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Heroes Community > Tavern of the Rising Sun > Thread: Let's talk about Maths!!!
Thread: Let's talk about Maths!!! This thread is 55 pages long: 1 10 20 30 40 ... 46 47 48 49 50 ... 55 · «PREV / NEXT»
ihor
ihor


Supreme Hero
Accidental Hero
posted November 06, 2010 05:39 PM

Aha I think I got it, those expressions confused me. Thats a subtraction in demand formula, right? I thought that dash means some correspondence.
Will try to solve when have time.

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 06, 2010 05:58 PM

You are right. It is subtraction. Just like in the demand at the entrance.
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Ecoris
Ecoris


Promising
Supreme Hero
posted November 06, 2010 06:09 PM
Edited by Ecoris at 18:09, 06 Nov 2010.

Alcibiades has already solved the first part of Corribus' problem. (One does not need to look at an infinite series to see this).

The answer to the other question, "Is there a value of "X" for which your chances of hitting a red area are maximized?", is yes. However, there is no angle that maximizes the yellow area (if we assume that the figure isn't defined for angles of 0 or 90 degrees).
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ihor
ihor


Supreme Hero
Accidental Hero
posted November 07, 2010 09:27 AM

Technically as slope approaches to 0 degrees (or 90 degrees), the chances to hit red go to 1/2.

1 - k + k*k - k*k*k + ... = 1/(1+k)

k is the ratio of two closest squares areas. k tends to 1 but less than 1 if slope approaches 0 deg.
1/(1+k) -> 1/2.

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted November 08, 2010 05:30 PM

Is it bad form to post a solution to one's own problem?

I suppose not, if it's a problem you made up.

Anyway, I got the same answer as provided above for part A.  The formula to calculate the percent area that is red that I came up with is:

Area = SUM[1/(2^(2n-1))] where the summation is from n = 1 to n = infinity.

If you evaluate this solution for the first few values of n, it is easily seen that the Area approaches 0.66 (or 66%).

B was a little more difficult, but I found that

Area = SUM[(Q^2-1)/(Q^(4n-2))], where the summation is over the same limits as above and Q = sin(theta) + cos(theta), theta being the angle of rotation.

I did not come up with a general analytical formula to express Area as a function of theta, but it is easily seen by evaluating the above expression over lots of values of theta that the case of theta = 45 degrees (shown in A) has the highest percentage of red area.  For theta values of <1 degree and >1 degree, the amount of area of red drops off significantly below 50% and approaches zero for the case of 0 and 90 degrees (where the figure is sort of undefined).  That might just mean my equation isn't valid for such small angles and Ihor's solution of approaching 50% at the limits may be more appropriate.

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ihor
ihor


Supreme Hero
Accidental Hero
posted November 08, 2010 08:33 PM

Quote:
At a zoo people have to enter through the west gate and exit through the east gate. John and Lillian sell balloons outside the zoo near the gates; John near the west gate and Lillian near the east gate. Every day exactly 500 kids visit the zoo with their parents. The parents are willing to buy at most one balloon per kid. The demand at the west gate is

500 - (price in cents of each balloon at the west gate).

The kids who did not buy a balloon at the entrance, as expected, clamor and cry while they are at the zoo since they do not have a balloon. Hence the parents tell their children that they will buy them a balloon at the exit, provided that the price is "fair". The demand at the east gate (exit) is

(The number of kids without a balloon) - (price in cents of each balloon at the east gate).

(a) If x and y are the prices at the two gates, what will be the gross income to John and what to Lillian ?

(b) Find the values for x = 200 and y = 150.

(c) If Lillian finds out that a balloon is priced 200 cents at the entrance, how should Lillian price the balloons at the exit gate ?

(d) If John knows that Lillian gets this information and acts accordingly, and if Lillian knows that John knows that Lillian acts this way, then what could Lillian expect from John ?

(If necessary, assume that both John and Lillian act selfishly; i.e. there is no co-operation between the two.)


Nobody have posted on Balloons problem, so I give my ideas. First of all let's assume the prices 0<=y<=x<=500, otherwise the demand functions will give us negative numbers, which not passes verification on common sense.

a) The price at west gates is x, and the demand is equal to 500 - x. John's gross income will be J(x,y) = x*(500-x).
500-x balloons have been bought at west gates, it means 500-(500-x)=x is the number of kids without a balloon inside. Lillian's price is y and demand consequently will be x-y. So Lillian's gross income is L(x,y) = y*(x-y)

b) J(200,150)=200*(500-200)=60000 cents = 600$
  L(200,150)=150*(200-150)=7500 cents = 75$

c) L(200,y)=y*(200-y) - thats the function of Lillian's gross income. She wants to maximize it, so taking a derivative and comparing to zero we see the maximum will be at point y=100. She should set price equal to 100.
L'y(x,y)=x-2y - diff. by y. See Lillian will maximize her income, when set the price, halved of John's price.

d) Nasty thing here. John could act in different ways. This question is much closer to Nash's Games theory.
1) John wants to maximize his own income.
As we can see here, John's gross income function is not dependent on Lillian's price ( J(x,y)=x*(500-x) ), so John will set price equal to 250. Similarly to c) Lillian will set price 125.
2) John wants to minimize Lillian's income
x - John's price, as I said Lillian's price should be always x/2 to maximize her income. So according to x her income will be
x/2*(x-x/2)=x^2/4. It's minimal when x=0 and equal to 0. However in this case John's income will also be 0 - all balloons will be free

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 09, 2010 04:04 AM

nice
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ihor
ihor


Supreme Hero
Accidental Hero
posted November 13, 2010 10:18 AM

Today I decided to recover my old materials from school Math olympiads and share some problems with you. I can't say all those
problems are very interesting but some of them may be really difficult. Personally I love the process of solving hard tasks. What about you ?
I think I'll post 5 problems now. Probably it will be a good idea not to post your answers for 3 days, so everyone who is interested may investigate carefully the problem(problems) he/she likes. Afterwards you may post your solution or maybe good ideas you found, but not more than 1 problem in one post, so after it we all could discuss or even improve ideas. Okay, lets move to problems. BTW those are intended for school graduates.

1. Peter has several sticks. Some of them are 5cm long, other - 6cm long. Total length of all sticks is 6 meters. Prove that it is possible to form a regular decagon from all his sticks.
Note 1meter = 100cm

2. Given function f(x) = lg[x] + lg{x}. For particular real number a, f(a)=2. Prove that f(a^2)>4.
Note lg(x) - is common logarithm (logarithm to base 10). [x] - is floor function of the number, {x} - is a fractional part of the number {x} = x - [x].

3. There is a table 11x11 with positive numbers in cells. It is known that product of all numbers in every row is equal to 1, the product of all numbers in every column is also equal to 1 and the product of all numbers in every 3x3 square is equal to 2. Find the product of first and second cells in third row.

4. Given a triangle ABC and points C1 and B1 selected on sides AB and AC correspondingly. BB1 and CC1 intersect at O. Point D is selected, so AB1DC1 is parallelogram. Prove that if D is inside the triangle ABC, then area of quadrilateral AB1OC1 is less than area of triangle BOC.


5. Sequence (x_n), n=0..infinity satisfies following conditions:
x_0 = 1
x_1 = 6
x_n+1 = x_n + sin(x_n), if x_n > x_n-1
x_n+1 = x_n + cos(x_n), if x_n <= x_n-1
Prove that x_n < 100 for all nutural numbers n.

Good luck.

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted December 03, 2010 04:42 AM
Edited by dimis at 06:30, 03 Dec 2010.

Medical researcher discovers integration, gets 75 citations

Hey ihor,

I didn't have time to even read what you posted yet, but I will do so soon. Meanwhile I received the titled email today and I simply have to share.

trialaree, trialaro

Apparently, it was written in 2007. As of today, the paper has received at least 137 citations.
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JoonasTo
JoonasTo


Responsible
Undefeatable Hero
What if Elvin was female?
posted December 03, 2010 06:51 AM
Edited by JoonasTo at 06:52, 03 Dec 2010.

Way to go doctors!



Conserning the darts question

I put together a formula for it. Might want to check if it's completely wrong. It's morning. I don't think well in the morning.

F=(A^2)*SUM(((-(sinb+cosb)^-2))^n),

where;
F=red area
A=wall of the square
b=angle, 0<b<Pi/2

When dF/db=0, b=Pi/4. Simply take lim b closes 0,Pi/2 and F(Pi/4) and you see F ranges from 1/2 when b is nearly zero to 2/3 when b=Pi/4 and back to 1/2 when b reaches Pi/2(as is symmetrically logical).

Base equation seems same as Corribus' but is easier(shorter) to comprehend(if you write it on paper, at least to me) than his.

And you're right, it isn't defined at b=0 and b=Pi/2. Think logically, if each one replaced the old area completely it can't be defined at infinity. It would depend on if infinity is even number or not.
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friendofgunnar
friendofgunnar


Honorable
Legendary Hero
able to speed up time
posted January 11, 2011 12:02 PM

A fun quickie:

A) There's a pair of rings drawn on a wall that divide it into three sections.  A center "bullseye" worth nine points, an outer ring worth 4 points, and a vast "out of bounds" area worth zero points.  You have an infinite number of darts.  What's the highest score that is impossible to get?

B) prove your answer.

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted January 12, 2011 04:31 PM

A possible score (y) is defined as one that satisfies

y = 4*n + 9*m

where n and m, the numbers of times that 4-point and 9-point areas are hit, respectively, both belong to the set of whole numbers (0, 1, 2, ... infinity).

Let x be a possible score and x-1 be an impossible score.

If x+1, x+2 and x+3 are also possible scores, then x-1 is the highest impossible score.  This is fairly easy to see, because if x is a possible score, then x+4 (or any multiple of four) is also a possible score).  Ditto for x+1 and x+5, x+2 and x+6, etc.

That is, if you have a chain of four consecutive possible scores, than every other number above those is also a possible score.  So all you have to do is find the lowest chain of four consecutive possible scores.

This is not so hard with a bit of trial and error.

As it turns out, the first set that I found is 24, 25, 26 and 27.  All of these are possible scores (24 = 6*4, 25 = 9*1 + 4*4, 26 = 9*2 + 4*2 and 27 = 9*3).  Note that 28, 29, 30 and 31 are all possible by simply hitting the 4-point target one more time over the previous four scores, and 32, 33, 34 and 35 are all posssible by simply hitting the 4-point target two more times, and etc.  So every value able 24 is possible just by additional multiples of four. 23 is an impossible score to get with multiples of 9 and 4, and because every score above 23 is possible, 23 becomes the highest impossible score.

By the way, you'll notice a pattern there with the way to get scores of 24, 25, 26 and 27.  Using that pattern, you should be able to see an analytical solution to any variation of the values for each region of the bullseye.

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted January 13, 2011 05:07 PM

Hey, I forgot how much I enjoyed this thread.  Let's do another.

New Puzzle: In two decks of cards, what is the least amount of cards you must take to be guaranteed at least one four-of-a-kind?
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg

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friendofgunnar
friendofgunnar


Honorable
Legendary Hero
able to speed up time
posted January 14, 2011 12:26 PM
Edited by friendofgunnar at 12:27, 14 Jan 2011.

For one deck it's 40 cards.  Wouldn't it be the same for 2?

btw the proof is correct although the trial and error bit is a bit unglamorous

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted January 14, 2011 03:38 PM

Quote:
For one deck it's 40 cards.  Wouldn't it be the same for 2?

Indeed it would be the same no matter how many decks of cards there are.

Quote:
btw the proof is correct although the trial and error bit is a bit unglamorous

I know. I've always taken the brute force approach to problem-solving.

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friendofgunnar
friendofgunnar


Honorable
Legendary Hero
able to speed up time
posted January 15, 2011 01:23 PM

I realized the answer would be different if you were also using the jokers.

For one deck it would be 42 cards.  For two decks and above it would be 43 cards.

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Jabanoss
Jabanoss


Promising
Legendary Hero
Property of Nightterror™
posted January 15, 2011 01:29 PM

If demanding a four of a kind with members from all the colors, you need to draw 79 cards for 2 deck.(ignoring jokers)

However I'm wonding, the next time someone posts a pussle or problem here will it be a requirement to post a solution together with a potential right answer?
(I assume not, unless stated otherwise)
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- Meroe

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Gnoll_Mage
Gnoll_Mage


Responsible
Supreme Hero
posted January 15, 2011 01:55 PM
Edited by Gnoll_Mage at 13:56, 15 Jan 2011.

Can I post something (logic problem)?
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Ecoris
Ecoris


Promising
Supreme Hero
posted January 17, 2011 11:19 AM

Yes, certainly. That should not fall outside the scope of this thread.

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Jabanoss
Jabanoss


Promising
Legendary Hero
Property of Nightterror™
posted January 21, 2011 06:39 PM
Edited by Jabanoss at 18:40, 21 Jan 2011.

Okay I have a serious problem...
I was just posting something in the Altar where I needed to calculate a not-so-hard discrete mathematical problem.

Let say we want to have 7 buildings, all named 1, 2, 3... and so on.
Now we want to create combinations with these buildings.
First we want to combine the buildings so that all the buildings are in combination with all the other buildings. For example combination of buildings involving building 1 is 1:2, 1:3, 1:4, 1:5, 1:6, 1:7.
For building 2 it is 2:3, 2:4, 2:5, 2:6, 2:7
and so on, Note that 1:2 == 2:1.
This should give us a total of 20 combination where 2 buildings are involved. (if my mental arithmetic is correct)

Now what I'm wondering is, How many combination will there be if we use a combination of 3 buildings in the same manner as we did will 2 buildings?
I know there is this fine formula for it, but I can't seem to remember it and my mental arithmetic just won't do here...
This should be easy, but like I said I just don't remember
(please tell me if anything is unclear, this is my first time posting problems here)
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