Thursday, May 29, 2014

Geometry Problem 1019: Circle, Diameter, Perpendicular, Chord, Tangent, Sum

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

Click the figure below to see the complete problem 1019.

Online Math: Geometry Problem 1019: Circle, Diameter, Perpendicular, Chord, Tangent, Sum

Posted by Antonio Gutierrez at 5:44 AM
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7 comments:

  1. Peter TranMay 29, 2014 at 10:40 AM

    Algebra solution :
    Let C’ is the point of sysmetry of C over AB => AC’=AC
    Let α=∠(BAD) , AB= 2.R and AH= 2p
    We have AD= 2R.cos(α) , DF=2R.cos(α) ^2 +2p
    Since AD=DF => 2.R.cos(α)= 2p+2Rcos(α)^2 => R.cos(α)^2-R.cos(α)+p=0
    Let α1 and α2 are solutions of this quadratic equation
    Summation of solutions= 1 => cos(α1)+cos(α2)=1=> AD/AB+AC’/AB=1
    Or AB=AD+AC

    ReplyDelete
  2. UnknownMay 30, 2014 at 6:28 AM

    Consider:
    R=DA, r = CA, 2a = AB(where O is the center)
    Join B and D ->Triangle BDA is right triangle
    Join B and C ->Triangle BCA is right triangle

    Draw a radius DF (DF is perenpedicular to MN)
    Draw a radius CE (CE is perenpedicular to MN)
    Draw an tangent to circle O threw A in that way that it cut DF at L and CE at M thus MEFL is rectangle.

    Triangle DAL is similiar to BDA and so MAC to BCA (because they has 90 angle of MEFL and becasue of the tangent : R/2a = DL/R (triangles DAL and BAD) - >DL = R^2/2a THUS LF = R - R^2/2a
    AC/BA = MC/AC -> r/2a = MC/r (triangles CAM and BCA) - >MC = r^2/2a THUS ME = r - r^2/2a

    Since MEFL is rectangle we get that ME = LF:
    r - r^2/2a = R - R^2/2a
    after short develop we get
    2ar - r^2 = 2aR - R^2
    R^2 - r^2 = (R - r)(R+r) = 2a(R - r)
    thus 2a = R+r - > AB = DA + AC

    ReplyDelete
  3. Ivan BazarovAugust 24, 2014 at 1:31 PM

    Choose midpoint of AH to be origin and BH as y-axis.
    If a=AH/2 and r=AB/2, parabola 4ay=x^2 cuts circle x^2+(y-(a+r))^2=r^2 at C and D==>
    4ay+y^2-2y(a+r)+(a+r)^2=r^2,
    y^2-2y(r-a)+a^2+2ar=0,
    CE+DF=(sum of y-coordinates of C and D)+2a=2(r-a)+2a=2r=AB

    ReplyDelete
  4. Sumith PeirisNovember 17, 2015 at 11:37 PM

    Let AC = a and AD = b and AB = c and AH = x

    Using Pythagoras

    EH^2 = a^2-(a-x)^2 = a^2(c^2-a^2)/c^2 since EH.AB = BC.CA

    So (a-x)^2 = a^4/c^2
    Hence a-x = a^2 /c......(1)
    Similarly b-x = b^2/c.....(2)

    (1) - (2) a-b = (a^2-b^2)/c

    Dividing both sides by a-b,
    c = a+b

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Sumith PeirisNovember 17, 2015 at 11:44 PM

    Note that E,A,D are collinear as well as F,A,C though this fact was not used in the above proof.

    ReplyDelete
  6. Sumith PeirisNovember 17, 2015 at 11:46 PM

    Another problem from the above.,,,

    Prove that 1/AH = 1/AC + 1/AD

    ReplyDelete
  7. Sumith PeirisMarch 16, 2016 at 3:49 AM

    Also EH/ BC + FH/BD = 1

    ReplyDelete

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