Chief Delphi - Math Quiz 9

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- - Math Quiz 9 (http://www.chiefdelphi.com/forums/showthread.php?t=149436)

Quote:

Originally Posted by Hitchhiker 42 (Post 1602388)

EDIT: Gosh darn I only did those with both points touching the the edges. Lemme work on that.

Also, don't forget to weight by the number of segments - the x coordinates near the center contain more segments than those near the left and right edges.

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle." We've got another build session this evening getting ready for Red Stick Rumble, so I don't know how much I'll get done on this tonight.

Addition:
Hitchhiker, thanks for answering a different question. While walking across the street to get lunch, I think I put some nails in this coffin. Consider these two questions:

What is the average length of a vertical chord of the unit circle?
What is the average length of a chord of the unit circle which passes through (-1,0)?

You answered the first - the area of the unit circle divided by its width: pi/2.
I answered the second in my paper a score or so posts back (first problem), in a form that is not incompatible with Ether's method, provided angle is checked before length: 4/pi.
If you answer the second using Ether's method but checking length first, then picking one of two possible angles, you will get an average value of 1.

Completing reducto ad absurdum is left to the reader - back to work!

Quote:

Originally Posted by Ether (Post 1602392)

I just can't trip you up, can I :)

Well, not that easily.

Quote:

Originally Posted by GeeTwo (Post 1602391)

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle."

I just can't trip you up, can I :)

Quote:

Originally Posted by FiMFanatic (Post 1597038)

...you can draw a ton more 0.000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 01 lines than you can 1 inch (or larger on a diagonal) lines.

Quote:

Originally Posted by FiMFanatic (Post 1597049)

...fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

Quote:

Originally Posted by FiMFanatic (Post 1597051)

Quote:

Originally Posted by z_beeblebrox (Post 1597050)

I think the problem is more "Choose two random points inside by a 1" square and measure the distance between them. Repeat infinitely. What's the average of those measurements?"

I agree then - your answer would make sense if that were the question. However, it was:

What's the average length of all the line segments which can be drawn inside a 1 inch square?

@FimFanatic: Are you still reading this thread? I've done the best "devil's advocate" job I can arguing your point. Gus has answered every challenge. Has this discussion been helpful for you?

Quote:

Originally Posted by Ether (Post 1602408)

I've done the best "devil's advocate" job I can arguing your point.

Whew! I've been torn among several ideas:

Ether needs coffee.
Ether has started trolling.
Ether's CD account has been hijacked by a technically competent troll.
Ether is up to a deep teaching moment.

I'm so glad to read that it was #4!

I'm not sure why CD wouldn't let me attach a file to a post on this thread, but I've posted my solutions to the cube at this "white paper": https://www.chiefdelphi.com/media/papers/3289.

For any who find this white paper a bit too much, let me simply note here that you don't want to see all of my blind alleys and solutions which worked but were much uglier than what I've shown here.

Edit: And OBTW, if anyone feels that my substitutions showed a bit too much prescience, that's ok, because I'm posting this on my birthday, and we should always expect prescience on our birthdays, no?