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Thread: Hard linear transformation question (corrected)

  1. #1
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    Hard linear transformation question (corrected)

    Hard linear transformation question (corrected)-scan.jpg

    The diagram above shows a tessellation of a plane of equilateral triangular tiles. A right-angled triangle is of such a shape that it can cover exactly half of any tile. Its initial position is marked t in the diagram.

    The operation of reflecting t in its shortest side is written A, and starting with the given position t the result is A(t), which is marked with the figure 1.

    Reflection in the second side is denoted by B, so that B(t) is 2.

    Reflection in the hypotenuse is C so that C(t) is 3.

    Position 4 is B(3) = B(t), so that the operation carrying t to 4 is BC.

    Questions...

    (i) Show that operations AB and BA are the same


    (ii) Describe the combined operation BC geometrically.


    (iii)Simplify C(CB), (CC)B and BC(CB)...using I to denote the identity operation if necessary.


    (iv) Obtain expressions, as combinations of A, B and C for operations which carry


    (a) t to x
    (b) t to y
    (c) x to t

    If you can answer some or all of this it is much appreciated.
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  2. #2
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    Re: Hard linear transformation question (corrected)

    (ii) if initial triangle is $mnp$ then

    $C(m)=m,C(p)=p,C(n)=q$

    and

    $B(m)=m,B(p)=r,B(q)=q$

    therefore

    $BC(m)=m,BC(p)=r,BC(n)=q$

    that's a rotation
    Thanks from Phavonic
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  3. #3
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    Re: Hard linear transformation question (corrected)

    My guess at (iV) is that

    (a) BC

    (b) ABC

    (c) CB
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