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 Quiz of the Month (July 2003)
Hector C. Parr
***
SOLUTION TO LAST MONTH'S QUIZ
1. 2.828 m
2. 4.123 m
3. 4.162 m
Notes
1. By symmetry, in the critical position the rod makes an
angle of 45^{o} with the corridors. So length = 2 * sqr(2).
2. In plan view the diagram will look the same as in Q.1, but the top
of the pole is 3 m above the bottom. So its length is the diagonal
of a rectangle with height 3 and base 2 * sqr(2).
3. By similar triangles, y/2 = 1/x
So y = 2/x
RO = 2 + x and SO = 1 + y
So RS^{2} = RO^{2} + SO^{2}
= (2 + x)^{2} + (1 + 2/x)^{2}
= x^{2} + 4x + 5 + 4/x + 4/x^{2}.


To find the shortest length RS, differentiate this expression and
put equal to 0. Solving, we get x = cube root of 2.
This is the x value in the critical position, and we find
RS, the length of the longest pole, from the above formula.

THIS MONTH'S QUIZ
1. How many times between noon and midnight do the two hands of a clock form an angle of exactly 90^{o}?
2. At what time exactly, between noon and 1 o'clock, is the minute hand of a clock 90^{o} ahead of the hour hand?
3. What time does the clock indicate if the minute hand and the hour hand each point exactly to one of the sixty divisions on the clock face, with the minute hand one division ahead of the hour hand?

***
(c) Hector C. Parr (2003)
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