 Posers and Puzzles

1. 31 Jul '08 07:07
At work we have a coffee machine. We use styrofoam cups. I noticed the liquid is delivered in equal volumes, three times. The bottom of the styrofoam cup is 3 cm wide. The first shot made a depth of 1 cm, the second shot made a depth of 0.8 Cm and the third made a depth of 0.64 Cm. What is the angle of the conical shaped cup?
2. 31 Jul '08 21:45
x = distance from actual bottom of the cup to where the bottom would be if the cone were completed
z = angle

The three (equal) volumes are:

{1/3 Pi [(1 + x)tan(z)]^2 (1 + x)} - [1/3 Pi (1.5)^2 (x)]
{1/3 Pi [(1.8 + x)tan(z)]^2 (1.8 + x)} - {1/3 Pi [(1 + x)tan(z)]^2 (1 + x)}
{1/3 Pi [2.44 + x)tan(z)]^2 (2.44 + x)} - {1/3 Pi [(1.8 + x)tan(z)]^2 (1.8 + x)}

Get two equations from that (it would be easier to just solve for x and tan(z), then find z from that).
Anyway, my answers for z are 21.27 or 66.24 degrees. I think I made a Mathematica error; my x's came out negative.
But I'm still pretty sure of the method.
3. 01 Aug '08 06:48
Originally posted by Jirakon
x = distance from actual bottom of the cup to where the bottom would be if the cone were completed
z = angle

The three (equal) volumes are:

{1/3 Pi [(1 + x)tan(z)]^2 (1 + x)} - [1/3 Pi (1.5)^2 (x)]
{1/3 Pi [(1.8 + x)tan(z)]^2 (1.8 + x)} - {1/3 Pi [(1 + x)tan(z)]^2 (1 + x)}
{1/3 Pi [2.44 + x)tan(z)]^2 (2.44 + x)} - {1/3 Pi [(1.8 + x)tan(z)]^2 (1.8 + ...[text shortened]... made a Mathematica error; my x's came out negative.
But I'm still pretty sure of the method.
This seems to be an unneccesary complicated method of solving the problem.
Mathematica may be a good program, but it cannot replace brain cells.
4. 03 Aug '08 06:09
Originally posted by sonhouse
At work we have a coffee machine. We use styrofoam cups. I noticed the liquid is delivered in equal volumes, three times. The bottom of the styrofoam cup is 3 cm wide. The first shot made a depth of 1 cm, the second shot made a depth of 0.8 Cm and the third made a depth of 0.64 Cm. What is the angle of the conical shaped cup?
I want to bump this interesting thread. The problem is clever, it is well defined, and it is realistic. No loopholes, very clean.

I don't solve the problem now, I just give a way of solving it.

We know the area of the inside of a circle. Let's form a function f(r) showing the area as a function to it's diameter.
A cup is a part of a cone where the tip is cut off. The volume of this cup is an integreation of f(x) from point pb centimeters from the cone's tip to point pt centimeters. pb < pt.
Now we know that there is 1 unit of volume if you integrate f(x) from pb to pb+1 cm (I1), 1 unit of volume from pb+1 to pb+1.8 cm (I2), and 1 unit from pb+1.8 to pb+2.44 (=pt) cm (I3)

Now we need to calculate pb by finding out for wich pb I1 = I2 = I3.
When we know pb, then we can construct a triangle to find the angle.

I think we can solve the problem only with two observations, I3 is not neccesary.

This is a way to solution, if I'm right, perhaps I'm not.
5. 03 Aug '08 14:28
Originally posted by FabianFnas
I want to bump this interesting thread. The problem is clever, it is well defined, and it is realistic. No loopholes, very clean.

I don't solve the problem now, I just give a way of solving it.

We know the area of the inside of a circle. Let's form a function f(r) showing the area as a function to it's diameter.
A cup is a part of a cone where the ...[text shortened]... vations, I3 is not neccesary.

This is a way to solution, if I'm right, perhaps I'm not.
yes,... I agree.... , the third observation is not needed. 🙂
6. 03 Aug '08 22:18
I started with the following equality..

(pi / 3) * [ (3 + 1.8k)^3 - 27] = 2 * (pi / 3) * [ (3 + k)^3 - 27]

Working from that, I arrived at a cubic equation with no constant term. Assuming a positive volume of coffe is found in each squirt, I divided both side by k, and found the followinf equality with 2 solutions, one positive, one negative.

479k^2 + 1,395k - 675.

The positive root is irrational, but approximately equal to 0.442560 cm.

The inverse tangent of this is 22.907 degrees, and that is the only angle of the cup that works for the first two parts of the problem as given.

I know my work in this post is very sketchy, but perhaps someone can follow it.