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TitaniumAlloy
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Professional
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posted November 19, 2008 02:04 AM |
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Lord PC is correct.
I didn't see that during the exam. *doh*!
I on instinct tried integration without looking at it geometrically I guess.
I tried finding the inverse function etc. blah blah must have screwed something up I got pi/2 but the worse thing is I wasted so much time on such a short question.
Pretty sure I got the followup questions, though.
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dimis
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posted November 20, 2008 05:39 AM |
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Edited by dimis at 16:03, 22 Nov 2008.
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Don't worry TA; things like that happen to everyone everyday. Problems you solve one day, you might not solve them another day, and so on ...
So, are you going to spend more time in your life on math, or is it all over now?
@Ecoris: I was amazed that you had an even simpler solution than my last one! I loved the problem! 
edit: We have another problem by Ecoris. Prove that the sum or product of R_alg numbers is R_alg. Any ideas? May be a proof for a simple example?
e.g. Let a = square_root(3) and b = third_root(2). Can we prove that a * b is also R_alg ? What about a + b?
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Nikita
Famous Hero
Meepo is underrated
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posted November 23, 2008 04:49 AM |
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what does R_alg stand for?
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dimis
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posted November 23, 2008 09:07 AM |
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R_alg = Real Algebraic Numbers = numbers that are roots of some polynomial with integer coefficients. So, why is square_root(3) a real algebraic number?
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Ecoris
Promising
Supreme Hero
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posted November 23, 2008 01:28 PM |
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Edited by Ecoris at 13:29, 23 Nov 2008.
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Consider the square below. B is the midpoint of the right edge. A is the lower left corner. You may place the point M anywhere on the top edge.
Where should you place M to minimize the total distance from A to B to M? (i.e. the total length of the two lines inside the square)
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TheDeath
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with serious business
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posted November 23, 2008 05:26 PM |
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Edited by TheDeath at 17:28, 23 Nov 2008.
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@Ecoris: So we have coordinates (chosen arbitrary for simplicity):
A(0,0)
B(1,0.5)
M has y=1, but the x varies:
M(x,1)
Then we calculate the distance (just basic vector stuff):
distance(AM) = sqrt((x-0)^2 + (1-0)^2) = sqrt(x^2+1)
distance(BM) = sqrt((x-1)^2 + (1-1/2)^2) = sqrt((x-1)^2+1/4)
So we take these two and sum them:
sqrt(x^2+1) + sqrt((x-1)^2 + 1/4) = dist
Then we have to find the minimum of this. Unfortunately I couldn't use the first derivative because if I equate the first derivative with 0, Maxima (a program) can't solve it for x. Then I tried to plot it in Maxima, and the graph looks like this:
(the minimum is somewhere around 0.6)
So I calculated it manually with Pari and it gave me around 0.666666666666666666666666666..... (infinite number of 6s)
which is:
2/3 (this means, it's to the right of the midpoint, 2/3 from the upper-left and 1/3 from the upper-right corners respectively).
this problem was interesting because I use this kind of vector math often in 3D algorithms
even though I didn't know how to do it 100% algebraically. In fact I'm not sure how to do it since the first derivative doesn't work (or am I missing something)?
(if there are any errors or I am completely off the point please tell me I did it in a hurry  )
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No jokes were harmed during the making of this signature.
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Gnoll_Mage
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posted November 23, 2008 06:14 PM |
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You have a grid of squares, X squares wide and X squares high, where X is a multiple of two. Suppose I point to a square in the grid, and ask you to cover all of the squares in the grid except the one I pointed to with pieces of the following shape:
X
XX
Prove that it is always possible to do this.
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Ecoris
Promising
Supreme Hero
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posted November 23, 2008 06:18 PM |
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Edited by Ecoris at 18:19, 23 Nov 2008.
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2/3 is the right answer. One should should be able to solve it using first derivatives: You will get two fractions with a squareroot in the denominators, if you add the fractions after having given them a common denominator you may discard the denominator since you have to find the x that makes the first derivative = 0.
The numerator is not nice, though:
x*sqrt((1-x)^2+1/4) + (x-1)*sqrt(x^2+1)
But this can be done in Maple or similar programs, which will give you 2/3.
Edit: The problem can be solved without use of differentiation and/or computers, it has a very simple solution.
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TheDeath
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posted November 23, 2008 06:21 PM |
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Well Maxima (a program I use) seems to be stupid to do this automatically and I probably had to do it manually by "splitting" the denominator first, but anyway Pari did the job with brute force solution 
(Pari mostly deals with numeric equations while Maxima with algebraic (that is with variables, not numbers))
I heard about Maple but it costs a lot (Pari & Maxima are free and open-source)
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No jokes were harmed during the making of this signature.
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Ecoris
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posted November 23, 2008 06:21 PM |
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Quote:
You have a grid of squares, X squares wide and X squares high, where X is a multiple of two. Suppose I point to a square in the grid, and ask you to cover all of the squares in the grid except the one I pointed to with pieces of the following shape:
X
XX
Prove that it is always possible to do this.
If you point out a square in a 6x6 board there are 35 squares left. That can't be done.
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Ecoris
Promising
Supreme Hero
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posted November 23, 2008 06:23 PM |
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Edited by Ecoris at 18:25, 23 Nov 2008.
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Quote:
I heard about Maple but it costs a lot (Pari & Maxima are free and open-source)
Ah yes, it probably does. I have a license through the university.
(Edit: Floodprotection? That's not something I experience often)
Edit: One can of course check that 2/3 solves the equation after having found the solution numerically.
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Gnoll_Mage
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posted November 23, 2008 06:31 PM |
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Whoops, sorry, X is a power of two, not a multiple, so two, four, eight etc.. (Observe that one less than a power of four is always divisible by three - you can prove that first if you want).
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Gnoll_Mage
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Supreme Hero
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posted November 23, 2008 06:34 PM |
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To restate:
You have a grid of squares, X squares wide and X squares high, where X is a power of two. Suppose I point to a square in the grid, and ask you to cover all of the squares in the grid except the one I pointed to with pieces of the following shape:
X
XX
Prove that it is always possible to do this.
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Ecoris
Promising
Supreme Hero
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posted November 23, 2008 06:38 PM |
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Edited by Ecoris at 18:38, 23 Nov 2008.
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I can prove that. Lets see what other people come up with.
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dimis
Responsible
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posted November 23, 2008 07:37 PM |
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Edited by dimis at 19:58, 23 Nov 2008.
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I am impressed again by the witty problem posed by Ecoris.
And it's funny what a Physicist would say ...
And we have one more problem by Gnoll_Mage.
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Ecoris
Promising
Supreme Hero
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posted November 24, 2008 11:22 AM |
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In fact, the Gnollmage's statement is true for any value of X as long as X is not 5 or divisible by 3.
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Ecoris
Promising
Supreme Hero
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posted November 27, 2008 04:46 PM |
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Edited by Ecoris at 16:48, 27 Nov 2008.
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As promised, a simple solution involving no differentiation of the minimizing problem.
If we reflect B in the top horizontal edge of the square we get the following picture:
It is now clear that the distance from A to M to B is the same as the distance from A to M to B'. This distance is minimal if M lies on the straight line between A and B'. The triangle with corners A, B' and the lower right corner of the square will then be three times as large as the triangle with corners M, B' and the upper right corner of the sqaure. Therefore the distance from M to the upper right corner of the square is 1/3.
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dimis
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posted November 28, 2008 10:32 PM |
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The pace has dropped probably because the semester is almost over. Let's see ...
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dimis
Responsible
Supreme Hero
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posted February 25, 2009 05:46 AM |
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IMs sent and a problem
Ok, I have to apologize for not doing this earlier.
I just sent IMs to Celfious, Domzilla, JoonasTo, homm3megejas, and Keldorn asking for their birthdays. What I don't know though is if Domzilla is an active member in the forum in general.
I am posting here so that I increase the chances of people logging in and answering to my question and consequently giving an end to this.
Getting along, does anyone have a nice problem or would like to suggest for problems targeting something specific?
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A funny thing happened to me during Christmas time. Somebody posed me the following problem:
Consider a 3x3x3 cube (like Rubik's cube) made of cheese. We have a mouse that starts eating the cheese from one of the corners. Each time the mouse eats a small cube of cheese, keeps on eating in a hortizontal or vertical direction (i.e. never diagonal) a small cube that is next to what has just been eaten.
Question: Is there a "way of eating" (i.e. a path) such that the mouse eats the "inner-most" small cube last ?
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Binabik
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posted February 25, 2009 07:00 AM |
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This thread is pages long: 
