[bonanova]

Four nested cups fit together with no space between.

The walls and bottoms are of equal thickness t.

The internal heights h_i and diameters d_i have the same ratio = h_i/d_i for all i.

The diameter d₄ of the largest cup is 1 foot.

The interior volumes V_i of the 3 smallest cups together equals the internal volume V₄ of the largest cup: V₁ + V₂ + V₃ = V₄

What is t?

[Guest]

it's the 20th letter of the modern English alphabet!

[Izzy]

it's the 20th letter of the modern English alphabet!

...

..

.

<_<

Still working on it.

[Guest]

Ahh..that made me laugh cause it true.

t is the thickness of each cup?

[Izzy]

1 ft = 73 mm

t = 6 mm

6/73 = 0.0822

So t = about 8.22% of 1 foot. Which is 0.9864 inches.

I'm gonna add in the rounding bit, and call it an inch.

Call it a flawed proof, but I bet I'm right.

Unless your image isn't drawn to scale... >_>

[Izzy]

Ahh..that made me laugh cause it true.

t is the thickness of each cup?

... <_<

t is the thickness of the cup, but you're supposed to solve for t.

[Guest]

it's the 20th letter of the modern English alphabet!

Ahh..that made me laugh cause it true.

t is the thickness of each cup?

perhaps BN should have asked us to solve for t instead?

I think what we're asked here is to find a value for t given the relationships of the dimensions.

edit: Izzy beat me to it. <_<

Edited January 13, 2009 by Grayven

[Guest]

1 ft = 73 mm

t = 6 mm

6/73 = 0.0822

So t = about 8.22% of 1 foot. Which is 0.9864 inches.

I'm gonna add in the rounding bit, and call it an inch.

Call it a flawed proof, but I bet I'm right.

Unless your image isn't drawn to scale... >_>

is right. The thickness of each cup is equal to 1 inch.

Spoiler for here comes the proof:

So I started with d₁+2t=d₂, d₂+2t=d₃=d₁+4t, and d₃+2t=d₄=d₁+6t=12 inches. Therefore, d₁=12-6t.

The volume of each cylinder is πd_i²h_i/4. Since the "π/4" term is common in the volumes of all four cylinders, it can be factored out and canceled. So what we have now is V₁+V₂+V₃=V₄, or d₁²h₁+d₂²h₂+d₃²h₃=d₄²h₄.

Now, since h_i/d_i is a constant, and conversely, d_i/h_i is also a constant, we can say

K=d₁/h₁=d₂/h₂=d₃/h₃=d₄/h₄.

Multiplying the last equation by K leaves us with d₁³+d₂³+d₃³=d₄³.

Substituting d₁+2t in for d₂, d₁+4t in for d₃, 12 in for d₄, and then going back in and substituting (12-6t) for all d₁, after expanding and reducing, we end up with

12-18t+7t²-t³=0.

Graphing that function, we can plainly see that it crosses the y axis at t=1 inch.

[Guest]

t = 1 inch

My proof is fairly longwinded, but by h_i/d_i = h₄ (as d₄ = 1) and, from the diagram of how the cups are placed:

d₁ = 1 - 6t

d₂ = 1 - 4t

d₃ = 1 - 2t

Without writing out all the working, from the above we can express h_i in terms of h₄ and t for i = 1 to 3.

We can express the volume of each cup as V_i = pi x d_i² x h_i / 4

Putting it all into V₁ + V₂ + V₃ = V₄ (making all appropriate solutions so we only have t and h₄), we find that h₄ cancels out and (after simplifying) we are left with the cubic equation:

144t³ - 84t² + 18t - 1 = 0

Solving for that only gives one real answer, being t = 1/12, or 1 inch.

Phew! It's been a long time since I've worked through equations like that - good to get the brain some exercise!!

[Guest]

is right. The thickness of each cup is equal to 1 inch.

Spoiler for here comes the proof:

So I started with d₁+2t=d₂, d₂+2t=d₃=d₁+4t, and d₃+2t=d₄=d₁+6t=12 inches. Therefore, d₁=12-6t.

The volume of each cylinder is πd_i²h_i/4. Since the "π/4" term is common in the volumes of all four cylinders, it can be factored out and canceled. So what we have now is V₁+V₂+V₃=V₄, or d₁²h₁+d₂²h₂+d₃²h₃=d₄²h₄.

Now, since h_i/d_i is a constant, and conversely, d_i/h_i is also a constant, we can say

K=d₁/h₁=d₂/h₂=d₃/h₃=d₄/h₄.

Multiplying the last equation by K leaves us with d₁³+d₂³+d₃³=d₄³.

Substituting d₁+2t in for d₂, d₁+4t in for d₃, 12 in for d₄, and then going back in and substituting (12-6t) for all d₁, after expanding and reducing, we end up with

12-18t+7t²-t³=0.

Graphing that function, we can plainly see that it crosses the y axis at t=1 inch.

t = 1 inch

My proof is fairly longwinded, but by h_i/d_i = h₄ (as d₄ = 1) and, from the diagram of how the cups are placed:

d₁ = 1 - 6t

d₂ = 1 - 4t

d₃ = 1 - 2t

Without writing out all the working, from the above we can express h_i in terms of h₄ and t for i = 1 to 3.

We can express the volume of each cup as V_i = pi x d_i² x h_i / 4

Putting it all into V₁ + V₂ + V₃ = V₄ (making all appropriate solutions so we only have t and h₄), we find that h₄ cancels out and (after simplifying) we are left with the cubic equation:

144t³ - 84t² + 18t - 1 = 0

Solving for that only gives one real answer, being t = 1/12, or 1 inch.

Phew! It's been a long time since I've worked through equations like that - good to get the brain some exercise!!

I liked Izzy's better. These suck all the mystique out of it. Interesting puzzle BN, you almost had me trying math again....

[Guest]

perhaps BN should have asked us to solve for t instead?

I think what we're asked here is to find a value for t given the relationships of the dimensions.

edit: Izzy beat me to it. <_<

Probably, but my brain is running on low at the moment anyways..*very tired* so I most likely wouldn't have answered it..not now anyways,

[Izzy]

I liked Izzy's better. These suck all the mystique out of it. Interesting puzzle BN, you almost had me trying math again....

Haha, really? I have a secret fondness for really cool math proofs. The kind that just make you go "Woah. You figured this out how?" As you can see from my ruler use, I'm not very good at coming up with them.

Too bad our exams don't have things drawn to scale. <_<

[Guest]

t = 1 inch

My proof is fairly longwinded, but by h_i/d_i = h₄ (as d₄ = 1) and, from the diagram of how the cups are placed:

d₁ = 1 - 6t

d₂ = 1 - 4t

d₃ = 1 - 2t

Without writing out all the working, from the above we can express h_i in terms of h₄ and t for i = 1 to 3.

We can express the volume of each cup as V_i = pi x d_i² x h_i / 4

Putting it all into V₁ + V₂ + V₃ = V₄ (making all appropriate solutions so we only have t and h₄), we find that h₄ cancels out and (after simplifying) we are left with the cubic equation:

144t³ - 84t² + 18t - 1 = 0

Solving for that only gives one real answer, being t = 1/12, or 1 inch.

Phew! It's been a long time since I've worked through equations like that - good to get the brain some exercise!!

Oooh, I beat you by a whole 6 minutes.

I must say it was much more exhausting typing the proof with all those _{s and ^{s, as I'm sure it was for you too.}}

[Guest]

Oooh, I beat you by a whole 6 minutes.

I must say it was much more exhausting typing the proof with all those _{s and ^{s, as I'm sure it was for you too.}}

^{_{but i thought the formula for the volume of a cylinder was v=(pi)r(squared)h, not v=(pi)d(squared)h... i don't know if it would make a huge difference. (by the way, sorry for my poor formatting; i don't know how to insert a pi symbol or type the exponent.)}}

[Guest]

Oooh, I beat you by a whole 6 minutes.

I must say it was much more exhausting typing the proof with all those sub]s and sup]s, as I'm sure it was for you too.

Absolutely - good practice for that too!

but i thought the formula for the volume of a cylinder was v=(pi)r(squared)h, not v=(pi)d(squared)h... i don't know if it would make a huge difference. (by the way, sorry for my poor formatting; i don't know how to insert a pi symbol or type the exponent.)

You are correct - but both lazboy and my answers also include a /4 in the formula (which is that r=d/2 so r²=d²/4

[bonanova]

Note that since height and diameter are proportional, volume scales as the cube of the diameter.

Let t=1/2 in some units, and ask Mr. Fermat to look away for a moment.

The beautiful little equation 3³ + 4³ + 5³ = 6³ tells us that d₄ = 6 in those same units.

Since d₄=12", we have t=1"