Thursday, May 19, 2016

Geometry Problem 1215: Circle, Diameter, Chord, Radius, Midpoint, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1215.


Geometry Problem 1215: Circle, Diameter, Chord, Radius, Midpoint, Measurement

3 comments:

  1. join A to C, draw DH altitude of ADC => DH meet AB at O
    HO midleline of ABC => OH=a/2. From tr AOH AH²=x²-a²/4 (1)
    From tr ADH AH²=d²-(a/2+x)² (2)
    from (1) & (2) x²-a²/4=d²-(a/2+x)² => the conlusion

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  2. Let CB and AD extended meet at E. Since AD = CD, BD bisects < ABE, hence DE
    = d and BE = 2x
    So 2d^2 = 2x(2x+a) from which
    x^2 + ax/2 + a^2/16 = d^2/2 + a^2/16 by completing the square
    So (x+a/4)^2 = (a^2+8d^2)/16 and therefore
    x = (1/4){sqrt(a^2+8d^2) – a)}

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. By Ptolemy's theorem, AB·CD+BC·AD=AC·BD
    √((2x)^2-a^2)√((2x)^2d^2)+ad=2dx
    √(((2x)^2-a^2)((2x)^2-d^2))=d(2x-a)
    ((2x)^2-a^2)((2x)^2-d^2)-d^2(2x-a)^2=0
    16x^4-8d^2x^2-4 a^2x^2+4 a d^2x=0
    4x(2x-a)(2x^2+ax-d^2)=0
    x=0 (rejected, ∵x>0) or a/2(rejected, ∵ΔABC is a right triangle∴ AB>BC,ie. 2x>a or x>a/2) or 1/4 (sqrt(a^2+8 d^2)-a) or 1/4 (-sqrt(a^2+8 d^2)-a) (rejected, ∵x>0)

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