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mjfsutherland
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Inscribe your name in my circular heart with approximation

by mjfsutherland Mon Mar 29, 2010 12:48 am

In the recent challenge problem:

A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
(A) 50%
(B) 55%
(C) 60%
(D) 65%
(E) 70%

I made it all the way through the calculations and ended up with

percentage coverage = (Pi x sqrt(3) ) / 9

The official answer ends up with:

percentage coverage = Pi / (3 x sqrt(3) )

Both answers are identical and I realise the exact calculation to both is 60.46 % (rounded to the nearest 100th).

I used 3.1 to approximate Pi and 1.7 to approximate sqrt(3) just as the official answer does, but with my answer you end up with

5.17 / 9 which is a closer approximation to 55% than 60%.

I guess my question is how do you know whether your approximation will give you the correct result and what if any guidelines would you advise when using approximated values. Is it always better to keep approximated values on denominator or should you try and keep them balanced on numerator and denominator?
sanyalpritish
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Re: Inscribe your name in my circular heart with approximation

by sanyalpritish Mon Mar 29, 2010 4:34 am

Can you mention the Source and mention the steps you got to the Answer which U got
sanyalpritish
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Re: Inscribe your name in my circular heart with approximation

by sanyalpritish Mon Mar 29, 2010 5:06 am

Did some Googling and found the below mentioned learning them will help.

Circumscribed Circle Radius R=a(3)^(1/2)/3

Inscribed Circle Radius r=a(3)^(1/2)/6
mjfsutherland
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Re: Inscribe your name in my circular heart with approximation

by mjfsutherland Mon Mar 29, 2010 5:54 am

Sorry I forgot that the challenge problems are not available to everyone. Its one of the ManhattanGMAT "challenge" problems (March 1).

Basically the way you solve the problem is draw a line that bisects each of the internal angles of the equilateral triangle. Due to symmetry and the nature of an equilateral triangle, each line will pass through the centre of the triangle and the centre of the circle and will bisect the opposite side perpendicularly. You end up with 6 sub triangles. You can work out that each of these triangles is a 30-60-90 triangle. From this you can work out the ratio between the radius of the circle and the length of the side of the triangle. Hence you work out the percentage of area covered by the circle within the triangle.

If you do this you will end up with Pi / (3 x sqrt(3) ) or (Pi x sqrt(3) ) / 9 and this is where my initial query starts off....
thoppae.saravanan
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Re: Inscribe your name in my circular heart with approximation

by thoppae.saravanan Mon Mar 29, 2010 6:03 am

Reason is very simple. 5.17/9 approximates to 57.4. Here you have approximated the numerator values pi and sqrt(3) as 3.1 and 1.7 which is in a way lesser than the real one namely 3.142 and 1.732. So this clearly indicates the actual should always be greater than 57.4 and definitely not lesser than 57.4. So you should go for 60%.

Caution: Here the option D is 65%. Hence you can safely select the answer as C thinking that your approximation will not be varied much. But again it depends. If there is an option of 62 then for sure we are in trouble and we should go for second digit namely pi=3.14 and sqrt(3)=1.73.

My advice is always go for 22/7 for pi and always have sqrt(3) at the numerator. This way it will be very easy for calculation purpose so that you can arrive at best answer.
mschwrtz
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Re: Inscribe your name in my circular heart with approximation

by mschwrtz Wed May 12, 2010 3:50 am

If you have two factors, like xy, round one up and the other down. If you have a fraction, round both up OR both down.

The reason that you ended up further off is that the two approximations you used compounded one another. You rounded down each of two factors. If you have have two factors, and you have to round one down, try to round the other up by a similar proportion.

Our approximation, though, rounded both parts of a fraction down, so the changes offset one another, rather than compounding.