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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    10 Mar '14 20:17
    This is trivial enough, but I'll probably use it in a high school someday.

    A rectangle has circumference length of p. The diagonal has length d. What is the area of the rectangle?
  2. Subscriber joe shmo On Vacation
    Strange Egg
    11 Mar '14 21:23 / 3 edits
    Originally posted by talzamir
    This is trivial enough, but I'll probably use it in a high school someday.

    A rectangle has circumference length of p. The diagonal has length d. What is the area of the rectangle?
    I don't know if I did this the hard way, because it didn't seem obvious to me...

    Let the sides of the rectangle be x, and y respectively.

    Perimeter = P = 2*x + 2*y ....eq (1)

    from the Pythagorean theorem

    d² = x² + y² ...eq(2)

    from eq(1)

    y = 1/2*P - x ...eq(1)'

    from eq(2)

    y = Sqrt ( d² - x² )

    Set eq(1)' = eq(2)

    1/2*P - x = Sqrt ( d² - x² )

    solve for x in terms of d, and P by completing the square

    x = Sqrt(d²/2 - 1/16*P² ) + P/4 ...eq(3)

    Sub eq(3) into eq(1)'

    y = 1/2*P - [Sqrt(d²/2 - 1/16*P² ) + P/4]

    = 1/4*P - Sqrt(d²/2 - 1/16*P² )

    then Area = A = x*y

    = (P/4 + Sqrt(d²/2 - 1/16*P² ))(P/4 - Sqrt(d²/2 - 1/16*P² ))

    which is of the form (a + b)(a - b) = a² - b² : giving after simplification

    A = P²/8 - d²/2 ***
  3. 11 Mar '14 22:46 / 2 edits
    Originally posted by joe shmo
    Let the sides of the rectangle be x, and y respectively.

    Perimeter = P = 2*x + 2*y ....eq (1)

    from the Pythagorean theorem

    d² = x² + y² ...eq(2)
    What you did works well.

    Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

    If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

    Then rearrange to get your result.
  4. Subscriber joe shmo On Vacation
    Strange Egg
    11 Mar '14 23:14
    Originally posted by Diapason
    What you did works well.

    Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

    If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

    Then rearrange to get your result.
    Wish I would have saw that,...by far less work!
  5. Standard member talzamir
    Art, not a Toil
    12 Mar '14 07:33
    Nicely done. And the squaring trick is what I was thinking too.

    2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

    x^2 + y^2 = d^2

    combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A
  6. Subscriber sonhouse
    Fast and Curious
    21 Mar '14 11:42 / 2 edits
    Originally posted by talzamir
    Nicely done. And the squaring trick is what I was thinking too.

    2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

    x^2 + y^2 = d^2

    combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A
    It's funny, the first numbers I chose showed me something about rectangles:

    I chose d=10 and p=20 where A = zero

    which makes sense

    It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

    How can you fix it so that doesn't happen?

    Exploding ink?
  7. Subscriber joe shmo On Vacation
    Strange Egg
    21 Mar '14 22:12 / 1 edit
    Originally posted by sonhouse
    It's funny, the first numbers I chose showed me something about rectangles:

    I chose d=10 and p=20 where A = zero

    which makes sense

    It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

    How can you fix it so that doesn't happen?

    Exploding ink?
    You happened to pick your variables outside of the domain of the equation.

    In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

    So;

    A > 0

    P²/8 - d²/2 > 0

    P²/8 > d²/2

    P² > 4*d²

    P > 2*d (this is the domain of the equation), which say that all perimeters must be greater than twice the diagonal.

    You picked a perimeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?
  8. Subscriber sonhouse
    Fast and Curious
    22 Mar '14 01:29
    Originally posted by joe shmo
    You happened to pick your variables outside of the domain of the equation.

    In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

    So;

    A > 0

    P²/8 - d²/2 > 0

    P²/8 > d²/2

    P² > 4*d²

    P > 2*d ...[text shortened]... imeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?
    I knew that, just pulling your chain