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- #1

- Feb 14, 2012

- 3,680

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,680

Does that count as a solution?

Is Wolframalpha making us lazy?

- Dec 25, 2012

- 42

Anyway, so I see that two sides of the hexagon are each 11 in length, so if I draw a line from where each of these points connect to the circle, then I have a triangle. And since each side is length 11 each, I can assume it's a 45-45-90 triangle. So then I can use pythagorean's theorem to find the hypotenuse of that triangle, then divide it by two to find the radius:

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle 11^2 + 11^2 = c^2\)

\(\displaystyle 121+ 121= c^2\)

\(\displaystyle 242= c^2\)

\(\displaystyle \sqrt{242}\) =c

\(\displaystyle \frac{\sqrt{242}}{2} =\) radius

\(\displaystyle 7.778174593052\) =approximate decimal notation solution of radius

Am I on the right track?

Can you?...I can assume it's a 45-45-90 triangle...

- Dec 25, 2012

- 42

Good point, I probably could NOT. Thanks for that! I'm still pretty new to doing this. I've got a lot of polishing to do!Can you?

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- #6

- Feb 14, 2012

- 3,680

Hi M R,

Does that count as a solution?

Is Wolframalpha making us lazy?

If you don't mind me asking, could you please tell me more about how exactly did you come up with that brilliant formula? It might be obvious to you but it isn't to me...

Thanks!

And I'm really sorry, your solution isn't

- Feb 13, 2012

- 1,704

$$\cos ^{-1} (1-\frac{2}{r^{2}}) + \cos^{-1} (1-\frac{49}{2\ r^{2}}) + \cos^{-1} (1-\frac{121}{2\ r^{2}}) = \pi\ (1)$$

... and its solution is $r=7$...

Kind regards

$\chi$ $\sigma$

,

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- #8

The law of cosines gives us:

\(\displaystyle L^2=53-28\cos(A)=4r^2+121-44r\cos(B)\)

Because $A$ and $B$ are opposite angles in a cyclic quadrilateral, we know they are supplementary, and using the identity \(\displaystyle \cos(\pi-x)=-\cos(x)\), we may write:

\(\displaystyle L^2=53-28\cos(A)=4r^2+121+44r\cos(A)\)

This gives us:

\(\displaystyle \cos(A)=\frac{53-L^2}{28}=\frac{L^2-4r^2-121}{44r}\)

Now, by Pythagoras, we have:

\(\displaystyle L^2=4r^2-121\)

Hence, we may state (after some simplification):

\(\displaystyle \frac{2r^2-87}{7}=\frac{11}{r}\)

Cross-multiplying, we get the cubic:

\(\displaystyle 2r^3-87r-77=0\)

This may be factored as:

\(\displaystyle (r-7)\left(2r^2+14r+11 \right)=0\)

Discarding the negative roots, we are left with:

\(\displaystyle r=7\)

- Dec 25, 2012

- 42

So what exactly lead you to use the law of cosines for this problem?

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- #10

The law of cosines gives us a convenient way to relate the three sides and one of the interior angles of any triangle.So what exactly lead you to use the law of cosines for this problem?

Also, knowing that $A$ and $B$ are supplementary, and the fact that we also have a right triangle in there helps too.