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Interesting problem with Geometry .. any idea ??

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by vishubn » Fri Aug 01, 2008 12:10 am
FOund this problem in manhattan ! can neone help me out ! with the easiest way i took more than three months to solve it :(

any short cut method would be of greta help


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by Stuart@KaplanGMAT » Fri Aug 01, 2008 12:39 am
There's some truly ugly math in this question, but the principles are all basic geometry.

Let's draw a line directly up from O to the top side of the square and call the point on the top of the line V.

Now let's look at triangle OVX. OV is a radius of the circle and VX is also equal to the radius of the circle, so we've created a 45/45/90 triangle with OX as the hypotenuse.

Let's call the radius r. Since we have a 45/45/90 triangle, the sides will be in the ratio of x:x:xroot2. In our triangle, x=r, so we can call the 3 sides r, r and r(root2).

If we look at the hypotenuse OX, we can see that it has two parts: radius OP and extra bit PX. From the question, we know that PX=1. So, line OX has length r + 1.

Using our 45/45/90 ratio, we now know that r(root2) = r + 1

Manipulating that equation, we get:

r(root2) - r = 1
r(root2 - 1) = 1
r = 1/(root2 - 1)

At this point, let's answer the question. We want the circumference of the circle, which is 2(pi)r. So, we can say that:

circumf = 2pi(r) = 2pi(1/root2 - 1)

If we know that root2 is approximately 1.4 (a good thing to know for test day), we can substitute in to get:

2pi(1/1.4 - 1) = 2pi(1/.4) = 2pi(10/4) = 2pi(5/2) = 10pi/2 = 5pi... choose (d).

Here are the big takeaways from this question (every time you do a practice question, you should ask yourself "what did I learn from this question that I can apply to future questions?"):

Almost all multiple shape questions on the GMAT require you to use what you know about simple shapes: circles, squares/rectangles and triangles. Accordingly, look for very familiar shapes inside the complicated ones.

Know your special right triangles, especially 45/45/90 and 30/60/90.
Last edited by Stuart@KaplanGMAT on Fri Aug 01, 2008 12:43 am, edited 1 time in total.
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by pepeprepa » Fri Aug 01, 2008 12:42 am
Radius: x
We know that: OX=x sqrt(2)
And with the graph we see that: OX=OP + PX= x + 1
We join both equations, it gives
x sqrt(2) = x + 1
So our radius is now x=1/(sqrt(2)-1)

sqrt(2) is 1,4 more or less so sqrt(2)-1=4/10 more or less and x=5/2 more or less

Circumference formula is: 2piR
2pi(5/2)=5pi
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by sudhir3127 » Fri Aug 01, 2008 12:55 am
hey how did u get OX as X sqrt 2..( are we drawing a line from O and assuming it to be a issoscles triangle?)
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by pepeprepa » Fri Aug 01, 2008 12:59 am
It is isocele because we have two sides which have the length of the radius.
T:point middle in the middle of XY
The triangle OTX has OT=TX=radius
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by Stuart@KaplanGMAT » Fri Aug 01, 2008 1:00 am
sudhir3127 wrote:hey how did u get OX as X sqrt 2..( are we drawing a line from O and assuming it to be a issoscles triangle?)
From my post above:
Let's draw a line directly up from O to the top side of the square and call the point on the top of the line V.

Now let's look at triangle OVX. OV is a radius of the circle and VX is also equal to the radius of the circle, so we've created a 45/45/90 triangle with OX as the hypotenuse.

Let's call the radius r. Since we have a 45/45/90 triangle, the sides will be in the ratio of x:x:xroot2. In our triangle, x=r, so we can call the 3 sides r, r and r(root2).
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