Melody

 - - 4 = +4

Hi Tony,

This question is long and tedious but reasonably straight forward.

We have points:  A(5, 3), B(– 2, 2) and C(2, – 6) and we trace the perpendicular bisectors  of segments AB and AC.

a) Find the coordinates of the intersection point of these perpendicular bisectors.

Find  the gradient of AB, the gradient of the perpendicular will be the negative reciprocal.

Find the midpoint of AB

Use the point and gradient to find the formual of the perpendicular bisector.

Do the same things for AC

Solve the two equations simultaneoulsy to find the point of intersection

b) Show that this point is at equal distance from all points (A, B and C)

Now use the distance formula 3 times to show this. 

I am sure that you can either do all of it, or do a chunk of it yourself :)

I probably do not have time right now to look at your logic.

My scenario was different from yours because I allowed for the scenario where some cookies were not allocated.

Still I can still use my method 

2 people 10 cookies

10 people 1 divider because all the cookies must be distributed = 11C1 = 11   (We both agree)

3 people 10 cookies

10 cookies 2 dividers = 12C2 = 66  ways

30 cookies 5 people no left over cookies = 34C4 = 46376 ways

Did you look at that stars and bars video.  I am almost certain that your problem can be done the same way.

You can use your formula and see if you get the same results.

If you respond you had better send me a private message, with the address, because otherwise I am not likely to see it.

Thanks :)

Hi Quazars,

I'm still stuck at your box .

If there are only 40 cookies and 5 people how can this happen

They get 40, 1,1,8,and 1 respectively.

I obviously do not understands what you mean. :///

maybe you can do this with the stars and bars method?

Say you have 6 cookies and 3 people (just to make it simpler)

1st               2nd        3rd          leftovers        

c                |c c       | c c           |c  

so that is 9C3 ways  =   84 ways to allocate the cookies

so is there were 40 cookies  and  5 people then maybe there is

45C5 ways = 1,221,754 ways to divide up the cookies  

(there can be some leftover cookies and/or some people who do not get a cookie)

I am not certain that you can do it this way but it seems logical to me....

It is a different scenario to this video clip but.... maybe...

here they are, you can choose the correct one.

I drew a right angled triangle to help me answer this question.  sides 1,x and sqrt(1+x^2)

If

\(if\;\;x\ge0\;\;and\;\;\theta=tan^{-1}x\\then\\sin\theta=\ \frac{x}{\sqrt{1+x^2}}\\ cos\theta = \frac{1}{\sqrt{1+x^2}}\\ cos(tan^{-1}x)\\=cos(2\theta)\\=cos^2\theta-sin^2\theta\\ =\frac{1}{1+x^2}-\frac{x^2}{1+x^2}\\ =\frac{1-x^2}{1+x^2}\)

Hi Kulki   

It is nice to meet you :)

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

Mmm  it is good that you use lots of brackets but maybe if some of them were square brackets it might be easier to make sense of.   Not to worry, lots of backets is good.

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\)

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{ln\left[ \frac{x+2}{x-2}       \right]^{x+1}}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{(x+1)ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(1+\frac{1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(\frac{x+1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]*(\frac{x+1}{x}) }\\ \text{The limit as x tends to infinity of (x+1)/x = 1 so}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]}\\ =e^\left[{\displaystyle\lim_{x\rightarrow\infty}\;{xln\left[ \frac{x+2}{x-2}       \right]}}\right]\\ \)

Now  I will look at the nw limit

\(\displaystyle\lim_{x\rightarrow\infty} \frac{ln\left[ \frac{x+2}{x-2} \right]}{\frac{1}{x}}\\ \qquad \text{The numerator and the denominator both tend to 0 so I can use L'Hopital's rule}\\ \qquad \frac{d}{dx}\;ln\left[ \frac{x+2}{x-2}\right]=\frac{(x-2)-(x+2)}{(x-2)^2}=\frac{-4}{(x-2)^2}\\ \qquad \frac{d}{dx}\;x^{-1}=-x^{-2}=\frac{-1}{x^2}\\ \qquad =\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{-4}{(x-2)^2}\div \frac{-1}{x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{4x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{x^2-4x+4} \;\;\;\frac{\div x^2}{\div x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{1}{1-\frac{4}{x}+\frac{4}{x^2}} \right]\\ \qquad=4\left(\frac{1}{1-0+0}\right)\\ \qquad=4\\ \text{so the original limit}=e^4 \)

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}=e^4\)

Hi Tony :)

7 (sqrt (10))  (sqrt (14) ) ) / (6 (sqrt (35) ) )

\(\frac{7 \sqrt {10} \sqrt {14}}{6 \sqrt {35} }\\ =\frac{7\sqrt{2} \sqrt {5} \sqrt {2}\sqrt{7}}{6 \sqrt {5}\sqrt{7} }\\ =\frac{7\sqrt{2} \sqrt {2}}{6 }\\ =\frac{7*2}{6 }\\ =\frac{7}{3 }\\ =2\frac{1}{3}\)

10C4nCr (10,4) = 210