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Difficult Math Question

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Difficult Math Question

by robk100 » Mon Jul 28, 2008 11:52 am
This isn't exactly a gmat problem but difficult I was hoping someone could help me here:

3 identical circles with radius 10 are inscribed in an equilateral triangle. I need to find the length of 1 of the sides of the equilateral triangle. The circles are touching, don't overlap, and touch the sides of triangle. Thanks
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by parallel_chase » Mon Jul 28, 2008 12:01 pm
Its better you post the entire question; a figure along with is appreciated.
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by rs2010 » Mon Jul 28, 2008 2:56 pm
Use this

side of triangle = 2* radius* (root 3 + 1)


If you want I can prove it.
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by parallel_chase » Mon Jul 28, 2008 3:03 pm
hemantsood wrote:Use this

side of triangle = 2* radius* (root 3 + 1)


If you want I can prove it.
Pls do so, I am very interested in this formula.

Thanks.
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by rs2010 » Mon Jul 28, 2008 3:47 pm
Ok here you go ...

Consider triangle ABC and circles inside it with centers a,b and c in such a way that circle with center a touches sides AB and AC, circle with center b touches sides BC and AB and circle with center c touches sides BC and AC.

Now draw perpendicular from center b to side BC. Now you will have a triangle Bbb' with bb' being perpendicular.

In triangle Bbb' <bBb' is 30 degrees.

So tan 30 degree = bb'/Bb'

root 3 = bb'/Bb'

Bb'= root 3 * radius of circle ( bb' is radius of circle)

Similarly perform this operation with center c.

you will get

Cc' = root 3 * radius of circle.

BC=Bb'+b'c'+c'C
BC=root 3*r + 2*r+ root 3 * r
=2*r*(root 3 +1)

I dont scanner other wise I would have attached the figure.

Please let me know if face any problem in understanding it.
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by parallel_chase » Mon Jul 28, 2008 4:00 pm
Can you do this without using trigonometry, its been almost 8 years since I have touched tan, cos or sin. To my knowledge GMAT does not tests you on trigonometry. Let me know your thoughts.
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by rs2010 » Mon Jul 28, 2008 5:28 pm
Instead of using tan you can consider a 30 60 90 triangle.

If hypotenous is a then side opposite to 30 degree would be a/2 and opposite to 60 degree would be root 3 * a /2.

bb'/Bb'= a/2 /root 3 * a/2.
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by malolakrupa » Tue Jul 29, 2008 6:41 am
Hi Hemant ,

I did not get the part where you said

"In triangle Bbb' <bBb' is 30 degrees. "

Thanks
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by rs2010 » Tue Jul 29, 2008 2:46 pm
I meant angle bBb' is 30 degrees.
This was the closest sign that I could think of :lol:
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