Monday, May 23, 2011

Geometry Problem 605: Circle, Square, Chord, Perpendicular, Metric Relations

Geometry Problem
Click the figure below to see the complete problem 605.

Geometry Problem 605: Circle, Square, Chord, Perpendicular, Metric Relations.

Posted by Antonio Gutierrez at 9:09 PM
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2 comments:

  1. PravinMay 24, 2011 at 9:05 PM

    Let 2a, 2b be the sides of the squares (a>b)

    Note a-b = 4 and

    (2a-x)^2 + a^2 = r^2 = (2b+x)^2 + b^2

    5(a^2-b^2)= 4(a+b)x,

    Hence x = 5(a-b)/4 = 5

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  2. Peter TranMay 25, 2011 at 2:42 AM

    http://img836.imageshack.us/img836/732/problem605.png
    Let R= radius of the circle
    Note that MN=2.LN and CF=2.CL
    So 3 points F, L, M are collinear
    Distances LN and LC are the solutions of system equations:
    1. equation y=2.t represent line FM
    2. equation t^2+(y+x)^2=R^2 represent circle radius R, center O
    2 positive solutions are LN= abs(t1)=1/5(-2x+SQRT(5R^2-x^2))
    And LC=abs(t2)=1/5(2x+SQRT(5R^2-x^2))
    DB=LC-LN=4/5.x=4

    So x=5

    Peter Tran

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