20050314, 15:31  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
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 (sa) (sb) (sc) ] 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. 
20050314, 16:09  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
1. is easy. x+y=a, xy=b, x=(a + sqrt(a^2  4*b))/2
Alex 
20050314, 16:38  #3  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
\problems of Antiquity
Quote:
Mally. 

20050314, 16:58  #4 
Jan 2005
Caught in a sieve
2×197 Posts 
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. 
20050314, 17:41  #5 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
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.

20050314, 22:55  #6 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
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 20050314 at 23:06 
20050315, 03:33  #7  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Problems of Antiquity
Quote:
Mally 

20050315, 07:31  #8 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Does this thread belong in the Puzzles section of the "Fun stuff" subforum?

20050319, 14:28  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Problems of Antiquity
Quote:
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 

20050321, 16:47  #10  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1E0C_{16} Posts 
Quote:


20050321, 16:56  #11  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Problems of Antiquity
Quote:
Mally 

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