BR Maths Corner1
Re: BR Maths Corner1
International Math Olympiad is taking place in Bath UK.
First paper is tomorrow (July 16th 2019) and the second paper is on July 17, 2019.
Good luck to all participants specially to team USA and team India.
Here is the Team India:
Soumil Aggarwal (Contestant 1) (17 years old)
Bhavya Agrawalla (Contestant 2) (17 years old)
Anubhab Ghosal (Contestant 3) (17 years old)
Ojas Mittal (Contestant 4) (15 years old)
Ritam Nag (Contestant 5) (16 years old)
Pranjal Srivastava (Contestant 6) (15 years old)
C. R. Pranesachar (Leader)
The Team USA:
Vincent Huang (Contestant 1) 18 years old
Luke RobitailleLuke Robitaille (Contestant 2) 15 years old
Colin Shanmo TangColin Shanmo Tang (Contestant 3) 17 years old
Edward WanEdward Wan (Contestant 4) 16 years old
Brandon WangBrandon Wang (Contestant 5) 17 years old
Daniel ZhuDaniel Zhu (Contestant 6) 17 years old
PoShen LohPoShen Loh (Leader)
First paper is tomorrow (July 16th 2019) and the second paper is on July 17, 2019.
Good luck to all participants specially to team USA and team India.
Here is the Team India:
Soumil Aggarwal (Contestant 1) (17 years old)
Bhavya Agrawalla (Contestant 2) (17 years old)
Anubhab Ghosal (Contestant 3) (17 years old)
Ojas Mittal (Contestant 4) (15 years old)
Ritam Nag (Contestant 5) (16 years old)
Pranjal Srivastava (Contestant 6) (15 years old)
C. R. Pranesachar (Leader)
The Team USA:
Vincent Huang (Contestant 1) 18 years old
Luke RobitailleLuke Robitaille (Contestant 2) 15 years old
Colin Shanmo TangColin Shanmo Tang (Contestant 3) 17 years old
Edward WanEdward Wan (Contestant 4) 16 years old
Brandon WangBrandon Wang (Contestant 5) 17 years old
Daniel ZhuDaniel Zhu (Contestant 6) 17 years old
PoShen LohPoShen Loh (Leader)
Re: BR Maths Corner1
Ramanujan machine automatically generates conjectures for fundamental constants
https://phys.org/news/201907ramanujan ... ental.html
https://phys.org/news/201907ramanujan ... ental.html
Re: BR Maths Corner1
Here is an easy problem 
Can you find at least one odd factor of (2019^8 +1) (Without using a calculator/computer).
Can you find at least one odd factor of (2019^8 +1) (Without using a calculator/computer).
Re: BR Maths Corner1
Here is problem # 6 of IMO 2019. (The exam has just finished a few hours ago  so it is not on official website but has been shared). Nice problem in geometry.
Let I be the incenter of acute triangle ABC with AB is not equal to AC. The incircle w of ABC is tangent to sides BC, CA, and AB at D, E, and F, respectively. The line through D perpendicular to EF meets w again at R. Line AR meets w again at P. The circumcircles of triangles PCE and PBF meet again at Q. Prove that lines DI and PQ meet on the line through A perpendicular to AI
The problem was proposed by Anant Mudgal, India.
Let I be the incenter of acute triangle ABC with AB is not equal to AC. The incircle w of ABC is tangent to sides BC, CA, and AB at D, E, and F, respectively. The line through D perpendicular to EF meets w again at R. Line AR meets w again at P. The circumcircles of triangles PCE and PBF meet again at Q. Prove that lines DI and PQ meet on the line through A perpendicular to AI
The problem was proposed by Anant Mudgal, India.
Re: BR Maths Corner1
^^^ From what I heard .. No one (except one partial) from India solved the above problem. .
(I think it is a neat problem proposed by India)..
(OTOH from partial scores  the person who got partial (IND  6) is doing quite good had has near perfect score on first 3 problems).. that's good ( This means, I think India has at least one gold).
*** (Scores are partial  not official and not all problems are graded .. Both US and India are at around 20th for total scoring but this will change
Results will be out on July 20, 2019 But it looks like First Gold for India since about 9 years is by a 9th grader 15 year old. (Pranjal Srivastava) . Very Impressive and congratulations to him.
Final Jury meeting in 9PM on July 20 ( See here: https://www.imo2019.uk/)
(I think it is a neat problem proposed by India)..
(OTOH from partial scores  the person who got partial (IND  6) is doing quite good had has near perfect score on first 3 problems).. that's good ( This means, I think India has at least one gold).
*** (Scores are partial  not official and not all problems are graded .. Both US and India are at around 20th for total scoring but this will change
Results will be out on July 20, 2019 But it looks like First Gold for India since about 9 years is by a 9th grader 15 year old. (Pranjal Srivastava) . Very Impressive and congratulations to him.
Final Jury meeting in 9PM on July 20 ( See here: https://www.imo2019.uk/)
Re: BR Maths Corner1
To me a strange/ironic thing is the last problem (which I posted here) was proposed by an Indian could be solved by none of the Indians (Except Pranjal who got 1 all else got 0)..
( Nice problem by Anant but may be he should also teach Indian team too .. )
( Nice problem by Anant but may be he should also teach Indian team too .. )
Last edited by Amber G. on 20 Jul 2019 05:12, edited 1 time in total.
Re: BR Maths Corner1
FYI: Pranjal got Silver last year, and at that time I was impressed with him. He is getting gold this time! (It is not official  we may have to wait for as much a day ..  but I am quite sure, seeing the partial results). He has two more years to represent India.
I posted this last year:
I posted this last year:
Amber G. wrote:^^^ For the record ( Name, score (out of 42) and Medal) (*** This is year 2018 **)
IND1 Sutanay Bhattacharya 21 Bronze
IND2 Spandan Ghosh 21 Bronze
IND3 Amit Kumar Malik 10 HM
IND4 Anant Mudgal} 26 Silver
IND5 Pulkit Sinha 26 Silver
IND6 Pranjal Srivastava 28 Silver
(I am specially impressed by Pranjal  He is youngest  had 4 perfect scores (Prob 3 and Prob 6 were missed)  and method used were nice).. I am sure he is going to do very well in the next olympiad and math in general.
***
Prob 3, (which I posted above) turned out to be rather difficult. None of Indian team solved it. Only 2 from USA did it.
I was kind of surprised as I posted above, when I checked, the problem has indeed came in Scientific American. (This is probably 3040 years late, but I knew how to solve it then ..)
Problem is posted above. (With the hint that solution is in SA one can do internet search)..
(Hint: Answer, is one can not find a solution for N>5 (so not solution for 2018))
One solution posted in SA came from 4th grader (for N=6)indeed quite simple.
Last edited by Amber G. on 20 Jul 2019 05:54, edited 1 time in total.
Re: BR Maths Corner1
Results of the International Math Olympiads are posted.
Top Teams: 12 USA / China
3  Korea
15 India.
USA (and China too) got 6 gold.
India got 1 Gold, 4 Silver, 1 HM.
Individual scores (Out of 42) (Cutoff for Gold was 31 points and Silver was 24)
Pranjal Srivastava 35 (Rel Rank = 97.26%) Gold medal
Ritam Nag .. 27 83.87% Silver medal
Anubhab Ghosal 27 83.87% Silver medal
Bhavya Agrawalla 27 83.87% Silver medal
Ojas Mittal 24 76.94% Silver medal
Soumil Aggarwal 16 51.29% Honourable mention. (Missed Bronze by 1 point)
Specially impressed by Pranjal  except for P6 he got near perfect score.
USA had 2 perfect scores and 6 Golds.
(China too had 2 perfect scores and 6 Golds and tied with USA)
Link: http://imoofficial.org/team_r.aspx?code=IND&year=2019
Top Teams: 12 USA / China
3  Korea
15 India.
USA (and China too) got 6 gold.
India got 1 Gold, 4 Silver, 1 HM.
Individual scores (Out of 42) (Cutoff for Gold was 31 points and Silver was 24)
Pranjal Srivastava 35 (Rel Rank = 97.26%) Gold medal
Ritam Nag .. 27 83.87% Silver medal
Anubhab Ghosal 27 83.87% Silver medal
Bhavya Agrawalla 27 83.87% Silver medal
Ojas Mittal 24 76.94% Silver medal
Soumil Aggarwal 16 51.29% Honourable mention. (Missed Bronze by 1 point)
Specially impressed by Pranjal  except for P6 he got near perfect score.
USA had 2 perfect scores and 6 Golds.
(China too had 2 perfect scores and 6 Golds and tied with USA)
Link: http://imoofficial.org/team_r.aspx?code=IND&year=2019

 BRFite
 Posts: 119
 Joined: 22 Mar 2017 06:19
Re: BR Maths Corner1
Amber G. wrote:Here is an easy problem 
Can you find at least one odd factor of (2019^8 +1) (Without using a calculator/computer).
I managed it with a hit and trial approach:
My first thought was to try and cast it in a form similar to Fermat's little theorem:
2019^8 = 1 (mod a)
and we are looking for a. To get a 1 on the RHS, I square:
2019^16 = 1 (mod a)
this equation may admit more solutions than the original one, but with that in mind, I thought at this point I had the solution: 17. But a quick check using modular arithmetic reveals this to be false. Still, we can try higher powers of 16 (or even powers of 8 from the original equation) as they would all produce 1 on the RHS. We need
2019^(a1) = 1 (mod a)
where a is prime, and (a1) = 0 (mod 16).
checking : 16*2 = 32 = 331 (not prime), 16*3 = 491 (not prime), 16*4 = 651 (not prime), 16*5 = 811 (not prime), 16*6 = 971 (prime)
So the next candidate is 97, and indeed, (2019^8)%97 = (79^8)%97 = (6241^4)%97 = (33^4)%97 = (1089^2)%97 = (22^2)%97 = 484%97 = 96 = 1 (mod 97)
is a solution.

 BRFite
 Posts: 119
 Joined: 22 Mar 2017 06:19
Re: BR Maths Corner1
Just looking at IMO results data, did a quick average rank calculation for the last 10 years (20102019)
Results are very interesting in that they seems to correlate neither to country size nor per capita income. East Asian countries dominate.
CHN 1.70
USA 2.10
KOR 4.30
RUS 5.20
PRK 7.43
SGP 8.30
THA 9.00
TWN 11.20
VNM 11.40
JPN 11.60
IRN 13.9
CAN 14.5
UKR 14.8
UNK 16.4
ROU 17.2
AUS 18.7
HUN 19.5
TUR 19.7
SRB 21.6
ITA 22
POL 22.8
GER 23.6
BGR 24
IDN 24.6
KAZ 25.2
PER 25.4
HKG 25.9
BRA 26.7
ISR 27.3
MEX 29.1
HRV 29.6
FRA 30.4
IND 30.4
Results are very interesting in that they seems to correlate neither to country size nor per capita income. East Asian countries dominate.
Re: BR Maths Corner1
Rahulsidhu wrote:Amber G. wrote:Here is an easy problem 
Can you find at least one odd factor of (2019^8 +1) (Without using a calculator/computer).
I managed it with a hit and trial approach:
My first thought was to try and cast it in a form similar to Fermat's little theorem:
2019^8 = 1 (mod a)
and we are looking for a. To get a 1 on the RHS, I square:
2019^16 = 1 (mod a)
this equation may admit more solutions than the original one, but with that in mind, I thought at this point I had the solution: 17. But a quick check using modular arithmetic reveals this to be false. Still, we can try higher powers of 16 (or even powers of 8 from the original equation) as they would all produce 1 on the RHS. We need
2019^(a1) = 1 (mod a)
where a is prime, and (a1) = 0 (mod 16).
checking : 16*2 = 32 = 331 (not prime), 16*3 = 491 (not prime), 16*4 = 651 (not prime), 16*5 = 811 (not prime), 16*6 = 971 (prime)
So the next candidate is 97, and indeed, (2019^8)%97 = (79^8)%97 = (6241^4)%97 = (33^4)%97 = (1089^2)%97 = (22^2)%97 = 484%97 = 96 = 1 (mod 97)
is a solution.
Nice!
***
From above  all prime factors for (any_number)^8+1 will be of the form 16k+1.
Historically this is how Euler found the factor 641 of (2^32+1) because he knew that all factors have to be of the type (64k+1) .. and ignoring nonprimes like 65, 129 etc, found 641.
(See here: http://eulerarchive.maa.org/hedi/HEDI200703.pdf
(Note  One can improve this method in the case of 2 and one can prove all factors of (2^(2^n)+1) have to be of the form (k*(2^(n+2))+1).
Re: BR Maths Corner1
A very entertaining lecture by Terence Tao
2007 lecture where Prof. Terence Tao starts from the definition of Primes and goes all the way to outlining the proof of Green_tao Theorem. I recommend it highly. It is just about 50 minutes long (including Q&A).
I suggest that you watch it on your desktop with a Wikipedia page on another tab. That way you can look up terms and definitions which you haven't come across previously.
2007 lecture where Prof. Terence Tao starts from the definition of Primes and goes all the way to outlining the proof of Green_tao Theorem. I recommend it highly. It is just about 50 minutes long (including Q&A).
I suggest that you watch it on your desktop with a Wikipedia page on another tab. That way you can look up terms and definitions which you haven't come across previously.
Re: BR Maths Corner1
^^^ It is indeed very good. Thanks for posting it.
***
Terry Tao  I have *many* posts about him in this math dhaga  is one of the greatest living mathematician. He is a great teacher, and his blog (*many links about interesting topics are here in this math dhaga ) . He is still the youngest ( he was 9 or 10 the first time he represented his country in International Math Olympics) IMO gold medal(s) winner.
***
Terry Tao  I have *many* posts about him in this math dhaga  is one of the greatest living mathematician. He is a great teacher, and his blog (*many links about interesting topics are here in this math dhaga ) . He is still the youngest ( he was 9 or 10 the first time he represented his country in International Math Olympics) IMO gold medal(s) winner.
Re: BR Maths Corner1
Amber G. wrote:^^^ It is indeed very good. Thanks for posting it.
You are welcome ma'am/sir.
Prof. Terence Tao started taking collegelevel courses at the age of 9. He first participated in IMO when he was 10. He had a bronze first year, silver next year, and then he finished with a gold when he was 13. He got a Masters by 16 and a PhD from Princeton by the "ripe old" age of 20. He joined UCLA and got promoted to a full professor by the age of 24. Extraordinary brain.
The other I like is Prof. Eric Demaine, EECS at MIT; originally from Canada.
Terence Tao's parents are from Hong Kong who moved to Australia. Tao himself is a SinoAustralian and now working in the US.
Goes to show that academia is sans state boundaries.
Re: BR Maths Corner1
Mathematician Wins $3 Million Breakthrough Prize for 'Magic Wand Theorem'
https://www.livescience.com/breakthroug ... nners.html
https://www.livescience.com/breakthroug ... nners.html
Re: BR Maths Corner1
The following post was a few months old, we now have an answer raised in that post!
Okay we now have an answer for 42!
Ramanujan would be proud.
Wow! Thanks to MIT's mathematician Andrew Sutherland, and a little help from worldwide computer network called Charity Engine. We now know:
(80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42
So all numbers less than 100, not of the form 9k+4, can be written as sum of three cubes.
It sure feel Glorious and Overwhelming to find the answer for 42. And thankfully unlike in Adams' search for the truth, the entire Earth wasn't destroyed in the process of finding the answer!
Amber G. wrote:In related matter  Interestingly, this is from recent math news  Ramanujan will be proud..
It was thought (but not proven one way or other) that was impossible to write 33 as sum of 3 cubes it was found that
Wow! 33 = 8866128975287528 ^ 3  8778405442862239 ^ 3  2736111468807040 ^ 3
Some theory: It was known that 
All numbers of the form 9k+4 or 9k+5 can not be written as sum of three cubes. (we can prove this)
for rest of the numbers the answer is not known.
Some numbers are easy. For example 29 = 3^3+1^3+1^3
or 1 = 1^3+0^3+0^3
also 1 = 10^3+9^31^3 (Ramanujan!)
Some numbers are not that easy.. and it required a computer to check many answers..
For example 29 is easy but 30 is hard.
***
Till recently only 33 and 42 were two numbers less than 100 which people did not know the answer..
For example see this youtube video; The Uncracked Problem with 33
Okay we now have an answer for 42!
Ramanujan would be proud.
Wow! Thanks to MIT's mathematician Andrew Sutherland, and a little help from worldwide computer network called Charity Engine. We now know:
(80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42
So all numbers less than 100, not of the form 9k+4, can be written as sum of three cubes.
It sure feel Glorious and Overwhelming to find the answer for 42. And thankfully unlike in Adams' search for the truth, the entire Earth wasn't destroyed in the process of finding the answer!
Re: BR Maths Corner1
SaiK wrote:Mathematician Wins $3 Million Breakthrough Prize for 'Magic Wand Theorem'
https://www.livescience.com/breakthroug ... nners.html
The mathematician honored (along with the winner of the prize) is none other than Maryam Mirzakhani.. The only woman (and Iranian) who have won a Fields Medal in Math.
I wrote about this in brf about a few times .. last time about his death which was mourned all over the world.
https://forums.bharatrakshak.com/viewtopic.php?p=2185948#p2185948
Amber G. wrote:From an Indian Newspaper Iran media mourns death of mathematician Maryam Mirzakhani, run front page pictures without hijab
(I don't know now but in 70's quite a few bright Iranian students used to come to India (IIT's etc). Iran always did good in International Math competitions  unlike most of the muslim countries)
Re: BR Maths Corner1
Just for fun, Here is a problem from last IMO, which is sort of not too difficult. Let 'f' be a function which applies on integers (and result is an integer) which has following property:
f(2a)+2f(b) = f(f(a+b))
Find all such functions which has this property.
f(2a)+2f(b) = f(f(a+b))
Find all such functions which has this property.
Re: BR Maths Corner1
Another fun  not too difficult if you hit the right idea  problem .. x, y are real numbers, can you find the values:
1/x + 1/2y = (x^2+3y^2)(3x^2+y^2)
1/x  1/2y = 2 (y^4x^4)
1/x + 1/2y = (x^2+3y^2)(3x^2+y^2)
1/x  1/2y = 2 (y^4x^4)
Re: BR Maths Corner1
Spectra in mathematics and in physics
I found this typed manuscript (11 pages long) to be a great read on how the mathematical concept of a spectrum got intertwined with the physical reality of the spectrum of EM waves as well as stability (or lack there off) of celestial mechanics. The last two pages touch on Quantum Mechanics and eigenpairs.
I found this typed manuscript (11 pages long) to be a great read on how the mathematical concept of a spectrum got intertwined with the physical reality of the spectrum of EM waves as well as stability (or lack there off) of celestial mechanics. The last two pages touch on Quantum Mechanics and eigenpairs.
Re: BR Maths Corner1
For people interested in modern physics, this from renowned Prof Sachdev is VERY nice.
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