Sunday, September 11, 2011

Problem 668: Right triangle, Incircles, Parallel, Common Tangent, Mind Map

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 668.

 Problem 668: Right triangle, Incircles, Parallel, Common Tangent, Mind Map.

4 comments:

  1. http://img819.imageshack.us/img819/5083/problem668.png
    Connect CO and AO
    Note that CO and CO2 are angle bisector of angle (BCA) so
    C, O, O2 are collinear
    Triangle OEC congruence with triangle OHC ( right triangles with common hypotenuse and CO is angle bisector)
    So OC is the angle bisector of angle (HOE) >>Incircle O2 tangent to OE also tangent to OH.
    Similarly AO is angle bisector of angle HOF) and incircle O1 tangent to both OF and OH
    Peter Tran

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  2. Let circle O₁ touch AB at J, AC at K & FG at L.
    Complete the square O₁JHM
    O₁M = KH = AH - AK = AF - AJ = KF = radius of O₁ (note FJ O₁K is a square)
    follows OH touches circle O₁.
    Similarly we can show OH touches circle O₂

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  3. Refer my solution to Problem 668:
    "FJO₁L is a square" (not FJO₁K as previously typed)

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  4. another point of vue
    in ABC r=s-b
    triangles AFG and ABC are homothetic,center A ratio (s-a)/c
    the incircle of AFG is the image of the incircle of ABC;r1=(s-a)r/c=(s-b)(s-a)/c
    line OH becomes the parallel line O1H1
    distance H1H=(s-a)-(s-a)²/c=(s-a)(s-b)c=r1
    incircle O1 tangent to OH
    The same way CDE and ABC are homothetic with center C and ratio (s-c)/a

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