Geometry

[Az]

Hey I was thinking of trying to reply but if you two are having a private conversation...

[Sirotnikov]

thank you everyone.

now i will give you all a funny link:

http://video.google.com/videoplay?docid=-3022812996105150373&q=hung+up

[Zkribbler]

Originally posted by Mercator
How about this...

AC=AB

Now, draw a line straight down from A, giving you point E.

AB²=AE² + EB²
AD²=AE² + ED²

ED < EB thus, AD < AC/AB

Damn! I was going to say that!

[Winston]

The picture shows a triangle.

[Solomwi]

Not that tere's a thing in the world wrong with the advice given, but out of curiousity, is it "legal" for you to use circles in the proof? My first thought was drawing a circle with the center at A and radius of AB. Then all that remains is showing that D is a point on a chord, and that the radius AC must therefore be longer than the line from center to chord AD.

[Mercator]

Originally posted by civman2000
Yeah, but I like your presentation of it better than mine.

Me too

Originally posted by Zkribbler
Damn! I was going to say that!

Tell me about it!

[Sirotnikov]

Originally posted by Solomwi
Not that tere's a thing in the world wrong with the advice given, but out of curiousity, is it "legal" for you to use circles in the proof? My first thought was drawing a circle with the center at A and radius of AB. Then all that remains is showing that D is a point on a chord, and that the radius AC must therefore be longer than the line from center to chord AD.

i'm not sure i'm supposed to do that...
i thought about it too.

[Sirotnikov]

Originally posted by Sirotnikov
thank you everyone.

now i will give you all a funny link:

http://video.google.com/videoplay?do...0373&q=hung+up

check this out though.

it is quite amazing.

[Winston]

That thing had the logo of one of our state-run TV stations in the corner. I think I recognized the guy at the keyboards as one Tommy Seebach, multiple time Eurovision Song Contest entrant in the 1970's and 80's, who is now deceased.

But I kept wishing they'd put on a proper ABBA song instead of the low-talented mumbo-jumbo that was pounding on.

[Az]

Not that tere's a thing in the world wrong with the advice given, but out of curiousity, is it "legal" for you to use circles in the proof? My first thought was drawing a circle with the center at A and radius of AB. Then all that remains is showing that D is a point on a chord, and that the radius AC must therefore be longer than the line from center to chord AD.

I attempted to use AB, AC as radii, but I haven't figured it out. I like your proof.

[Solomwi]

Only thing is that, while it works beautifully as a real world type solution, I suspect that following the rules of geometric proof will just lead you back down the paths illustrated above, with the water muddied a bit by the inclusion of the circle. It's been 15 years since I took geometry, though, so my memory of what tools are available for a proper proof is not to be relied upon at all.

[Az]

Yeah, well, you have to prove the OP to get to use it in the theorem we just used, so it's all very vague. Generally, all of the theorems of euclidian geometry are assumed, I think.

[Elok]

I can't put it in terms of a proof, only intuitively. Not that I need to, since Mercator did. But, if the shortest distance between A and the bottom BC is a straight perpendicular, it stands to reason that the longest line would be obtained by drawing a line from A to a point as far off to the side as possible. Since the given limits for the placement of such a point are the sides of the triangle, AC must be longer than AD no matter where D is. Which is basically the same principle already employed. But is it impossible to write a formal proof without squaring the sides?

Oh, wait. Duh. If the longest side of triangle ABD must be AB, given that AB=AC, AC must be longer than AD. You could just write that up formally, however that's done. Been years and years since I took geometry. But there you go, unnecessary alternate solution.

[Andemagne]

take a compass(divider, a toll that you circles with) and draw a circle, so that point A is the centre of the circle and B and C are on the circle. thus AB (and AC) becomes the radius of the circle. AD doesn't reach the circle, so it is shorter than AB or AC.


EDIT: and yes, I do think that using circles in this proof is perfectly legal.


EDIT2: why are we asked this kind of question anyway?

[KrazyHorse]

The "circle proof" as presented here is completely illegitimate.

showing that D is a point on a chord

What is really needed is to show that D is a point on a radius, and this is nontrivial. Actually, it might be harder to prove than the original statement.