I have five weeks to prove that, if the area of a parallelogram is half the square of the shorter diagonal, and the sides and area are of rational measure, the parallelogram MUST be a square. My actuary dad and I have been at it for a week, and we can prove a few inequalities, but nothing conclusive. Since the instructions say we can get help from ANY SOURCE except Lipscomb faculty, so long as we eventually present it in person without notes, I figured it couldn't hurt to post it here. Any hints or advice (or a full blown proof) would be wonderful. Posted by Mighty Blogger Snay (Member # 411) on :
I'd love to help you, but I'm going to have a hard enough time figuring out my taxes.
Posted by B.J. (Member # 858) on :
Are you sure you're saying this right? Is it half the square or half the square root? And when you say diagonal, are you referring to the line between two opposite corners?
Just trying to be clear about things!
B.J.
Posted by Krenim (Member # 22) on :
It would have to be half the square.
Say you have a square with sides of length s.
The diagonal of such a square would be d = SQRT(2)*s.
Switch it around for s = d/SQRT(2).
The area of the square would be A = s2.
Substituting in the diagonal gives A = (d/SQRT(2))2.
Which simplifies to A = d2/2.
Posted by Sol System (Member # 30) on :
It's all triangles somehow.
Posted by Siegfried (Member # 29) on :
A parallelogram is simply a rectangle that was built by the lowest bidder. Probably the same construction outfit that rebuilt the major road leading to my subdivision.
Posted by PsyLiam (Member # 73) on :
"Actuary dad"?
Posted by Daryus Aden (Member # 12) on :
Ok, this may be a retarded question, but how can a square have a shorter diagonal?
Posted by Cartman (Member # 256) on :
A square with sides s has two diagonals d of equal length. A parallelogram with base and height s has two diagonals dl and ds of unequal length, one of which is always longer and the other of which is always shorter than d. You are seeing now?
Posted by AndrewR (Member # 44) on :
I hated proofs too - infact I was talking about proofs with a friend at uni the otherday who did his first degree in mathematics!! Hah!
I guess I sorta understand now, but during higher maths as school in year 11 and 12 - I was always frustrated and annoyed that the bloody teacher could never 'explain' where he or HOW he STARTED the proof. It was as if he just pulled some formula or "If such and such is this, then that is that" out of his arse!
I'm guessing you have to use equations that are 'part of the "area"' of the equation or problem you are trying to proove? Doesn't that mean that you need a fair understanding of a specialise area of that mathematics branch you are working on at that time? How long does a proof have to be?
So you use "knowns" to "prove" an "unknown"? Sorta like doing experiments, only your tools are equations?
Anyway goodluck.
Posted by AndrewR (Member # 44) on :
Oh, an actual help Omega - have you tried some of the sci.* newsgroups - I'm sure there is a maths one there. Try looking and posting through Google Groups - I personally like Forte Free Agent to peruse my newsgroup messages (although I haven't in a while - I used to post there WAY more than I do here now - It was a little excessive!) Posted by Cartman (Member # 256) on :
So you use "knowns" to "prove" an "unknown"?
Erh, a real mathematician would probably cringe at this explanation, but you use something that you know and is proven (that the diagonals of a square are equal in length) to deduce something else that you also know but isn't proven (that, if half of the square of the shortest diagonal of a parallelogram is equal to its surface area, the parallelogram is in fact a square). Or... something.
Posted by Sol System (Member # 30) on :
Re image: huh.
Re geometrical proofs: they were my favorite thing, actually, in high school or junior high, prior to that point in school where I just sort of gave up on things (Algebra II).
Re nerds: I'm totally reading David Foster Wallace's Everything and More right now. (It is about Cantor and infinity and, I don't know, ordinals or something. Math is hard. I like shopping!)
Posted by AndrewR (Member # 44) on :
quote:Originally posted by Sol System: Math is hard. I like shopping!)
I didn't realise you were a blonde, 14 year old girl! Posted by PsyLiam (Member # 73) on :
Why else do you think he's so popular here? His sparkling personality? I laugh heartily.
Posted by TSN (Member # 31) on :
I thought it was because of his gigantic cock.
Oh, wait, subtlety...
Posted by Siegfried (Member # 29) on :
I LOVE TEH C OCK!!1!
Posted by The Mighty Monkey of Mim (Member # 646) on :
I, too, love the cock.
Posted by Cartman (Member # 256) on :
Re huh re image: what?
Posted by Sol System (Member # 30) on :
It's just not something you see every day, is all.
Posted by AndrewR (Member # 44) on :
quote:Originally posted by The Mighty Monkey of Mim: I, too, love the cock.
Tastes like chicken? Posted by AndrewR (Member # 44) on :
duo posto
Posted by Nim' (Member # 205) on :
Ipso facto
Posted by Siegfried (Member # 29) on :
Pax Cock ad absurdum ad maiorem Dei gloriam.
Posted by TSN (Member # 31) on :
I think that does it. If the area of a pythagorean triangle can't be square, then 1/2eh must not be square. That means 2eh must not be square, being a square times a non-square. But 2eh MUST be a square, otherwise b = e+h +-sqrt(2eh) has no rational solution, which it must. Therefore, we're not dealing with a pythagorean triangle, e=0, rectangle, square, QED.
Of course, I have to be able to reproduce and explain Fermat's original proof, and that explanation makes some jumps I'm thus far unable to follow. But there may yet be hope. Posted by Jason Abbadon (Member # 882) on :
The proof is in the pudding.
Hmmmm....pudding.
Posted by TSN (Member # 31) on :
If you have to replicate Fermat, just scrawl out "I have discovered a marvelous proof which is too large to fit on this page" and turn it in. If the teacher calls you on it, just die without explaining yourself.
Posted by AndrewR (Member # 44) on :
quote:Originally posted by Siegfried: Pax Cock
Peace cock? Posted by Jason Abbadon (Member # 882) on :
Just scrawl E=Mc2 on the page with a yellow post-it attached explaining "It's all relative! The voices were right!"
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