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$\begingroup$

I've got some trouble...

IJKL is a square and B, I, J, C are aligned (alternatively, |IJ| is confounded with |BC|.

h is the height of acute $\triangle$ ABC from A to side BC.

C1 is the red square.
C2 is the green square.
C3 is the blue square

a= |BC|
b= |CA|
c= |AB|

  1. How to find IJKL using a and h
  2. a< b< c. Classify C1, C2, C3 from the largest.
  3. Find ABC triangles (their area, S) when the square IJKL has the largest area.

squares in triangle

Thank you in advance.

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    What is this??.2012-11-30
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    You may want to upload the picture somewhere else, because I have to register to Math Help Forum to be able to see the picture to which you are linking.2012-11-30
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    Can you see the picture ?2012-11-30
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    @Dan, I separated the statements as you did in your post (use two "enters" to separate lines, or when you want to begin another line, use
    . I assumed that, e.g., [BC] means the length |BC|? I'm not clear what you mean by [IJ] is *confunded with* [BC]. Do you mean confounded? Or that |IJ| is proportional to |BC|, and varies as |BC| varies?
    2012-11-30
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    @amWhy, I mean that B, I, J and C are aligned. Thank you for editing my post.2012-11-30
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    What do you mean with 'How to find IJKL using a and h'? Do you wish to express the perimeter of the square IJKL in a and h? Or the coordinates of the points, somehow relative to the triangle's points?2012-11-30
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    In fact, I want to express the lenght of the sides of IJKL using a and h. Did you understand ?2012-11-30
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    All right, that's clear. Are you sure you mentioned all available data?2012-11-30
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    Yes, I'm sure, I need to obtain for the first question something like (a*h)/a+h...2012-11-30
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    Yep, that's correct for the first part.2012-11-30
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    But how do you find that result @JonathanChristensen ?2012-11-30
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    I'm writing it up now.2012-11-30
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    I forgot to add (I don't know if it will help) : three acute angles for ABC.2012-11-30
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    That fact is crucial for the second question.2012-11-30
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    Sorry I didn't realize2012-11-30
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    Do you think you can do it with what I added ?2012-11-30
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    I updated my answer with a trigonometric proof of 2., though I would still like a geometric one. The triangle being acute isn't as crucial as I thought--it just can't be a right triangle.2012-11-30

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