Friday, May 20, 2016

Geometry Problem 1216: Circle, 90 Degrees, Tangent, Common Chord, Collinearity

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

This entry contributed by Peter Tran, California, U.S.A.

Click the figure below to view more details of problem 1216.


Geometry Problem 1216: Circle, 90 Degrees, Tangent, Common Chord, Collinearity

3 comments:

  1. Problem 1216
    The common tangents at points C and D intersect at E. The point E belongs to the radical axis of the circles with centers O1 and O2 as whereby the points A, B, E, are collinear.Is <ECO=
    <EAO=<EDO=90,therefore the points O,A,C,E,D are concyclic.But CE^2=EB.EA=ED^2.So
    Triangle CBE is similar with triangle ACE and triangle DBE is similar with triangle ADE.
    Therefore <EBC=<ACE and <DBE=<ADE. But <DBE+<EBC=<ADE+<ACE=180. Therefore
    The D,B,C are collinear.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE

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    Replies
    1. Excellent solution.
      See below for the sketch:
      http://s32.postimg.org/c1xovx2ut/pro_1216.png

      Peter Tran

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  2. Пусть D' пересечение BC и окружность O, P основание высоты О1 до AB, <BCO1=a, и <ACO1=b.
    O, O1, и C лежит на одной прямой, так как окружностей O и O1 прикоснется друг друга.
    <PO1O=(<PO1A=a+b)-(<OO1A=2b)=a-b.
    <O1OA=<PO1O=a-b потому что ОА и PО1 параллельны.
    <OD'C=a,и <OAC=180-(a-b)-(b)=180-a, значит D', O, A, и C принадлежит одной окружность.
    Тогда мы знаем <D'AC=<D'OC=180-2a.
    <D'AB=<D'AC-<BAC=(180-2a)-(90-a)=90-a, значит описанная окружность D'AB касается окружность O в D', которые является самой точка D.

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