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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Black Puzzle Cards
#205 Whipsmart Hot Fudge
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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anansi
Boot


Joined: 11 Mar 2006
Posts: 51
Location: Leeds, UK

 		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.


PostPosted: Thu Aug 03, 2006 10:22 am
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Muffin
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Joined: 30 Oct 2003
Posts: 306
Location: UK, Leicester
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 Smile 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
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PostPosted: Thu Aug 03, 2006 12:17 pm
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fretty
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Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England
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


PostPosted: Thu Aug 03, 2006 12:22 pm
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doublecross
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Joined: 25 Jul 2005
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Location: London, UK

 		Sorry, it isn't.

Incidentally, I don't see why
Spoiler (Rollover to View):
the angle should be 7.5

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xx

PostPosted: Thu Aug 03, 2006 1:03 pm
Last edited by doublecross on Thu Aug 03, 2006 1:11 pm; edited 1 time in total
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fretty
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 		I wish I did know the answer!!

PostPosted: Thu Aug 03, 2006 1:07 pm
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doublecross
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Location: London, UK

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

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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').

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PostPosted: Thu Aug 03, 2006 1:40 pm
Last edited by doublecross on Thu Aug 03, 2006 2:16 pm; edited 1 time in total
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fretty
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Joined: 19 Nov 2004
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 		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?

PostPosted: Thu Aug 03, 2006 1:48 pm
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Generica
Greenhorn

Joined: 01 Aug 2006
Posts: 5
Location: Leeds, UK

 		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. Embarassed

PostPosted: Thu Aug 03, 2006 2:01 pm
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doublecross
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Joined: 25 Jul 2005
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Location: London, UK

 		Thanks Generica - in fact my calculations above were correct!
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PostPosted: Thu Aug 03, 2006 2:16 pm
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anansi
Boot


Joined: 11 Mar 2006
Posts: 51
Location: Leeds, UK

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

PostPosted: Thu Aug 03, 2006 4:52 pm
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devjoe
Boot

Joined: 24 Jul 2004
Posts: 21

 		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
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ppc205.gif

PostPosted: Thu Aug 03, 2006 8:35 pm
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fretty
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Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England

 		I've done all of the calculations and get:
Spoiler (Rollover to View):
1.14"

as my diameter.
Can anyone confirm this?

PostPosted: Fri Aug 04, 2006 5:01 am
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uptheblades
Boot

Joined: 13 Dec 2005
Posts: 19
Spoiler (Rollover to View):
fretty, I can confirm your maths are correct


PostPosted: Fri Aug 04, 2006 5:18 am
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fretty
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Joined: 19 Nov 2004
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Location: South Yorkshire, England

 		That's good to know.

PostPosted: Fri Aug 04, 2006 5:34 am
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crovax1234
Boot

Joined: 20 Jul 2006
Posts: 46

 		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.


PostPosted: Fri Aug 04, 2006 5:39 am
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