"Squaring a rectangle"

locum-motion

I reckoned I should start a new thread rather than derail the other one.

In this thread, user Armelodie mentioned the new (to me!) concept of "squaring a rectangle", that is, constructing with straight edge and compass a square whose area is equal to that of a given rectangle.

The concept is shown in this diagram:


Now, I understand the procedure - find M such that |AM| is half of the sum of the lengths of two adjacent sides of the rectangle, draw circle of radius |AM|, find F the point where this circle intersects the extrapolation of BC, and then the square on BF is the square you want.

What I don't understand is the proof. I don't recall ever being taught this theorem/construction/whatever-it's-called in school, so I've only got Armelodie's word for it that it's true. I've been trying to prove it to myself, without much success so far.

My question is this: can it be proven by manipulation of Pythagoras' Theorem regarding the triangle MBF? That's the tack I've been taking, but I'm getting bogged down in lots of brackets, fractions and squares. I'm going to bed now, so I'm not going to keep going with it now. I might have another go tomorrow.

Don't give me the whole thing, just let me know if I'm on the right track. Thanks.

CJC86

locum-motion wrote: »

My question is this: can it be proven by manipulation of Pythagoras' Theorem regarding the triangle MBF? That's the tack I've been taking, but I'm getting bogged down in lots of brackets, fractions and squares. I'm going to bed now, so I'm not going to keep going with it now. I might have another go tomorrow.

Don't give me the whole thing, just let me know if I'm on the right track. Thanks.

You're on the right track here. To prove the areas are equal you need to show that AB*BC=BF^2, so try writing AM and MB in terms of AB and BC.

MathsManiac

Or, have you ever encountered the "intersecting chords theorem"?

locum-motion

MathsManiac wrote: »

Or, have you ever encountered the "intersecting chords theorem"?

Possibly, but I don't recall it if I did or not!

But to use it to prove the Squaring the Rectangle thing, I'd need to prove it first.

Whereas I know Pythagoras is true, so I'll use that!

MathsManiac

There is of course an entire edifice leading up to most useful theorems. The "intersecting chords" theorem is easily proved using the "similar triangles" theorem (If two triangles are similar, then the lengths of corresponding sides are in proportion; you probably used this theorem on your way to proving Pythagoras' theorem in the first place) and the theorem that states that two angles standing on the same arc are equal.

Along similar lines, you could try to prove your result directly by showing that triangles ABC and FBE are similar (i.e., equiangular). The resulting proportionality of the sides then gives the result you want.

locum-motion

Describing the lengths of the various lines in terms like |AM| is very tedious to type/write, so I tried to describe them in terms of x, y and z.

Let x = |AB| = the long side of the rectangle
Let y = |BC| = the short side of the rectangle
Let z = |BF| = the side of the square

The hypotenuse of the triangle MBF is the radius of the circle, so it's (x + y)/2
The other two sides are z and x-[(x+y)/2]

Therefore:

{(x + y)/2}^2 = z^2 + {x-[(x+y)/2]}^2

Rearranging gives:

z^2 = {(x + y)/2}^2 - {x-[(x+y)/2]}^2

So basically that as far as I got last night. I know I have to reduce that down as far as

z^2 = xy

but - as I said last night - I got bogged down in all the brackets, fractions and powers. It's gotten late on me again, so I'm not going to try again now, but I wanted to let you know how far I got.

Thanks for your answers.

MathsManiac

Well, you're practically there. Just use the identity A^2 - B^2 = (A - (A + , and Bob's your uncle.

locum-motion

MathsManiac wrote: »

There is of course an entire edifice leading up to most useful theorems. ...

I sort of knew the lower floors of that edifice 25 and 24 years ago when I did my Leaving Certs (yes, plural)! Now, I can't remember which floors are closer to the ground, which ones are halfway up, or even whether I visited some of them at all. There are some floors of that edifice that my Ordinary Level elevator didn't have access to!

ps Nice analogy, by the way. Never heard it described thus before.

locum-motion

z^2 = {(x + y)/2}^2 - {x-[(x+y)/2]}^2

= < {(x + y)/2} - {x-[(x+y)/2]} > * < {(x + y)/2} + {x-[(x+y)/2] }>

(< & > aren't less/greater than in this case; I just needed a 4th different type of bracket)

= <2 {(x + y)/2} - x > * < {(x + y)/2} - {(x + y)/2} + x>
= <(x+y) - x> * <x>
= y*x


Thanks MathManiac. Your post #7 is what got me through it.

Mwalimu

MathsManiac wrote: »

Along similar lines, you could try to prove your result directly by showing that triangles ABC and FBE are similar (i.e., equiangular). The resulting proportionality of the sides then gives the result you want.

I've just come back to this problem and something doesn't ring true. I don't see that triangles ABC and FBE are similar. If they were, then AB/BC would be equal to FB/BE. However, in the construction, BE is equal to BC, which would mean that AB = FB, which it isn't.

MathsManiac

Mwalimu wrote: »

I've just come back to this problem and something doesn't ring true. I don't see that triangles ABC and FBE are similar. If they were, then AB/BC would be equal to FB/BE. However, in the construction, BE is equal to BC, which would mean that AB = FB, which it isn't.

I beg your pardon. You are correct. I was combining two proofs in my head and wasn't careful enough. Extend BC down until it cuts the circle at C'. It is ABC' that is similar to FBE.

When you then write the ratio and cross-multiply it, you get the result you want (since BC' = BF, as the diameter drawn perpendicular to a chord bisects the chord).

Mwalimu

MathsManiac wrote: »

I beg your pardon. You are correct. I was combining two proofs in my head and wasn't careful enough. Extend BC down until it cuts the circle at C'. It is ABC' that is similar to FBE.

When you then write the ratio and cross-multiply it, you get the result you want (since BC' = BF, as the diameter drawn perpendicular to a chord bisects the chord).

Now it all makes sense! I really like the way it combines various bits of knowledge about triangles and circles to establish the result, by contrast with the algebraic solution, which can be mesmerising for some students as well as being more prone to mistakes in the expansion/calculation.:)