Monday, January 11, 2010

Problem 417: Tangent circles, Tangent lines, Angles

Proposed Problem
Click the figure below to see the complete problem 417 about Tangent circles, Tangent lines, Angles.

Problem 417: Tangent circles, Tangent lines, Angles.
See also:
Complete Problem 417
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:39 PM
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7 comments:

  1. c .t . e. oJanuary 12, 2010 at 5:52 AM

    join E to A, D to B
    draw DH perpendicular to EA

    in ABDH
    ang( HAB+ABD+BDH+DHA)= 164 + BDH + 90 (HAB+ABD=2∙82)
    => ang BDH = 106
    => ang HDF = 164

    => x = 16

    ReplyDelete
  2. AnonymousJanuary 18, 2010 at 9:05 PM

    solution needs to be improved
    how did u get HAB + ABD = 2.82

    ReplyDelete
  3. c .t . e. oJanuary 23, 2010 at 7:16 AM

    thanks, I saw it today

    ang ECD = 1/2 ( arc EC + arc CD ) (build common tang)

    ang EAC + ang CBD = arc EC + arc CD

    ReplyDelete
  4. APOSTOLIS MANOLOUDISJune 17, 2016 at 9:18 PM

    Problem 417
    I design the common external tangent of two circles intersecting tinEF
    in point K and DF intersects the KC at the point M. If <MCD=a=<MDC, then <KMF=2a.
    But <KCE=82-2a=<KEC, so <FKC=184-2a.Then x+2a+184-2a=180 therefore x=16.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  5. Sailendra TJuly 28, 2017 at 2:08 PM

    Extend FD to meet EC at G and let m(FEC) = x and m(GDC) = y

    Join AE, AC and form the isosceles triangle AEC in which
    m(AEC) = m(ECA) = 90-x

    Similarly join CD, DB and form the isoceles trianlge CDB in which
    m(CDB) = m(DCB) = 90-y

    Since A,C and B are collinear we have
    (90-x)+82+(90-y) = 180
    => x+y=82------------(1)

    Now consider the triangle CGD. Since m(GCD) = 82 and m(GDC) = y, we have m(EGD) = 82+y
    In the triangle EGF, we have
    m(GEF) = x
    m(EGF) = 82+y
    Hence m(F) = 180-(x+82+y) = 180-(164) = 16 degrees (Substitute value of x+y from (1))

    ReplyDelete
  6. Sumith PeirisSeptember 11, 2017 at 9:47 AM

    Let the common tangent at C meet EF at P and FD extended at Q.

    PE = PC and DQ = QC so we realize that
    < FPQ + < FQP = 2 X 82 = 164

    Hence x = 180 - 164 = 16

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. AnonymousMarch 19, 2025 at 3:02 PM

    note that (AE) and (BD) are perpondicular to (FE) respectively (DF)
    from B extend line BD to meet EF at H
    we have AE=AC and CB=BD
    now for notation let < AEC= θ and <BCD= β
    since A,C and B are colinear then θ+β+82°=180° and so far θ+β=98° (1)
    we have also < CEH=90-θ and < DHB=90-x and so far <BHE=90+x
    finally focus on the quadrilateral ECDH we have:
    (90°-θ)+(90°+x)+82°+180°-β=360
    we conclude using (1) that x=16°

    ReplyDelete

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