1. ## proportions of circle

Prove that there is a magnitude p such that the area of a circle of radius r is to the area of a square of side r as p is to 1.

2. Originally Posted by rmpatel5
Prove that there is a magnitude p such that the area of a circle of radius r is to the area of a square of side r as p is to 1.
Is "p" supposed to be $\pi$? If so then we can solve this. The area of a circle of radius r is $\pi r^2$. The area of a square of side r is $r^2$, so the ratio of these two areas is
$\frac{\pi r^2}{r^2} = \frac{\pi}{1}$

-Dan

3. Originally Posted by topsquark
Is "p" supposed to be $\pi$? If so then we can solve this. The area of a circle of radius r is $\pi r^2$. The area of a square of side r is $r^2$, so the ratio of these two areas is
$\frac{\pi r^2}{r^2} = \frac{\pi}{1}$

-Dan
no..
The proof depends on three earlier results in the Elements:
1. Given two magnitudes AB > CD, if from AB a quantity more than half is removed, and from the remainder a a quantity more than half is removed, and so on, eventually what is left will be less than CD.
Proof One of the axiomatic properties of magnitudes is that there is a positive integer n such that n CD > AB, and a property of intgers is that there is a positive integer m such that 2m > n. Thus there is an m such that CD : AB > 1:2m. Then the remainder after the m -th cut off is < 1/(2m)AB < CD.

2. Let P and Q be regular similar polygons with correspnding side lengths p and q . Then Area P : Area Q = p2 : q2.
Since this is true for similar triangles, by the construction of Thales, it is true for all similar polygons by decomposition.

3. The Law of Trichotomy (not explicitly stated, but used throughout): If x and y are magnitudes, then x < y or x = y or x > y . To us, this is an axiom of the real numbers; to the Greeks it was an unstated axiom of magnitudes.
Circles C and D are to one another as the squares on their diameters c and d .
Either Area C : Area D = c2 : d2 or < c2 : d2 or > c2 : d2.
Suppose Area C : Area D < c2 : d2, so Area D > Area C.(d2/c2) . Let S = Area C.(d2/c2) , so S < Area D .
By Proposition 1 of Book X, we can continually remove from Area D larger and larger polygons inscribed in D until we get a polygon P such that S < Area P < Area D, so Area P > Area C. d2/c2.
Now inscribe in C a polygon Q similar to P so Area Q < Area C and Area Q : Area P = c2:d2. Thus Area P = Area Q . (d2/c2) < Area C. (d2/c2), a contradiction.
A similar contradiction arises if Area C : Area D > c2 : d2 and we inscribe a polygon in C .