Problems of Antiquity

[wblipp]

2. (Only rectangles with area = perimeter have area 16 and 18)


Let the height and width be x and y. We need xy=2x+2y. There is no solution
for y=2 because that would require 2x=2x+4. If y is not 2, then

x = 2y/(y-2).

The only possible common factor of y and y-2 is 2. Since (y-2) divides 2y, the
only possible positive values for y-2 are 1, 2, and 4. Substituting these in
results in rectangles of 6x3, 4x4, and 3x6.

Checking the negative values of y-2 (-1, -2, -4) gives no traditional solutions,
but does reveals some non-traditional solutions: (-2)x(1), 0x0, 1x(-2)

[Ken_g6]

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.

The volume of the pyramid VP(a,H) = 1/3*a^2*H

Now, the frustrum, or frustum, is:

1. frustum -- (a truncated cone or pyramid; the part that is left when a cone or
pyramid is cut by a plane parallel to the base and the apical part is removed)

So, first, we need to find the height of the full pyramid, and then remove the
"apical part" (the top). Well, as you go up cutting the pyramid with a plane parallel to
the base, clearly the side length of the square is a linear function. And we have two
points for that function.

l(0) = a
l(h) = b

This leads to:

l(H)=a+H*m (m = slope = rise/run) so:
l(h)=b = a+h*m, so m = (b-a)/h, and:
l(H)=a-H*(a-b)/h

So let's find the zero of this function, the top of the pyramid:

l(H) = 0

a=H*(a-b)/h

H=h*a/(a-b)

The height of the full pyramid is h*a/(a-b), so the height of the apical part is
h*a/(a-b)-h = h*(a/(a-b)-1)

Then their respective volumes are:

vp = 1/3*a^2*h*(a/(a-b))
va = 1/3*b^2*h*(a/(a-b)-1)

Note that:
a/(a-b)-1 = (a-a+b)/(a-b) = b/(a-b)

So:
va = 1/3*b^2*h*(b/(a-b))

vp = 1/3*h*a^3/(a-b)
va = 1/3*h*b^3/(a-b)


and the total volume is:

v = vp-va = 1/3*h*(a^3-b^3)/(a-b)
Now the difference of cubes a^3-b^3 factors to (a-b)*(a^2+a*b+b^2).

So:

v = 1/3*h*(a^2+a*b+b^2), for a not equal to b, and v=0 for a=b.

[mfgoode]

Originally Posted by wblipp

2. (Only rectangles with area = perimeter have area 16 and 18)


Let the height and width be x and y. We need xy=2x+2y. There is no solution
for y=2 because that would require 2x=2x+4. If y is not 2, then

x = 2y/(y-2).

The only possible common factor of y and y-2 is 2. Since (y-2) divides 2y, the
only possible positive values for y-2 are 1, 2, and 4. Substituting these in
results in rectangles of 6x3, 4x4, and 3x6.

Checking the negative values of y-2 (-1, -2, -4) gives no traditional solutions,
but does reveals some non-traditional solutions: (-2)x(1), 0x0, 1x(-2)

Thank you wblipp for your very powerful and excellent deduction of the area = perimeter problem. It exhibits a touch of a master with its deep insight of the Theory of numbers. The economy in words and the sheer paucity of sweeping and elegant statements, each a gem in itself, is dazzling.

I say this because you have tackled the problem head –on and skilfully steered clear of the inevitable quadratic eqn. (the way I solved this problem) which could lead one into deep waters.

I would like to clarify that there are an infinite number of rectangles in which area = perimeter with rational numbers. But the only possible positive INTEGER values for ( y – 2 ) are 1 , 2, and 4

I also discovered that this problem is basically in disguise and similar to the 1st. problem I gave viz: Given ( x + y ) and ( x y ) find the numbers.
The ancients knew this solution (refer problem #1) from which this particular problem follows as an off shoot.

Mally.

[mfgoode]

Originally Posted by Ken_g6

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.

The volume of the pyramid VP(a,H) = 1/3*a^2*H

So, first, we need to find the height of the full pyramid, and then remove the
"apical part" (the top).
l(H)=a+H*m (m = slope = rise/run) so:
l(h)=b = a+h*m, so m = (b-a)/h, and:
l(H)=a-H*(a-b)/h

l(H) = 0

a=H*(a-b)/h

H=h*a/(a-b)

The height of the full pyramid is h*a/(a-b), so the height of the apical part is
h*a/(a-b)-h = h*(a/(a-b)-1)

Then their respective volumes are:

vp = 1/3*a^2*h*(a/(a-b))
va = 1/3*b^2*h*(a/(a-b)-1)

Note that:
a/(a-b)-1 = (a-a+b)/(a-b) = b/(a-b)


v = vp-va = 1/3*h*(a^3-b^3)/(a-b)
Now the difference of cubes a^3-b^3 factors to (a-b)*(a^2+a*b+b^2).

So:

v = 1/3*h*(a^2+a*b+b^2), for a not equal to b, and v=0 for a=b.

A bit tedious but splendid and rigorous. A work of sheer cold beauty and elegance isn't it?

It may be simpler to get the height (H) of the pyramid by similar triangles knowing that the height of the frustrum is given and bearing in mind what was known and possibly not known by the Ancients.

Now that we assumed the volume of a pyramid was known and given, the question is how did these people work out the volume, not knowing the Integral calculus ?
I havent worked this out myself but I believe Archimedes had the solution.
Please take a shot at it Ken_g6 :wink

Mally

[Ken_g6]

According to Mathworld, calculus is necessary to work out the solution. Although I don't see where I used it in my proof.

Anyway, who said Archimedes didn't know the Integral calculus? The so-called Archimedes Palimpsest contains a diagram where he basically integrates the volume of a wedge-shaped slice of a cylinder.

[mfgoode]

Originally Posted by Ken_g6

According to Mathworld, calculus is necessary to work out the solution. Although I don't see where I used it in my proof.

Anyway, who said Archimedes didn't know the Integral calculus? The so-called Archimedes Palimpsest contains a diagram where he basically integrates the volume of a wedge-shaped slice of a cylinder.

Well when Mathworld says that calculus is necessary to work out the solution (which solution?) If its the volume of the truncated pyramid
they are right as we assumed the volume of a pyramid which requires some form of summation of infinitisimals to get the result.
It is true that Archimedes was one of the early originators of the Integral calculus which came before Differentiation though these days its taught first.
The area of a circle etc are obtained by him by summation.
What I asked is how did he logically do it?
Mally

[cheesehead]

Originally Posted by Ken_g6

The so-called Archimedes Palimpsest contains a diagram where he basically integrates the volume of a wedge-shaped slice of a cylinder.

PBS recently had a program about that. IIRC the palimpsest was "lost" for several hundred years. Suppose that, even if Archimedes hadn't generalized his method, someone else had invented integral calculus after reading his work, about 1000 AD, six hundred years before Newton? :surprised

[mfgoode]

You have made an important point Cheesehead but I’d put the figure at 2000 yrs.
Tho’ we are fortunate for the discovery of the palimpsest, which has been overwritten, we derive much of our knowledge of ancient maths from the two important mathematical papyri– viz: the Papyri Rhind and the Moscow Papyrus ( 2 centuries older). These are untouched and are not overwritten.

History records that Eudoxus (408 B.C. --) a contemporary of Plato preceded both Archimedes (287.B.C. to 212 B.C.) of Syracuse and Euclid (330 B.C.-275 B.C.)
He discovered the theorem of the Golden Section and also established the ‘Method of Exhaustion’.
Thus he calculated the volume of the pyramid or a cone is 1/3rd. the Volume of a cylinder on the same base and same altitude.
This was the beginning of the Integral Calculus.

Archimedes by similar reasoning took up the baton and calculated the value of Pi ,lying between 3(1/7) and 3(10/71). This was a very great leap forward because it embodied the elements of Calculus: infinite processes and limits.
One more step and he would have been inside the door that was opened almost 2 thousand years later By Sir Isaac Newton.
Moreover he believed, unlike his contemporaries, that infinity was beyond our reach. :surprised

With the tragic death of Archimedes math progress came to a virtual standstill. He had developed the subject to the point where no further advances could be made without Algebra and Analytic Geometry.
History had to wait 17 centuries for the next major step with the advent of Rene Descartes (1596-1650A.D.)
Ref: ‘A Short History of Mathematics’ By W.W. Rouse Ball.
mally