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Inspired from a recent post: How does one obtain the radius

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young_gun
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Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 09:18
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?



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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 09:45
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:)

for me:
\(\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}\)

\(r=\frac{1}{sqrt{3}}x\)
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 09:40
Jack.Zhou wrote:
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?


I get (√3) * x / 6


the answer is x sqrt(3)/3
actual solutions, as opposed to answers, would be greatly appreciated [walker, what's taking you so long to respond??].
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 10:16
walker wrote:
:)

for me:
\(\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}\)

\(r=\frac{1}{sqrt{3}}x\)


any non-trig way to do this?
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 10:34
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young_gun wrote:
walker wrote:
:)

for me:
\(\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}\)

\(r=\frac{1}{sqrt{3}}x\)


any non-trig way to do this?


\(r^2=(\frac{x}{2})^2+(\frac{r}{2})^2\)

\(r=\frac{1}{sqrt{3}}x\)
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 10:44
walker wrote:
young_gun wrote:
walker wrote:
:)

for me:
\(\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}\)

\(r=\frac{1}{sqrt{3}}x\)


any non-trig way to do this?


\(r^2=(\frac{x}{2})^2+(\frac{r}{2})^2\)

\(r=\frac{1}{sqrt{3}}x\)


thanks, but do you think you could post a diagram or explain more...i'm not seeing it. seems like you used pythag? where does (r/2)^2 come from? is it known that in an equil. triangle, the mid point to the middle of the sides is r/2?
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 11:12
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hope this help :)
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 13:56
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?


:P knew that question would spark some interest.

Walker has a nice picture, but essentially the long side of the smaller triangle is x/2 then small side is xsqrt3/6

So the hypt. is xsqrt3/3 Which is the radius. or as walker puts it x/sqrt3
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Re: Inspired from a recent post: How does one obtain the radius [#permalink] New post   17 Jan 2008, 15:51
Great analysis of the question. You guys deserve some kudos.

GMATBLACKBELT wrote:
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?


:P knew that question would spark some interest.

Walker has a nice picture, but essentially the long side of the smaller triangle is x/2 then small side is xsqrt3/6

So the hypt. is xsqrt3/3 Which is the radius. or as walker puts it x/sqrt3




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Re: Inspired from a recent post: How does one obtain the radius [#permalink]
17 Jan 2008, 15:51
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