Tuesday, September 10, 2013

Geometry Problem 922: Circle, Tangent Lines, Tangency Point, Chord, Secant, Harmonic Mean

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 922.

Online Geometry Problem 922: Circle, Tangent Lines, Tangency Point, Chord, Secant, Harmonic Mean

8 comments:

  1. http://img59.imageshack.us/img59/9554/7fts.png
    Draw circle diameter PO
    Let M is the midpoint of AB and PO cut DE at N
    We have PA.PB=a.b= PE^2 ( power from P to circle O)
    PM=(a+b)/2
    O,N,C,M are cocyclic => PC.PM=PN.PO=PE^2 ( relation in right triangle)
    Or c. (a+b)/2 = PE^2 => a.b= c.(a+b)/2 => c= 2.a.b/(a+b)

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  2. Anonymous Erina NJ

    PD^2 = A*B PD^2 - PC^2 = DC*CE = AC*CB

    a*b-c^2 = (c-a)(b-c)-> c = 2ab/(a+b)

    ReplyDelete
  3. Anonymus Erina Nj

    Even the previous solution is from Erina NJ

    For Pravin

    Sinx = 2 sin10 cos (30-x)
    sinx = sin (40-x)- sin (20-x)
    sin (40-x)- sinx= Sin (20-x)
    2sin (20-x) cos 20-sin (20-x) =0
    sin(20-x) [2cos20-1] =0
    2cos20-1 different from 0 => sin(20-x) = 0
    X=20

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    Replies
    1. Erina: This solution has been moved to problem 920. Thanks.

      Delete
  4. Sorry about that Antonio(your system seems to have cut out a few details):

    In the quandrangle ABCD, angles A, B, C, are obtuse. Prove that BD is greater than AC

    Erina-NJ

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  5. Let PD = p, DC = m and CE = n and let X be the mid point of DE

    Then PX^2 = p^2 - (m+n)^2/4 = c^2 - (n-m)^2/4

    So p^2 -c^2 = mn = (c-a)(b-c)

    But p^2 = ab

    Solving c = 2ab/(a+b)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete