1. 0

## For Ptichka

Kasparov Goes Undefeated!

Twenty-four 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 sixty-nine 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?

2. 0

## 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=24-a

Since everybody has to play with everybody in his group, the number of games played in group A is:
(a-1) + (a-2) + ... + 1
(Same logic for group B);
To make writing it out it easier, let's designate this function
f(a-1)
(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(a-1) + 69 = f(b-1)
f(a-1) + 69 = f[(24-a) - 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=8-d

Draws are 0.5, wins are 1
d*0.5+w*1 = 5.5
0.5d + (8-d) = 5.5
0.5d - d =5.5-8
-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

3. 0

## Re: For Ptichka

Aaarrrrgggghh!!!

BTW, the equation (24-a)(23-a)/2 - (a)(a-1)/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

4. 0

## Re: For Ptichka

Hints:

1. Recall the Gauss-Bonnet 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.

5. 0

## 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

6. 0

## For CBorsky, or anyone

:D

And now, if you <em>really</em> want to make Jeff Buttle proud:

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 three-space, the formula for the surface area of the sphere of radius r is

A = 4pi*r^2.

In negatively curved three-space 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^x-e^(-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 2-sphere was A = 12.5663800 (billion light-years)^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

7. 0

## 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

8. 0

## Re: For CBorsky, or anyone

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 three-dimensional 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 -- Calabi-Yau manifolds, and things like that. Cool stuff.

Mathman

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