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This is one of Akamai's .. nice one...

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CEO
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This is one of Akamai's .. nice one... [#permalink] New post 10 Oct 2004, 00:15
This is one of Akamai's .. nice one...

http://www.manhattangmat.com

Praetorian
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 [#permalink] New post 10 Oct 2004, 05:07
7/5.

Draw a radius from B to BZ (to from diameter YZ).

r = 360/pi (from the length-of-an-arc formula).

Draw a radius from B to A, results in a 'small' semicircle with diameter of 360/pi. Area of this 'small' circle is 90(180)/pi. You have to add it to the area of the sector YBC in the denominator of the ratio, and subtract it from the area of the semicircle in the numerator (together with the area of sector ABZ).

Now, angle YXA = 105, draw a line from X to Z, forming angle YXZ inscribed in a semicircle, = 90 degrees. Then angle AXZ is 105-90=15, and results in arc AZ of 30, with angle ABZ = 30 degrees.

From there, get the area of sector ABZ = 30/360pir^2 = (3*360)/pi

For the ratio, in the numerator, you have Area of semicircle - area of ABZ - area of 'small' semicircle

in the denominator, you have area of ABZ + area of small 'semicircle'

solve it out, and you should get 180*210/180*150 = 210/150=7/5.

After this one, I guess I will retire for today!
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 [#permalink] New post 10 Oct 2004, 07:08
Praet,

Could you please repost it I guess there is a pb ...
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 [#permalink] New post 10 Oct 2004, 07:26
twixt wrote:
Praet,

Could you please repost it I guess there is a pb ...


repost? ok, this might work...

http://www.manhattangmat.com/index.cfm?cid=11&sethome=1

Praetorian
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 [#permalink] New post 10 Oct 2004, 12:30
Thanks Praet,

Same method, same pain...

Answer is 7/5 after some nice plays with isoceles

Took me 9 mn...
Joined: 31 Dec 1969
Location: India
Concentration: Strategy, Human Resources
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GMAT 4: 710 Q45 V41
GPA: 3.31
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Re: Manhattan GMAT problem of the week [#permalink] New post 10 Oct 2004, 18:00
Knowing this theorem can help a lot:
The opposite angles of quadrilaterals in circles are equal to two right angles.

If you complete the quadrilateral YCAX, then according to above theorem, ang YXC + ang YCA = 180

so,ang yca = 180-105=75

Another theorem to remember:
Angle at the centre is double the angle at the circumference if the angles have the same circumference as base.

Based on the above thm, ang YBA is double of the ang YCA.
So, ang YBA = 150

So ang YBC=180-150=30

Area of arc ybc = 0.5*r^2*pi/6

since 30 deg = pi/6

Area below = 0.5*r^2*pi/6 + pi*r^2/8

Area above = 0.5*r^2*5*pi/6 - pi*r^2/8

so ration above/below=7/5






Praetorian wrote:
This is one of Akamai's .. nice one...

http://www.manhattangmat.com

Praetorian
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Manhattan GMAT problem of this week [#permalink] New post 11 Oct 2004, 20:28
EDITED: anish, Manhattan gmat sells the archives of these questions. We may not want to be in copyright trouble :) thats why we usually only post the link, not the problem.
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 [#permalink] New post 11 Oct 2004, 23:19
yes, the answer is indeed 7/5. go to the website to see the legend akamai's explanation. thanks all :-)
  [#permalink] 11 Oct 2004, 23:19
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