Round 260

Date Released: May 1st, 2008.
Date Rewarded: May 8th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm³, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s², what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.

Prize


Steel Negg

Click to show/hide solution

Using the stress equation [1]:

rho = F/A

where rho = 400 MPa = 400,000,000 kg/m·s² is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.

The crossed area is a regular pentagon can be calculated by the following equation [2]:

A = 5/4 t² tan(54°) = 0.000688190960236 m²

where t = 2 cm = 0.02 m is the side of the pentagon.

Now, we can calculate the force that the steel bar can withstand:

F = rho * A = rho * 5/4 t² tan(54°)

Using Newton's gravity law [3], we have:

F = m * g

Therefore, the mass that the steel bar can widthstand is:

m = F /g = rho * 5/4 t² tan(54°) / g = 28060.793485650833 kg

where g = 9.81 m/s² is the gravitational acceleration.

Don't round anything yet. The mass is contributed by 3 components: the big sphere m₁, small spheres m₂, and the steel bar m₃. Since all of them are steel, they have the same density. We have:

m₁ + int(h/3) * m₂ + m₃ = m
<=>[V₁ + int(h/3) * V₂ + A * h] * d = m

where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm³ = 7,800 kg/m³ is the density of steel, V₁, V₂ are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.

We can calculate V₁, V₂ by the sphere volume formula [4]:

V₁ = 4/3 * pi * (r₁)³ = 0.065449846949787 m³
V₂ = 4/3 * pi * (r₂)³ = 0.014137166941154 m³

where the radius r₁ = d₁/2 = 25 cm = 0.25 m, r₂ = d₂/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:

V₁ + int(h/3) * V₂ + A * h = m/d
<=>int(h/3) * V₂ + A * h = m/d - V₁

We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:

h/3 * V₂ + A * h = m/d - V₁

<=>(V₂/3 + A) * h = m/d - V₁
<=>h = (m/d - V₁)/(V₂/3 + A) = 654.0200901109209 m

Now round h down to an integer. We then have h = 654 m. Next, test h into the equation int(h/3)* V₂ + A * h = m/d - V₁. If the left hand side is smaller than the right hand side, the steel bar can withstand the gravitational forces. The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer.

Using h = 654 m, the left hand side: int(h/3)* V₂ + A * h = 3.5320. The right hand side: m/d - V₁ = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V₂ + A * h = 3.5327. The right hand side: m/d - V₁ = 3.5321. (not good). Therefore, the answer is h = 654 m.

Results: 986 people guessed the correct answer earning themselves 2029 NP each.

References:
[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume

Round 258

Date Released: April 17th, 2008.
Date Rewarded: April 24th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.

Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s². Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)

Prize


Underwater Tour

Click to show/hide solution

First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:

P = rho * g * h <=> h = P / (rho * g) = 5.096839959225281 m

where P = 50k Pa = 50,000 kg/m·s² is the maximum hydrostatic pressure, rho = 1000 kg/m³ is density of water, g = 9.81 m/s² is the gravitational acceleration. Remember that do not round h yet.

Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):

V = 1000 * pi * h² * (r - h/3) = 212276.11

where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.

We know that: 1 m³ = 1000 dm³ = 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and we get V = 212,280 liters.

Results: 462 people guessed the correct answer earning themselves 4330 NP each.

If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.


We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:
Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r

References:
[1] Hydrostatic pressure
[2] Pythagorean theorem

Round 23

Date released: August 29th, 2001.
Date rewarded: September 12th, 2001.

The Fontaine sisters are seamstresses by trade. They have been commissioned to create as many 2 by 3 feet rectangular Jolly Roger flags as they can for the pirates in their village, so Loretta Fontaine has gone out to buy cloth to embroider.

Dark cloth costs 150 Neopoints a foot and comes in a roll five feet wide. It can be bought in any length (even down to the inch). She has a total of 1,260 Neopoints to spend on the cloth.

How many WHOLE Jolly Roger flags will they be able to make???

Prize
Unknown

Click to show/hide solution

Number of feet of the Dark cloth that the Fontaine sisters can buy:

1260 / 150 = 8.4 ft

The area of the bought cloth:

8.4 * 5 = 42 ft²

Number of possible whole flags with that area of the cloth:

42 / (2 * 3) = 7

However, if we check the below figure, the is no way to make 7 flags from the cloth (each flag must have 2 ft x 3 ft dimension). The sisters can make maximum 6 flags from the cloth.


Therefore, the Fontaine sisters can make 6 whole Jolly Roger flags.

Round 21

Date released: August 10th, 2001.
Date rewarded: August 14th, 2001.

'Excuse me, Slave', said Princess Vyssa, 'I have a job for you'. 'What is it?', Burlak groaned??? Princess Vyssa handed the sculptor a big ball of clay. 'I would like some decorations for my new palace. I want you to build me lots of little pyramids out of this clay. The base has to measure 8 centimeters by 8 centimeters, and they have to be 6 centimeters high..

Burlak looked at the ball of clay, and measured it. It's diameter was 20 centimeters, and it seemed to be a perfect sphere. He wondered how many pyramids he could make... do you know?

Prize
Unknown

Click to show/hide solution

Volume of a pyramid [1]:

V₁ = B * h /3 = 8² * 6 / 3 = 128 cm³

where B is the area of the base of the pyramid, h is the height of the pyramid.

Volume of the sphere [2]:

V₂ = 4/3 * pi * r³ = 4/3 * pi * 10³ = 4000/3 * pi cm³

where r is the radius of the sphere.

Total number of pyramids that Burlak can make:

N = V₂ / V₁ ~ 32.72

Therefore, he can make 32 pyramids from the clay sphere.

References:
[1] Pyramid volume
[2] Sphere volume