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DensetsuNo
Joined: 11 Oct 2015
Last visit: 05 Aug 2023
Posts: 88
Given Kudos: 38
Status:2 months to go
Schools: HBS 2+2 '22 MIT Sloan MSMS"18
GMAT 1: 730 Q49 V40
GPA: 3.8
26 Jul 2016, 06:46
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In the diagram above, the triangle is equilateral with a side of 2.
All three circles are of equal size and all are tangent to each other and to two sides of the triangle.
Which of the following is the radius of one of the circles?
A) 1+ √3
B) 2 - √3
C) √6/4
D) 1 - √3/2
E) (√3-1)/2)
Attachment:
GMAT_PS_Manhattan_380.png [ 9.89 KiB | Viewed 7771 times ]
DensetsuNo
Joined: 11 Oct 2015
Last visit: 05 Aug 2023
Posts: 88
Given Kudos: 38
Status:2 months to go
Schools: HBS 2+2 '22 MIT Sloan MSMS"18
GMAT 1: 730 Q49 V40
GPA: 3.8
26 Jul 2016, 08:57
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Here's how I'd solve it.
As in the first 30secs I didn't reckon any pattern (there are, I just didn't see them), I estimated.
We can find the height of the equilateral triangle which is √3, and we can see that r has to be between √3/4 and √3/5, so approximately 0,43<r<0,34.
The solutions are:
A) 1+ √3 ≈ 2,7
B) 2 - √3 ≈ 0,29
C) √6/4 ≈ 0,6
D) 1 - √3/2 ≈ 0,25
E) (√3-1)/2) ≈ 0,35
The only one included in the range is E), so the answer has to be E). (And in fact it is).
There are other ways to approach it, if enough of you guys like the problem I'll post Magoosh official solution (although recopying everything is a pain :D).
Den
As in the first 30secs I didn't reckon any pattern (there are, I just didn't see them), I estimated.
We can find the height of the equilateral triangle which is √3, and we can see that r has to be between √3/4 and √3/5, so approximately 0,43<r<0,34.
The solutions are:
A) 1+ √3 ≈ 2,7
B) 2 - √3 ≈ 0,29
C) √6/4 ≈ 0,6
D) 1 - √3/2 ≈ 0,25
E) (√3-1)/2) ≈ 0,35
The only one included in the range is E), so the answer has to be E). (And in fact it is).
There are other ways to approach it, if enough of you guys like the problem I'll post Magoosh official solution (although recopying everything is a pain :D).
Den
That was hard. It actually took me a good time.
I thought in dividing the figure in more geometric figures:
If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.
If smaller side is \(r\) - of the new (30-60-90) triangle - than its hypotenuse is \(2r\) and its longer leg is \(r\sqrt{3}\).
Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.
Now you have \(r\sqrt{3}\), then \(2r\) between the 2 (30-60-90) triangles and another \(r\sqrt{3}\).
These measurements summed up must be equal to the side, which is 2.
\(r\sqrt{3}+r+r+r\sqrt{3} = 2\)
\(r = \frac{sqrt(3)-1}{2}\)
ps: I tried to use square root inside the fraction, but it didn't work :/
I thought in dividing the figure in more geometric figures:
If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.
If smaller side is \(r\) - of the new (30-60-90) triangle - than its hypotenuse is \(2r\) and its longer leg is \(r\sqrt{3}\).
Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.
Now you have \(r\sqrt{3}\), then \(2r\) between the 2 (30-60-90) triangles and another \(r\sqrt{3}\).
These measurements summed up must be equal to the side, which is 2.
\(r\sqrt{3}+r+r+r\sqrt{3} = 2\)
\(r = \frac{sqrt(3)-1}{2}\)
ps: I tried to use square root inside the fraction, but it didn't work :/
