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Quiz of the Month (June 2003)

Hector C. Parr

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SOLUTION TO LAST MONTH'S QUIZ

    1. 6 2/3 cm
    2. 4 cm
    3. 4 cm
Notes 
  On the actual fence the quadrilateral ABFE is a rectangle, so
  its diagonals must meet at the mid-point of CD. The same must be
  true on the perspective drawing.
    Now triangles CQE and ABE are similar.
    So CE/AE = CQ/AB   i.e. y/(x+y) = b/2a      -     (i)

    Similarly,              x/(x+y) = b/2c      -     (ii)

  Adding (i) and (ii):  1 = b/2a + b/2c
  Whence 2/b = 1/a + 1/c
  
  Dividing (ii) by (i):  x/y = (b/2c)/(b/2a) = a/c

  So for any three consecutive posts we have:
  (a) the drawn length of the middle post is the Harmonic
        Mean of the drawn lengths of the other two posts.
  (b) the position of the middle post divides the distance
        between the other two posts in the ratio of their lengths.
 
  The solutions follow easily from these results.

THIS MONTH'S QUIZ

1. A thin, straight rod is being carried along a corridor 1 m wide and 3 m high. If the corridor has a right angled corner, what is the length of the longest rod that can be taken round the corner, if the rod must be kept horizontal?

2. If the rod may be tilted, what then is the longest rod that can be taken round the corner?

3. Another corridor has a right angled corner, but the two sections which meet at the corner have widths of 1 m and 2 m respectively. What is the length of the longest rod that can be taken round this corner, if the rod must be kept horizontal?

***

(c) Hector C. Parr (2003)


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