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Jun 14th, 2006, 10:29 AM
#1
Thread Starter
Hyperactive Member
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Jun 14th, 2006, 11:51 AM
#2
Re: Rectangle Problem
one possible way would be to make a drawing, another is to use a table. Each time a side of the rectangle is mowed, so it gets 1m smaller in one of the two directions (he either mows a horizontal or a vertical strip), so you get something like:
after 0 sides: 20x40
after 1 sides: 19x30
after 2 sides: 19x39
after 3 sides: 18x49
...
after n sides: 0xSomething
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Jun 14th, 2006, 03:08 PM
#3
Re: Rectangle Problem
To clarify, that is:
20x40
19x40
19x39
18x39
18x38
etc
Perhaps another way to look at it is that he mows 40m^2, 19m^2, 39m^2, 18m^2, 38m^2, 17m^2, 37m^2, etc. The pattern is fairly obvious and is just a simple sum. When this sum exceeds half the area of the lawn, namely 400m^2, then he has done his bit.
zaza
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Jun 15th, 2006, 10:29 AM
#4
Addicted Member
Re: Rectangle Problem
20x40 = 800 m2
19x40 = 760 m2
19x39 = 741 m2
18x39 = 702 m2
18x38 = 684 m2 (one time)
17x38 = 646 m2
17x37 = 629 m2
16x37 = 592
16x36 = 576 m2 (two times)
15x36 = 540 m2
15x35 = 525 m2
14x35 = 490 m2
14x34 = 476 m2 (three times)
13x34 = 442 m2
13x33 = 429 m2
12x33 = 396 m2 (< 400 m2, therefore appr. 3.7 times)
12x32 = 384 m2
Todd will have to go around approximately 3.7 times until he finishes mowing his half.
Rui
...este projecto dos Deuses que os homens teimam em arruinar...
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Jun 15th, 2006, 11:17 AM
#5
Thread Starter
Hyperactive Member
Re: Rectangle Problem [resolved]
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