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### Re: BR MathsCorner-1

Posted: 27 Jul 2008 21:48
zulu wrote:There are 9 pearls. 8 weighing equal and 1 pearl weigh slightly more than others.
We have a balance scale but can use it only twice.Find out the odd pearl.
(solvable in a min)

Yes that was easy
1. Divide the nine pearls in three sets of three each A, B and C
2. Balance set A and set B against each other.
3. If set A equals set B in weight do step 4 with set C else do step 4 with heavier set between A and B.
( IF A equals B THEN R = C
ELSE IF A heavier than B THEN R = A
ELSE R = B )
4. Call three pearls of the chosen set R1, R2 and R3. Balance R1 against R2.
5. If pearl R1 equals R2 then R3 is the odd one, else the heavier between R1 and R2 is the odd one.

Kya mein paanchvi pass se tez hoon ?

### Re: BR MathsCorner-1

Posted: 27 Jul 2008 22:05
SK Mody wrote: .. Move to the rotating frame of the earth. We can pretend that earth is an inertial frame by adding centrifugal force - which is radial so it does not affect the direction of downward force. ....

To add to Nandu - to pretend (rotating) earth is an inertial frame, you need to add centrifugal force ( (omega)x(omega)x (r)) as well as another factor (-2 (omega)x(velocity vector) )). The other factor can (most of time) be neglected for small velocities (because omega is small)) .. But .. it does produces trade winds and effects on ballistic missiles (And it also produced lot of messages in nukkad thread about which direction one should shoot a satellite)

(Here omega is angular velocity vector (=(2*pi/(24*60) sec^(-1)) and "x" in normal vector cross multiplication).

### Re: BR MathsCorner-1

Posted: 27 Jul 2008 23:09
derkonig wrote:AoA math mujahids,
mdhoat,
the case of atleast 3 heads out of 5 tosses can easily be solved by using the bernoulli trails method...

Derkonig ji.. It was stated in the math problem itself, that it is supposed to be solved by using Inclusion/Exclusion formula and venn diagrams. That was where I got derailed putting the numbers in venn diagrams and then applying the formula

### Re: BR MathsCorner-1

Posted: 27 Jul 2008 23:16
Nandu wrote:
SK Mody wrote:One can either (a) solve the eqn of motion in radial field or (b) argue as follows: Assume that centre of earth is an inertial frame (Newton's laws hold). Move to the rotating frame of the earth. We can pretend that earth is an inertial frame by adding centrifugal force - which is radial so it does not affect the direction of downward force. Then the initial velocity of the ball is zero in this frame and it can can only move along the lines of force. The result then follows. Thus for example a ball thrown upwards will land exactly at the point from which it was thrown.

No, your first approach was correct. You can't just change to a rotating frame of reference except locally.

In 10 seconds, the angular displacement of the points under consideration will be too small to be significant to the solution of the problem.

IMHO, of course.

Amber G wrote:
To add to Nandu - to pretend (rotating) earth is an inertial frame, you need to add centrifugal force ( (omega)x(omega)x (r)) as well as another factor (-2 (omega)x(velocity vector) )). The other factor can (most of time) be neglected for small velocities (because omega is small)) .. But .. it does produces trade winds and effects on ballistic missiles (And it also produced lot of messages in nukkad thread about which direction one should shoot a satellite)

Yes, I see. My changing the reference frame to the rotating one already assumed that the ball had only radial velocity in the rotating frame. So the argument was circular. Anyway, it prompted me to work out the exact equation of motion in 2-D, which is:

dA/dt = k/(r^2)
d2r/dt2 = k^2/(r^3) - GM/(r^2)

Where A is the angle, r is the radial distance, k is an arbitrary constant, M is mass of earth.
From the first equation we can see that the only solution for which the angular velocity dA/dt is constant is the one for which r is also constant - that is the ball orbits round the earth. All other solutions have non-constant dA/dt. Thus the ball can never land at the original ground point.

### Re: BR MathsCorner-1

Posted: 27 Jul 2008 23:48
Yes, I see. My changing the reference frame to the rotating one already assumed that the ball had only radial velocity in the rotating frame.

Sorry, I have not read/understood the whole post, so you already know that, but, radial velocity will experience the other (Coriolis) force. This will actually make it move "eastward". You normally ignore it in physics classes because it is quite small and can be ignored. (Only if the velocity is in the same direction as a line joining south/north pole - will have no second term or Coriolis force would be zero).

about 24 cms to the east.
The correct formula for eastward shift (t=time of fall, g=gravitational acc., L = latitude (0 at equator), w= angular velocity = 2*pi/(24*3600) ... t is given by 500 meters = (1/2) gt^2) )
x= g (cos L) (2wt - sin (2wt))/(4*w^2)
(More details can be seen in any book of physics or classical mechanics eg Goldstein etc)
(This assume that I have not made any mistake . .. I will double check and correct it, if need be)

### Re: BR MathsCorner-1

Posted: 28 Jul 2008 00:10
mdhoat wrote:
derkonig wrote:AoA math mujahids,
mdhoat,
the case of atleast 3 heads out of 5 tosses can easily be solved by using the bernoulli trails method...

Don't want to repeat my self, but for the case of at least 3 heads out of 5 (or at least 11 heads out of 21 for that matter) the probability is exactly 1/2.. as explained before.

### Re: BR MathsCorner-1

Posted: 28 Jul 2008 11:11
For math geeks here

Worms Do Calculus to Find Food

Sun Jul 27, 10:21 AM ET

Like humans with a nose for the best restaurants, roundworms also use their senses of taste and smell to navigate. And now, researchers may have found how a worm's brain does this: It performs calculus.

Worms calculate how much the strength of different tastes is changing - equivalent to the process of taking a derivative in calculus - to figure out if they are on their way toward food or should change direction and look elsewhere, says University of Oregon biologist Shawn Lockery, who thinks humans and other animals do the same thing.

This research could one day benefit some of the more than 200,000 Americans who detect a foul smell or taste that is actually pleasant or have a weakened or depleted ability to appreciate the scent of a lilac or savor the flavor of a juicy burger.

"The more we know about how taste and smell function - not just at the level of primary sensory neurons, but downstream in the brain - the better prepared we will be to understand when the system is broken," Lockery says.

With the aid of salt and chili peppers, Lockery reached the calculating-worms conclusion by studying two anatomically identical neurons from the worm's brain that collectively regulate behavior. These two neurons function like "on" and "off" gates in a computer in response to changes in salt concentration levels. This dubiously delicious discovery, detailed in the July 3 issue of the journal Nature, hints at the method for smelling and tasting that is thought to be common among a wide variety of species, including humans.

Like human visual systems that respond to the presence and absence of light, Lockery and colleagues found that when the left neuron fires as salt concentrations increase, the roundworm continues crawling in the same direction. The right neuron responds when salt concentrations decrease, and the worm turns in search of a saltier location.

Lockery said this is similar to a game of hot-and-cold with a child. But there is one key difference: the worm doesn't need an observer to say if it's getting closer to or farther from the target - the worm calculates the change by itself.

Observing the worm responding to changes in concentration suggested an experiment to see if the worm's brain computes derivatives. The mathematical concept of a derivative indicates the rate at which something, such as salt concentration, changes at a given point in time and space. So Lockery tried to verify that these neurons recognize changes in salt concentration and then tell the worm where food is and where it is not.

To do so, he artificially activated each neuron with capsaicin, the spicy component in chili peppers, which worms naturally cannot detect. Worms with capsaicin applied to the left neuron crawled forward. When the worm's brain indicated that the current motion leads to increasing salt concentrations, it continues moving in its original direction. But when the worm's right neuron is activated by capsaicin, it is duped into thinking the salt levels are decreasing. So the worm changes direction, hoping to find salt elsewhere.

"We found a new way to do calculus with neurons," Lockery told LiveScience.

Previous studies have identified "on" and "off" cells in the brains of other chemosensory animals such as fruit flies, cockroaches, frogs, lobsters and rats. Given the strong similarities between the olfactory regions of the brains in rats and other mammals, Lockery says that humans should also be included in this list. So his work suggests that this circuit may be a universal derivative for smelling and tasting.

In response to the lingering mystery of why worms go toward salts in search of food, Lockery offers an untested theory that the decaying carcasses of invertebrates, like snails and earthworms, provide a common source of bacteria. Since animals are very salty inside, he thinks there could be a link between salt and bacteria in the wild.

### Re: BR MathsCorner-1

Posted: 28 Jul 2008 12:48
Amber G. wrote:
Yes, I see. My changing the reference frame to the rotating one already assumed that the ball had only radial velocity in the rotating frame.

Sorry, I have not read/understood the whole post, so you already know that, but, radial velocity will experience the other (Coriolis) force. This will actually make it move "eastward". You normally ignore it in physics classes because it is quite small and can be ignored. (Only if the velocity is in the same direction as a line joining south/north pole - will have no second term or Coriolis force would be zero).

about 24 cms to the east.
The correct formula for eastward shift (t=time of fall, g=gravitational acc., L = latitude (0 at equator), w= angular velocity = 2*pi/(24*3600) ... t is given by 500 meters = (1/2) gt^2) )
x= g (cos L) (2wt - sin (2wt))/(4*w^2)
(More details can be seen in any book of physics or classical mechanics eg Goldstein etc)
(This assume that I have not made any mistake . .. I will double check and correct it, if need be)

Thanks. (In your initial question you had said to assume that building was at equator). I had assumed that when you shifted to a coordinate system rotating with the earth, the velocity of the ball would be purely radial (equivalent to assuming that the coriolis term in the rotating frame simply did not exist). Hence by definition, no coriolis force. But the equations show that moving to such a coordinate system does not remove the angular velocity of the ball (unless the ball is in stationary orbit). Anyway no point discussing what was an error.

By the way, the coriolis force solution is also approximate (but probably pretty accurate). You need to solve the differential equation to get the "exact" solution.

The equations I got (in non-rotating frame fixed at earth center) were:

rd2A/d2t = -2(dr/dt)(dA/dt)
d2r/dt2 = r(dA/dt)^2 - GM/(r^2)

which simplify to:

dA/dt = k/(r^2)
d2r/dt2 = k^2/(r^3) - GM/(r^2)

I should have written the equations in a rotating frame to see the coriolis force more clearly, but the end result should be the same.
There are several approximations needed to reduce this to the solution you got.

### Re: BR MathsCorner-1

Posted: 28 Jul 2008 20:18
I had assumed that when you shifted to a coordinate system rotating with the earth, the velocity of the ball would be purely radial (equivalent to assuming that the coriolis term in the rotating frame simply did not exist)

Just to clarify, I did not change/shifted the coordinate system, just gave the solution for general latitude (which BTW, just introduces a factor of (cos L) which is MAX (=1) at the equator, not zero)... that is horizontal displacement would be more at the equator than other places.

To reiterate, Coriolis force is NOT zero for a 'radial' motion. At equator, it is zero, for a north(or south direction) horizontal direction, but for falling body (or body going up) it is not zero. (That was the whole debate in nukkad, why one should launch satellite in east direction etc).
(Direction of the Coriolis force is perpendicular both to velocity vector and axis of earth's rotation)

You are right, of course, one can solve, the equation in non-rotating frame of reference (in this the trajectory of the ball is a narrow ellipse (thrown with a zero radial velocity at the highest point, and with a tangential velocity = '(r+h)w' etc) and, the numbers do come out the same. Beautiful thing is, one can find the exact solution as the diff. equations could be solved analytically.
There are several approximations needed to reduce this to the solution you got.

Not really, the solution is/can_be quite exact (IOW can be solved exactly without introducing numerical way to solve) (as long as you assume point masses etc..) the approx. solution, if you expand it terms of 'wt' which is very small indeed, one gets:
(1/3) w g t^3 (at equator)
(w is angular velocity, t= sqrt(2h/g) etc... )

Cheers.

### Re: BR MathsCorner-1

Posted: 30 Jul 2008 12:20
Amber G maam. No questions about the numerical part of the answer. That of course is correct. But the question is about direction.

It is like this. w acts as a vector along the z axis (earth spinning around north south axis, ), object along the equator is the X axis ,so the cross product 2 rho dot X w should act on the Y axis.. That will translate to right /left of right under the building..

Yeah, ok. since it is on the equator , it will fall on the equator and doesn't move north or south (cos angle between omega and r dot is 90deg), so falls either due east or due west (like Bangalore /Chennai on same latitude..). I guess if not on equator, it will wont translate nicely as 24cms east/west etc.

### Re: BR MathsCorner-1

Posted: 30 Jul 2008 19:09
Vinaji - Its due east. (You/I may have missed a '-' sign ) Using normal convention - Coriolis force is given by
(-) 2 (omega) x (velocity) ) but in any case it is easy to guess which direction it will fall.
(Easiest to look at conservation of angular momentum - use non-rotating frame of reference - you get both direction/velocity/displacement etc.. easily. (and then calculate displacement relative to fixed point on earth)

For normal latitudes (Bangalore /Chennai) the displacement is still virtually in east direction (magnitude is a little less than equator - at Banglore it will be about 90% of that 24 cms) - max deflection about half a degree from pure east-west direction in any case- anywhere on earth)

If curious: the north-south (southward in northern hemisphere) deflection is about (lowest term in wt) = (1/6) g w^2 t^4 (Sin L cos L) (If I am not making a mistake ( which comes out to be a millimeter or two for 500m fall)

### Re: BR MathsCorner-1

Posted: 31 Jul 2008 00:35
Amber G, do you have the questions of the recent IPO ??
thanks.

### Re: BR MathsCorner-1

Posted: 31 Jul 2008 01:39
Do you mean IPhO (Physics), if so the results are at:
http://ipho2008.hnue.edu.vn/Competition/Results/tabid/82/Default.aspx
The problems may not have been posted yet. The problems, this year, from what I heard/seen were too long, and hard (In MO).. American team solved (had time for) about 60% of the problems.

### Re: BR MathsCorner-1

Posted: 31 Jul 2008 01:50
even so, I would be obliged if you could post a couple, especially if it covers mech or EM.

TIA.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 00:11
Rahul - Believe the problems (and official solutions) will be posted on the above site, soon. I'll wait for a little while.
BTW
lakshmic wrote:Amber-G-saar,

approx sqrt(1000/g)*500*2*pi/(60*60*24) meters or approx 0.3 meters away in the east ? or am I missing something

I think, your answer is about 50% more (If I am not mistaken).. needs a factor of (2/3) in the front, which you may have missed.
FWIW: Using conservation of angular momentum gives the answer without much math:
(Assuming w= angular velocity of earth, h= height, r=radius of earth, g=acc due to gravity)
Assume non rotating frame:
tangential velocity = (r+h)w ===> angular momentum = (r+h)^2 w
tangential velocity at the surface = (r+h)^2 w / r = rw + 2hw + ... (higher order terms)
Now assume rotating frame:
tangential (relatvie) velocity = 2hw = wgt^2 (assume h=(1/2) gt^2 does not change much)
Integrating we get the tangential displacement = x = (1/3) w g t^3

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 00:34
Two Not so Hard Problems:
1. (Old problem from Hindu Math sources)
A dairy farm has exactly 75% cows, and the rest 25% are buffaloes. Also exactly 80% of all these animals are producing milk (rest 20% are non-milk producing). There are exactly 76 milk producing cows in the farm. What is the total number of animals in the farm.

2. "Ankan" of a number is defined as the total number of distinct digits one has to use to write that number (in normal decimal notation, without leading zeros). For example, ankan of 245 is 3 , ankan of 7272722 is 2 (Only 2 and 7 are used), ankan of 1001 is 2 and ankan of 32456782 is 7 and ankan of 888888 is of course 1.

Prove that for any number given, you can always find a multiple of that number which has ankan of exactly 2.
For example if you start with 286 (which has a ankan of 3) but its multiple
286*7 = 2002 has an ankan of 2 (only '2' and '0' are used to write this number)

Good luck.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 00:49
Also let me post an easy looking Geometry problem.
ABC is an isosceles triangle with Angle A=20 degrees and angle B= Angle C = 80 degrees.
Let D be a point on AB such as angle BCD = 60 degrees.
Let E be a point on AC such that angle CBE = 50 degrees.
Find the angle AED.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 03:56
Amber G. wrote:Two Not so Hard Problems:
1. (Old problem from Hindu Math sources)
A dairy farm has exactly 75% cows, and the rest 25% are buffaloes. Also exactly 80% of all these animals are producing milk (rest 20% are non-milk producing). There are exactly 76 milk producing cows in the farm. What is the total number of animals in the farm.

2. "Ankan" of a number is defined as the total number of distinct digits one has to use to write that number (in normal decimal notation, without leading zeros). For example, ankan of 245 is 3 , ankan of 7272722 is 2 (Only 2 and 7 are used), ankan of 1001 is 2 and ankan of 32456782 is 7 and ankan of 888888 is of course 1.

Prove that for any number given, you can always find a multiple of that number which has ankan of exactly 2.
For example if you start with 286 (which has a ankan of 3) but its multiple
286*7 = 2002 has an ankan of 2 (only '2' and '0' are used to write this number)

Good luck.

Here is my solution to problem 1.
C&C welcome

Total number of animals is 120. 90 cows & 30 buffaloes.

My solution: (Not the most elegant one!!)

a) 75% cows --> Number of Cows is a multiple of 3 --> Total number of animals is a multiple of 4.
b) 80% of total animals are milch --> Total is a multiple of 5.
from a & b, Total number of animals is a multiple of 5*4=20.

76 cows are milch --> least number of cows is 78(multiple of 3) --> least total of animals is 4/3*78=104 -->
--> nearest multiple of 20 is 120.

Try total number of animals = 120. So 90 cows and 30 buffaloes. 96 animals are milch -> 76 cows & 20 buffaloes

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 05:47
Amber G. wrote:Also let me post an easy looking Geometry problem.
ABC is an isosceles triangle with Angle A=20 degrees and angle B= Angle C = 80 degrees.
Let D be a point on AB such as angle BCD = 60 degrees.
Let E be a point on AC such that angle CBE = 50 degrees.
Find the angle AED.

Never thought I'd run into matrix algebra for this one...but in the end I was able to avoid it with a bit of trial and error.

Essentially, you draw all the lines and do the usual [angle x + angle y + angle z = 180] for triangles and straight lines and end up with 4 simultaneous eqns with 4 variables.

The 2 lines CD and BE intersect at a point, call it O. Angle BOC is 70 degrees (since BOC is a triangle and COB=50, BCO=60. Angles DOE = 70= BOC. So, angle CEO = 50 (since CEO is a triangle=sum of angles 180), Similarly, Angle BDO= 40. So, the 4 unknown angles are ODE, EDA, OED and DEA ( = a, b, c, and d respectively). You get a+b = 140, c+d = 130, a+c=110 and b+d=170). 4 eqns. 4 unknowns. Solve. After that the problems started, but with a little pushing one gets the angle AED = 60 and ADE= 100

Added later: The plot thickens....still need to work things out.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 06:17
chandraS -
Ok, least number may be 120 but why it can't be greater than 120?

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 06:25
Srikumar: from your
a+b = 140, c+d = 130, a+c=110 and b+d=170

If you add first two, you get a+b+c+d = 270
and if you add last two, you get a+b+c+d = 280
I think there is something wrong

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 06:28
Amber G. wrote:Srikumar: from your
a+b = 140, c+d = 130, a+c=110 and b+d=170

If you add first two, you get a+b+c+d = 270
and if you add last two, you get a+b+c+d = 280
I think there is something wrong

Sorry, my last eqn. has a typo: it should read b+d= 160 and not 170. Then the objection raised above is avoided and my answer does satisfy the eqns.... but there is the issue of multiple solutions....I'm definitely missing an extra condition.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 09:50
Amber G. wrote:chandraS -
Ok, least number may be 120 but why it can't be greater than 120?

Well, the number of milch cows is fixed at 76. So any increase will cause the buffaloes to violate the 3:1 ratio. The number of milch buffaloes for a total of 140 is (112-76) = 36 > 35 which is 25% of 140.

Full Disclosure: I brute forced the calc for 140.!!

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 21:58
Amber G. wrote:Two Not so Hard Problems:
1. (Old problem from Hindu Math sources)
A dairy farm has exactly 75% cows, and the rest 25% are buffaloes. Also exactly 80% of all these animals are producing milk (rest 20% are non-milk producing). There are exactly 76 milk producing cows in the farm. What is the total number of animals in the farm.

Let total = n, # of milk producing buffaloes = m
Then 76+m= 0.8n
with 0.75n >= 76 and 0.75n must be an integer
m < 0.25n and 0.25n must be an integer.

In other words must have
1) 4n-5m=380
2) n > 101
3) m <= n/4
4) 0.75n integer,
5) 0.25m integer.

so as suggested by ChandraS n=120, m=20 is a solution satisfying all conditions.

We can try the others:
n=125, m=24 fails (4)
n=130, m=28 fails (4)
n=135, m=8=32 fails (4)
n=140 m=36 - fails (3)
increasing n further fails (3), so cant go anymore this way.
n=115 m=16 - fails (4)
n=110 m=12 - fails (4)
n=105 m=8 - fails (4)
decreasing n further fails (2)

Thus n=120, m=20.

### Re: BR MathsCorner-1

Posted: 02 Aug 2008 22:16

There are also other possibilities that you can consider
4n-5n=380 = (1*2*10*19)

First solve 4n-5m=1
n=9, m=7 is a solution.
Therefore 380*9 = 3420, 380*7 = 2660 is a solution.
We can for example also solve 4n-5m=2 and then multiply the solution by 180.
and so on ...
But all these solutions are related to ChandraS's solution by 120+5i, 20+4i with i=...-2,-1,0,1,2...
The reason is simple. If (n1,m1) and (n2,m2) are any two solutions, then taking the difference of the two solutions gives:
4(n2-n1)-5(m2-m1) = 0
Since 4 and 5 have no common divisors, the only possible solutions to this are:
m2-m1=4i
n2-n1=5i
(i=...-2,-1,0,1,2...)
As seen in the previous post only one of these solutions 120, 20 satisfies _all_ the conditions.

### Re: BR MathsCorner-1

Posted: 03 Aug 2008 03:30
Amber G. wrote:Two Not so Hard Problems:
1. (Old problem from Hindu Math sources)
A dairy farm has exactly 75% cows, and the rest 25% are buffaloes. Also exactly 80% of all these animals are producing milk (rest 20% are non-milk producing). There are exactly 76 milk producing cows in the farm. What is the total number of animals in the farm.

Good luck.

My solution

if D is total, 0.8D are milk producing. if y is milkable buffaloes, (76+y) = 4/5*D

==> D = (76+y)*5/4

0.75D = cows ==> (76+y)*15/16 ==> this has to be a natural number

so (76+y) is a multiple of 16, possible values 4,20,36 etc

4 wont work, 20 works, 36 and greater makes no sense.

so y = 20. and the rest follows.

D = 120 etc

### Re: BR MathsCorner-1

Posted: 03 Aug 2008 04:17
Chalam, SK Mody, ChandraS:
FWIW; Perhaps something along this line may be a little bit clearer.
Since (1/4) are buffalos ==> total number must be divisible by 4
Similarly (1/5) are non-milk producing ==> total number must be divisible by 5
This means total number of animals must be divisible by 20, so let the total number be 20n (where n is positive integer)

Now Total # of cows = 15n and 76 or them producing milk
Total # of Buffalos = 5n and (16n-76) of them producing milk.

So 15n >= 76 ==> n > = 76/15 > 5
And 5n >= 16n – 76 ==> 76 >= 11n ==> n <=76/11 <7
So n is greater than 5 and less than 7, ==> n=6 or
total number = 20n = 120

### Re: BR MathsCorner-1

Posted: 03 Aug 2008 06:51
Amber G.: Thats a pretty elegant way of solving the problem.
Thanks for being the prime mover for this thread. Hopefully, I will find my feet with maths once again

### Re: BR MathsCorner-1

Posted: 03 Aug 2008 22:02
Amber G. wrote:ABC is an isosceles triangle with Angle A=20 degrees and angle B= Angle C = 80 degrees.
Let D be a point on AB such as angle BCD = 60 degrees.

Man, the new admins act fast....a quick flash of steel and all traces of haram activity are gone.... no signs left of Mullah Nayakuddin's overnight infiltration.

For the problem above, I had to resort to trigonometry. I think you may be looking for a geometry-based solution, hence I am only posting my approach/method below:

There are 2 triangles that can be formed with the base of the main triangle: BCE and BCD. We want to find out angle AED. This is possible if we know angle DEB. To determine this, we need to find the length of the sides BE and DB.

Take triangle BCE: It is isosceles. (Angles CBE and CEB are 50 degrees)
Assume length of BC = 1 unit.
Therefore, BE = 2 x BC cos 50 = 2 cos50. ------ Eqn 1.

Take triangle BCD, and use the sine formula: a/sin(A) = b/sin(B)
we have: DB/sin (60) = BC/sin(40). Since BC = 1, we get:
DB = sin(60)/sin(40) since BC = 1 .................... Eqn. 2

Now, to determine DE take triangle BED, and using the cosine formula in triangles: you get DE^2= BE^2 + DB^2 -2*DB*BE*cos30.

Plug in values for BE and DB from eqns (1) and (2) above to get DE and use the sine formula to get angle BED. Angle AED = 130- BED.

I tried to simplify the eqns to minimize the need to refer to tables, but I could not fully eliminate cos(50).

### Re: BR MathsCorner-1

Posted: 03 Aug 2008 23:05
ChandraS - Thanks.
Thanks also to admin(s)) for cleanup.
Sri Kumar - Your approach should work. Resorting to trig is okay, although, as you said, "simple approach" may be more fun. Just wanted to add that (Maha Hint! ) everything does simplifies in a neat and cool way.! .(even without resorting to trig tables , (or to trig for that matter -if one uses a non-trig method) )..

### Re: BR MathsCorner-1

Posted: 04 Aug 2008 10:39
Amber G. wrote:Also let me post an easy looking Geometry problem.
ABC is an isosceles triangle with Angle A=20 degrees and angle B= Angle C = 80 degrees.
Let D be a point on AB such as angle BCD = 60 degrees.
Let E be a point on AC such that angle CBE = 50 degrees.
Find the angle AED.

50 degrees.
Proof took me a while. I attach a figure with my proof here. Note my paint-fu sucks so the figure may not reflect reality as much, especially the area around DHGE, but the logic makes sense.

Given
BAC = 20 degrees, BCA=80, CBA = 80,
BCD = 60, CBE = 50 degrees

From the figure
CBD = CBA = 80
BCE = BCA = 80

First, since sum of angles in triangle is 180, therefore:
BDC = 180 - CBD - BCD = 180 - 80 - 60 = 40 degrees

Hence, CDA = 180 - BDC = 180 - 40 = 140 degrees

Next we consider BCE
CBE = 180 - CBE - BCE = 180 - 80 - 50 = 50 degrees
Therefore BCE is an isoceles triangle and BC = CE

Next draw DG parallel to BC.
Therefore CBG = BCD
Since ABC is isoceles and DG is parallel to BC, therefore BD = CG
Therefore triangles BCD and CBG are also congruent (since BC is common, CBD = BCG and BD = CG)

Now (this part took me a while to work out ), since DG is parallel to BC, BCD = CDG (alternate angles)
Since BCD = 60 (given), therefore CDG = 60 as well.
Similarly, we can work out that BGD = 60.
Now consider triangle DGH. Since HDG = 60 and HGD = 60, therefore we have DHG = 180 - 60 - 60 = 60
Therefore triangle DGH is equilateral triangle
Since DG is parallel to BC and DGH is equilateral, BHC is also equilateral

Therefore BH = CH = BC

Earlier we determined that BCE is isoceles and BC = CE. Hence BH = CH = CE

Now consider triangle CEH
We just proved CH = CE above, hence triangle CEH is isoceles.
Now angle HCE = BCA - BCD = 80 - 60 = 20
Therefore, CHE and CEH = 80 degrees

Now DHG = 60 degrees (proved above) and CHE = 80 degrees
Therefore GHE = 180 - DHG - CHE = 180 - 60 - 80 = 40 degrees

Now since triangle BDC AND BGC are congruent, therefore BDC = BGC
We already determined BDC = 40 degrees above, hence BGC = 40 degrees as well
BGC = EGH, hence EGH = 40 degrees

So now we have GHE = 40 and EGH = 40 degrees, so triangle EGH is isoceles (doesn't look that way in the figure)
therefore GE = EH
Now we already proved that DH = GH (since DGH = equilateral)
Therefore DHEG is a kite shape and line DE bisects the angle HDG
Hence EDH = half of HDG = 1/2 * 60 = 30 degrees
Since EDH = EDC, therefore EDC = 30 degrees

Now we're cooking, since we already determined that CDA = 140 degrees very early on above
Therefore, EDA = CDA - EDC = 140 - 30 = 110 degrees
Therefore AED = 180 - DAE - EDA = 180 - 20 - 110 = 50 degrees

Hence AED = 50 degrees

### Re: BR MathsCorner-1

Posted: 05 Aug 2008 04:43
Rahul M - the problems and solutions of IPhO are now here:
http://ipho2008.hnue.edu.vn/Competition/ProblemsSolutions/tabid/80/Default.aspx

ArmenT - Nice work. Answer is correct. The problem (and few of its variations) is classic (IOW it has been asked quite a few times in contests) and , of course, I have seen many different ways to solve this. SriKumar and others - now since answer is there, you may be able to simplify the trig (or other solution) solutions.
BTW - I gave this (geometry) problem to one of my "nephew" (friend's son), the "nephew" his father (an IIT grad) and grandfather (a math prof) worked on it quite a while. They had a long trig solution, I sent them my solution which had just one picture (and no words)..so sometimes if you are lucky enough to see the problem right way it may be better .

### Re: BR MathsCorner-1

Posted: 05 Aug 2008 17:06
Amber G. wrote:Two Not so Hard Problems:
1. (Old problem from Hindu Math sources)
A dairy farm has exactly 75% cows, and the rest 25% are buffaloes. Also exactly 80% of all these animals are producing milk (rest 20% are non-milk producing). There are exactly 76 milk producing cows in the farm. What is the total number of animals in the farm.

Good luck.

Let me give a rigorous solution to this cow and buffalo problem. A common mistake here is to assume that the milk producing animals are homogeneous in two groups (cows and buffaloes), even though one may get the right solution.

# cows = x;
# buffaloes = y.

Total number of cattle = (x + y)
Using the 75% and 25% ditribution of cows and buffaloes, we have : x = 3y
Number of milk producing cattle = 80% of the total = (4/5)(x+y)
Number of milk producing cows = 76. So, x must be greater than 76.
Number of milk producing buffaloes = z (say) So z must be between 0 and y.
We have (76 + z) = (4/5)(x + y) = (4/5)(3y +y);
i.e., 380 + 5z = 16y, where x > 76, 0<= z <= y, and x, y, z are all integers.

We need to solve the last equation within the feasible set. Note that, z can never be odd since the right size
is always even. So now try with z = 0, 2, 4, 6, etc. With z = 4, we get y= 25, but this is impossible since this
would imply x = 75, smaller than 76. The next possible value of z is 20, which yields y = 30, and x = 3y = 90.
(Other higher possibilities also may exist. Essentially we need (380 + 5z)/16 to be an integer, that's all.)
So, the total number of cattle = 90 + 30 = 120.

### Re: BR MathsCorner-1

Posted: 05 Aug 2008 17:59
Kati wrote:
Let me give a rigorous solution to this cow and buffalo problem. A common mistake here is to assume that the milk producing animals are homogeneous in two groups (cows and buffaloes), even though one may get the right solution.

# cows = x;
# buffaloes = y.

Total number of cattle = (x + y)
Using the 75% and 25% ditribution of cows and buffaloes, we have : x = 3y
Number of milk producing cattle = 80% of the total = (4/5)(x+y)
Number of milk producing cows = 76. So, x must be greater than 76.
Number of milk producing buffaloes = z (say) So z must be between 0 and y.
We have (76 + z) = (4/5)(x + y) = (4/5)(3y +y);
i.e., 380 + 5z = 16y, where x > 76, 0<= z <= y, and x, y, z are all integers.

We need to solve the last equation within the feasible set. Note that, z can never be odd since the right size
is always even. So now try with z = 0, 2, 4, 6, etc. With z = 4, we get y= 25, but this is impossible since this
would imply x = 75, smaller than 76. The next possible value of z is 20, which yields y = 30, and x = 3y = 90.
(Other higher possibilities also may exist. Essentially we need (380 + 5z)/16 to be an integer, that's all.)
So, the total number of cattle = 90 + 30 = 120.

You also need the number of buffaloes to >= (80% of total - 76), since they have to make up the rest of the milch animals. 120 is the only answer.

BTW, I solved no equations. I just set up the constraints as given, in Microsoft Excel. Shame on me.

### Re: BR MathsCorner-1

Posted: 06 Aug 2008 00:26
Nandu wrote:
Kati wrote:
Let me give a rigorous solution to this cow and buffalo problem. A common mistake here is to assume that the milk producing animals are homogeneous in two groups (cows and buffaloes), even though one may get the right solution.

# cows = x;
# buffaloes = y.

Total number of cattle = (x + y)
Using the 75% and 25% ditribution of cows and buffaloes, we have : x = 3y
Number of milk producing cattle = 80% of the total = (4/5)(x+y)
Number of milk producing cows = 76. So, x must be greater than 76.
Number of milk producing buffaloes = z (say) So z must be between 0 and y.
We have (76 + z) = (4/5)(x + y) = (4/5)(3y +y);
i.e., 380 + 5z = 16y, where x > 76, 0<= z <= y, and x, y, z are all integers.

We need to solve the last equation within the feasible set. Note that, z can never be odd since the right size
is always even. So now try with z = 0, 2, 4, 6, etc. With z = 4, we get y= 25, but this is impossible since this
would imply x = 75, smaller than 76. The next possible value of z is 20, which yields y = 30, and x = 3y = 90.
(Other higher possibilities also may exist. Essentially we need (380 + 5z)/16 to be an integer, that's all.)
So, the total number of cattle = 90 + 30 = 120.

You also need the number of buffaloes to >= (80% of total - 76), since they have to make up the rest of the milch animals. 120 is the only answer.

BTW, I solved no equations. I just set up the constraints as given, in Microsoft Excel. Shame on me.

Nandu, many many thanks. That settles the answer uniquely. With that restriction (which I overlooked),
we have now: y >= (4/5)(x+y)-76, which finally gives the restriction on y as: 0<y<= (380/11)<35.
Therefore, the even-integer valued solution of z must exist within the range (0, 35), and there is only one
solution, that is z = 20. This gives the unique values of x and y as x = 90, y = 30.

### Re: BR MathsCorner-1

Posted: 06 Aug 2008 00:35
Kati - You may like to look at one solution (in tiny fonts ) posted around 6:47 pm (Aug 2)- which is quite rigorous and may be simpler.
(In your equations, x (and y) has to be multiple of 5, setting them as such in the beginning may make it simpler.

(Also there is standard mathematical technique to reduce, say eq. like
380 + 5z = 16y .. (called Linear Diophantine equations)
z has to be multiple of 4, so take it as say 4c, y has to be divisible by 5 so make it 5b and you have 19+c=4b a much easier eq. to work with)

(Actually if you apply y<35 so that b<7 it is becomes even simpler ....)

(If you are interested, you can look up Linear Diophantine equations (in wiki) - (Some name it after Aryabhata) something Ancient Hindu Mathematician were very good at... actually very advanced at (Aryabhata has written quite advance topics in this .. dealing with nonlinear stuff too)

Nandu - One version of this problem, was a little harder (one can not solve by brute force that easily) but I chose to put an easier version here.

Anyway, it looks like we still have a few unsolved problems ("ankan" and factoring).

### Re: BR MathsCorner-1

Posted: 06 Aug 2008 19:06
Many years ago in 1985 I proved the following lemma that I used in proving a major theorem about
the stability of matter when I was at the Institute in Princeton as a post-doc. Can anyone reprove it? I have another lemma that I proved lets see if you guys can prove that too. Here is the first lemma:

Take a square it works for a cube in any dimension. Now play the following game. Lets take a square for example.
Bisect each side you get now 4 congruent mini squares. Call this Step 1. Each congruent mini square can be bisected further to get now 16 mini squares in step 2. Proceed so on. One gets at step n, 4^n squares.

Now remove N DISJOINT squares in any way, that is remove a step 10 square, a step 25 square etc. The remaining
figure is a big square with holes like Swiss cheese Prove you can cover the remaining figure; the Swiss cheese looking part by exactly C N squares. C is a universal constant that only depends on the dimension. I forget the value I found for C for dimension 2.

By cover I mean nothing must penetrate the holes where squares were taken out.

The paper I wrote appears in Communications in Partial Differential Equations towards the end of the paper.

Another cute geometric lemma I proved in 1991 appears in a paper in Journal of Differential Geometry.
Only recently has more progress been made on it. Here is the lemma, actually can anybody improve the lemma?

Here is the lemma. Take ANY set E in n-dimensions, n>1. Assume it is covered by finitely many balls of various radii, radii is all arbitrary. Covered means every point in E is in some ball. Now prove one can find a sub family of the balls that have the following property:

(1) For any POSITIVE number delta<1/2, if I magnify the radii of each ball of the sub-family by 1+\delta, i.e.
new radius is (1+delta)r, then E is still covered by the balls of the sub-family with the radius of each magnified by 1+delta

(2) FURTHERMORE, a point does not belong to more than C (delta)^{-n} balls of the sub-family, where C is a UNIVERSAL constant depending only on the dimesnion n. That is in other words the sub-family without the radius magnification is more or less disjoint, there is overlap between balls but the overlap is controlled, i.e no more than C{delta}^{-n} balls of the sub-family overlap.

This lemma is used by me to obtain a powerful result a conjecture of ST Yau. Point is can you improve statement
(2) to (delta)(- {\square root n}} for example, any improvement?

By the way there is a new paper by Dan Mangoubi on arxiv.org who is claiming to have improved my technique.
I have not checked his computations. There is a huge amount of work that came out of the innocuous lemma 2.
Here is one which eventually appeared in the American Journal of Mathematics one of the leading journals of the world in 2004.

http://arxiv.org/abs/math/0402412

You will be able to do better than what the authors above do if you improve the second lemma since you will improve my result and its ensuing major consequences.

### Re: BR MathsCorner-1

Posted: 06 Aug 2008 21:17
In my previous post, lemma 1, the Swiss cheese lemma, when you cover the Swiss cheese part by squares nothing
of the covering squares must leak into the holes AND I forgot to add NOTHING of the covering squares must leak out of the big square you started with in the bisection process.

I have many more of such geometric puzzles I found that lead to powerful uses in Physics, Geometry and other branches of mathematics. It is heartening to see some of these problems begin to move 20 years after I discovered them, they are non-trivial. Even a Fields medallist Charles Fefferman has contributed see the refs of the paper I linked. No cookie points for guessing who yours truly Vsunder is from the refs hehehehe!!!

### Re: BR MathsCorner-1

Posted: 09 Aug 2008 20:47
vsunder wrote:Many years ago in 1985 I proved the following lemma that I used in proving a major theorem about
the stability of matter when I was at the Institute in Princeton as a post-doc. Can anyone reprove it? I have another lemma that I proved lets see if you guys can prove that too. Here is the first lemma:

Take a square it works for a cube in any dimension. Now play the following game. Lets take a square for example.
Bisect each side you get now 4 congruent mini squares. Call this Step 1. Each congruent mini square can be bisected further to get now 16 mini squares in step 2. Proceed so on. One gets at step n, 4^n squares.

Now remove N DISJOINT squares in any way, that is remove a step 10 square, a step 25 square etc. The remaining
figure is a big square with holes like Swiss cheese Prove you can cover the remaining figure; the Swiss cheese looking part by exactly C N squares. C is a universal constant that only depends on the dimension. I forget the value I found for C for dimension 2.

By cover I mean nothing must penetrate the holes where squares were taken out.

It's not April 1, but I will bite. If this is not a joke, could one please write it in a way so it makes sense mathematically by being precise in definition.
For example: What exactly the sentence mean "Take a square it works for a cube in any dimension." ?? If rest of the part is particular example in D2, then state so, if not, then were exactly, say 4 in 4^n comes from?
Also what is "N" is is different than "n"? for example, when n=1, we have only one choice (3 squares) making C=3 (Assuming N and n are same) while for n=2, one can easily get an example where you need only 2 squares to cover.

In short it makes no sense.

If this was a joke, can the spam be removed.

(I am not even commenting on the rest of the post.. as it makes even less sense to me)
***********************************************

Okay - speaking of squares, and covering .. here is an old classic problem (If you have not heard it before, try to solve it without googling it etc... if there is an interest, someone can post a solution in tiny format. - I will do so if there is interest).

There is a carpet 9(units) x 16 (units), and you want to cover 12 x 12 square. In how many (minimum) pieces you have to cut the original carpet (and resew them) to make it fit 12 x 12.?

(No trick question, but the cut pieces need not be rectangle .. they can be any shape)

### Re: BR MathsCorner-1

Posted: 10 Aug 2008 05:33
Amber G. Here is shot for the puzzle you provided involving your friend born in Auburn on Feb 3, 1964. This sort of exercise is generally frowned on. All mistakes are due to my own lack of skill/amateur understanding of Jyotish. Please do not view Jyotish as an absolute set of rules for prediction, but more like a search program that gets better with a little feedback to reduce error. Could you provide a little feedback esp. regarding points 2 and 3? Then I can try to provide more specific info, perhaps dates. Let’s try this for a few cycles and then you can evaluate.

It is outside my ability to comment on the factual data you have requested. I am unsure of unsure of gender, probably (90%) same as yours.

1. The individual is a short witty argumentative person. He is attracted by mysteries and knowledge. He may be fond of some wind-related sport.

2. He is a very forceful, even fierce speaker, very ethical, but maybe cutting and sarcastic. He has problems with his mouth/teeth and has undergone some dental surgery. This person is a boss of several educated folks and powerful. He has huge expenditure but is not poor at all.

3. He is very well educated (definitely post-graduate) business oriented, and emotionally close to his mother. He lives in a very fine, spacious upscale house (is it light-yellow/tan?), that has water features in it.

There is something unusual (not necessarily too negative) about his forearms. Tattoos? birthmark? muscular/shapely? siblings did something?
He has several unusual siblings, at least one older and younger, I think. They are accomplished. The person is very courageous and has stood up for contentious issues. He communicates very extensively through short write-ups and heavy use of e-mail.

His child(ren) maybe belligerent and confrontational.

His spouse is likely coarse/materialistic and there is probably more than one marriage.

He has very very successful career wise (but he himself may think he was thwarted in achieving his success). Involves something with water, fluids? The choice was clearly influenced by his father and mother.

He is rather spiritually motivated.

If he survived past 1969 he will likely die in 2016. May 1987 (+ 2 months) marked an important transition to his current professional life.

Added info (such as his height, speech, and forearms) if different, could reveal much more. Highest confidence in points 2 and 3.