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Assuming  that   letters can  be  repeated....we  can  choose   any of the 26 letters for the  first  three places  and  any of the remaining 25  for the last one

So

26^3 * 25   =  439400

LOL!!!

Never forget   the  "pie"   !!!!!

Yep that's what I got too

Thank you. CPhill!!!

I got it!!!!!!

I forgot to type in the pi...

Its pi/4!!!!

Oops...I put in the incorrect volume....see my corrected answer  !!!!

Thank you for the answer CPhill, it still seems to be wrong...

I put my work below, to show what I got before

Im still confused on what is wrong ...

It is against Web2.0 policy to answer aops questions.    Sorry,

It is called  'Art of Problems Solving' becasue they want you to learn to problem solve - without outside help.

They provide a notice board with some hints.

They do not want you to get outside help.

It is not  called  'Art of getting people to help on the net'.   

I am not trying to be harsh but AoPS have shown their frustration enough times.

I can most definitely understand their point of view.

Original post: https://web2.0calc.com/questions/consider-rhombus-abcd-with-side-length-6

My work: 

I got that the circumference would be 6, so the area of the base would be 9. Then, I got that the height is 3/2 because 9*3/2 = 27/2. But, when I typed in 1/4 for the answer which is (3/2)/6 it was wrong.

Here's my best attempt

Since A  joins C   (without overlap)   the circumference of the  cylinder  =  6

So   the radius  can  be figured     as   

6  =  2pi * r

r  = 3  / pi

And since we know the volume of the  cylinder  we  can  figure the height as

(27/2) =  pi  (3/ pi)^2  * h

(27/2)  =  9 / pi  * h

h =  ( 27pi)  / (2 * 9)  =    27pi /18  =   3 pi /2

So.....this also   =  the  rhombus height

And the sin BAD  =   rhombus height  / rhombus side   =   (3 pi/2) / 6  =   pi  / 4

Corrected   !!!!

please give us the link to the original post.

It should be answered on the original.

You do not nedd to repeat the question,  Just give the link and ask for more help.

 lol

sure

CPhill --

I got your answer ??? but a different number.

Number Eugene starts with:  E

Number Frank starts with:     F

Eugene gave away 1/3 of his stamps to Frank ...

Now Eugene has:      (2/3)E

and Frank has:          F + (1/3)E

Frank then gave 1/5 of his stamps to Eugene ...

Now Eugene has:      (2/3)E + (1/5)( F + (1/3)E )

and Frank has:           (4/5)( F + (1/3)E )

They now have the same number of stamps:

        (2/3)E + (1/5)( F + (1/3)E )  =  (4/5)( F + (1/3)E )

          (2/3)E + (1/5)F + (1/15)E  =  (4/5)F + (4/15)E

multiply by 15:     10E + 3F + E  =  12F + 4E

                                   11E + 3F  =  12F + 4E

                                             7E  =  9F

Since  E + F  =  772      --->     F  =  772 - E

                                             7E  =  9(772 - E)

                                             7E  =  6948 - 9E

                                            16E  =  6948

                                                E  =  434.25

Final result:  ?????

LOL!!!....not always   !!!!!

If both have the same  number at the end  then  each  has   (772/2)   = 386

Let  E  be the number that Eugene started with and F what Frank started with

And   F  =  772  - E        (1)

Eugene gives away  (1/3) of his

And from Frank he gets   (1/5) ( F  + (1/3) E ]  =   (1/5) ( 772  -E + 1/3)E )     (2)     

So  at the end,   what  Eugene started with  - what he gave away + what he got from Frank  = 336

So

E  - (1/3)E  +  (1/5) (772 - E + (1/3)E  ]  = 336   

(2/3)E +  772/5 + (1/5)(-2/3)E  = 336

(2/3)E - (2/15)E  =   336   - 772/5

[ 8/15] E  =  908 / 5

8E  =  2724

E = 340.5   ?????

Are your numbers correct  ???  (it's also possible that I made an error !!! )

Lets see.

\(\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\ =\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\ =\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\ =\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\ =\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\ =6\frac{7}{9}\)

LaTex:

\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\
=\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\
=\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\
=\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\
=\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\
=6\frac{7}{9}

Would you care to look at that again Geno ?

Because of the   leading x^4 term,  the max  is unbounded    =    inf

Using some Calculus  to find the min

q'(x)  = 4x^3 + 8x - 24x

q'(x)  =  4x^3 - 16x

Set this  =  0

4x^3 -16x   = 0

4x (x^2 - 4)  = 0 

x=  0     or  x   =2

Taking the second derivative

12x - 16

When x = 0     this is negative .....so....not a min here

When x =  2  this is positive, so  x is a min here

Petting 2 back into  the  function and evaluating  gives us    -30   

So

Range  =   [ -30 , inf )

Very nice, BuilderBoi   !!!!