Bharat Rakshak

AmberG,

From the diagram above, r1 is the initial position of the rabbit, f1 is the initial position of the fox. The base of the triangle represents the X axis.

The fox adopts the following strategy -
If the rabbit moves away from r1 towards r2, the fox moves in a direction perpendicular to the direction of the rabbit i.e. from f1 towards f2.
Let us say at r2, if the rabbit reverses and moves towards r3, the fox moves in a direction perpendicular to the opposite side of the triangle i.e. from f2 towards f3.
In both cases, the fox moves with a minimum speed such that its x coords always matches the x coords of the rabbit.

The fox will always keep moving closer to the rabbit along the y axis until it finally catches it (the distance can increase when the rabbit is reversing, however for a given point of reference, it will keep reducing e.g. if the rabbit backs up all the way to its starting point, the fox would have been closer than it had been earlier). If the rabbit keeps still, the fox moves along the y-axis .

What remains is to trivially determine this minimum speed at which the x coords remain equal. If the rabbit keeps going to the left towards e, the fox will catch up with it if it travel at 1/sqrt(3) times the speed of the rabbit. Motions in the opposite direction are similar i.e. the fox always has to travel at 1/sqrt(3) times the speed of the rabbit so that the x coords match while it comes closer along the y-axis until it finally catches up with it.

So, if the aim is for the fox to catch the rabbit, the rabbit does not require to travel @ less than 2x but only at less than sqrt(3)x. To put it another way, the fox is guaranteed to catch the rabbit if it is faster than 1/sqrt(3) times the rabbit. Indeed, it is trivially possible to calculate the angle at which the fox should travel for a given ratio of speeds.

I think this problem as given is simpler than the duck-fox lake problem but it might be tougher to prove the minimum ratio at which the duck can never be caught, especially for the range between sqrt(3) and 2.

Gakakkad, my (older) sister went into medicine, and she still enjoys math. In those days to get MBBS/MS/MD one need not even know calculus but she kept her interest. (I wish I learned more bio-science from her *smile* ). My soon to become DIL is finishing residency in neurology but is extremely good in, and always enjoyed math/physics (majored, and has taken graduate level math course)...So please keep up with physics/maths .. we need good doctors with strong math physics background.

You are welcome. Hope the information was useful.

Brilliant analysis and clear figure. (See my comments below)

For more on this problem look at IMO compendium (Around page 100, prob 52-YUG3 ) here is the link from google books:
IMO compendium



All one has to do, (looking at your diagram) is for fox to go horizontally (parallel to x axis) (Instead of perpendicular to the side) the first time when rabbit chooses a side, and if rabbit reverses (which it has to) fox turns in upward slant (easy to calculate the angle of the upward slant).

Rest is simple.


(This way the minimum speed (for fox) is "improved" from (1/(sqrt(3)) to 0.5. )
Added later: For clarity, see my next post for few more comments.

I agree with that.. Biological sciences are nothing without a backing from physical science and math.. I often have an argument with my American colleague's who want the India/UK/Australia type system of getting into medicine straight from school ..

Pure neuroscience(not to be confused with neurology) is an example of biology that is greatly dependent of Mathematics...
In fact one of the tenured prof of Neuroscience in columbia univ is a PhD in Mathematics..

Eric Lander , one of the founding members of broad institute at MIT (involved in the human genome project) too has a PhD in Mathematics... the guy even taught in Harvard Business .. He now teaches basic biology at MIT ... In fact it was neuroscience that got him interested in Mathematics..

Continued from my post above:

For clarity, see Skumar's figure above, key strategy for the fox is to to keep the x-component of its velocity exactly equal to to the rabbit's while having the maximum possible y-component (vertical velocity) (That is, if the rabbit is stationary, fox moves straight up, else on a slanted line). As long as the fox speed is > 1/2 of rabbit's, there is always a positive movement in y-direction so fox will eventually catch the rabbit.

(Conversely, if the fox is slower that 1/2 of the rabbit, the rabbit just waits till the last moment till the fox makes a move in the horizontal direction, and the rabbit escapes on the other side - if fox does not make a move in the horizontal direction, rabbit can choose any side)

The same method works with duck-fox and circular lake problem. The duck starts towards 180 degrees opposite from the fox and then keeps it's angular velocity same as fox (so it remains on the opposite side) while having maximum radial velocity .. It can do so as long as it is inside the "smaller" circle (see gakakkad''s picture) as the radial component of the velocity will be positive. Once it reaches that point it simply takes a straight line path to the shore...

Here is a similar old classic calculus problem...

.

I guess one can see farther on clearer days .

Thanks.

I'd like to raise my knowledge of applied math in general. i.e. learn techniques of problem solving primarily for engineering problems. I know the field of math is vast but what book can you recommend that starts at a basic level (and i emphasize basic) yet leaves me with a reasonably good knowledge of math by the end of it.

Due to time constraints, I'd prefer doing an online course watching a video or watching a DVD of some guy explaining the concepts. Anything like that out there that goes topic by topic?

Fellow BRF members:

I am probably the only brf oldie, for whom, Math dhaga comes as one of the most active dhaga. Thanks to Ramana who started this thread. I too, kind of took a lead here and posted many problems, commented on Maths recent issues, posted math related news items etc. In all I enjoyed (and continue to enjoy) posting and reading this thread. From the feed back we got, I think similar views are shared by many.

Yet, I have to be honest. I don't know in which direction BRF is going. Personally I don't know if I can even continue posting here , even if I want to, as powers to be here, it seems to me, are clearly telling me that BRF is no longer a place for diverse views or technical discussion.

Of course, I am not the first, or will be last, who has articulated this. Many of valuable contributors, at least it seems to me, have articulated this.

To underscore that I am not overreacting, let me just point out one example, in this Math dhaga itself.
One of the moderator gave me a board warning (my first board warning!) for my comments in math thread, and called my comment “idiocy”. He rebuked me as someone “who wanted to test math ability in nuke discussion thread” :?
See: this for details

Please also see my reply to that:
(Reply)

What is noteworthy to me is

1- No admin ever (in last 3 years) made a comment about uncivil language used against me. (pointed out in my reply above)

2. Many of the other members, including admins, who since then made no contribution to math thread, jumped in to pass a few more insults.

3. Even after 3 years, no one has answered or gave clarification to the question I raised, namely, how the original problem/observation posed there made sense or why some seemingly random, and sometimes broken links given there, were relevant?

I do not think I am over reacting . I have tried to give honest feed back to forum admins with little result. Others are free to give feedback to BRF admins.

******
Let me end this with a less known but a beautiful theorem in Geometry.

Take any triangle, trisect each angle. The inner triangle thus formed (try to draw it and you will know what I mean) will always be a equilateral triangle. (If outer triangle is ABC, and inner PQR while angles
CBP = PBR = RBA , and angles BAR=RAQ=QAC and angles ACQ=QCP=PCB – triangle PQR will always be a equilateral triangle no matter what shape is ABC)

AmberGji,
You have a legitimate grievance,IMVVHO, against the moderators on that issue in this dhaaga.In my view,the other poster was just showing off.

Don't you mean "bisect" instead of "trisect"? It would seem that I'd get a six-sided figure if I divided each angle into 3 parts? (I'll admit I haven't drawn an actual figure yet, so I may be wrong here). Bliss to clarify onlee?

I did mean Trisect..
Eg:


The Green triangle above is always an equilateral riangle, irrespective of the shape of the blue triangle. ( The theorem is simple and beautiful, yet first time it got mentioned was in 20th century ..with newer proofs given as late as 1980's or in 2000's - First time when I heard it (in 1980's) I was surprised that virtually no one knew about it - it took me some time, and lot of trigonometry/algebra calculation to prove its validity before I can believe it)

Whoa! The maths dhaga is one BIG reason I visit BR Forum. Don't leave!

Neshant ji

May I suggest "An Introduction to Applied Math" by Gilbert Strang of MIT? He also has an on-line video course on Linear Algebra based on his other classic "Linear Algebra and Its Applications" which is another book that deals with that part of mathematics which is arguably the bed-rock of Applied Mathematics and Engineering.

Added later: Forgot to mention, the video lectures on Linear Algebra are at OCW.

Some of you - esp. AmberG - might get a kick out of this.

Kiran Kedlaya was a contestant on Jeopardy today and won. So he will be on tomorrow as well.

UK's Intelligence agency posts an online cryptographic challenge to attract recruits

One who also brought math to a broad audience...
Robert Osserman, noted Stanford mathematician, dies

Methods of computing square roots. Bakhshali approximation

http://en.wikipedia.org/wiki/Methods_of ... roximation

By the way, Fibonacci aka Leonardo of Pisa gives all the credits to Indians in his book Liber Abaci under Modus Indorum - the methods of Indians:

Here is one just using law of sines and some trigonometry.
Let angles PAB =PAQ=QAB=a; RBC=RBP=PBA=b and RCB=RCQ=QCA=c.
a+b+c = 60 ---- 1

In Triangle ABC
AB/Sin3c= BC/Sin3a
BC = AB * Sin3a/Sin3c

Since
Sin3a = Sina(3cos^2a – Sin^2a)
Sin3a = Sina * (sqrt(3)cosa + sina) * (sqrt(3)cosa – sina)
Sin3a = Sina * 4 * (sqrt(3)cosa/2 + sina/2) * (sqrt(3)cosa/2 – sina/2)
Sin3a = Sina * 4 * (sin60cosa + cos60sina) * (sin60cosa – cos60sina)
Sin3a = Sina * 4 *sin(60+a)*sin(60-a))

Hence

BC = AB * Sina * sin(60+a) * Sin(60-a) / Sinc * sin(60+c) * Sin(60-c) ---- 2

In Triangle ABP

BP/Sin(a) = AB/Sin(a+b)
BP = AB*Sin(a)/Sin(a+b)
BP = AB*Sin(a)/Sin(60-c) ---- 3

In Triangle BRC

BR/Sin(c )=BC/Sin(b+c)
BR = BC*Sin( c )/Sin(b+c)
BR = BC*Sin(c )/ Sin (60-a)

Replacing BC from 2

BR = AB * Sina * Sin(60+a) / sin(60+c) * Sin(60-c) ---- 4

Now from 3 and 4

BP/BR = Sin(60+a)/Sin(60+c)

Hence Angle BRP = 60+c and angle BPR = 60+a in Triangle BPR.

We can similarly prove for triangle APQ
Angle APQ = 60+b & angle AQP = 60+c

And for triangle QRC
Angle CQR = 60+a & angle AQP = 60+b

We can now calculate angles PQR, RPQ and PRQ

PQR = 360-CQA-CQR-PQA = 360-(180-a-b)-(60+a) – (60+b) = 60
RPQ = 360-RPB-BPA-APQ = 360-(60+a)-(180-a-b)-(60+b) = 60
PRQ = 360-BRC-BRP-QRC = 360-(180-b-c) – (60+c) – (60+b) = 60

^^^ Very nice.
As said before, what I find very remarkable (and thus beautiful) is that such a simple theorem is still not that well known, and I have seen quite a few serious mathematician sharing the same view when they first saw this.
(They could not believe that that they have not heard about this before.)

The first famous paper was in 1929. (People still publish new proofs of this or its generalization)...Nice stoy about how the theorem was discovered while one was studying cardioids (those who don't know it is heart shaped curve - locus of a point on a circle if one rolls this circle inside a bigger circle)

I didn't follow this step. Just because the ratio of the sides is expressible as the ratio of two sines, are they necessarily the angles of that triangle? Am I missing something obvious?

Yes, since third angle of triangle is known and fixed (b in this case) there can only be 1 such solution for remaining 2 angles. For any other Sine ratio which will meet this requirement (BP/BR), third angle will not be b.

To put it other way if you have 2 sides of a triangle as AB and BC (AB/BC is known), one can create as many triangles as one wants but 3rd side CA (And corresponding angle as well as height of triangle which gives law of sines) would be different for each of these combinations. Since 3rd Angle in this case is fixed, only 1 solution is possible.

^^^
Also, one of interesting (and easy to understand with using very elementary geometry - ) way is to work backwards.. we need scissors and very little geometry.

First, for simplicity let the three angles of our triangle ABC be 3a, 3b, and 3c (Angle CAB = 3a, etc)..
(We have here sum of three angles = 180 = 3a+3b+3c or a+b+c= 60)

First cut a triangle say PQR where PQ=QR=RP=1 and each angle is 60)
Then cut 3 triangle(s), let the first of these three triangles, one side is same as PQ and call this triangle PQZ... such that side PQ=1 and angle PQZ= a+60 and angle QPZ=b+60 (so that third angle (PZQ=c)

The other two triangles are similarly symmetric.. (Let us call them QRX and third triangle as RPY)

(Very easy to use scissors and cut the triangles than to describe it here in words..... with very little work one can see (and prove) that the triangle XYZ is similar to triangle ABC...

QED!

To add to amit's post. (or my last post..).

I don't think it is obvious on the face of it.. (that is just because the ratio is (sin(60+a)/sin(60+c)) it is not obvious (IMO) that the angles have to be.. 60+a and 60+c)..

The critical part is, that we do know that the third angle is b and ((60+a)+(60+c) +b) happens to be 180 degrees (because a+b+c=60) hence we can deduce the angles are indeed 60+a and 60+b)
(IOW if the ratio of two sides is (sin x / sin y) and the angle between them is (180-x-y) then the two angles would be x and y)

In this forum just the outline of proof is obviously okay but if a detailed proof is required, I could also use the cos theorem (You know the two sides (or ratio of them) (PB and RB) and the angle between them (b) just calculate PR and see that it is symmetric in abc.. (IOW PQ=QR=RP).) to make the proof a little easier...
(Hope it is helpful)

matrimc: Thanks, I did know and indeed watched Kiran Kedlaya on Jeopardy.

Thanks both! Having two sides whose ratio is known and an included angle doesn't fix the triangle. But all the triangle solutions are obviously similiar, so the other two angles are the same in all the solutions.

May be of interest to high school students (or even earlier grades - 6th and upwards)... In USA (and similar programs in India) there are interesting contests coming up where every one, if interested in math, can take part and start ...

AMC (American Math Competition) open to all (Just ask your high school to register etc) as first step towards selection process for math olymnpiad..

Feb 7 ( or alternate day is Feb 22) - AMC 10/AMC 12
(If you do well the next step is AIME in March , USAMO in April and IMO in July..)
(In India similar contests RMO, InMO etc..)

Also some other math contests are in February too ..starting at school level. (and later district level, and then state /national level at later dates). One such program is Mathcounts where district level competitions are in February)

May be of interest. One of OCW's most popular course, Linear Algebra, is now available in a version designed to support independent learning.
There is a press release for this but of course, one can check out ocw.mit.edu for details.
Press release at:http://ocw.mit.edu/about/media-coverage/press-releases/linearalgebra/


Also IIT's indeed have lot of gems in their OCW .. One can check out, for example, their product or NPTEL channels .. ( a joint venture by seven Indian Institutes of Technology (IITs) and Indian Institute of Science (IISc))..

http://www.youtube.com/profile?user=nptelhrd#grid/playlists

Here is a Jingo ancient problem - Hawk and Rat problem

A hawk is sitting on a pole whose height is 18 meters. A rat which had gone out of its dwelling, at the foot of the pole, to a distance of 81 meters, while returning to its dwelling, is killed by the cruel hawk on the way. Say how far has it gone towards the hole and also the horizontal motion of the hawk (the speeds of the rat and the hawk being the same).

One more:

There is a reservoir of water of dimensions 6 X 12. At the north-east of corner of the reservoir there is a fish; and at the north-west corner there a crane. For the fear of the crane the fish, crossing the reservoir, hurriedly went towards the south in an oblique direction but was killed by the crane who came along the sides of the reservoir. Give out the distances travelled by them (assuming their speed is same)

Disclaimer: I am not a mathematician but i found some mathematical problems interesting enough to share here.

Is there a formula to divide X hours into Y tasks (total X hours = sum(y1, y2...yn)) based on Z resources? Each Z resource have different rate of work like z1-100% of his work hours, z2-50%, etc..?

Hungarian Scientist Receives Top Mathematics Prize
Endre Szemeredi, is this year’s laureate of the Abel Prize for mathematics, an award often compared to the Nobel Prize.

His guru Paul Erdos is one of my hero.

Budapest /Hungary remains one of the top place were best mathematics students all over the word try to spend some time.

Amber G. ji

Hope you have read "The man who loved only numbers". If not, please do. It is a bio of Erdos.

Regards

might be trivial for folks here, but stumbled upon a coarse approximation to pythogoras theorem. c^2 = a^2 + b^2 ~ max(a,b) + min(a,b)/3. works better for slivers than it does for a=b, but still a good rough approximation.

(Assuming the above meant c ~ max(a,b) + min(a,b)/3)
Above is not that helpful or accurate unless b is zero (which is uninteresting) or the sides are (3,4,5) (or have the same ratio)

When a=b, the actual value is about 1.41a and approximation 1.33 a... besides sqrt(a^2+b^2) is simple enough...
except for the special case (of ratio 3:4:5 triangle), for thin slivers (that is if one side is relatively much smaller than the other).. just using, for example, max(a,b) alone will often give better approx than including the second( min(a,b)/3)) term.
(for example a=40, b=9 value of c (=41) is closer to 40 than 43)

Thanks. Agree. The book was a best seller and fun to read (even non-mathematicians will like it)
If any one is interested, here is an online copy. http://www.jcu.edu/math/Faculty/Shick/erdos.pdf
(It is easy to buy too, and has been translated in many languages)

Erdos has written more papers than anyone else in mathematics. He is well known and was loved by many students because he regularly posed extremely interesting problems for his (world wide) students. He (his institution) will have award/prize for such problems. More often than not, the institution did not have to dole out those prizes because winners never cashed the checks. (Erdos himself signed the checks, and students will just frame the signed checks.. A signed check was worth much more than the prize...)

(Not surprisingly some of the problems I have posted in the past, have come from ( (or inspired by) him )

Here is one of his very famous problem. (It is very easy to understand even for an elementary school student)

- Start with any number (say n) .
- If the number is even, take the half of the number (n/2)
- If the number is odd, multiply by three and add 1. ( 3n+1)
Now repeat this with the new number you got by above.
If you repeat enough times, you will always get a 1. ( The problem was to prove this, of find a number where you will never get a 1)

Example, start with 6
Next - (6 is even so we get) 3
Next - (3 is odd so we get 3*3+1) 10
Next - (10 is even so we get) 5
Next - (5 is odd so 3n+1 gives) 16
Next - 8
Then you get 4,2,1
You reach 1.
(For example if you started with 7 the sequence would have been (7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1 ....)
****

Try it with different numbers.. (trivial to write a computer program)
Here is a challenge for BRFites... see who can find a starting number which has the longest (among BRFitest) chain before the sequence reaches 1.

Really stupid stuff.