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Math Help - Circle Geometry: Shortest Chord

  1. #1
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    Circle Geometry: Shortest Chord

    AB, CD and XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove that XY is greater than either AB or CD.

    I hav no idea how to approach this problem
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  2. #2
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    Extreme value problem...

    Quote Originally Posted by Lukybear View Post
    AB, CD and XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove that XY is greater than either AB or CD.

    I hav no idea how to approach this problem
    Look at it as an extreme value problem like this:
    Since both AB and XY are chords of the same circle that intersect in the midpoint L of AB, you have that XL\cdot LY = AL\cdot LB =: k and that XY=XL+LY.
    Since XL\cdot LY=k, independently of the exact choice of XY, you'll find that the case where XY=XL+LY attains its smallest value is exactly the case where XL=LY=\sqrt{k}. Since we know that AL=LB=\sqrt{k}, we can infer from this that XY\geq AB=2\sqrt{k}. Similarly for XY\geq CD.
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  3. #3
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    Sorry about this, but i havent learnt half of this. Could you just explain to me why XL.LY = AL.LB =k?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Lukybear View Post
    Sorry about this, but i havent learnt half of this. Could you just explain to me why XL.LY = AL.LB =k?
    This is called the Chord Theorem (aka. Chord-Chord Power Theorem): it can be shown to hold by considering similar triangles, for example.
    However, if what I wrote seems like so much gibberish to you, then, maybe, another approach is needed than what I have suggested. I just cannot think of a fundamentally different approach at the moment.
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  5. #5
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    Perhaps a simple geometrical proof using circular properties would do? I have no idea, but if anyone could help me out on this that would be good.
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