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For Ptichka
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Kasparov Goes Undefeated!
Twentyfour players competed in a recent chess tournament held in Moscow. The players were divided into two sections. In each section each competitor played one game against every other player. There were sixtynine more games played in Section B than in Section A. Kasparov, unbeaten in Section A, scored five and a half points, where a win counts as 1 point, a draw is half a point, and a loss is zero points. How many of Kasparov's games ended in a draw?

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Re: For Ptichka
I believe, MM, that <strong>5 of Kasparov's games ended in a draw</strong>.
First, we need to find out how many games Kasparov has played. Let's try to figure out how many players are in group A.
a is the number of players in group A
b is the number of players in group B
a+b=24, so b=24a
Since everybody has to play with everybody in his group, the number of games played in group A is:
(a1) + (a2) + ... + 1
(Same logic for group B);
To make writing it out it easier, let's designate this function
f(a1)
(BTW, f(x) is calculated as (x+1)/2 * x)
We also know that (# of games in group A) + 69 = (# of games in group B).
So, we get an equation:
f(a1) + 69 = f(b1)
f(a1) + 69 = f[(24a)  1]
We try plugging in the different numbers into the equation, and the only one that makes it work out is 9. So Kasparov has played 8 games.
Now it's easy.
d is # of games ended in draw
w is # of games K. won
He played 8 games total, so
d+w = 8, so w=8d
Draws are 0.5, wins are 1
d*0.5+w*1 = 5.5
0.5d + (8d) = 5.5
0.5d  d =5.58
0.5d = 2.5
d = 2.5 / 0.5 = 5
Edited to add: MM, I know you can solve the equation that I solved by "plugging in numbers" algebraically; I was just too lazy to do it

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Re: For Ptichka
Aaarrrrgggghh!!!
BTW, the equation (24a)(23a)/2  (a)(a1)/2 = 69 appears to be quadratic, but the a^2 terms cancel, so it is really only linear.
OK. Now the gloves come off.
Farmer Xeke lives on the perfectly spherical planet Xeton. He owns a 40 acre farm shaped like a triangle. One day he decided to measure the three angles of the boundaries of his farm. He found them to be:
Angle A = 30.00000102 degrees
Angle B = 60.00000115 degrees
Angle C = 90.00000102 degrees
What is the diameter of Xeton?
Mathman

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Re: For Ptichka
Hints:
1. Recall the GaussBonnet Theorem: On a curved surface the sum of the three angles of a triangle (measured in <strong>radians</strong>) satisfies the formula
A + B + C = pi + the integral over the triangle of K dS,
where K is the Gaussian curvature and where dS is the element of surface area. For a sphere of radius R the curvature is 1/R^2 (a constant).
2. The Eltamina Principle: If you get the units right, everything else is bound to be right, too. Radians are unit free. K carries the units of "inverse area"  "per square mile" would be appropriate for this problem. There are 640 acres in a square mile.
Your pal, Mathman

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Re: For Ptichka
Well Ptichka, I hope you don't mind that I'm answering this question addressed to you. Mathman gave me permission.
So here is my solution: Diameter = 2119 miles
First to convert degrees to radians you mulitiply by pi and divide by 180 degrees.
Using Mathman's formula and plugging in the values we know, we get
pi*[(30.000 001 02+60.000 001 15+90.000 001 02)/180]=pi+integral(R^(2)dS)
Since R^(2) is constant, the integral of R^(2)dS = S/R^2, where S is the area in square miles.
So after moving the pi's to the left hand side we have,
pi*[((30.000 001 02+60.000 001 15+90.000 001 02)/180)1]=S/(R^2)
The area of the 40 acre farm in square miles = 40acres*1mile^2/640 = 0.625 square miles
Now rearranging and simplifying the formula and substituting in the value for the area we get
R^2=0.0625/(pi*(1.77222x10^8)
R=square root(1122567.621)
R=1059.5 miles
Diameter = 2*R = 2119 miles
If anyone finds any mistakes, please correct me.
~Cassie

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For CBorsky, or anyone
:D
And now, if you <em>really</em> want to make Jeff Buttle proud:
Recall that the Quadrivium comprises Arithmetic, Astronomy, Geometry and Music. On these "quadrivia" threads we have had several questions about arithmetic and about music, and now one about geometry.
So here is a question about astronomy.
Recent observations of distant galaxies support the hypothesis that the universe is either flat with a substantial cosmological constant (dark energy), or else is negatively curved. In flat threespace, the formula for the surface area of the sphere of radius r is
A = 4pi*r^2.
In negatively curved threespace the formula is
A = (4pi/k^2)*(sinh(kr))^2,
where k is the curvature and where sinh is the hyperbolic sine function, sinh(x) = (e^xe^(x))/2.
We can estimate the surface areas of large imaginary spheres in the heavens, centered at the Earth, by observing the density of galaxy clusters at distance r, where r can be estimated by red shift.
I did that yesterday with my handy Mersenne telescope and found that at a distance of r = 1 (billion light years), the surface area of the corresponding 2sphere was A = 12.5663800 (billion lightyears)^2. What is the curvature of the Universe?
Extra credit. You can do this without a computer by using the Taylor series for the hyperbolic sine function: sinh(x) = x + (1/6)x^3 + ... The first two terms are an excellent approximation if x is small.
Mathman

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Re: For CBorsky, or anyone
I wish I could see the answer to this one as easily as I did the last, but unfortunately this one is beyond my knowledge today. Hopefully in the future (after a few more math courses at UBC) I'll be able to get it!
Good Challenge Mathman, I hope you put up a detailed solution after a few days if no one else answers it.
Thanks,
Cassie

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Re: For CBorsky, or anyone
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Come on, Cassie, I know you can do it! Here's what you know: first, the formula
A = 4pi/k^2*(sinh(kr))^2.
We know that A = 12.56638 when r = 1.
(BTW, this right away tells us that the universe is negatively curved, because if it were flat then we would get
A = 4pi*r^2 = 12.56637
Thus the surface area of a sphere is <em>bigger</em> than we would expect using Euclidean geometry. In observational terms, this means that as we look further and further out, we see <em>more</em> galaxy clusters than we expected. Einstein thought that we should see <em>fewer</em> than expected, which would be the case if the universe were positively curved, like a threedimensional sphere.)
OK, never mind that. Back to the equation. Since we know A and r, all we have to do is solve the equation for k. You could do this with a computer, or even just by using the "Equation Solver" key on your calculator, but that's cheating. So...
Write sinh(k) = k + (1/6)k^3. The resulting equation is quadratic in k^2.
If you want to learn about this, sign up for a class with Prof. Jim Bryan at UBC. He is not exactly a cosmologist per se, but he is an expert on the geometric underpinnings of string theory  CalabiYau manifolds, and things like that. Cool stuff.
Mathman
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