [..........]. and y = -[........ ]if x>y

Mally
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Old 2005-03-15, 07:31   #8
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Does this thread belong in the Puzzles section of the "Fun stuff" subforum?
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Old 2005-03-19, 14:28   #9
mfgoode
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Quote:
Originally Posted by cheesehead
2a) square number 16.

For each square number n^2, the area is n^2 and the perimeter is 4n. n^2 = 4n => n=4.

But it could be that n^2 = 16 falls out as a special case of rectangular number, which proof I haven't completed.
[Quote=cheesehead] Does this thread belong in the Puzzles section of the "Fun stuff" subforum]


This theorem was quoted by Plutarch [40A.D. - 120A.D.] who blended sacred history with mathematical theorems. It was well known by the ancients Pythagoras included.

It was recommended by Davis and Hersh to try to prove it.
"This is a nice theorem, Prove it."

I am still waiting for you to complete it Then decide if it should be shoved into. the Puzzles section of the "Fun Stuff" subforum!!!

Mally
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Old 2005-03-21, 16:47   #10
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Quote:
Originally Posted by mfgoode
I am still waiting for you to complete it Then decide if it should be shoved into. the Puzzles section of the "Fun Stuff" subforum!!!
Advisory: Completion is way down on my priority list, like ... end-of-summer-ish.
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Old 2005-03-21, 16:56   #11
mfgoode
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Cool Problems of Antiquity

Quote:
Originally Posted by cheesehead
Advisory: Completion is way down on my priority list, like ... end-of-summer-ish.
Suit yourself buddy!. Im waiting for Christmas

Mally
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