Monday, May 19, 2008

Elearn Geometry Problem 70



See complete Problem 70
Squares Inscribed in a Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. Proof.

    If n = 1 then s↓1= bh/(b+h), as showed in the problem proposed 69 (with proof posted in September 9, 2009). If n = 2 we have s↓2= s↓1h↓2/(s↓1+h↓2), where h↓2 is the height of the triangle with vertex B and base s↓1.
    From h↓2= h - s↓1 we have h↓2= h - bh/(b+h) = h↑2./(b+h). So,

    s↓2= s↓1h↓2/(s↓1+h↓2) = (bh/(b+h))(h↑2./(b+h))/((bh/(b+h)+(h↑2./(b+h)) = bh↑2./(b+h)↑2.

    Therefore, by induction we have s↓n= bh↑n./(b+h)↑n.

    QED, Ianuarius.

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  2. In addition to the above proof by Unknown on September 11, 2009, find the finalization of the proof by induction here.

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  3. With the help of Problem 69, or just simply by similat triangle, it is easily found that s1=bh/(b+h)

    By similar triangle again, (h-s1-s2)/h=s2/b
    hs2=bh-bs1-bs2
    (h+b)s2=bh-bs1
    By putting back s1=bh/(b+h) and with simplification, the desired s2 is achieved
    The rest can be deducted by MI easily

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