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# A castle is at the center of the several flat paths which su

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OA B

let distance from the castle to the circular moat is x

distance from 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat ( addition of lenght of diagonals of sqr)= 4x

Thus side of inscribed sqr = x*sqrt2 (use pythagoras 2x^ = (side of sqr)^2.....hence perimeter of sqr = 4*sqrt2*x

circumference of perimeter = pi * (2*x)

Thus our eqn becomes => q= pi * (2*x) + 4*sqrt2*x + 4x

on simplification we get x = q / 2(2+2*sqrt(2)+pi) km...OA B
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Thanks for the solution...
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Castle and the path - [#permalink]
A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?

1) q/4(2+2\sqrt{2}+pie)
2) q/2(2+2\sqrt{2}+pie)
3) q/(2+2\sqrt{2}+pie)
4) 2q/(2+2\sqrt{2}+pie)
5) 4q/2(2+2\sqrt{2}+pie)
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CastlePaths1.gif [ 2.33 KiB | Viewed 15322 times ]

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bhushan252 wrote:
OA B

let distance from the castle to the circular moat is x

distance from 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat ( addition of lenght of diagonals of sqr)= 4x

Thus side of inscribed sqr = x*sqrt2 (use pythagoras 2x^ = (side of sqr)^2.....hence perimeter of sqr = 4*sqrt2*x

circumference of perimeter = pi * (2*x)

Thus our eqn becomes => q= pi * (2*x) + 4*sqrt2*x + 4x

on simplification we get x = q / 2(2+2*sqrt(2)+pi) km...OA B

Strange, this problem seems to have a different answer when we take the side of the square to be a ( or r) or may be I am doing something wrong!
Here, what I am doing -

Let the side of the sq be a, then sum of four sides = 4a;
Using formula, diag of a sq is sqrt(2)a, we get sum of the paths four paths from castle to moat ( i.e, sum the diagonals) as = 2a sqrt(2) ;
Circumference of the circle as 2pie a sqrt(2)/2

Putting it all together we
4a +2asqrt(2) +2pie sqrt(2)/2 = q
=> 4a +2asqrt(2) +pie asqrt(2) =q
=> sqrt(2)a (2 sqrt(2)+2 +pie) = q
=> a = q/sqrt(2) (2 sqrt(2)+2 +pie)

Can anybody shed some more light on this..?
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Re: Castle and the path - [#permalink]
A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?

1) q/4(2+2\sqrt{2}+pie)
2) q/2(2+2\sqrt{2}+pie)
3) q/(2+2\sqrt{2}+pie)
4) 2q/(2+2\sqrt{2}+pie)
5) 4q/2(2+2\sqrt{2}+pie)

Let take radius of circular path is R
so distance from the castle to the circular moat = R

total length of all of the pathways = total length of 4 straight paths that travel from the castle to its circular moat + length of circular path which borders the moat + total length of square path which has its corners at the ends of the 4 straight paths

(Diagonal of square is 2R, (90-45-45 right triangle with hypotenuse 2R) so side of square would be $$R\sqrt{2})$$

$$q = 4R + 2R*pie + 4 (R \sqrt{2})$$
$$q = 2R (2+pie+2 \sqrt{2})$$
$$R = q/2 (2+2 \sqrt{2}+ pie)$$

Originally posted by swatirpr on 18 Nov 2009, 07:36.
Last edited by swatirpr on 19 Nov 2009, 21:11, edited 1 time in total.
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Re: Castle and the path - [#permalink]
swatirpr wrote:
A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?

1) q/4(2+2\sqrt{2}+pie)
2) q/2(2+2\sqrt{2}+pie)
3) q/(2+2\sqrt{2}+pie)
4) 2q/(2+2\sqrt{2}+pie)
5) 4q/2(2+2\sqrt{2}+pie)

Let take radius of circular path is R
so distance from the castle to the circular moat = R

total length of all of the pathways = total length of 4 straight paths that travel from the castle to its circular moat + length of circular path which borders the moat + total length of square path which has its corners at the ends of the 4 straight paths

(Diagonal of square is 2R, (90-45-45 right triangle with hypotenuse 2R) so side of square would be R\sqrt{2})

q = 4R + 2R*pie + 4 (R \sqrt{2})
q = 2R (2+pie+2 \sqrt{2})
R = q/2 (2+2 \sqrt{2}+ pie)

Hi Swatipr,
have a look at - castle-paths-82093.html#p653668
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Thanks a lot swatipr..
oblivious to obvious..strange manifestation of alcohol after effect!
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can we solve this in more simpler way..say plugging smart nos
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This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes?
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TooLong150 wrote:
This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes?

You can actually do it in a minute after you understand that all you need is the radius of the circle given that sum of lengths of all lines and curves here is q.

So q is made up of the circle, the inscribed square in it and the 4 radii of the circle. Assume radius of the circle is r.
Each side of the square will be $$\sqrt{2}r$$ because each side makes an isosceles right triangle with two radii. Each radius is r and sides of isosceles right triangle are in the ratio $$1:1:\sqrt{2}$$

q = Circumference of circle + Perimeter of square + 4*radius
$$q = 2 \pi r + 4\sqrt{2}r + 4r$$

So $$r = q/2( 2 + 2\sqrt{2} + \pi)$$

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Re: A castle is at the center of the several flat paths which su [#permalink]
VeritasPrepKarishma wrote:
TooLong150 wrote:
This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes?

You can actually do it in a minute after you understand that all you need is the radius of the circle given that sum of lengths of all lines and curves here is q.

So q is made up of the circle, the inscribed square in it and the 4 radii of the circle. Assume radius of the circle is r.
Each side of the square will be $$\sqrt{2}r$$ because each side makes an isosceles right triangle with two radii. Each radius is r and sides of isosceles right triangle are in the ratio $$1:1:\sqrt{2}$$

q = Circumference of circle + Perimeter of square + 4*radius
$$q = 2 \pi r + 4\sqrt{2}r + 4r$$

So $$r = q/2( 2 + 2\sqrt{2} + \pi)$$

Hi Karishma,

The part that threw me off was the $$\sqrt{2}r$$

I understood that the hypotenuse was $$\sqrt{2}$$ but I couldn't connect the R there, and more importantly, I couldn't tell that the problem was asking us to solve for R.

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Re: A castle is at the center of the several flat paths which su [#permalink]
wow..the wording is very confusing...
total pathways : circumference, the square, and 4 radii.
circumference is 2*pi*r
square - since we have 4 equilateral triangles, and all right triangles, we can apply 45-45-90 property, and get one side of the square equal to 4*r*sqrt(2)
then we have 4r
we are then told that:
q=2*pi*r + 4*r*sqrt(2) + 4r
factor r
q=r(2pi+4*sqrt(2)+4)
r= q/(2pi+4*sqrt(2)+4)

we can eliminate D and E right away
from the above, we can factor out a 2;
q/2(pi+2*sqrt(2)+2)

so B.
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Re: A castle is at the center of the several flat paths which su [#permalink]
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Re: A castle is at the center of the several flat paths which su [#permalink]
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