Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circle's slightly harder challenge

  1. Dec 23, 2006 #1
    Alright, well seeing as my last challenge lasted a whole 30 minutes before Jimmy Snyder solved all my brain teasers, I dug deep to find more challenging ones. :rolleyes: :smile:

    Please use and and the closing tag [/ color] to make your text unreadable to those who don't want to see the solution, thanks.

    If you recognize any of these puzzles from old threads please either ignore them or link to them and I may remove those puzzles from this thread to ensure that these are all new. Thanks.

    Good luck!!

    1.Three distinct points with integer coordinates lie in the plane on a circle of radius r > 0. Show that two of these points are separated by a distance of at least r^(1/3).

    2.What chemical compound is represented by the following: HIJKLMNO?

    3. On his way to work each day, a man living on the fifteenth floor of an apartment building takes the elevator to the first floor from the fifteenth floor. On his return, he is forced to take it to the seventh floor and walk the remaining eight floors to his apartment. Why?

    4.Three criminals just robbed a bank and go back to their hideout. They put the money behind a high tech security door. There are 3 locks on the door, each activated/deactivated by a button next to it. All locks are originally deactivated, and once a lock is activated it is impossible to tell whether it is activated or not. The three criminals want to work out a system so that any two of them can access the money but a single criminal cannot. The 2 criminals accessing the money must be assured that all locks are deactivated, otherwise an alarm will sound, and built-in lasers will shoot them. Also, each criminal may only give information about which locks he toggled to one other criminal. Figure out how their system will work.

    5.Punctuate the following so it makes sense: Alice while Matthew had had had had had had had had had had had a better effect on the teacher.

    Last edited: Dec 23, 2006
  2. jcsd
  3. Dec 23, 2006 #2
    2. H2O

    Last edited: Dec 23, 2006
  4. Dec 23, 2006 #3
    edit: Yup, :tongue:
    Last edited: Dec 23, 2006
  5. Dec 23, 2006 #4
    3. A probable answer would be that the man has his office also on the first floor of the building, and the office extends from floors 1 to 7 (say) and the elevator is privately owned by the office for the employees . Note that the first statement of the question doesn't say from which floor the elevator is boarded, only the destination (1st floor). So the man may well have his residence on the 15th floor, walk upto the 7th floor and then take the elevator to the 1st floor.
    Who would construct such a building anyway ?

    5. Can we delete spaces, so that we may combine all but the last two hads into a very big surname for Matthew

  6. Dec 23, 2006 #5
    Lol, nice try on both :biggrin:.

    Highlight to see :

    I am afraid that you are not quite right on either, but your thinking on Q#3 is on the right path, it is something quirky like that.

    For Q#5, no you can't get rid of spaces, nice try :biggrin:
  7. Dec 23, 2006 #6
    Well, I only said that my answer to question 3 is only "possible", don't you agree. Maybe you could add a few more details so that you may contradict this case ?


  8. Dec 23, 2006 #7
    True enough I suppose, I shouldn't have said you were incorrect. There is another answer though. heh. I'll post it in a few days.
    Last edited: Dec 24, 2006
  9. Dec 23, 2006 #8
    I also changed it so that he takes it from the 15th floor in the morning.
  10. Dec 23, 2006 #9
    He is too short to reach the buttons above the 7.

    Alice, while Matthew had had "had had" had had "had". "Had had" had had a better effect on the teacher.

    I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted.
  11. Dec 23, 2006 #10
    the first problem is hard
  12. Dec 23, 2006 #11
    For Q#3, that is the "original answer" good job.

    Correct on Q#5 as well, that is the answer (pretty close to "original" answer but its right anyway)

    The "original answer" is::

    Alice, while Matthew had had "had," had had "had had." "Had had" had had a better effect on the teacher.
    Last edited: Dec 23, 2006
  13. Dec 24, 2006 #12


    User Avatar
    Gold Member

    the first two criminals activate or deactivate the first lock. then the first and third ones together either activte or deactivate the second one. then the second and third one either activate or deactivate the third lock. this way any given criminal knoes the position of only two locks.
    am i missing something?
  14. Dec 24, 2006 #13
    Given the difference between my answer and the original answer, you should be able to punctuate the sentence at the bottom of my post.

    I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted.
  15. Dec 24, 2006 #14
    "I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted." - Jimmy Snyder

    I can't see any other way to correctly punctuate "except that I while she"

    Last edited: Dec 24, 2006
  16. Dec 26, 2006 #15
    I showed this to my wife and she gave a similar answer to this except that I, while she had had 'had had "had" had had "had had"', had had 'had had "had had" had had "had"'. 'Had had "had"' had been in the answer I posted.

    By iterating in this fashion, any number of had's can be strung together and meaningfully punctuated.
  17. Dec 27, 2006 #16
    Yea definitely, it still takes some effort to do it though (although it is not a very creative process) I tried to select a few puzzles for different types of people. :smile: I already knew that any number of had's can be strung together in that fashion, it's not the kind of puzzle I enjoy doing but some may.

    Although of course you still realize that your answer to your question is not grammatically correct.
  18. Dec 27, 2006 #17
    It seems ok to me. What's the problem?
  19. Jan 2, 2007 #18
    This one's still bugging me. I've asked a few people I know, and so far everyone's stumped.

    Just so I'm clear, though, can the question can be re-written as the following?

    Three distinct points A, B, and C lie in the x,y plane, and each have exact integer coordinates. Each of the points also lies on a circle of radius R. Prove that at least one pair of these points is separated by a distance greater than or equal to the cube root of R.

    Any hints on the method of solving it?

  20. Jan 2, 2007 #19
    Had to ask a friend how to do this as I sorta forget how he showed me, use the points to form a triangle / circumcircle. If you need more help I have more hints. And yes I believe that your wording is fine.

    What is the equation for the Circumradius if the sides are a,b,c, and D = largest distance between two points, and A = area?
    Last edited: Jan 2, 2007
  21. Jan 5, 2007 #20
    Alright. Uncle!

    After trying this for a couple days, I asked a few of my more geeky friends, but nobody could figure it out. So I sent the problem to a math whiz at MIT, a college math professor, and someone who majored in math a long time ago. Nothing.

    So far, I think the best working theory someone had was that there were only a few quasi-rational coordinates on the unit circle. Not that the coordinates are actually *rational* per se, but that the radius could be adjusted to a particular value that would *make* them be rational. For example coordinates sqrt(3)/2, 1/2, which, while irrational, can be made to *be* rational with a radius that's a multiple of sqrt(2).

    Anyway, the assumption was that there's a fixed number of possible exact-integer coordinates on any given circle. And knowing that, you can calculate the circumradius of the points and compare it to the distance of the furthest points, and get an answer.

    But otherwise, I dunno. The sticky wicket in all cases is somehow establishing equasions wherein the coordinates of the three points are necessarily integers.

    Could you post (or PM me) the proof?

  22. Jan 5, 2007 #21
    Here it is. The question appeared on a putnam exam (Had to find hard ones!) :-)

    We are given three points in the plane. Let D be the largest distance between any pair. If we view these points as vertices of a triangle, then we may view r as the radius of the circumcircle of a triangle. If the lengths of the sides are a,b, and c, and the area is A, then the circumradius is given by r = abc/(4A). Since D = max(a,b,c), we have r < D^3/4A, whence D > (4A r)^(1/3)

    On the other hand, the area of the triangle may be determined from the coordinates of its vertices; it's half the area of a certain parallelogram, which is in turn computed by a determinant, showing that the area of such a triangle must be at least 1/2. This gives D > (2 r)^(1/3), a slightly stronger result than was asked for.
    Last edited: Jan 5, 2007
  23. Jan 5, 2007 #22
    Hm. Ok, I still don't quite get it.

    I understand how we get to D > (4Ar)^(1/3), that's fine. Although, I don't quite see:

    1) How do we know the area is at least 1/2? I may figure that one out, cuz... that looks like it might be something I could prove... But I'm still fuzzy on why that's true.

    2) So, if the area is 1/2 or greater, I understand how we get a minimum value and it turns into D > (2r)^(1/3). But aren't we looking for D > r^(1/3)? Should the problem have been stated as "the cube root of the diameter" instead of "the cube root of the radius"?

  24. Jan 5, 2007 #23
    if D > (2r)^(1/3) then D must also be greater than r^(1/3). right?
    Last edited: Jan 5, 2007
  25. Jan 5, 2007 #24
    Do we need to pay attention to the fact that the problem says the point coordinates are integers? I was trying to solve it a totally different way.
  26. Jan 5, 2007 #25
    Yes it is very important because the points become lattice points and Pick's theorem can be applied, and in general allows you to easily compute the minimum area of the triangle who's vertices are the 3 points.

    Davee, you can use Pick's theorem to prove that the area of a triangle whose vertices have integer coordinates is at least 1/2.

    Actually, the fact that a primitive triangle has an area of 1/2 can be used to prove Pick's theorem.
    Last edited: Jan 5, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook