[Generica]

Don't forget the answer they're looking for is the diameter, not the radius, which is probably what you'll be solving for. I wasted two solve attempts by typing in the radius.

[fretty]

Another point to bring up is whether you are allowed to round up. Obviously you can't have a diameter that is bigger than the maximum but on the solve page it says "to the nearest inch". Does this mean you can round up?

[doublecross]

But I think that you can work out the angle as follows:

Spoiler (Rollover to View):

At your point 3 you know that you are dividing the 15 degrees in two. On one side you have the angle you want, say theta, and on the other side you have another angle, say phi. You know that theta plus phi is 15. Also, sin(phi)=x/(the line you have just drawn) and sin(theta)=x.sqrt(2)/(the line you have just drawn). I calculate theta to be 8.79395 and phi to be 6.20605 (approx!). Plugging this back into your a+b=c, which gives a horrible x.sqrt(2)/tan(theta)-sqrt((1+x)^2-2x^2)=1/sin(15) (I think), but this gives x as *removed* which is right! (Using Excel 'goal seek').

[fretty]

I wish I did know the answer!!

[doublecross]

Sorry, it isn't.

Incidentally, I don't see why

Spoiler (Rollover to View):

the angle should be 7.5

[fretty]

Spoiler (Rollover to View):

All of the work with the top I think is complete. Now for the sides: I've managed to get a little further: 1. using the vanilla sphere draw a triangle with the 1 inch being perpendicular to the cone. Use trig along with the 15 degree angle to find the hypotenuse. Call it a 2. draw the right angled triangle as described before, with x root 2 and x+1. Use pythagorus to find the other side in terms of x. Call is b 3. Using the same x root 2 line in number 2, draw lines from each end to meet with the bottom point of the cone. You should end up with an angle of 7.5 at the bottom. Use trig to find the adjacent side in terms of x. Call this c. 4. Notice now how a+b=c. This forms an equation that can be solved to find x. Solving this equation gives the maximum diameter to be 0.87" Please say that this is right

[Muffin]

Spoiler (Rollover to View):

Okay, if we assume the angle is 15°, which I'm not convinced of but can't disprove or prove as yet then I agree with the calculation. But remember this works out the radius, and that the cards asks for a diameter! Giving the answer as 0.45 inches

This is a very good card!

M

[anansi]

fretty,

Spoiler (Rollover to View):

I don't agree that the angle is 15 degrees. I'm using the same triangles to calculate it but if the bottom angle were 15 deg, it would mean that the hypotenuse (1+x) and the side of the cone are parallel and I don't think we can assume that.

[fretty]

Sorry, yeah that's what I get now, bloody google.

[Sophiecat]

I got the same equation, but when I solve it I get

Spoiler (Rollover to View):

0.22" to 2d.p.

Unfortunately I'm locked out until 10pm (not helped by trying my first two solves in cm!)so can't check either.

[fretty]

Can someone check if this is correct:

Spoiler (Rollover to View):

Calling x the radius Making a triangle from above in a horizontal plane with hypotenuse 2x (from two adjoining chocolate spheres), the distance between the centre of the cone and the centre of one of the chocolate spheres can be worked out by using pythagoras to be x root 2 as almagest said. Then by making a triangle with angle of 15 degrees, an opposite side of length x root 2, and a hypotenuse of length x+1 (as the two radii combine), you can say that: sin 15 = (x root 2)/(x+1) Solving gives x=0.224... or 0.22 (to 2 d.p.)

EDIT: changed the end value

[almagest]

Well, I said it was messy! I used more than one triangle.

[anansi]

almagest, are you meaning

Spoiler (Rollover to View):

that you work out x from a triangle with two (opposite) chocolate centres and the vanilla centre as its vertices, or a triangle with two (opposite) chocolate centres and the point of the cone as its vertices (so that the centre of the vanilla scoop lies on the plane formed by the triange? Either way, I'm struggling to find the angles in the riangle. Am I missing some really simple way to calculate it? I should really get this, my highest mark at uni was in Geometry!!

Jack

[almagest]

Unless I am missing some elegant solution, this is a messy geometry problem. Note first the slight ambiguity in the meaning of apex angle. It means that the angle between the axis of symmetry and a line in the surface through the apex is 15 deg.
If you look at the four choc spheres (each radius x), then the distance of their centres from the axis of symmetry is x root 2. (Just take a section through all their centres). Now take a section through the vanilla centre, the apex and two of the choc centres. You know the angles and can easily get an equation for x.

[Sammykat]

D: Dammit.

I'm gonna have to hope someone can tell me the answer then. I really am that bad at maths. I dont even know how to do trig, at all. I can't even remember studying it.