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        riddles >> medium >> Market Day
        
(Message started by: ThudanBlunder on Feb 1^st, 2009, 2:57pm)

Title: Market Day
Post by ThudanBlunder on Feb 1^st, 2009, 2:57pm

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.

Title: Re: Market Day
Post by Azgard on Feb 1^st, 2009, 6:55pm

What is the length of each hurdle? Or is that not relevant to the problem?

Title: Re: Market Day
Post by ThudanBlunder on Feb 1^st, 2009, 7:39pm

on 02/01/09 at 18:55:33, Azgard wrote:

What is the length of each hurdle? Or is that not relevant to the problem?

Sorry, they are each one metre long.

Title: Re: Market Day
Post by towr on Feb 2^nd, 2009, 4:24am

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.)

Title: Re: Market Day
Post by ThudanBlunder on Feb 2^nd, 2009, 6:39am

on 02/02/09 at 04:24:26, towr wrote:

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. :P

Title: Re: Market Day
Post by towr on Feb 2^nd, 2009, 12:12pm

Huh, I hadn't expected that they'd all have the same area. Hard to tell on a drawing.

[hide]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=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(9²+29²-cos(acos(3/5)+acos(-3/5)+acos(21/29))*2*9*29)
S_A=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((A+9+29)*(-A+9+29)*(A-9+29)*(A+9-29))/4=90

B=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(9²+25²-cos(acos(3/5)+acos(-3/5)+acos(-3/5))*2*9*25)
S_B=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((B+9+25)*(-B+9+25)*(B-9+25)*(B+9-25))/4=90

C = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(17²+25²-cos(acos(3/5)+acos(77/85)+acos(-3/5))*2*17*25)
S_C=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((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[/hide]

Title: Re: Market Day
Post by ThudanBlunder on Feb 2^nd, 2009, 12:47pm

on 02/02/09 at 12:12:28, towr wrote:

Huh, I hadn't expected that they'd all have the same area. Hard to tell on a drawing.

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

Title: Re: Market Day
Post by Immanuel_Bonfils on Feb 8^th, 2009, 11:05am

on 02/02/09 at 12:47:38, ThudanBlunder wrote:

It is possible to avoid the messy trig at the end by considering the proportions 5:2 and 3:2

Could you be more explicit?

Title: Re: Market Day
Post by Immanuel_Bonfils on Feb 12^th, 2009, 7:05am

on 02/08/09 at 11:05:30, Immanuel_Bonfils wrote:

Could you be more explicit?

Thanks, guess I've got it... mistake doing arith by hard