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jerethdagryphon
alternate route

i got the answer correctly by using a cad program c4d to be presise
theres was no need for trig or anything just basic maths

anyways nice card

Posted: Fri Feb 02, 2007 6:14 pm
Platinumflux
But drew the solution on Coreldraw and got the answer in 5 minutes!

 Spoiler (Rollover to View): 1.135

But they wanted two decimal places so rounded up to
 Spoiler (Rollover to View): 1.14

I put the inaccuracy down to me!

Posted: Sat Aug 26, 2006 10:26 am
fretty
That's a good point, rosemary, because I use these facts often, I take them as obvoius. That's why there are lots of people stuck with this.

There is one thing though, how come you couldn't solve the end equations, I only ended up with simple trig equations and one quadratic at the end (which was easily solved using the formula).

Posted: Sat Aug 26, 2006 3:18 am
rosemary
for anyone who is still struggling, I don't know if this will help but the key to the drawings is

 Spoiler (Rollover to View): a line that meets a curve (circle) at only one point is a tangent to that curve. ie it meets it at 90 degrees.

And

 Spoiler (Rollover to View): angles in a straight line must add to 180 degrees

Also

 Spoiler (Rollover to View): for two triangles, if they have the same base point, and the lines for the two sides from the base extend outward to form the second triangle (ie the angle at the base is the same), the ratio of the sides opposite the base equals the ratio of the equivalent sides of the original triangle compared to the extended triangle.

having said that, after I'd formed the equations, I don't think I would have solved them without mathcad. This was harder than I expected for a black, compared to some of the others or indeed some of the silvers
My apologies if anyone thinks I'm trying to teach them to suck eggs, but this was intended for those who didn't understand the maths behind the drawings so beautifully illustrated previously.

Just read the post below and realised I made this far harder than I needed to. I ended up with weird equations in terms of inverse sine and cosine functions.

Posted: Fri Aug 25, 2006 6:53 pm
crovax1234
Noone ever said it was a pretty equation. Also, please spoiler tag that equation.

Posted: Fri Aug 04, 2006 3:37 pm
almagest
No, I don't think this is a nice little problem at all! The exact solution is:

 Spoiler (Rollover to View): (194/121) - 52t/121 - (116 Sqrt(3+t))/121 + (76/121) Sqrt(9+3t), where t = Sqrt 3.

You call that nice!

Edit - Added spoiler tags. -Cass

Posted: Fri Aug 04, 2006 12:35 pm
Dragonscales
Bloody hell... Whipsmart don't care about their customers, do they? Making us figure out the exact measurements of ice-cream. I wouldn't know - I would have eaten it before the question started.

Ben

Posted: Fri Aug 04, 2006 6:48 am
Curlytek
This nice little problem can be done using two scale diagrams, one of the side on view similar to devjoes, and one of the top down view showing the max size of four circles in a bigger circle....and then you only need to do two simple pieces of maths: 1) work out the dimensions of the cone (or, the triangle in the 2D diagram) with a bit of pythag, and 2) work out how far from the centre line the chocolate scoop needs to be using a simple ratio from the top down view.

Be precise using a nice drawing package that lets you define the dimensions of the objects nicely, line up the edges well and presto, you can take your answer straight from the picture.

Posted: Fri Aug 04, 2006 6:09 am
crovax1234
Devjoe - your diagram was quite useful to me, though

 Spoiler (Rollover to View): I had to enlarge the upper spheres to help me visualize it better. I realize that it'd be a good deal further up the cone from the 2-inch ball, and worked with the idea in mind that the ball had to be greater than an inch wide. Then I did some somewhat wonky square math, taking the radius of the cone at certain heights and dividing it by root 2 to get the diameter of the balls, and just did a lot of plugging in at 0.01 inch increments. Sloppy, but got the job done.

Posted: Fri Aug 04, 2006 3:39 am
fretty
That's good to know.

Posted: Fri Aug 04, 2006 3:34 am
 Spoiler (Rollover to View): fretty, I can confirm your maths are correct

Posted: Fri Aug 04, 2006 3:18 am
fretty
I've done all of the calculations and get:
 Spoiler (Rollover to View): 1.14"

as my diameter.
Can anyone confirm this?

Posted: Fri Aug 04, 2006 3:01 am
devjoe
When I solved this, I drew lines as shown in the figure attached. There are three distinct angles on each side of the center line at C. One of them is 75 degrees, one of them is calculated from the known triangle CFG in terms of x, and the third one I solved for by extending CF until it meets the cone at H and working with the two similar right triangles HEF and HBC to find the length of the hypotenuse. The last two must add to 105 degrees, and now you need to solve numerically for x to make this true. I don't have the card to check my answer, but the approximation to two decimal places is quite close to the exact number.
 ppc205.gif Description Filesize 1.44KB Viewed 1360 Time(s)

Posted: Thu Aug 03, 2006 6:35 pm
anansi
It's not a very nice calculation though!! Will have a go at solving it now...

Posted: Thu Aug 03, 2006 2:52 pm
doublecross
Thanks Generica - in fact my calculations above were correct!

Posted: Thu Aug 03, 2006 12:16 pm
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