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Maths Quiz!

  1. Sep 3, 2003 #1

    jcsd

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    Okay Maths quiz! normal rules apply (first to answer question correctly gets to ask the next question, etc.):

    1) √(5 + √(24)) = ? in exact surd form
     
    Last edited: Sep 3, 2003
  2. jcsd
  3. Sep 3, 2003 #2

    Hurkyl

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    That font doesn't work for me. :frown:
     
  4. Sep 3, 2003 #3

    jcsd

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    Font corrected :)
     
  5. Sep 3, 2003 #4

    Hurkyl

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    You looking for something like

    √2 + √3

    ?
     
  6. Sep 3, 2003 #5

    jcsd

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    Correct!
     
  7. Sep 3, 2003 #6

    Hurkyl

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    Hrm, my favorite math puzzle is a picture puzzle, but that'd be too hard to describe in words. :frown:


    Ok, here's one:

    A carpeting company has taken an order to carpet an annulus-shaped room (the region between two cocentric circles)... however their drunken surveyor only measured the length of the longest straight line that can be drawn in the room! The manager complained to the resident mathematician about the problem, because this measurement didn't seem to be enough to find the area of the floor, but the mathematician said "But there's a formula for that..." to which the manager responded "Oh, there is? I know what it is then" and walked away.

    What is the formula?
     
  8. Sep 3, 2003 #7

    jcsd

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    πl2/2 ?
     
  9. Sep 3, 2003 #8

    Hurkyl

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    not quite!
     
  10. Sep 3, 2003 #9

    jcsd

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    3πl2/4?
     
  11. Sep 3, 2003 #10

    Hurkyl

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    nope!
     
  12. Sep 3, 2003 #11

    Integral

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    Here is how I see it.
    R= Radius of Large circle
    r= radius of small concentric circle
    D = measurement.

    The measured distance must be tangent to the smaller circle so we have
    ;
    The Radius of the larger circle is the Hypotenuse of a right triangle with sides r and D/2 so

    (D/2)2 = (R2 -r2)= D2/4

    The Area we need is:
    A = [pi] R2-[pi]r2
    A = [pi](R2-r2)
    so
    A = [pi]D2/4
     
  13. Sep 3, 2003 #12

    ahrkron

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    Since (according to the local mathematician) D is enough information to obtain the area, we have to conclude that the result is independent of R and r, which means that we can choose whichever values we like. Using r=0, we have that R = D/2, and the area is [pi]R2, or [pi](D/2)2, and should be the result for all cases.
     
  14. Sep 4, 2003 #13

    marcus

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    If Maths Quiz is played according to similar rules to Astronomy Q/
    A then Hurkyl should speak up and say it is Integral's turn because he was the "first to answer question correctly"

    And if anybody forgets to do what theyre supposed to do, according to the Nicool rules after "2 or 3 days" someone can
    do what's needed to keep the game going. Like, just go ahead and ask another question.

    Because JCSD started the gamethread he can decide on how to adapt Nicool's rules as necessary (which are in Nicool's post at the start of the Astronomy gamethread)

    Those were both neat quiz questions I hope Hurkyl shows up presently, and awards Ahrkron an honor point for elegance
     
  15. Sep 4, 2003 #14

    jcsd

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    It's annoying because I had the right answer but I worked it out using a very roundabout method (using chords and tangents) and I made a mistake :(
     
  16. Sep 4, 2003 #15

    Integral

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    I agree with this. Mine is a brute force solutoion. Hurkyl arrived at the same conclusion with simple logic. Much nicer solution. Hurkyl, if you have a problem on the par with the ones posted so far, please post it. I would have trouble comming up with a good one.
     
  17. Sep 4, 2003 #16

    Hurkyl

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    Yep, Integral got the right answer, and Ahrkron got the clever solution!

    I don't have any other good ones off hand, though...
     
  18. Sep 4, 2003 #17

    chroot

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    As a Math Doctor on Drexel's "Ask Dr. Math" website, I see a whole lot of clever math puzzles come through the door. I will try to post some soon!

    - Warren
     
  19. Sep 4, 2003 #18

    marcus

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    referree call

    Hurkyl asked, Integral answered (seconded by Ahrkron)
    The turn goes to Integral, who passes back to Hurkyl
    and Hurkyl declines the turn.

    Therefore it is Ahrkron's turn. Ask a good one, Ahrkron!
    If you dont happen to come up with something subtle then
    ask something
    straightforward. This could be a good game.
     
  20. Sep 5, 2003 #19

    ahrkron

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    OK, here it is:

    You probably know the following riddle:


    John leaves home one morning and does the following:
    - Stands in front of his house, facing South
    - walks straight ahead for 1 mile,
    - turns left and walks 1 mile, sees a bear
    - turns left and walks 1 mile,
    at which point he finds himself back home.

    Riddle's question: What is the color of the bear?


    The solution (the color of the bear) is unique. My question is: if we take the bear out of the question, how many different ways can such a walk be achieved? explain!

    I will give some hints if necessary.
     
    Last edited: Sep 5, 2003
  21. Sep 5, 2003 #20

    Integral

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    Couple of things I would like to see.

    A derivation of some sort, or at least an indication of the logic used should be a part of the solution..... There may be some who are wondering how to get the solution to that first problem (think quadratic !)

    Also I see quality problems as important as who posts them. The winner gets first shot, but if he does not have a good proplem I would rather it be passed to someone who has one.
     
  22. Sep 5, 2003 #21
    White.

    He was at the north pole therefore the only bear he saw would have been a polar bear.
     
  23. Sep 5, 2003 #22
    Oh, I didn't see the real question!

    If I understand the question correctly then there would be an infinite number of places on Earth where someone could move 1 mile forward, 1 mile east then 1 mile backward and be back to where he started.

    Take a point on Earth that is very close to the north-pole. (say a couple of meters, It only has to be close enough). Understand that this parallel of latitude at our chosen point circles the north pole and is 1/n miles long.

    Situate John's house at a point which is exactly 1 mile south of this parallel.

    Now, John walks forward (north) to our chosen line of latitude (call point P). Turns and walks east 1 mile, walking n times around that latitude and ending up back at P. Now he simply walks one mile backward to where he started.

    Simply, take any point on Earth (Nth hemisphere) walk north x miles and stop at the right place before you reach the north pole. Turn east and walk that x miles again (if you stopped at the right spot your x miles should end up back at that spot) ending where you first turned east. Walk south x miles again.

    Therefore there are an infinite number of ways this can be acheived. NOTE: This is possible because we are on a polar surface!
     
  24. Sep 5, 2003 #23

    marcus

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    Oxymoron! please add to your solution that the house can be near the south pole. This will keep nitpickers quiet (of course at PF everybody is nice and there are no nitpickers)

    The house is facing south. According to the original wording he must set out walking south. Then after a mile he will turn left and (presumably walk one mile East going around and around the south pole) then he will turn left again and walk north to his house.

    To remind you of the precise wording I will "take the bear out of the question" and recopy the puzzle Ahrkron gave:



    John leaves home one morning and does the following:
    - Stands in front of his house, facing South
    - walks straight ahead for 1 mile,
    - turns left and walks 1 mile,
    - turns left and walks 1 mile,
    at which point he finds himself back home.
    ...how many different ways can such a walk be achieved?


    As I understand it, Oxymoron, you are an "Ozzie" that is an Australian, so probably you really mean south pole when you say north pole. We understand that Ozzies are upsidedown in some respects. And if I understand you correctly you are saying that the answer is MANY different ways

    For every positive integer N there is a circle around the south pole which has circumference 1/N mile, and so one can walk for one mile to the east along this circle going N times around and be back at the same place.

    This is a brilliant idea. Please edit your answer so that it will be letter-perfect in the eyes of Ahrkron and then we will get to hear what problem you can come up with!
     
  25. Sep 5, 2003 #24

    HallsofIvy

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    Actually, I doubt that there are any polar bears (or any other) land animals within a mile of either the north or south poles.
     
  26. Sep 6, 2003 #25

    ahrkron

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    Good job guys!

    Yes, there is an infinite number of solutions. I just want to point out one family[\i] of solutions that has not been explicitly mentioned.

    So far the following initial points have been described:
    - The north pole,
    - Points that are at such a distance from the south pole that, upon walking one mile due South and changing your heading to East, one mile walking makes you circle around the south pole an integer number of times.

    The third family would be:
    - Points such that the first 1-mile segment touches and passes the south pole, and deposits you in one of the (1/N mile) circles around the south pole.

    After crossing the South pole, you are not waking due South anymore, but that is allowed by the wording of the problem, since it only states the heading at the begining of the walk.

    (Btw, the original wording should have said "... walks one mile, keeping his (N/S/E/W) heading").

    Regarding HallsofIvy's comment, polar bears have been reported as far north as the North Pole, but they only live on the northern polar region, which is why I "took the bear out" of the original riddle.

    So,

    Oxymoron got the gist of the solution, and gets to post the next problem. Marcus gets an honor point for accuracy.
     
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