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animeshsaxena
Posts: 520
Joined: June 19th, 2008, 2:56 pm

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May 8th, 2011, 7:55 pm

From the diagram I guess it's an incircle for the smaller triangle.The angle bisectors will hit the incenter, so first I need to find the two angles.One of them is obviously 45degree. Other one after some trig...can be found to be 63.43(approx)Tan[45/2] = r/xTan[63/2] = r/[a-x]a-x + x = r/tan[45/2] + r/tan[63/2]I had drawn a perpendicular from circle centre to opposite side...so r = tan[45/2]tan[63/2]/(1+tan[45/2]tan[63/2])
 
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wileysw
Posts: 593
Joined: December 9th, 2006, 6:13 pm

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May 8th, 2011, 7:56 pm

a*2/(sqrt(5)+sqrt(8)+sqrt(9)) or a*(1+sqrt(2)+sqrt(5)-sqrt(10))/6any other interesting form?----- ----- ----- ----- -----hmm... scooped:(
Last edited by wileysw on May 7th, 2011, 10:00 pm, edited 1 time in total.
 
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animeshsaxena
Posts: 520
Joined: June 19th, 2008, 2:56 pm

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May 8th, 2011, 8:01 pm

so who gets the two points....answers match
 
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animeshsaxena
Posts: 520
Joined: June 19th, 2008, 2:56 pm

*puzzle quiz*

May 8th, 2011, 10:06 pm

Made a very very silly mistaker = tan[45/2]tan[63/2]/(1+tan[45/2]tan[63/2])should have beenr = tan[45/2]tan[63.43/2]/(tan 45/2 + tan [63.43/2]) = 0.248instead of writing 1 + in the denominator....must be slope formula swirling in my head....Had verified the answer...but instead of typing it...wrote the silly formula I agree damn stupid Definitely 0 points for this!
 
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Traden4Alpha
Posts: 23951
Joined: September 20th, 2002, 8:30 pm

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May 8th, 2011, 10:47 pm

Hmm... Technically, the problem statement is ill-formed. What is the height of the outer rectangle?
 
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animeshsaxena
Posts: 520
Joined: June 19th, 2008, 2:56 pm

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May 9th, 2011, 7:01 am

I think it's a square
 
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Traden4Alpha
Posts: 23951
Joined: September 20th, 2002, 8:30 pm

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May 10th, 2011, 12:04 pm

No. This problem is a 3-D variant on a knights tour problem in which each move is down one unit in two of the dimensions and up two units in one dimension.This movement pattern means that the differences in the population values remain constant modulo 3 after any change. Achieving a monochromatic population requires a modulo 3 difference of 0 between two elements in the initial condition and the given initial condition of [20,18,16] has modulo 3 differences of 2.
Last edited by Traden4Alpha on May 9th, 2011, 10:00 pm, edited 1 time in total.
 
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katastrofa
Posts: 8730
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

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May 10th, 2011, 7:17 pm

QuoteOriginally posted by: outrunNo, its not a square, if so,, then the solution would only be correct for a=1 Actually "a" stands for "Area", I forgot to mention that. It is definitely a square and a is its side, unless wileysw measured the radius in the units of area, which would be very wrong :-)a~
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