BR Maths Corner-1

Amber G. · Post by **Amber G.** » 22 Sep 2012 06:23

To add to

Akashdeep, Rijul and Mrudul got Silver
Shubham got HM

Excellent. 2 Golds and 3 Silvers... One of the best result in last 10 years.. congrats.

So India makes it in top 20.. (11th - and IMO excellent result)

The final results were a little surprising..seeing Korea and Thailand and Canada on the top, and
Lim from Singapore to score a perfect score.

1) Republic of Korea
2) People's Republic of China
3) United States of America
4) Russian Federation
5) Canada & Thailand

To keep up with Brf tradition, here is one of the easier (relatively) problem from the competition ..
See if you can do as well as other high-schoolers ..(all in Indian team were able to do this problem)

Problem slightly changed to make it easier to type it here...
..
Let (a,b,c,d,e,f,g,h,i ) are (9) positive real numbers such that (a*b*c*d*e*f*g*h*i = 1)

Prove that (1+a)^2 (1+b)^3 (1+c)^4 (1+d)^5 .... (1+i)^10 > 10^10

svenkat · Post by **svenkat** » 13 Oct 2012 22:59

Calling AmberGji,
Any clues for the mathematics olympiad problem.The dhaaga is very quiet.

Amber G. · Post by **Amber G.** » 14 Oct 2012 01:33

One Hint for the above problem.... can one solve

Given a is positive real number, can one show ... for example ..

(1+a)^5 > 12 a

(If one can show this the same method can be used for the first post..)
(Or can one prove that (1+a)^10 > 25 a ) (see if one can see the pattern.. to see that (1+a)^n > ??(a))
(BTW the Olympiad problem was just a little different, instead of just 10 numbers, one has to prove this for n numbers - Small tidbit all Indian kids got the correct solution for this)

Amber G. · Post by **Amber G.** » 14 Oct 2012 20:33

I will put my simple solution (outline) for the above problem here below. If you don't want to see it please do not read the tiny print..This solution is just a one line or extremely short and requires really middle school math,, (There are many other standard ways to solve the above too, for example, using calculus it is a simple max-min problem)

You use the fact that arithmetic mean is always greater or equal to geometric mean.
Hence:
(a+1/4+1/4+1/4+1/4)/5 >= (a/4^4)^(1/5)
and you have it
(1+a)^5 >= 5^5/4^4

Going back to original problem, RHS is reduced to
(10^10/9^9)(9^9/8^8)..... abcdefghi which is nothing but 10^10

Vayutuvan · Post by **Vayutuvan** » 14 Oct 2012 23:02

AmberG
For n numbers one have to show the product of the increasing powers to be greater than n^n, not 10^n.

Amber G. · Post by **Amber G.** » 15 Oct 2012 05:35

^^^ Yes, and that was the original problem in IMO, I simplified it a little.

Vayutuvan · Post by **Vayutuvan** » 15 Oct 2012 06:55

^ One could get the easier bound
PRODUCT OF THE POWERS > n!

Amber G. · Post by **Amber G.** » 16 Oct 2012 03:17

But n^n is always greater than n! right?
***
Back to the original Q..
One interesting point, if one used the method I used above, one must add one more step before the solution can get full marks...

(Hint: I used ">=" sign in the shorter Q while ">" in the longer Q - so one still has to prove that the product is actually greater than and not equal to)

Murugan · Post by **Murugan** » 18 Oct 2012 20:34

A good small video on Madhava

http://www.youtube.com/watch?v=TiKoB8nl ... re=related

Vayutuvan · Post by **Vayutuvan** » 19 Oct 2012 01:17

Amber G. wrote:But n^n is always greater than n! right?
***

That's why I said it is a easier bound in the sense that a weaker bound that is easy to get, at least for me, in a limited time. Of course, it is not a as good as the greater lower bound of n^n.

Amber G. · Post by **Amber G.** » 20 Oct 2012 21:57

^^^ Thanks, I understood that, was just curious what was your method to get n! bound.

Vayutuvan · Post by **Vayutuvan** » 21 Oct 2012 00:42

Vayutuvan · Post by **Vayutuvan** » 21 Oct 2012 00:42

I haven't put it on a paper. Let me see if my method works - not sure.

Added later: I realized that I am implicitly using calculus and the bound is directly the tighter one. I haven't looked at your hint yet. Will try when get some time.

Amber G. · Post by **Amber G.** » 21 Oct 2012 22:05

Okay, this is a little different type of post. It is to help out a problem posted in GD Thread. Please post your answer and see if we get the same answer from various posters.

Here is the problem:

There is a town called Arundhati, its longitude is 165.88 degrees and latitude is 56.6 degrees (N)
There is a near by town called Vasistha and its Long = 165.70 and lat = 56.4 degrees.

We are also given that magnetic north pole moves around arctic circle making one round in 26000 years in uniform motion. Arctic circle is 23.5 degrees south of north pole. At present (Yr 2012) the longitude of the magnetic north pole is exactly zero (of course the latitude is 66.5 degrees N)

Now all I have a magnetic compass, using this which direction (shown by magnetic compass) should I take if I want to go from Vasistha to Arundhati. Does answer depends on the location of the magnetic pole? Do I have to move slightly eastward or westward? Does that answer depend on the location of the magnetic pole?

I found that at certain time in the past Muni X, going from Vasistha to Arundhati went in the direction shown exactly magnetic north (using a magnetic compass). What era did he live in?

(You may use a graph paper, calculator or other computing device of your choice

)

Vayutuvan · Post by **Vayutuvan** » 22 Oct 2012 07:54

AmberG ji and others interested in the IMO problem:

Here is a rough outline. I used big-oh notation for simplicity but the algebra can be worked out in detail but would not add much to understanding the method.

(added later: the case of all numbers == 1 can be trivially disposed off.)
1. n numbers sorted in non-increasing order and let x1 be the largest and x_n the smallest. We want to show that the given product is (call it P) > (n+1)^(n+1)
2. Given x_1...x_n = 1.0 there is at least one number greater than 1.0 and at least one number less than 1.0, i.e. x1 > 1.0 and x_n < 1.0
3. P >= (1+x1)^2 (1+x_n)^(m-3) where m = (n+1)(n+2)/2 = O((n+1)^2)
4. But we know x_n = 1/(x_1...x_(n-1)) > 1/x1
5. P > (1+x1)^2 (1+(1/x1))^(m-3) > ((x1+1)/x1)^(m-3)
6. x1 > 1.0 implies, for an epsilon > 0.0, P > (1+epsilon)^(m-3) = (1+epsilon)^O((n+1)^2) = ((1+epsilon)^(n+1))^(n+1) > (n+1) ^ (n+1), where there exists a C such that (n+1) > C. To find C, for example, one can expand (1+epsilon)^(n+1) using binomial theorem and drop all terms beyond the third term as it is second degree polynomial in n and would dominate the degree one polynomial beyond some n.

ArmenT · Post by **ArmenT** » 22 Oct 2012 14:56

Amber G. wrote:Okay, this is a little different type of post. It is to help out a problem posted in GD Thread. Please post your answer and see if we get the same answer from various posters.

Here is the problem:

There is a town called Arundhati, its longitude is 165.88 degrees and latitude is 56.6 degrees (N)
There is a near by town called Vasistha and its Long = 165.70 and lat = 56.4 degrees.

We are also given that magnetic north pole moves around arctic circle making one round in 26000 years in uniform motion. Arctic circle is 23.5 degrees south of north pole. At present (Yr 2012) the longitude of the magnetic north pole is exactly zero (of course the latitude is 66.5 degrees N)

Now all I have a magnetic compass, using this which direction (shown by magnetic compass) should I take if I want to go from Vasistha to Arundhati. Does answer depends on the location of the magnetic pole? Do I have to move slightly eastward or westward? Does that answer depend on the location of the magnetic pole?

I'm going to say, what does going from Vasistha to Arundhati have to do with the magnetic pole anyway? The lat/long of the two towns should not change irrespective of where the magnetic pole moves, so I'd say you'd have to move in a northeast course (in a true north sense, assuming you're not using the compass) to get from one town to other. The only thing you should need to watch out for when using the compass to navigate is what is true north vs. magnetic north (which moves), because the compass may indicate you're moving slightly northwest when you're really moving slightly northeast and vice versa, depending on where the magnetic north is at the moment. So you should correct for true north vs. magnetic north in order to get to your destination.

Amber G. wrote:I found that at certain time in the past Muni X, going from Vasistha to Arundhati went in the direction shown exactly magnetic north (using a magnetic compass). What era did he live in?

(You may use a graph paper, calculator or other computing device of your choice )

Does the magnetic north pole rotate clockwise or anticlockwise?

Vayutuvan · Post by **Vayutuvan** » 22 Oct 2012 20:48

Amberg ji

Re. the n! lower bound - I remember now. I was going to sleep, got the bound, fell asleep and promptly forgot. Here it is.

Expand each individual term using binomial theorem. The product of the second terms.i.e those terms with an exponent of 1 for x_i w in each expansion is
n! (x1...x_n-1) = n!.

Amber G. · Post by **Amber G.** » 23 Oct 2012 01:42

matrimc wrote: 2. Given x_1...x_n = 1.0 there is at least one number greater than 1.0 and at least one number less than 1.0, i.e. x1 > 1.0 and x_n < 1.0

IMO graders are ruthless ..

there is a little sloppiness here..
First: Following need not be be true: "

Given x_1...x_n = 1.0 there is at least one number greater than 1.0 and at least one number less than 1.0"

I can take x_1=x_2=...=1, not a single number is greater than 1.

Second: x_1 and X_n do not appear in symmetric in the expression.. there powers are different, so you do not have liberty to choose which is greater than and which is less than 1.

****
If you want to prove n! case, one easy way I can think of...
It is easy to see (1+x)^n > nx (because x is positive and LHS = 1+ nx+ n(n-1)/2*x^2 and other terms..
This case, one line proof becomes .. LHS > 2.3.4...n (product of all x's) = n! qed.

***

Amber G. · Post by **Amber G.** » 23 Oct 2012 01:50

ArmenT wrote:
Now all I have a magnetic compass, using this which direction (shown by magnetic compass) should I take if I want to go from Vasistha to Arundhati. Does answer depends on the location of the magnetic pole? Do I have to move slightly eastward or westward? Does that answer depend on the location of the magnetic pole?

I'm going to say, what does going from Vasistha to Arundhati have to do with the magnetic pole anyway? [/quote]
As said before, one is measuring direction by using magnetic compass only.. so the question here is not true north/south (which as you said does not change) but wrt to magnetic north/south direction....

So you should correct for true north vs. magnetic north in order to get to your destination.

Yes that is what being asked, we need numerical answer .. how many degrees of correction?

***

Your second question (which direction -- clock wise vs anti-clock wise) you can choose any direction you want.. answer does not depend on that.

HTH

P.S the question is based on Nilesh Oak's problem about seeing these two stars at the time of Mahabharat war.. (see the GD thread)

Vayutuvan · Post by **Vayutuvan** » 23 Oct 2012 03:59

Amber G. wrote: I can take x_1=x_2=...=1, not a single number is greater than 1.

Second: x_1 and X_n do not appear in symmetric in the expression.. there powers are different, so you do not have liberty to choose which is greater than and which is less than 1.

I did put all 1s as a spacial case (later as I remembered that case has to be handled).

As for choosing x_1 and x_n as the largest and smallest - yes, we do have the liberty to choose. With a non-increasing order the given product achieves the least value. For the lower bound to work, the least value of the product, i.e. product with this permutation, has to be greater than the asked lower bound. If one doesn't show this then one has not shown the asked lower bound.

I am unable to find a more elementary method yet.

Added later: The method I outlined doesn't depend on the ordering I choose, though it is a convenient way to label the n numbers. Essentially

P > (1+x)^(m-1) > (1+x)^(n^2 )
Case 1: x => 1.0, it is easy to see P >= 2^(n^2) > n^n
Case 2: x < 1.0. Then we have to show sqrt((1+x)^n) > n for all n > 1. Simple binomial theorem application will show this.
QED I haven't checked this. Will try.

SriKumar · Post by **SriKumar** » 23 Oct 2012 09:30

Amber G. wrote:There is a town called Arundhati, its longitude is 165.88 degrees and latitude is 56.6 degrees (N) There is a near by town called Vasistha and its Long = 165.70 and lat = 56.4 degrees.
We are also given that magnetic north pole moves around arctic circle making one round in 26000 years in uniform motion. Arctic circle is 23.5 degrees south of north pole. At present (Yr 2012) the longitude of the magnetic north pole is exactly zero (of course the latitude is 66.5 degrees N)

Now all I have a magnetic compass, using this which direction (shown by magnetic compass) should I take if I want to go from Vasistha to Arundhati. Does answer depends on the location of the magnetic pole? Do I have to move slightly eastward or westward? Does that answer depend on the location of the magnetic pole? [/b]

Spent some time thinking about it but not sure if I understood this correctly. Clarification # 1: It says 'magnetic north pole (MNP) moves *around* the arctic circle. I assume it means: MNP moves ON the arctic circle (and not in a circle that lies outside the arctic circle).

Next thing that is not super-clear: the longitude for A is 165.88 and for V it is 165.7. So, V is already west of A to begin with. Now, the MNP is at 0 longitude, at 66.5 N. This is at the 'far side' of the arctic circle from the towns of A and V, which are close to 180. I would think that the magnetic direction starting from V would take it even further west of A, so one would have to travel more east to correct for the location of Magnetic north (somehow, I dont think this is what you were driving for, but certainly following magnetic north would take one further west than if one followed geographic north. Either that, or I got my diagram wrong. The diagram's basis is explained below).

My approach was to look at the projection of the earth as viewed from, say, 30,000 miles above the geographic north, i.e. along the axis of rotation. Then, all latitudes appear as concentric circles (and longitudes as straight lines). The center of these three circles is at geographic north.The three latitudes of interest are: 66.5 N (magnetic North sits here), latitude 56.6 N ('A' sits here) and 56.4 ('V' sits here).

We know some distances: radius of earth = 6400 km. Radius of each circle of latitude = radius_earth*cosine(latitude). With this approach, radius of arctic circle = 2552 km. Radius of 'Vashishta circle/latitude' = 3542 km, for 'Arundhati circle/latitude = 3523 km. One thing that strikes me is how close V and A are: just 19 km north-south- however, this is a projection and there might be a bit of error associated with it. The longitudes uniquely locate each place- V, A and MNP, on each circle, respectively. We have 3 circles, 3 points and two angles (longitudes of V and A w.r.t. 180 degrees).

Using the above locations, distances and some angles (longitudes) plus some trig., it seems like we could work out the unknown angles and vectors. But before I rush headlong into that, I wanted to see if any of this is in the right direction and if the assumptions are warranted.

Amber G. · Post by **Amber G.** » 23 Oct 2012 19:59

SriKumar - I think (or at least hoped) that the problem was simple and well defined. Basically it asks a problem and wants to find a numerical answer to verify calculation posted in GD thread about when MBH war took place.

The answer to when Arundhati and Vasistha will be exactly "magnetic north" (north according to magnetic compass)
is same as when the stars by the same name in the sky and same long/lat in skyglobe will be "north/south" per the precessing "north pole"

For detail see the thread in GD, and my analysis. Nilesh got the answer by doing lot of hard work and using software which shows star positions in the past, my suggestion was, using a simple model one can get the same answer by simple calculation.

I was hoping the numerical answer posted here matches with what Nilesh had this in his book. (Hint: The calculation is not that hard using spherical geometry..

)... Anyway I will wait for some time to see if any answers are posted here.

Getting back to SriKumar - Yes, A &V are outside (when one looks from north pole) the Arctic circle. What the problem is essentially asking: is if we join A&V and draw a "straight line" (= great circle on the earth) where it will meet the arctic circle. see this picture from Nilesh:
http://forums.bharat-rakshak.com/viewto ... 6#p1354646

Or the post just below that:
http://forums.bharat-rakshak.com/viewto ... 3#p1354653

Dan Mazer · Post by **Dan Mazer** » 24 Oct 2012 05:41

There's probably a simpler way to do it if you know spherical trig. which I don't. This is how I tried to do it:

The longitudes of A and V need to obtained in the globe with the magnetic poles as the new North and South poles.

In a cartesian system with its x-axis along the direction from the earth's centre to the 0 N, 0E point and z-axis along the direction from the earth's centre to the north pole, the coordinates of a location with latitude l and longitude m, are (cos(l)cos(m),cos(l)sin(m),sin(l)). So the longitude is given by arg(x,y).

In the new globe with the magnetic poles as its N and S pole, the 0 N, 0 E shifts to l0,m0. The new unit vectors in terms of the old ones are given by

i' = cos(l0)cos(m0) i + cos(l0)sin(m0) j + sin(l0) k
j' = -sin(m0) i + cos(m0) j
k' = -sin(l0)cos(m0) i - sin(l0)sin(m0) j + cos(l0) k

So the coordinates in the new system,

x' = cos(l)cos(m)cos(l0)cos(m0) + cos(l)sin(m)cos(l0)sin(m0) + sin(l)sin(m0)
y' = -cos(l)cos(m)sin(m0) + cos(l)sin(m)cos(m0)

and the new longitude is given by arg(x',y').

m0=-23.5 here and the latitude l0 that results in the same longitude for A and V needs to be solved for. I got latitudes that would have occurred in 11080 BC and 2180 BC as solutions.

Looking at the Archaeo-astronomy thread, I seem to have made a mistake somewhere since only the 11080 BC answer matches the solution there.

Nilesh Oak · Post by **Nilesh Oak** » 24 Oct 2012 08:22

Dan Mazer wrote:
m0=-23.5 here and the latitude l0 that results in the same longitude for A and V needs to be solved for. I got latitudes that would have occurred in 11080 BC and 2180 BC as solutions.

Looking at the Archaeo-astronomy thread, I seem to have made a mistake somewhere since only the 11080 BC answer matches the solution there.

Dan Mazer ji,

I will leave it to Amber G to coment on your approach and accuracy of calculations. All I can tell you is that I am very impressed by your calculations. Since I took a mathematically less rigorous route of simulations and have generated these over a period of 40,000 years, I wanted to make a comment where one of your solutions (2180

seems to deviate from number (4380 BC or 4508 BC quoted by me).

The region of 4380/4508 BC is very sensitive... i.e. small change (error) can lead to bigger variation in estimated solution, which is not the case on the other side (i.e. 11091 BC ). My point being this could be the reason why your answer is different than intended 4380/4508 BC. And keep in mind, that while my simulation took into account 'Nutation (Zigzag perturubation of the assumed smooth circle of precession), your calculations is not accouting for it. Such small change can make up for the difference. Great job.

I will of course leave final grading to discretion of teacher (i.e. Amber G).

I have enclosed graph of 'Right ascension Delta between AV' ( or delta between longitudes of AV..as formulalated problem ...by Amber G) for a period of over 40,000 years. Here it is..

Dan Mazer · Post by **Dan Mazer** » 24 Oct 2012 09:33

Nileshji,

My plot looks the same as yours, only its left-right inverted because I've taken the opposite direction of rotation for the precession (magnetic pole in this problem). My plot is also shifted a bit because the starting longitude is 0 in this problem. I'd need to know Polaris' longitude in the ecliptic coordinate system to shift my plot by the right amount to match yours. So the 11081 BC figure matching was just a coincidence

, but since the plots match I think my solution isn't wrong.

Dan Mazer · Post by **Dan Mazer** » 24 Oct 2012 10:19

I used the north star longitude on the ecliptic system - 90 E and the plot now matches yours nicely. The times when the A and V longitudes match up are 13400 BC - 4500 BC.

ArmenT · Post by **ArmenT** » 24 Oct 2012 12:11

AmberG, I think I've partially solved some parts. Here's some code I whipped together to do some distance and bearing calculations (I got the forumlae from another website by googling for "haversine distance". I've referenced the website in my comments).

Code: Select all

import math

# Geo calculations written by Armen Tanzarian (c) 2012
# Released into the public domain
# Written for python 3.0. For python 2.xx, remove parens around the print() statements
# Formulae shamelessly filched from http://www.movable-type.co.uk/scripts/latlong.html

def sqr(x):
    return x*x

def haversine_distance(start_coords, end_coords):
    """Calculates great circle distance between two pairs of coordinates"""
    (lat1, lon1) = start_coords
    (lat2, lon2) = end_coords
    earth_radius = 6371 # Radius in km, used by the FAI for aviation purposes.

    # First convert all degrees to radians
    (lat1, lon1, lat2, lon2) = map(math.radians, (lat1, lon1, lat2, lon2))

    delta_lat = (lat2 - lat1)
    delta_lon = (lon2 - lon1)
    
    a = sqr(math.sin(delta_lat/2)) + \
        math.cos(lat1) * math.cos(lat2) \
        * sqr(math.sin(delta_lon/2))
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))
    distance = earth_radius * c

    return distance

def bearing(start_coords, end_coords):
    """Calculates the bearing course between two pairs of coordinates"""
    (lat1, lon1) = start_coords
    (lat2, lon2) = end_coords

    # First convert all degrees to radians
    (lat1, lon1, lat2, lon2) = map(math.radians, (lat1, lon1, lat2, lon2))
    delta_lon = (lon2 - lon1)

    a = math.sin(delta_lon) * math.cos(lat2)
    b = math.cos(lat1) * math.sin(lat2) \
        - math.sin(lat1) * math.cos(lat2) * math.cos(delta_lon)
    bearing_radians = math.atan2(a, b)
    bearing_degrees = math.degrees(bearing_radians)    
    final_bearing = (bearing_degrees + 360) % 360

    return final_bearing        

if __name__ == "__main__":
    Arundhati = (56.60, 165.88)
    Vasistha = (56.40, 165.70)
    print(bearing(Vasistha, Arundhati))

Answer prints as 26.34056057216918, which is actually incorrect because I've made a BIG assumption here: I've assumed that magnetic north = true north in Vasistha. In reality, I'd have to add/subtract the difference between the two to get the actual bearing. Here's where I'm really stumped and want to get some clarification:

Let's say in 2012, magnetic north is pointing to a spot 66.5 N, 0 E. So if I happen to be somewhere in Siberia at say 66.5 N, 90 E, my compass should say that the north pole is due west of me, correct?

If this is so, I'd have to calculate what the magnetic declination would be from Vasistha's lat/long to current lat/long of magnetic north to figure out how much to correct my bearing by. So when the position of the magnetic north keeps changing year by year, I would get a different magnetic declination value. And this is where I'm kinda hitting a wall. How do I calculate that bit, given that I know the magnetic zero's coordinates for any year and also the coordinates for Vasistha? Any handy pointers or clues here to find this formula people?

Dan Mazer · Post by **Dan Mazer** » 26 Oct 2012 02:24

As I'd suspected, found out that it's trivial working in spherical coordinates:

Equation of a great circle is: tan(l)=Asin(m-m0)

Use the latitude, longitude values for A and V and solve to get A=-3.948 and m0=-36.71 deg. This is equation of the great circle connecting A and V.

We need to find the longitudes, m1 and m2 at which this great circle intersects the arctic circle on which the latitude, l is 66.5 deg.

They're 178.92 deg and 287.66 deg. Longitude of the north pole currently is 90 deg E and increased further east as you go back in time. So these two longitudes would have occurred in 4422 BC and 12275 BC.

Nilesh Oak · Post by **Nilesh Oak** » 26 Oct 2012 03:09

Dan Mazer wrote:As I'd suspected, found out that it's trivial working in spherical coordinates:

Equation of a great circle is: tan(l)=Asin(m-m0)

Use the latitude, longitude values for A and V and solve to get A=-3.948 and m0=-36.71 deg. This is equation of the great circle connecting A and V.

We need to find the longitudes, m1 and m2 at which this great circle intersects the arctic circle on which the latitude, l is 66.5 deg.

They're 178.92 deg and 287.66 deg. Longitude of the north pole currently is 90 deg E and increased further east as you go back in time. So these two longitudes would have occurred in 4422 BC and 12275 BC.

Very nice!, Dan Mazer ji.

ArmenT · Post by **ArmenT** » 26 Oct 2012 10:06

Would someone please explain the math above? I don't get the missing steps. For instance:

tan(l)=Asin(m-m0)
What is l, m and m0?

Dan Mazer · Post by **Dan Mazer** » 26 Oct 2012 10:51

ArmenT wrote:Would someone please explain the math above? I don't get the missing steps. For instance:

tan(l)=Asin(m-m0)
What is l, m and m0?

l is the latitude and m is the longitude. Sorry for not making it clear, I was using the notation from my earlier post. A and m0 are just constants of the equation. Solve for them by using the (l,m) values for A and V. Once you do that, use the equation to get the two longitudes for which the latitude is 66.5 deg i.e. m = (m0 + asin(tan(l)/A)), (m0 + pi-asin(tan(l)/A))

Amber G. · Post by **Amber G.** » 05 Nov 2012 22:07

Thanks to all who have provided solutions or commented.

Here is my comment.

First, the problem was, in my opinion, relatively simple and straight forward. For perspective I asked this problem to to my son (a relatively good math student, post-doc in Physics) over phone and got the answer while I was on the phone. I also asked this problem to some bright middle school students (I volunteer to coach math teams) and the students who had no background in spherical trigonometry were able to solve this using ordinary trigonometry and algebra.
(in about 15 minutes using scientific calculators)

Here is one of those methods.

Work in rectangular coordinates..
Assume the radius is 1, we can easily find (x,y,z) coordinates of A (Arunadhati) and V (Vasistha).
Now all you have to do is to find the equation of the plane which passes through center of earth (0,0,0) and the points V & A
Find the line where it intersects the plane of arctic circle (z=sin (66.5)).
You are done!

****
This is even simpler ( really a one-two liner) using vectors... or matrices/determinants..

For those who may like to see here are some details... (Actually doing it takes much less time than writing this post

First: X (any point V or A) = cos (lat) cos (long)
Y = cos (lat) sin (long)
Z= sin (lat)

For arctic circle, we have Z= sin (66.5 deg) ... and if we find the long(let us call it x) we are done..
as all we have to do is to convert it to in years knowing 360 degrees is 26000 years etc)

Equation of the plane going through (0,0,0)(x1,y1,z1)(x2,y2,z2) is
det |x, y, z |
|x1,y1,z1| = 0
|x2,y2,z2|
Where this meets the arctic circle is now trivial, we put x=cos(66.5)cos (x), y=cos(66.5) sin (x) and
z = sin (66.5)
So we know every value in above except "x", solve for it !!!

(Actually it boils down to - simplifying by dividing each row by cos (lat) etc.) we get
det | cos(x) ; sin(x) ; tan(66.5) |
| cos (long-V); sin (long-V); tan (lat-V)| = 0
| cos (long-A); sin (long -A) ; tan (lat -A)|
equate this to zero, and you can get the value of x..

A simple trick, favorite of mine when a trigonometric equation has to be solved, is to assume:
tan(x/2)= t , we can get
cos(x) = (1-t^2)/(1+t^2) and
sin(x) = 2t/(1+t^2)
The above equation is quadratic in t, giving two values of t...

Hope this helps.

Nilesh Oak · Post by **Nilesh Oak** » 05 Nov 2012 22:33

Amber G. wrote:Thanks to all who have provided solutions or commented.

Here is my comment.
--------------------------------

Hope this helps.

This helps a lot. Amber G, you are dearly missed on Archeo-astronomy thread. On the other hand, I won't blame you for staying away from it, considering non-mathematical discussions that occur (and are occuring) on that thread.

Thank you.

Vayutuvan · Post by **Vayutuvan** » 10 Nov 2012 05:30

Quoting my earlier post -

Case 2: x < 1.0. Then we have to show sqrt((1+x)^n) > n for all n > 1. I haven't checked this. Will try.

Since 0.0 < x < 1.0, sqrt(1+x) is (1+e) where 0.0 < e < 1.0 and application of binomial theorem will give us a polynomial in with degree > 1 in n that has all positive coefficients which asymptotically would be greater than n. I am unable to come up with the simpler proof Amber G ji is talking about yet. Obviously those IMO kids are really bright.

Amber G. · Post by **Amber G.** » 10 Nov 2012 22:25

^^^ Sorry if I am repeating something ... but the proof, as said before (and posted)is extremely simple...

(If you want to prove that it is greater than n!, all you have to know that (1+x)^n > 1+nx)

Rest is simple ... as posted above..

If you want to prove n! case, one easy way I can think of...
It is easy to see (1+x)^n > nx (because x is positive and LHS = 1+ nx+ n(n-1)/2*x^2 and other terms..
This case, one line proof becomes .. LHS > 2.3.4...n (product of all x's) = n! qed

For the original question in IMO, you use the relationship (1+x)^n > n^n x /((n-1)^(n-1)))
The above is easy to prove, by AM-GM inequality or with Calculus - which again is simple)
AM-GM was posted before, if you want to use calculus...
Take log of ((1+x)^n /x ) we get, n log(1+x) - log x, , putting d/dx=0 we again get the simple result....

(The difficulty I have with matrimc's work is that it is not clear, at least to me right away, why one can assume x < 1 .. in any case, this is not necessary because (1+x)^n > n^nx/((n-1)^(n-1))>nx... all the time, as long as x>0.)

Vayutuvan · Post by **Vayutuvan** » 11 Nov 2012 04:16

Ambre G ji, I have not looked your proof because I wanted to try when I get time. As I said, greater than n! is not a problem. I will look at your short proof shortly. By the way, I am not using calculus more than the concept of limit.

As for x < 1, look at it this way - there are only two cases. Either all x_i are equal to 1.0 (in this case the bound is obvious) or there must be at least one x that is less than 1.0. What needs to be proved is that the product for any permutation of the x_i's should be greater then n^n. The worst case permutation (i.e. the permutation which gives the least value for the product) is when the x_i are sorted in non-decreasing order. Let x be the least number and has to be less than 1. Now the product as defined in the problem be P.

P > (1+x)^2 (1+x)^3 (1+x)^4 ... (1+x)^(n+1)
P > (1+x)^{n(n+1)/2-1}
For all n > 1, P > (1+x)^{n^2/2} = {sqrt(1+x)^n}^n

To prove P > n^n, we need to prove (the term in bold above) {sqrt(1+x)}^n > n which asymptotically is true (i.e. for all values of n greater than an n_0, which is dependent on x) which in turn proves that P > n^n, for all n > n_0.

Hopefully the above is clearer. I may be missing something obvious which will make some steps redundant.

Added Later:
OK. Now I see why I am not clear. There exists an x <= 1 because of the condition \Pi x_i = 1.

Amber G. · Post by **Amber G.** » 11 Nov 2012 22:31

Nilesh Oak wrote:
Amber G. wrote:Thanks to all who have provided solutions or commented.

Here is my comment.
--------------------------------

Hope this helps.
This helps a lot. Amber G, you are dearly missed on Archeo-astronomy thread. On the other hand, I won't blame you for staying away from it, considering non-mathematical discussions that occur (and are occuring) on that thread.

Thank you.

Thanks, Here is above even in more simpler form..

Cross posted ,

For a layman .. here is a formula which you can use to fine out at any given time how "east" any star is, thus you can tell if a given star is "ahead" of "behind" any other given star... (You can do your own calculation aka arundhati or vashistha..)

The formula, of course, is very simple (probably first thing you learn in astronomy when you want to change coordinates - it is also discussed in math dhaga)

Let a = longitude of a star (For Vasistha = Long = 165.70)
b = Lattitude of a star (For Vasistha = 56.4 degrees --- see math dhaga)

Given, earth is tilted 23.4 degrees, let C = Cos (23.4) ~ 0.92 , and S= sin(23.4) ~ 0.40
Also assume cycle of precession = 26,000 year, and if you want a epoch T years (from now.. use negative sign for earlier events)

Let x = a + 360*(T/26000)

So at any future (or past time) you want to get the RA of that star (to see how east or west it is), .. Let RA of the star be R

Then
tan (R) = C*tan (x) - S tan(b) / cos (x)

(Here, C is about 0.92, S is about 0.40 and x is calculated by adding precession correction to longitude)

***
So basically if you want to see if A is "ahead" of behind of V ... and MORE IMPORTANTLY by HOW much? ..
All you have to do is to find out the value of C*tan (x) - S tan(b) / cos (x) for each star, subtract is and see if it is positive or negative...

Hope this helps.

Amber G. · Post by **Amber G.** » 13 Nov 2012 20:58

Dan Mazerji - Forgot to thank you.. Very nice work.

(Also, your formula tan(i)= A sin(m-m0) is equivalent to tan(l)=A*sin (m)+ B*cos(m) and that way it may be a little simpler to calculate A&B.. but, of course, sin(m-m0) is more standard... This is essentially same as what I posted using one of the student's method).

Will be interesting to add "delta" for error in time .. (uncertainty is about .005 for long, and .05 for lat for both A&V) and resolving power (say 0.01 degrees (also for .1 degree - more reasonable) in your graph.

Regards.

Amber G. · Post by **Amber G.** » 04 Dec 2012 09:22

Continued from: http://forums.bharat-rakshak.com/viewto ... 4#p1370954

Part II of the article above is going to be published in the January 2013 issue. (It will be posted online in a few days)..

BTW the notices (http://www.ams.org/notices) is available to everyone online (without subscription)

I did like The men who knew infinity

Also nice to see (if you have not seen it yet) the 1987 documentary "Letters from an Indian Clerk which can be watched on youtube:

Amber G. · Post by **Amber G.** » 04 Dec 2012 09:43

In honor of Ramanujan, Here is one problem posed by him... A *VERY* simple looking problem, see if we have a solution here...

(1+n!) is a perfect square when n=4 or 5 or 7, can you find another number for which this is true.

(n! means 1x2x3...n, for example 4! = 1x2x3x4 = 24
5! = 1x2x3x4x5 = 120
and 7! = 120x6x7 etc...)

Ok computer gurus .. or math gurus ... find another n...