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Topic: Market Day (Read 1287 times) |
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ThudnBlunder
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The local village is rather quiet, except on the annual Market Day, at which time lines of hurdles are erected from the flag-pole to each of the corners of the quadrilateral that is the market-place. They also run along three of its four sides and enclose three triangular pens of different sizes. The remaining triangular area is occupied by dealers and spectators. Each pen is filled with animals, one to each square metre, and the number of animals in each pen is equal to the number of hurdles forming that pen.
What is the area of the market-place?
EDIT: Each hurdle is one metre long.
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| « Last Edit: Feb 1st, 2009, 7:40pm by ThudnBlunder » |
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Azgard
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Re: Market Day
« Reply #1 on: Feb 1st, 2009, 6:55pm » |
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What is the length of each hurdle? Or is that not relevant to the problem?
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ThudnBlunder
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Re: Market Day
« Reply #2 on: Feb 1st, 2009, 7:39pm » |
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on Feb 1st, 2009, 6:55pm, Azgard wrote:| What is the length of each hurdle? Or is that not relevant to the problem? |
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Sorry, they are each one metre long.
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towr
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Re: Market Day
« Reply #3 on: Feb 2nd, 2009, 4:24am » |
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Is one side of the market longer than either of the diagonals? (And I'll assume you only consider convex quadrilaterals; because otherwise I'll probably get complaints about how the fourth vertex doesn't really count as a corner or something.)
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ThudnBlunder
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Re: Market Day
« Reply #4 on: Feb 2nd, 2009, 6:39am » |
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on Feb 2nd, 2009, 4:24am, towr wrote:| Is one side of the market longer than either of the diagonals? (And I'll assume you only consider convex quadrilaterals; because otherwise I'll probably get complaints about how the fourth vertex doesn't really count as a corner or something.) |
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I see what you are driving at, but I don't believe any further information is necessary in order to arrive at the unique solution.
However, it will probably simplify your options greatly to know that there are no compactified dimensions involved. 
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towr
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Re: Market Day
« Reply #5 on: Feb 2nd, 2009, 12:12pm » |
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Huh, I hadn't expected that they'd all have the same area. Hard to tell on a drawing.
The three animal pens have sides 6-8-10, 9-10-17 and 6-25-29, with areas 24, 36 and 60 respectively.
There's three ways to put them together to form the market, depending on how you attach the other two triangles to the 6-8-10 one. Which forms one of the triangles A-9-29, B-9-25 and C-17-25 for the merchants.
A= (92+292-cos(acos(3/5)+acos(-3/5)+acos(21/29))*2*9*29)
SA=((A+9+29)*(-A+9+29)*(A-9+29)*(A+9-29))/4=90
B=(92+252-cos(acos(3/5)+acos(-3/5)+acos(-3/5))*2*9*25)
SB=((B+9+25)*(-B+9+25)*(B-9+25)*(B+9-25))/4=90
C = (172+252-cos(acos(3/5)+acos(77/85)+acos(-3/5))*2*17*25)
SC=((C+17+25)*(-C+17+25)*(C-17+25)*(C+17-25))/4=90
So in total the market has an area of 24+36+60+90=210
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ThudnBlunder
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Re: Market Day
« Reply #6 on: Feb 2nd, 2009, 12:47pm » |
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on Feb 2nd, 2009, 12:12pm, towr wrote:Huh, I hadn't expected that they'd all have the same area. Hard to tell on a drawing.
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Well done, towr. Neat, eh? As Kevin Costner might say.
It is possible to avoid the messy trig at the end by considering the proportions 5:2 and 3:2
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| « Last Edit: Feb 4th, 2009, 1:58pm by ThudnBlunder » |
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Immanuel_Bonfils
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Re: Market Day
« Reply #7 on: Feb 8th, 2009, 11:05am » |
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on Feb 2nd, 2009, 12:47pm, ThudanBlunder wrote:
It is possible to avoid the messy trig at the end by considering the proportions 5:2 and 3:2 |
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Could you be more explicit?
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Immanuel_Bonfils
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Re: Market Day
« Reply #8 on: Feb 12th, 2009, 7:05am » |
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on Feb 8th, 2009, 11:05am, Immanuel_Bonfils wrote:
Could you be more explicit? |
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Thanks, guess I've got it... mistake doing arith by hard
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