[jerethdagryphon]

 alternate route

[Platinumflux]

Had to admit, I panicked when I saw the answer would involve trig!
But drew the solution on Coreldraw and got the answer in 5 minutes!
My first answer was

Spoiler (Rollover to View):

1.135

But they wanted two decimal places so rounded up to

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1.14

I put the inaccuracy down to me!

[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).

[rosemary]

for anyone who is still struggling, I don't know if this will help but the key to the drawings is

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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

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angles in a straight line must add to 180 degrees

Also

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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.

[crovax1234]

Noone ever said it was a pretty equation. Also, please spoiler tag that equation.

[almagest]

No, I don't think this is a nice little problem at all! The exact solution is:

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(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

[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

[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.

[crovax1234]

Devjoe - your diagram was quite useful to me, though

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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.

[fretty]

That's good to know.

[uptheblades]

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fretty, I can confirm your maths are correct

[fretty]

I've done all of the calculations and get:

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1.14"

as my diameter.
Can anyone confirm this?

[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.

[anansi]

It's not a very nice calculation though!! Will have a go at solving it now...

[doublecross]

Thanks Generica - in fact my calculations above were correct!