Thursday, July 9, 2009

Problem 317: Right triangle and Inscribed Squares

Proposed Problem
Click the figure below to see the complete problem 317.

 Geometry Problem 317: Right triangle and Inscribed Squares.
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Complete Problem 317
Level: High School, SAT Prep, College geometry


  1. Denoting AH & PC by c & d resply. and noting the fact that all rt. angled triangles within Tr. ABC are similar to it, we can say that c/(c+a)= a/x while d/(d+b)= b/x. Further, c/a = AB/BC and d/b = BC/AB or c/(c+a)= AB/(AB+BC) and d/(d+b)= BC/(AB+BC). This gives us, a/x + b/x = c/(c+a) + d/(d+b)= AB/(AB+BC) + BC/(AB+BC) = 1. In other words, x = a + b QED.

  2. tr JKE and tr EBF are similar ( an JEK = an BFE )

    => EB/BF = a/x-a

    tr EBF and MFN are similar ( an BFE = an FNM )

    => EB/BF = x-b/b

    from above => a/x-a = x-b/b

    => x = a + b

  3. May I ask a question?
    ABC is a right angled Triangle with right angle at A. How to inscribe a square PQRS such that P. Q lie on the sides AB, AC respectively and both R, S lie on the hypotenuse BC ?.
    Vijaya Prasad Nalluri (Pravin)
    Rajahmundry, INDIA.

    1. Geometry solution:
      Let BC= a and AH= h ( see sketch)
      1. Draw altitude AH= h
      2. Extend BT such that BT= AH= h
      3. Connect AT
      4. Draw BQ//AT . Q is on AC
      5. Draw QP//BC. P is on AB
      6. Draw QR ⊥BC. Q is on BC
      7. Draw PS⊥BC. S is on BC
      Perform calculation we will get PQ=QR= a.h/(a+h)
      And PQRS is a Square