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Math Help - Puzzle I made up

  1. #1
    Super Member Matt Westwood's Avatar
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    Puzzle I made up

    Take a rectangle ABCD.

    Draw the line AC dividing the rectangle diagonally in two.

    In triangle ABC, inscribe a circle. Let its centre be E.

    Draw perpendiculars from E to AB and BC, meeting CD at F and DA at G

    What is the area of ABCD divided by the area of EFDG?


    How neat is that?
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  2. #2
    Newbie Coffee Cat's Avatar
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    I think it makes a pretty fine problem



    Let's say the length of rectangle ABCD is L, and the width is W. If r is the radius of circle E, then the length of rectangle EFDG is (L-r) and the width is (W-r).

    We can split triangle ABC into three smaller triangles: AEC, AEB and BEC. Each has an altitude of r.



    area(ABC) = area(AEC) + area(AEB) + area(BEC)

    <br />
\frac {LW} {2} = \frac {r \sqrt {L^2 + W^2} }{2} + \frac {Lr} {2} + \frac {Wr} {2}<br />

    <br />
r = \frac {LW} {L + W + \sqrt {L^2 + W^2} }<br />

    So the area of rectangle EFDG is

    <br />
A_{EFDG} = (L - \frac {LW} {L + W + \sqrt {L^2 + W^2} } ) (W - \frac {LW} {L + W + \sqrt {L^2 + W^2} } )<br />

    Whoa. So I suppose unless we are provided with the L/W ratio, we will not have a pretty answer for this problem.
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  3. #3
    Super Member Matt Westwood's Avatar
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    There's a neater way of doing it. Hint: use Pythagoras.
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  4. #4
    MHF Contributor Quick's Avatar
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    Sorry, your Web browser is not Java compatible. This prototype will not be of much interest to you.
    Quote Originally Posted by Matt Westwood View Post
    Take a rectangle ABCD.

    Draw the line AC dividing the rectangle diagonally in two.

    In triangle ABC, inscribe a circle. Let its centre be E.

    Draw perpendiculars from E to AB and BC, meeting CD at F and DA at G

    What is the area of ABCD divided by the area of EFDG?


    How neat is that?
    I cheated and didn't do the algebra. Although making the incenter took a good deal of time so I wouldn't recommend using the java [draw] [/draw] tags.

    Anyway, you can drag points "A" and "C" around and you'll find that the ratio is always 2
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  5. #5
    Newbie Coffee Cat's Avatar
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    To Quick: Amazing! How did you make this fine applet?

    To Matt Westwood: Still thinking about it ^^

    Edit: Ah, I'm retiring for the night. I'm so curious about the answer. ^^
    Last edited by Coffee Cat; August 10th 2008 at 05:30 AM.
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  6. #6
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Coffee Cat View Post
    To Quick: Amazing! How did you make this fine applet?

    To Matt Westwood: Still thinking about it ^^
    I did this:

    [draw]Point(20,60) [label ('A')]; UnitPoint(1,230) [hidden]; Line(1,2) [hidden]; Point(230,240) [label ('C')]; Perpendicular (3, 4) [hidden]; Intersect (3, 5) [label('B')]; Perpendicular (3, 1) [hidden]; Perpendicular (5, 4) [hidden]; Intersect (7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw]

    That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.
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