Elegance Revisited (1 Viewer)

:: ck ::

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oh yeh crap i didnt read properly!! arghhh >.<

ok so the coordinates of S and T are S(a,-b) and T(b,a)

mST = (-b-a)/(a-b)
Pt Grad Formula [using S]... equation ST is...

y+b = [ (-b-a) / (a-b) ] (x-a)

Ok so coordinates of Q must lie within the x axis, so sub y=0
b = [ (-b-a) / (a-b) ] (x-a)
x = b(a-b)/(-b-a) + a
= (ba-b<sup>2</sup>) / ( -b-a ) + a

So coordinates of Q [ (-b<sup>2</sup>+ab) / ( -b-a ) + a, 0]

R lies on the line, y=x
Thus sub y =x (vice versa)

x + b = [ (-b-a) / (a-b) ] (x - a)
x(a-b) + b(a-b) = x(-b-a) -a(-b-a)
x(a-b) + x(b+a) = -b(a-b) -a(-b-a)
x(a-b+b+a) = -ba + b<sup>2</sup> +ab + a<sup>2</sup>
x = (b<sup>2</sup> + a<sup>2</sup>) / 2a

and since y = x,
Coordinates of R is [ (b<sup>2</sup> + a<sup>2</sup>) / 2a, (b<sup>2</sup> + a<sup>2</sup>) / 2a ]

.'. Coords of R and Q such that PQ + QR + RP is minimised is [ (b<sup>2</sup> + a<sup>2</sup>) / 2a, (b<sup>2</sup> + a<sup>2</sup>) / 2a ], and Q [ (-b<sup>2</sup>+ab) / ( - b-a ) + a, 0]

man i confused myself with this typing in this little quick edit box.... i mite've stuffed up something... T_T
 
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OLDMAN

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Ryan.cck : Method is right. Reflecting across y=x and y=0 seems to be an elegant approach. Nice careful work.

Being careful is another requirement for doing well in Ext2. Practise your care on the following problem :

Let 0 <= a <=4. Prove that the area of the bounded region enclosed by the curves with equations y=1-|x-1| and y=|2x-a| cannot exceed 1/3.
 

:: ck ::

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can u just simply say the greatest area is when a = 1... find the interesection between the two lines, which happens to be 1/3 of the distance away from the point x = 1...

split up into two triangles and using area of one of the triangles x 2 = 1/3 [by symmetry] ?
 

OLDMAN

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____________________________________________________
:: ryan.cck :: can u just simply say the greatest area is when a = 1... find the interesection between the two lines, which happens to be 1/3 of the distance away from the point x = 1...

split up into two triangles and using area of one of the triangles x 2 = 1/3 [by symmetry] ?
____________________________________________________

Did you mean greatest area at x=2?

Maximization by symmetry? I'd pass it if I was the marker but...
 

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