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BMAD

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Posts posted by BMAD

  1. Suppose we have the following system

    x^2+y^2=r; x+y=r, such that the line crosses the circle at exactly two places. 

    Obviously with two equations and three variables, we have a solution set of answers that can satisfy the given conditions.  What I want to know is of the given solutions that satisfies this problem, what is the smallest and largest values x can possibly be?

  2. On 4/17/2019 at 6:02 AM, rocdocmac said:
      Hide contents

    116 seconds (or possibly even 106 seconds) presuming first two pieces are already in place.

    Toasting alone takes 90 seconds for slices numbered 1, 2 and 3 with sides A and B.

    Stages: (1) 1A  and 2A under grill (30 sec), followed by (2) 1B and 3A under grill (30 sec); finally (3) 2B and 3B under grill (30 seconds)

    Normal additional time for single movements:

    End of 1st stage: Slice 1 flipped from 1A to 1B (3 sec), 2A removed (5 sec), 3A inserted (5 sec)

    End of 2nd stage: Slice 1 removed (5 sec.), slice 3 flipped from 3A to 3B (3 sec), slice 2 (2B) re-inserted with flip side up (5 sec)

    End of 3rd stage: Done

    Normal additional time = 26 seconds

    Total time = 116 seconds.(1:56)

    Exchange in one go:

    If one is allowed to do simultaneous movements (exchanging in one switch of 5 sec), 10 seconds can be deducted, i.e. a total time of 106 seconds (1:46)

     

     

     

    14 hours ago, Pirate8 said:

     

    Left

    side

     

    Right

    side

    1-5

    A1 in

     

    6-10

    B1 in

    6-35

    A1 toast

     

    11-40

    B1 toast

    36-38

    A flip

     

    41-45

    B out

    39-68

    A2 toast

     

    46-50

    C in

    69-74

    A out

     

    51-80

    C1 Toast

    75-79

    B2 in

     

    81-83

    C flip

    80-109

    B2 toast

     

    84-113

    C2 toast

    110-114

    B out

     

    114-116

    C out

             

    Manual operations must not overlap.

     

     

  3. Three slices of bread are to be toasted under a grill. The grill can hold two slices at once but only one side is toasted at a time. It takes 30 seconds to toast one side of a piece of bread, 5 seconds to put a piece in or take a piece out and 3 seconds to turn a piece over. What is the shortest time in which the three slices can be toasted?

  4. 2 hours ago, Thalia said:

     

      Hide contents

    12 cuts, 42 pieces? Unless you want to line up the sausages and do one giant cut...

     

    You have the right idea with your comment. I an designing a cut as a swipe of the knife,  so lining them up and then cutting with one swipe would  count as one cut. 

  5. 20 hours ago, EventHorizon said:

    I'm sure I know what BMAD was saying.  Yes, ignore the units / squared units difference.  The problem doesn't involve a circle though.  You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function.  Here the arc length is like the length of a string representing the function cut at the two points.

    How to find the arc length of a function:

      Hide contents

    If the function was a straight line, you could use the Pythagorean theorem to find it.

    sqrt( (y2-y1)^2 + (x2-x1)^2 )

    If you assume the arc length was a straight line, and slowly bring the points on the function closer together, eventually you get the change in y approaching the slope of the line times the difference in x's.

    The arc length for that infinitesimally small section is then sqrt( ( f'(x)dx )^2 + dx^2 ) = sqrt( dx^2 * (f'(x)^2 + 1 ) = sqrt(   f'(x)^2 + 1 ) dx.  Where f'(x) is the derivative of f(x).

    Integrate that between two points to get the arc length between those two points.

    Example: f(x)=2x

    f'(x) = 2

    integrate( sqrt( 2^2 + 1 ) dx ) = integrate( sqrt(5) dx) = x*sqrt(5) + c

    If you evaluate the integral between 0 and 1 you end up with arc length =  sqrt(5).  Which is the same as if you just used the Pythagorean theorem.  Yeah, I'm too lazy right now to find a function that ends up with a nice arc length function that isn't just a straight line.

    Restatement of problem using previous spoiler:

      Hide contents

    Find a function f(x) such that F(x) = integrate( sqrt( f'(x)^2 + 1) ), where F(x) is the indefinite integral of f(x) and f'(x) is the derivative of f(x).

    One step further:

      Hide contents

    Taking the derivative of both sides gives

    f(x) = sqrt( f'(x)^2 + 1)

     

    One answer is simple, the other is less so.

    yes, this is what I mean.

  6. James and Mike are running in a race.  They both walked and ran for part of the rate.  They each walked and ran at the same speed.  James ran for half the distance and walked for half the distance.  Mike ran for half of his time and walked for half of his time.  Who finished first?

  7. On 12/15/2018 at 4:16 PM, CaptainEd said:

    I think not: please check my counting

      Hide contents


    In what follows, the rubber band’s path goes on the outside of the nails marked “x

     o o o o o o o
     o o o o o o o o x
     o o o o o o o o
     o o o o o o o
     o o o o o o
     o o o o o x
    N = 43
    P = 21 + sqrt(5)
    S = 62: (5x5:1, 4x4:4, 3x3:10, 2x2:18, 1x1:29)

    Compare with

           o o o o o o o 
     o o o o o o o o x
     o o o o o o o o
     o o o o o o o
     o o o o o o
     o o o o o x
    N = 43
    P = 21 + sqrt(5)
    S = 56: (4x4:2, 3x3:8, 2x2:17, 1x1:29)

     

    All you showed is the bound is wider than one thought.

  8. On 12/12/2018 at 4:52 PM, CaptainEd said:

    I question whether the information given about a figure is sufficient to uniquely determine the number of included squares:

      Hide contents


    If I understand right, these two figures have the same N and P, but don’t have the same number of squares.

        o o o o
     o o o o o
     o o o
    Nails = 12, perimeter = 7+sqrt2 + sqrt5
    Squares =5


     o o o o
     o o o o o
     o o o
    nails=12, perimeter = 7 + sqrt2 + sqrt5
    Squares = 6

     

     

    So then if we know N and P we should be able to bound  squares x by a two values. Are those values always consecutive?

  9. On 12/11/2018 at 9:00 PM, CaptainEd said:

    While this doesn’t answer the question with randomly placed 1s, I think it does give an upper bound to the number of columns that can be removed.

     

      Hide contents

     

    What is the smallest number of remaining columns that would allow (a) the 30 rows to be distinct, (b) each row to have 5 1-bits, and (c) all remaining columns to be distinct?

    7 columns: 7!/2!5! = 21, too few distinct rows
    8 columns: 8!/3!5! = 56, that’s enough.

    The first 12 columns in each of the 30 rows should be one of these patterns:
    11000 00000 00 
    01100 00000 00
    00110 00000 00
    00011 00000 00
    00001 10000 00

    00000 11000 00
    00000 01100 00
    00000 00110 00
    00000 00011 00
    00000 00001 10

    00000 00000 11
    10000 00000 01

    The last 8 columns should be chosen from the C(8,5) possible patterns.

    The resulting matrix satisfies the requirements because:
    (a) the last eight columns differentiate the 30 rows
    (b) the first twelve columns are distinct by design
    (c) the last eight columns are distinct because of the distinct patterns chosen from the C(8,5)
    (d) all row sums are 7 before column removal, and 5 afterwards

     

     

    I agree

  10. Suppose you have a triangle that has 2-1 inch lengths. Divide this triangle into half by drawing a line from vertex between the two identical sides,  choose one of the sides randomly and shade it.  The non-shaded side is cut in half again.  Choose one of these sides randomly and cut it in half again shading one random piece.  If this pattern of cut, shade, cut, cut, shade, cut, cut, shade cut, cut,.... was to be continued forever, what would be the area of the shaded region?

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