Mathematics marathon

[Rinfiyks]

Here's my initial stab at it.

Spoiler

Suppose the radii are \( r \). Then, connecting the spheres' centers, we have a square with every side length equal to \( 2r \) (because each sphere is tangent to one another, two centers of two such spheres will be no further than \( 2r \)). Since all faces (except the base) are equilateral triangles, this is called the first Johnson solid (\( J_1 \)). Not that that is relevant, but neato trivia.

Anyway to find the height, consider the distance to the center of the view given in the problem. The distance from the center of the image and any one of the four circle's centers is \( 2r/\sqrt{2} = r\sqrt{2} \) (why? construct a right triangle). Now we have the distance from the center of the pyramid's base, to one of its base vertexes. Now we have a right triangle if we travel along the edge of the pyramid to the apex. This is a right triangle, with base \( r\sqrt{2} \) and hypotenuse \( 2r \). Since \( \mathrm{base}^2+\mathrm{height}^2=\mathrm{hypotenuse}^2 \), we have

\( (r\sqrt{2})^2 + h^2 = (2r)^2 \)
\( 2r^2 + h^2 = 4r^2 \)
\( h^2 = 2r^2 \)
\( h = r\sqrt{2} \)

This is only the height from the plane which the centers of the four base spheres form, which is itself at a height of \( r \). So the center of the top sphere is at \( (1+\sqrt{2})r \). But if we want to know the distance the bottom of the sphere is from the ground, we just take off \( r \) from this, giving \( r\sqrt{2} \).

Not sure if this is correct, but it's my initial intuition.

You are correct
You might want to try it for 3 spheres on the base instead of 4.
Answer (I think):

Spoiler

\(
\frac{2\sqrt{6}}{3}r
\)

For something less proof-centered than quad's last problem:

For what values of M, N can an MxN checkerboard be tiled with dominoes? Explain.

Spoiler

My guess:
M and N must both be greater than 1, and at least one must be greater than 2.
Each domino will occupy 1 light square and 1 dark square. So both M and N cannot be odd.

Another problem:

Spoiler

You may have heard the riddle about the man who walks 1 mile south, 1 mile east, 1 mile north, and ends back where he started. The question is, where did he start? The celebrated answer is the North Pole.
However, there is another answer in which he starts slightly north of the South Pole. How far north does he start? And how does it work?

 

[Kynit]

Another problem:

Spoiler

You may have heard the riddle about the man who walks 1 mile south, 1 mile east, 1 mile north, and ends back where he started. The question is, where did he start? The celebrated answer is the North Pole.
However, there is another answer in which he starts slightly north of the South Pole. How far north does he start? And how does it work?

Spoiler

He is one mile north of the point where the distance around the earth is 1 mile. I'm not sure how to figure this out exactly, though

 

[keemy]

lewis:

Spoiler

wlog when M is even. with M even you can just place dominoes in a MX1 row N times. if both are odd then there are an odd # of tiles and since placing dominoes can only cover an even # of tiles wont work

quad (incomplete)

Spoiler

let \( f(x)= \frac {P(x)}{Q(x)} \) were P and Q are polynomials an Q not 0.

assume for sake of contradiction
\( \Delta f(x) = \frac{P(x+1)Q(x) - P(x)Q(x+1)}{Q(x+1)Q(x)} = \frac{1}{x} \)

so either 1.\( Q(x)=xR(x) \) or 2.\( Q(x+1)= xR(x) \) R being another non 0 polynomial

1. \( \Delta f(x) = \frac{xP(x+1)R(x) - (x+1)P(x)R(x+1)}{(x(x+1)R(x+1)R(x)} = \frac{1}{x} \)
so you can see that \( P(x+1)R(x) \) must have a factor of \( x+1 \) basically you can keep going down this path showing stuff needs more factors over and over so the polynomials can't be finite at least

2. same as above basically.

 

[Christopher Mowla]

\( \lim_{x \to 0} \frac{ \ln (1-x) -\sin {x} } {1 - \cos^2{x}} \) oMg

Answer

Spoiler

\( \underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\ln \left( 1-x \right)-\sin \left( x \right)}{1-\cos ^{2}\left( x \right)} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\ln \left( 1-x \right)-\sin \left( x \right)}{\sin ^{2}\left( x \right)} \right] \)

\( \text{This has the form }\left( \frac{0}{0} \right),\text{ so we can use L }\!\!'\!\!\text{ hospital }\!\!'\!\!\text{ s rule:} \)

\( \overset{L}{\mathop{=}}\,\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{d}{dx}\left[ \ln \left( 1-x \right)-\sin \left( x \right) \right]}{\frac{d}{dx}\left[ \sin ^{2}\left( x \right) \right]} \right] \)

\( =\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{1}{x-1}-\cos \left( x \right)}{2\sin \left( x \right)\cos \left( x \right)} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{1}{x-1}-\cos \left( x \right)}{\sin \left( 2x \right)} \right] \)


Clearly this limit does not exist due to the denominator.

 

[LewisJ]

You guys got that, now for the interesting version with the restriction which I forgot previously:
For what values of M, N can you cover an MxN checkerboard with an equal number of horizontal and vertical dominoes?

 

[keemy]

cmowla is the new state champion!!!!

lewis

Spoiler

when N and M r both even whoops

 

[einstein00]

i'm just wondering... how old are you guys? I think i need to step up my math a bit...

 

[Rinfiyks]

You guys got that, now for the interesting version with the restriction which I forgot previously:
For what values of M, N can you cover an MxN checkerboard with an equal number of horizontal and vertical dominoes?

Spoiler

Both M and N must be even, and at least one must be a multiple of 4?

Edit:

i'm just wondering... how old are you guys? I think i need to step up my math a bit...

I'm 18, so I have little experience with high level maths involving calculus and limits. That's why I'm posting the style of puzzles that only requires a small understanding of maths.

 

[maggot]

maggot: your problem is wrong y=x isn't a solution (unless you mean 2xy' for second term) in that case the solution is really nasty because it ends up being something like

Spoiler

\( x \int x^{-2} e^{x^2} dx \)

so pretty lame (someone can try for themselves it's pretty easy redux of order then solve the first order DE with an integrating factor or as a separable equation but then integral of that is as stated above).

and a problem (not that hard)

Spoiler

let \( a_0=c \) and \( a_1=k \) solve the recurrence \( a_{n+2} = \frac {-a_n}{2(n+2)} \) (if you want you can give separate solutions for even and odd)

thx for pointing out the mistake in the problem! it was late when i wrote the post lol. . .

i broke it down for the approx for fun. .
\( y(x) = -1 + x^2 + \frac {x^4}{6} + . . . + \frac {x^{2n}}{n!(2n-1)} + . . . \)

and for the above limit, i got it wrong lmao! good problem! i had to actually think about differentiating the trig functions lmao.. . you guys are making me think so hard! this thread is a good test of my memory!

 

[LewisJ]

keemy, rinfiyks:

Spoiler

They don't have to both be even; 3x8 is coverable.

einstein:

Spoiler

I'm 16, took multivar calc last semester, ODE next semester

 

[cyoubx]

i'm just wondering... how old are you guys? I think i need to step up my math a bit...

I finished linear algebra and differential equations freshman year, and multivariable calc as a sophomore in High School.

 

[einstein00]

:O i'm taking single variable calc this year (sophomore) and i guess i'm behind lolz

i got an 8 on the AIME last year though

 

[dbax0999]

A spherical oil tank with a radius of 10 feet is half full of oil that weighs 60 pounds per cubic foot. Its top is 6 feet under ground. How much work is needed to pump the oil out of the tank through a hole in its top?

 

[Keroma12]

i'm just wondering... how old are you guys? I think i need to step up my math a bit...

17. All I've been taught from school is up to the first half of high school calculus, derivatives. But I've already done some stuff on my own, not much though. I'm reading On Numbers and Games by John Conway right now. Anybody else here read it? What do you think?

 

[LewisJ]

A spherical oil tank with a radius of 10 feet is half full of oil that weighs 60 pounds per cubic foot. Its top is 6 feet under ground. How much work is needed to pump the oil out of the tank through a hole in its top?

Woo, I haven't done a problem like this in nearly a year. Let's see how well my memory holds.

I will assume the oil is to be pumped to ground level, so a constant 6 will be added to pumping distance.

I will let the bottom of the tank be y=0 and integrate wrt y
Our limits of integration are thus 0 and 10 since the tank is half full
We must integrate the work required to pump out a "slice" of the tank over this region

Work = Force * displacement
The displacement at height y is 20+6-y feet
The force is 60 * volume of "slice"
Volume is dy * pi * r^2 (r being slice radius)
Radius is found with a right triangle with hypotenuse being sphere radius, one leg along y-axis, other leg being slice radius, and we get
slice radius = sqrt(100 - y^2)

Plug everything back in and we see what we must integrate:
60pi(26-y)(100-y^2)dy
from 0 to 10.

To hell with integration by hand.

And we get 890000pi foot-pounds of work required.

Edit: I totally just did someone's math homework. Meh.

 

[cmhardw]

I always liked this problem.

There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?

 

[LewisJ]

I always liked this problem.

There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?

Spoiler

1/3; there are three cases with red on top; each side of the all-red card, plus the one side of the red/black card. Only one has black on the other side.

 

[Rinfiyks]

Spoiler

1/3; there are three cases with red on top; each side of the all-red card, plus the one side of the red/black card. Only one has black on the other side.

Spoiler

Really? I thought it would be 1/2. There are 2 cards with red on them. You've either got one or the other.

 

[cmhardw]

Spoiler

Really? I thought it would be 1/2. There are 2 cards with red on them. You've either got one or the other.

Spoiler

1/3 actually is the correct answer tu LewisJ). To convince you that this is true, it's very similar to the Monty Hall problem, another very interesting probability question.

 

[Rinfiyks]

Spoiler

1/3 actually is the correct answer tu LewisJ). To convince you that this is true, it's very similar to the Monty Hall problem, another very interesting probability question.

Spoiler

Ah yeah, you're given a random face, not a random card! I should've thought "what's the probability that the card has the same colour on both sides", which is effectively what it's asking! Awesome problem more!