Maths Problem
This is the beginning of a weird maths puzzle story which I found in a notebook under my bed.
Your beads look like this
David (you know him, don?t you?) also has a bag. He won?t let you touch it. He remembers that he was given them by a mysterious person who he is unable to describe. His label says, "175 excellent round glass beads. Diameter not known. 1cm long holes. Fine quality. Possibly cursed."
His beads look like this
You and David are discussing how to dispose of the beads when you are interrupted by a weird dwarf. He offers to buy the beads, and says he will pay by weight.
Assuming that the label on your bag is correct (you didn't count and measure them, unlike David), who will get the most money?
By the way, you missed the train. And you should answer your phone.
That?s ? funny I wonder how you solved it. The very fact that it?presumably?can be solved might be a good clue to get the right answer quickly in your head bypassing the calculations needed to actually work out each bead?s volume. I only realized this after doing it, though.
It seems like you?re doing justice to your nickname
EDIT ? Well, I?m assuming the density of David?s beads is the same as that of yours. Who knows if the difference between ?fine? and ?excellent? quality, or curses, mean a difference in density!
_________________
The red lake has been forgotten. A dust devil stuns you long enough to shroud forever those last shards of wisdom. The breeze rocking this forlorn wasteland whispers in your ears, “Não resta mais que uma sombra”.
The glass is all the same density. I'm rewriting the story to make things like that clearer, and then I'm going to give it to my friends and watch them struggle.
Some useful formulae (only one is actually needed, but the others are useful anyway)
Volume of a sphere (radius R):
Volume of a cylinder (radius r, height h):
Volume of a spherical cap (height h) in a sphere of radius a:
Edit: Forgot to say what R, r and h mean.
Last edited by MathematicalOwl on 25 Aug 2014, 10:32 am, edited 1 time in total.
How does the possible curse fit into the equation?
I think it?s actually less cumbersome to calculate a bead?s volume as ?
http://pastebin.com/EDKkhZ6H (encrypted so it doesn?t spoil the problem).
That seems a case of ?My normal approach is useless here?.
_________________
The red lake has been forgotten. A dust devil stuns you long enough to shroud forever those last shards of wisdom. The breeze rocking this forlorn wasteland whispers in your ears, “Não resta mais que uma sombra”.
Who ever has the most "tonnage" of beads gets the most money.
The problem is that you have a smaller number of heavier beads, but David has a bigger number of lighter beads.
Glass is glass (we can assume). So the density of the beads can be assumed to be the same. The curse is probably a red herring.
Thats the set up of the problem. Pretty simple.
But doing the calculations to answer it is quite complicated. You need to calculate the volume of the spheres, minus the volumes of the cylinders cut out of them, and the volume of the caps taken off the tops.
Just looking at the diagrams it looks like each of your beads has more than twice the volume of glass that each of David's have. So even though he has twice as many beads you would get the most money. But thats just a guesstimate.
How do you guesstimate that?
Each bead?s shape is entirely determined by the radius of the ball it was carved from and the height of the cylindrical hole. You?re only given the latter, but this, in itself, is the strong clue I mentioned before.
You could make a rough estimate forgetting about the different shapes and considering that a linear dimension of your beads is 3/2 times that of David?s. As an object is scaled up or down, its volume changes proportionally to the cube of the scaling factor, so each of your beads should have roughly 3³/2³, i.e., 27/8 times the volume of one of David?s, which is very comfortably greater than 175/85.
Nevertheless, since MathematicalOwl posted the formulae for the volumes, the problem is very much solved. If you use them, you can get an accurate calculation to improve the rough estimate above. I think the result is funny, and then you can see how the simple hint that the problem can be solved is a big clue.
_________________
The red lake has been forgotten. A dust devil stuns you long enough to shroud forever those last shards of wisdom. The breeze rocking this forlorn wasteland whispers in your ears, “Não resta mais que uma sombra”.
How do you guesstimate that?
Each bead?s shape is entirely determined by the radius of the ball it was carved from and the height of the cylindrical hole. You?re only given the latter, but this, in itself, is the strong clue I mentioned before.
You could make a rough estimate forgetting about the different shapes and considering that a linear dimension of your beads is 3/2 times that of David?s. As an object is scaled up or down, its volume changes proportionally to the cube of the scaling factor, so each of your beads should have roughly 3³/2³, i.e., 27/8 times the volume of one of David?s, which is very comfortably greater than 175/85.
Nevertheless, since MathematicalOwl posted the formulae for the volumes, the problem is very much solved. If you use them, you can get an accurate calculation to improve the rough estimate above. I think the result is funny, and then you can see how the simple hint that the problem can be solved is a big clue.
Hmmm..well..
In cross section David's beads appear as one square unit of air, and two tiny slivers of glass totaling maybe a third of the air. But your's appear as one half by one-and-a-half (ie 3/4 of a square unit) of air, plus about an equal amount of glass. So thats 33 percent of a square unit vs 75 percent of a square unit. Eighty five times 75 percent is about 64, and 175 times 33.3 percent is 58. So even in two dimensions "You" have slightly more. In three dimensions the ratio between the two totals would be squared making it more lopsided in your favor. I would think. thats how I guesstimated it.
But when I get a chance Ill roll up my sleeves and do the equations.
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