You are correct, but only so far as the time portion, so there is a "trick".

Another way to look at is Man A is running 3x faster than man C, and Man B is running 2x faster than Man C. Taking your previous calculations, man A runs three laps, Man B 2 laps and Man  C 1 lap, but they all meet at the start/finish line.

This is, of course, assuming they run identical paths, which breaks the laws of physics (three people cannot occupy the exact same space at the same time). Running different paths, as in a real race, presents a much more challenging equation, and switching paths mid race adds in variables that fry even my nerd brain.  :D

Ant
sounds like you might be ready for the wriggly worm problem i was reading about on the metro back from work today. it's the idea that a worm gets restless at night so it wriggles randomly, and its parents need to find a blanket that will cover it up from the cold, but blanket material is expensive so they're trying to find the smallest blanket possible to cover the wriggling worm.

first solution is that they assume the wriggling worm wriggles from one of the endpoints on its body, so a circle with a radius equivalent to the fully stretched length of the worm is assumed to be the blanket. this is actually a very bad solution, but it has the advantage of being simple.

then some wiseacre sez if you assume the worm is stationary in the middle then you come up with an ellipse defined by both endpoints of the wriggling worm, head and tail. yeah? i followed as long as that but then it went into this whole bit about "Cantor sets," which emerge from this repetitive operation of trisecting line segments, leaving the middle segment empty, and then trisecting the two resulting line segments on the ends and repeating the result iteratively. you end up with this line that's all endpoint and no segment, or something. it got really weird at that point and i just started staring out of the window wondering when i'd reach my metro stop. interesting book though.

http://www.amazon.com/Game-Set-Math-Conundrums-Mathematics/dp/0140132376

My brain just exploded.

My brain just exploded.
quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.

please show your work.

You are correct, but only so far as the time portion, so there is a "trick".

Another way to look at is Man A is running 3x faster than man C, and Man B is running 2x faster than Man C. Taking your previous calculations, man A runs three laps, Man B 2 laps and Man  C 1 lap, but they all meet at the start/finish line.

This is, of course, assuming they run identical paths, which breaks the laws of physics (three people cannot occupy the exact same space at the same time). Running different paths, as in a real race, presents a much more challenging equation, and switching paths mid race adds in variables that fry even my nerd brain.  :D

(unless they happen to run in what looks like a 3-phase power wave (i.e., like braiding hair), in which case they conveniently wind up having travelled the same distance despite their 120* offsets. This still doesn't solve the physics portion of the equation as two people would occupy the same space at the same time, but close enough... )

http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/Semi/03269.png

Ant



This is all well and good but either you are over thinking the problem or I'm under thinking  it. I assume the men starting as the
same point....90 yards from this finish line. So this means the starting postions would be at different locations just like a regular
track meet. They start in a straight line in a dash rash, any race around the track is scattered to allow a fair race for the outer
racers. Horses, on the other hand, are SOL.

quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.

please show your work.


Is there enough slack in the blanket to allow for differing paramedic velocities?

Is there enough slack in the blanket to allow for differing paramedic velocities?
yes but it must be absolutely held to a minimum because of the aforementioned blanket-being-expensive provision.

i cant friggin solve word problems to save my life but i think i've managed to actually come up with one that would get albert einstein himself going, um, dude, no clue here. :P

karaokelgebra! i think i can make it catch on.

This is all well and good but either you are over thinking the problem or I'm under thinking  it. I assume the men starting as the
same point....90 yards from this finish line. So this means the starting postions would be at different locations just like a regular
track meet. They start in a straight line in a dash rash, any race around the track is scattered to allow a fair race for the outer
racers. Horses, on the other hand, are SOL.


Yes, they do start at the same point. If it were a 90 yard dash, the problem of concentricity would not be an issue: each runner runs 90 yards in a straight line (in that case it's irrevelant what lane they start in because they each travel 90 yards).

There also would be a different answer: they would never meet (until sometime after all three crossed the finish line and drank Gatorade togther: an undefined/undeterminable answer).

Seeing as how they run in a circle, the runners in the outside lanes have to run a longer distance than those on the inside, which is why real races always have runners taking the shortest path (the inside lane). Staying in the outermost lane of a typical 8-lane run would put a runner at an extreme disadvantage (auto racing, like NASCAR, is a different animal altogether, but that's an algebra problem for another day).

As a purely mathematical problem, Kin and my answers make the whole problem solved. Bring in physics, and this hypothetical race becomes a nightmare to correctly calculate.


sounds like you might be ready for the wriggly worm problem i was reading about on the metro back from work today. it's the idea that a worm gets restless at night so it wriggles randomly, and its parents need to find a blanket that will cover it up from the cold, but blanket material is expensive so they're trying to find the smallest blanket possible to cover the wriggling worm.

first solution is that they assume the wriggling worm wriggles from one of the endpoints on its body, so a circle with a radius equivalent to the fully stretched length of the worm is assumed to be the blanket. this is actually a very bad solution, but it has the advantage of being simple.

then some wiseacre sez if you assume the worm is stationary in the middle then you come up with an ellipse defined by both endpoints of the wriggling worm, head and tail. yeah? i followed as long as that but then it went into this whole bit about "Cantor sets," which emerge from this repetitive operation of trisecting line segments, leaving the middle segment empty, and then trisecting the two resulting line segments on the ends and repeating the result iteratively. you end up with this line that's all endpoint and no segment, or something. it got really weird at that point and i just started staring out of the window wondering when i'd reach my metro stop. interesting book though.

http://www.amazon.com/Game-Set-Math-Conundrums-Mathematics/dp/0140132376


I'm not familiar with "Cantor Sets" either, but assuming the blanket can hold down the worm so that it can't escape, I imagine the answer would be close to the area of two circles with a combined area of 2 x 1/2L2 x Pi, as opposed to Pi x r2 (plus a little undetermined extra for the middle point of the worm, which might not be covered by the circles)

IOW, if the worm if 2 ft long, pi r2 would be a blanket of ~12.56 sq ft  (the worm moves as in the first solution you presented: length = radius).

My solution would be 2 x (1ft2 x pi) + (undetermined negligable amount for middle section), or ~6.28 sq ft.

Of course, that again is assuming the blanket can hold down the worm. As it wiggles randomly, I don't think there is a true solution to this problem.


My brain just exploded.


Lemme guess: you divided by zero?  ;D

Ant

Lemme guess: you divided by zero?  ;D

Ant


Her issue appears to be undefined.

Her issue appears to be undefined.


TTotM? 3-7 days of variables no mathematician will ever completely solve for...  ;D

*ducks sharp and heavy objects thrown at me by Jess*

Ant

http://i123.photobucket.com/albums/o283/Nerdprincess1980/Funny/catnip.jpg

quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.

please show your work.



I like to consider myself more the artistic sort, but the equation is fool proof.

I'm not familiar with "Cantor Sets" either, but assuming the blanket can hold down the worm so that it can't escape, I imagine the answer would be close to the area of two circles with a combined area of 2 x 1/2L2 x Pi, as opposed to Pi x r2 (plus a little undetermined extra for the middle point of the worm, which might not be covered by the circles)
http://en.wikipedia.org/wiki/Cantor_set

now it's clear as mud! you're welcome.

the upshot of what the book was saying, i think, is that questions asking to determine the least possible area of something are often hard to solve, i think for reasons kinda like why it's hard to prove a negative. you can establish that something is non-zero (that's sorta what the cantor set is, i think? nonzero but as close to zero as makes no odds) but beyond that it's easy to propose a solution that might be the least possible area, but hard to prove that there isn't some undiscovered alternative that could be less still.

"Let the regions within or on the indicated rectangles in the adjoining diagram represent the subsets R, S and T of a set U. Redraw and shade the region representing the subset of U specified in each exercise."

http://i28.tinypic.com/5dut6c.gif

1.

http://i31.tinypic.com/zmdrub.gif

http://i27.tinypic.com/10cvreb.gif

http://i31.tinypic.com/2rpzmhl.gif

http://i31.tinypic.com/eqv3wy.gif

that's a difficulty C. seemed pretty easy to me.

What exactly is Cramer's rule?

I think, probably, the best way to answer this question would be using the following method:

http://upload.wikimedia.org/math/e/9/6/e96379eeb8832176cbe89b39245f22a4.png
http://upload.wikimedia.org/math/d/d/e/dde2390688d8f36ec284e0ab274e3ad8.png
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://upload.wikimedia.org/wikipedia/en/6/60/CentralTendencyLV.jpg
http://performingarts.ufl.edu/wp-content/uploads/2007/05/parsons-dance-company-man-flying-cropped.jpg
http://earthscience.files.wordpress.com/2007/05/tornado.jpg
=4

that must be new math!

I think, probably, the best way to answer this question would be using the following method:

http://upload.wikimedia.org/math/e/9/6/e96379eeb8832176cbe89b39245f22a4.png
http://upload.wikimedia.org/math/d/d/e/dde2390688d8f36ec284e0ab274e3ad8.png
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://upload.wikimedia.org/wikipedia/en/6/60/CentralTendencyLV.jpg
http://performingarts.ufl.edu/wp-content/uploads/2007/05/parsons-dance-company-man-flying-cropped.jpg
http://earthscience.files.wordpress.com/2007/05/tornado.jpg
=4


It's going to take me a few hours to get it.  Eventually I will.

why does math hate babboons? :\'(

why does math hate babboons? :\'(


Because math have a Superiority complex, anything less is nonexistant.

Because math have a Superiority complex, anything less is nonexistant.
http://i32.tinypic.com/34e8b2r.gif

:(

Good math.

http://static.pyzam.com/img/funnypics/misc/1442.jpg


http://static.pyzam.com/img/funnypics/misc/fun32.jpg


http://static.pyzam.com/img/funnypics/misc/evilgirls.jpg