Geometry

[Urban Ranger]

Alternative proof:

Drop a line from A to BC so that AOC is a right angle.

AO² = AC² - CO² = AD² - DO²

Since CO² > DO², AC² > AD²

[KrazyHorse]

They already did that one

[Urban Ranger]

Did? Bummer.

[notyoueither]

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

Who are you, and what have you done to Burton Cummings?

[Solomwi]

Originally posted by 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.

I see the logic of proving it by showing D as a point on a radius, but in what way is the "proof" as presented illegitimate? If you're speaking of rigorous mathematical proof, I can buy that. If you're just speaking of the logic behind it, though, I need a little more explanation, please.

I do see one flaw in what I said, which is remedied by changing what you quoted to "showing that D is a non-endpoint on a chord," and showing D as a point on a radius suffers from the same nitpick.

[KrazyHorse]

Originally posted by Solomwi


I see the logic of proving it by showing D as a point on a radius, but in what way is the "proof" as presented illegitimate?

I mean that you haven't done what is by far the hardest part of the "proof"

[KrazyHorse]

Secondly, why are you talking about chords? Chords are simply line segments connecting two points on a circle. The distance we are interested in is that between what would be the centre of your circle (A) and some point which is not on the arc of the circle (D). Chords which go through the centre are called diameters, and half of a diameter is a radius.

[KrazyHorse]

Now, D is obviously on the chord BC, but while it is quite obviously true that the distance from any point on a chord of a circle to the centre of the circle is less than the radius of the circle, the proof of this requires at least as much as the proof of the original statement which Siro provided.

[Solomwi]

First post: gotcha, and don't disagree at all. That's why I used quotes around proof. I was simply throwing out the logical progression for it. Doing the hard part would require overcoming my laziness.

Second post: because any point that lies on a chord, and isn't an endpoint of the chord, is necessarily inside the circle, i.e., not on the arc of the circle, regardless of whether the chord passes through the center. The line from the center to any point inside the circle will necessarily be shorter than the radius. Of course, any point inside the circle is on a diameter, as well as an infinite number of chords that aren't diameters. Any one of them will do for Siro's purpose. I don't think we're saying different things at all on this point.

[Solomwi]

Originally posted by KrazyHorse
Now, D is obviously on the chord BC, but while it is quite obviously true that the distance from any point on a chord of a circle to the centre of the circle is less than the radius of the circle, the proof of this requires at least as much as the proof of the original statement which Siro provided.

X-post with the one above.

Right, and that's what I was referring to in my response to Az.

[Solomwi]

dp

[KrazyHorse]

inside the circle

Be very very careful here. There is no real definition of inside in classical geometry.

a) You have not proven to me that all points on chords in the circle are "inside" the circle

b) You have not proven to me that being "inside" the circle means that I'm closer to the centre that a radius

[KrazyHorse]

The proper definition of inside and outside had to wait for the Jordan Curve Theorem, which was finally proved in 1905, so it hardly forms part of classical geometry

[Solomwi]

Indeed, hence my CYA follow-up.

Works if for some reason you come across this on a jobsite, but not in the classroom.

[Az]

KH pwned us.