Two from Africa!
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3 жыл бұрынWe solve two questions from the 2004 Pan-African Math Olympiad.
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@onealjerry 3 жыл бұрын
For the first problem, there's a faster and simpler way to get the items under the square roots: it's to determine each as the square of something, using the "factoring" process we learned in Algebra I/II. So we want to find x and y such that 4 - 2*sqrt(3) = (x + y)^2 = x^2 + 2xy + y^2. In other words, we want x^2 + y^2 = 4 and 2xy = -2*sqrt(3) => xy = -sqrt(3). It's obvious that x = sqrt(3) and y = -1 works, so we get 4 - 2*sqrt(3) = 3 - 2*sqrt(3) + 1 = sqrt(3)^2 + 2*(-1)*sqrt(3) + (-1)^2 = [sqrt(3) - 1]^2. Thus sqrt[4 - 2*sqrt(3)] = sqrt(3) - 1. (Note: x = -sqrt(3) and y = 1 also works, but we want x+y>0 since we seek the positive square root at the end.) The second one is trickier, but in this case we find that x = 7 and y = -4*sqrt(3) works, so 97 - 56*sqrt(3) = [7 - 4*sqrt(3)]^2 => sqrt[97 - 56*sqrt(3)] = 7 - 4*sqrt(3). (Again, choose the positive root). Putting it all together, we find that 4*sqrt[4 - 2*sqrt(3)] + sqrt[97 - 56*sqrt(3)] = 4*[sqrt(3) - 1] + 7 - 4*sqrt(3) = 3.
@GDPlainA 3 жыл бұрын
Teacher: what is 1+2? Students: 4(sqrt(4-2sqrt(3)))+sqrt(97-56sqrt(3))
@freepimaths9698 3 жыл бұрын
what's funniest about that answer is it has two 3's within the expression, meaning the student clearly knew what 3 was in the first place.
@JordHaj Жыл бұрын
@@freepimaths9698 then define 3 as an infinite nested square root
@freepimaths9698 Жыл бұрын
@@JordHaj What? How was that related to my comment? Also, because you asked anyways... 3 = sqrt((1 + 2)sqrt((1 + 3)sqrt(1 + 4)sqrt(... The proof is left as an excersise to the reader.
@JordHaj Жыл бұрын
@@freepimaths9698 I was (jokingly) suggesting to replace both 3s in 4(sqrt(4-2sqrt(3)))+sqrt(97-56sqrt(3)) with the sum of roots itself, as it equals to 3 kind of back substitute into itself; it will look horrible but now we can define 3 as an infinite square root without the use of "3" Yeah, looking back my comment was pretty out of the blue and extremely vague so sorry for that
@freepimaths9698 Жыл бұрын
@@JordHaj Oh, no actually, looking back I see I clearly just misinterpreted when I shouldn't have 😅 the sorry is from my direction
@gorillajock 2 жыл бұрын
Your reasoning in that second question especially is amazing to me. I couldn't imagine even how to start answering it. Thanks!
@mach2570 3 жыл бұрын
The second problem is a lot of fun if you don't use the number theory trick, takes some time, but definitely worth it.
@andreamarino95 3 жыл бұрын
Love your solution. When dealing with a+b sqrt(3) numbers, i like to do it in a number theory fashion: Look for n, m such that (n-m sq(3)) ^2 = c-d sq(3) Which implies n^2 +3m^2 = c 2nm = d For c=4, d=1, m is forced to be 1 by first equation, and by second equation we get n=1. For c=97, d= 56 is a bit trickier. Looking at the first equation modulo 4, one sees that a must be odd and b even. Since ab=28, we have that b=4 and a=7. Then conclude by taking the right sign so that squared numbers are positive :)
@RexxSchneider Жыл бұрын
When trying to "decompose" (as Michael Penn puts it) something like √(x + y√p), it's a fact that the result can only be of the form u + v√p when p is a square-free product of primes. For the first part, we're looking for something like a + b√3, which when squared makes 4 - 2√3. So a^2 + 3b^2 + 2ab√3 = 4 - 2√3. This shows that ab = -1 and a^2 + 3b^2 = 4. That pretty obviously has a solution a=-1, b=1, which gives √(4 - 2√3) = √3 - 1. The other solution (a=1, b=-1) would give 1 - √3 which is negative, not respecting the convention that √ implies a positive value. So 4√(4 - 2√3) = 4√3 - 4. To simplify √(97 - 56√3), we're again comparing it with √(a^2 + 3b^2 + 2ab√3), but because we're expecting an integer, we should look for a simplification that looks like a - 4√3, so that the irrational part cancels that of the previous simplification, i.e. b=-4. So we would have a^2 + 48 = 97 and -8a = -56, leading us to a=7 in both cases, showing that √(97 - 56√3) = 7 - 4√3. Finally we have 4√(4 - 2√3) + √(97 - 56√3) = (4√3 - 4) + (7 - 4√3) = 3.
@andreivila7607 3 жыл бұрын
I know that the first problem is extremely easy, but I would like to point out a trick that can be used even in composed radicals that do not factor nicely. That is if I have to calculate sqrt(a+-sqrt(b)) I can simply write it as sqrt((a+c)/2) +- sqrt((a-c)/2) where c=sqrt(a^2 - b). This helps you out as I said when you can’t find any easy factorization and gets you rid of the composed radical. I hope this helps you in your problem solving!! P.S. by ,,+-“ I mean plus-minus, as in either plus, either minus...
@evanev7 3 жыл бұрын
Oh cool. Thanks Stalin!
@shatishankaryadav8428 3 жыл бұрын
First is easy make combination of 2ab then it become (a-b)^2
@andreivila7607 3 жыл бұрын
@@shatishankaryadav8428 I know what you are talking about, but some aren’t so good at that. That is the reason I left this trick here.
@benextinction__144 3 жыл бұрын
@@madhukushwaha4578 the first video that appears is literally why odd x odd is even
@V-for-Vendetta01 3 жыл бұрын
@@evanev7 lmfao
@caiotmz 3 жыл бұрын
I swear the I read 'Toto from Africa' in the title
@xevira 3 жыл бұрын
Dammit, now I got that song stuck in my head again... again. :P
@doctorb9264 3 жыл бұрын
Love playing that tune on bass !
@DanielWalvin 3 жыл бұрын
I bless the math!
@jamirimaj6880 3 жыл бұрын
Legendary song!
@ellenmarch3095 2 жыл бұрын
I think everybody did, lol.
@kozokosa9289 3 жыл бұрын
First question is just substitute squares into the roots.... no complications required. 49 +48 is 97, 7×4×2 is 56, 1+3 is 4 2×1 is 2.
@gameguy8101 3 жыл бұрын
It really boils down to be solving a two variable system of equations. You assume you can write 4-2sqrt(3) = (a+bsqrt(3))^2 = a^2 + 3b^2 + 2absqrt(3) and then align coefficients so that a^2+3b^2 = 4 and 2ab = -2. Since a+3b^2 is an integer, it is obvious that a and b both have magnitude 1 and opposite signs. Since 1-sqrt(3) is negative its clearly -1 + sqrt(3) then we do the same thing with the next root expression, and 97-56sqrt(3) = (a+bsqrt(3))^2 a^2+ b^2 + 2absqrt(3) since we know this new b has to be equal the other coefficient of sqrt(3) (in order to non integer terms to cancel in the end), b = 4 (since our first term is -4 + 4sqrt3) then its a simple calculation to solve for a = - 7 Then you know an integer solutions exists and you can check it yourself. So yes, substituting into squares works and by using the necessary details of the puzzle it is made straightforward to do by hand
@GreenMeansGOF 3 жыл бұрын
I agree with gameguy
@user-mo5rq4ct4k 3 жыл бұрын
@@gameguy8101 hey gameguy, are u genius?
@gfest1119 3 жыл бұрын
Yeah man, i watched it very fast, and dont understand how the author hasnt watched it
@pshr2447 3 жыл бұрын
@@gameguy8101 love you game guy
@mathwithjanine 3 жыл бұрын
Awesome video! :) Enjoyed these fun math problems!
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@NavyBlueMan 3 жыл бұрын
for solving a, you can write it as sqrt(1 - 2 sqrt(3) + 3) which is (1 - sqrt( 3 ) ) ^ 2) or (sqrt( 3 ) - 1) ^ 2 )
@DavidSavinainen 3 жыл бұрын
Similarly for solving b, you can write it as (49-56sqrt(3)+48) which is the same as (7-4sqrt(3))^2 or (4sqrt(3)-7)^2
@zanti4132 3 жыл бұрын
Expanding on your approach, there is a quick check to see if √(a ± b√c) can be reduced to the format √c ± √d, which can save you the trouble of trying to do something that won't work: Check if a² - b²c is a perfect square. This is true for both √(4 - 2√3) and √(97 - 56√3): 4² - 2² * 3 = 4 = 2² 97² - 56² * 3 = 1 = 1²
@V-for-Vendetta01 3 жыл бұрын
many have already pointed this out, but the first problem can easily be solved by converting the terms into (a+b)^2 form
@chhabisarkar9057 3 жыл бұрын
I did like that lmao and took me almost 4 steps
@tuffleader2448 3 жыл бұрын
Instead of the guess-and-check method for factoring in Q2, we can complete the square and then use the difference of squares to factor.
@shadowsky 3 жыл бұрын
I know a way to solve a=√4-2√3 it's a quadratic formula where 4 is a^2+b^2 and 2ab is 2√3, if you solve it you will get a=√(√3-1)^2 meaning that a is √3-1
@madhukushwaha4578 3 жыл бұрын
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@songguo2243 3 жыл бұрын
First question is straight forward: 4-2*sqrt(3)=sqr( sqrt(3) -1) and 97-56*sqrt(3)= sqr(7-4*sqrt(3)) hence 4*sqrt(4-2*sqrt(3)) + sqrt(97-56*sqrt(3)) = 4*( sqrt(3) -1) + 7-4*sqrt(3) =3
@madhukushwaha4578 3 жыл бұрын
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@siulibasak3804 3 жыл бұрын
First problem was an easy one but got to know a new NT fact from the second one.... You can pls solve few more questions from Indian National Mathematical Olympiad....
@jeanf6295 2 жыл бұрын
To reduce expressions of the form sqrt(a-sqrt(b)) you can use the following trick : if a²-b is a square then P(±) = [sqrt(a-sqrt(b)) ± sqrt(a+sqrt(b))]² are integers and thus sqrt(a-sqrt(b)) = [sqrt(P(+))-sqrt(P(-))]/2
@Downloader77 3 жыл бұрын
You did it your way and I like that.
@Chess-ks8lk 3 жыл бұрын
You're great mathematician!
@tomatrix7525 3 жыл бұрын
Great video, as usual.
@tomatrix7525 3 жыл бұрын
@@madhukushwaha4578 alright , I’ll def. have a look. Michael Penn on here also has more challenging problems, like some math olympiad ones can be a bit tough or usually I like the putnams. They’re the real deal
@edusoto91 3 жыл бұрын
Here is an analytic approach for the first one. 1. prove that 2 < a < 3 and 0 < b < 1 to obtain the candidate a + b = 3. 2. Prove that a+b = 3 by cleaning sqrt's carefully.
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@franckquessette9198 3 жыл бұрын
For all a, b positive, sqrt( (a+b) - 2sqrt(ab) ) = sqrt(a) - sqrt(b), just because in x^2 - 2xy +y^2 = (x-y)^2 set a = sqrt(x) and b = sqrt(y) and take sqrt of both parts.
@ireallydontknow3299 3 жыл бұрын
Here's two ways to deal with nested radicals. Let's say we want to find a different way to write sqrt(a +- b*sqrt(c)). there most definitely are x and y such that sqrt(a +- b*sqrt(c)) = sqrt(x) +- sqrt(y). But we want to find a _nice_ pair of x and y, preferably with integers or rationals. Well, let's work with that expression a bit... a +- b*sqrt(c) = x + y +- 2sqrt(xy) If we set x + y = a and 2sqrt(xy) = b*sqrt(c), and find x and y, we'll find the "nice" expression if there is one. This is sort of infallible and doesn't rely on guessing; as long as there's a nice way to express it you'll probably find it. Another faster way but that relies on a bit of guessing would be to expect that there is some way to write sqrt(a +- b * sqrt(c)) as something like sqrt((x +- y*sqrt(c))^2), so you can cancel out the square root. In which case it's usually easy enough to guess what x and y should be. For example with 4 - 2sqrt(3), since 2sqrt(3) = 2 * 1 * sqrt(3), a good guess would be x = sqrt(3) and y = 1, which does work. This one might take a few tries, but I'd say it's pretty reliable and a lot faster: 97 - 56sqrt(3). Since 56sqrt(3) = 2 * 28 * sqrt(3) = 2 * 14 * 2sqrt(3) = 2 * 7 * 4sqrt(3) and some others, there's a few things to try. But it wouldn't take to look to see x = 7 and y = 4sqrt(3) works, because 7^2 + (4sqrt(3))^2 = 97. So 97 - 56sqrt(3) = (7 - 4sqrt(3))^2.
@malawigw 3 жыл бұрын
These problems were quite simple but fun!
@recklessroges 3 жыл бұрын
Those are both beautiful. Thanks.
@HagenvonEitzen 3 жыл бұрын
I would have tried to *assume* that a and b are of the form u + v*sqrt(3), where we can readily find u,v for the a case and then in the light of the desired result already know the v for b. -- Then one just needs to verify for the two cases that u+v sqrt(39 is positive and squares to the right expression
@pierineri 3 жыл бұрын
The first radicand is very clearly the square of √3-1 since 4=3+1, so the first term is *4√3-4* . Now recall: *They ask to check if the sum is an integer* This is only possible if the other term is *n-4√3* . That is, we want *97-56√3=( n-4√3)²* which immediately implies 56=8n and *n=7* , that actually produces the integer *3* .
@matthieumoussiegt 3 жыл бұрын
for the first solution we can find directly the value of the coefficient by looking at the coefficient of the internal square root. To ensure that a number √(a+b√n) is simplifiable, you need to have a factorisation of a+b√n in a square (a'+b'√n)^2 meaning that a=a'^2+n*b'^2 and b=a'b'/2 leading to a system of equations
@Pablo360able Жыл бұрын
Second question seemed straightforward to me, but that's because I've used all the logic involved to solve cryptic sudoku.
@spartacus8875 3 жыл бұрын
Thank you for the movies ...the best
@ankitbhattacharjee280 3 жыл бұрын
For the first time, I was able to solve one of the problems put forward in these videos. I was able to calculate the answer as 3 for Q2.
@rmschad5234 3 жыл бұрын
I denested the radical with 97 by a more brute force method. I ended up having to solve x^2-97x+3*28^2=0, but happened upon (x-48)(x-49) pretty quickly after noting 28=2*2*7.
@CaradhrasAiguo49 2 жыл бұрын
3:18 that can be re-grouped as a^4 - 8a^2 + 4 = a^4 - 4a^2 + 4 - 4a^2, which is a difference of 2 squares; (a^2 - 2)^2 - (2a)^2, which results in the same factorisation. No need for guesswork Same for 6:55: b^4 - 194a^2 + 1 = b^4 + 2b^2 + 1 - 196b^2 = (b^2 + 1)^2 - (14b)^2
@burhan8795 3 жыл бұрын
at 6:00 after you determined that 4a = -4+4sqrt(3) you stipulate that b has to be of the form x-4sqrt(3) {since 4a is a mixed radical and we add together correspending types when we sum these sorts of numbers } which is only possible if b is the square root of a square. so sq(x-4sqrt(3)) = 97-56sqrt(3) after which you can quickly obtain x to be 7. you save yourself from factorizing a quartic twice.
@kristianwichmann9996 3 жыл бұрын
#2 was a really nice little problem :-)
@orchestrain88keys 3 жыл бұрын
The first problem is fairly straightforward, but solved in a convoluted way. The second problem is interesting.
@jlhjlh 3 жыл бұрын
I want to buy some merch, because this channel is really awesome. Can someone explain the "Normally ordered platypus" shirts? I don't want to wear one without understanding it.
@eliaskar5084 3 жыл бұрын
The second problem should be phrased as "...so that the sum of ANY 20 consecutive numbers is 75". I was pretty sure there was no solution to the problem until I saw the beginning of the solve, only then did I realize what the question meant.
@CM63_France 3 жыл бұрын
Hi, For fun: 1 "So, if you want to go ahead and", 2 "let's may be go ahead and", 2 "let's go ahead and", 1 "great", 2 "ok, great", 1 "the next thing that we want to notice", 2 "and so on and so forth', 1 "all the way up to".
@VideoFusco 3 жыл бұрын
In the first problem you can observe that 4-2sqrt(3)=3-2sqrt(3)+1=(sqrt(3)-1)^2. And something like that even for b.
@hamdiel-sissi7760 2 жыл бұрын
Super complicated when the solution is trivial!!
@ericzgrey 3 жыл бұрын
You can just solve it using exhaustion rotating by 20s through all of the points you have until you get every point that isn’t a -1. Then simply find that 210 is -1
@himanshigoyal7471 3 жыл бұрын
I solved the nested radicals without using the quadratic formula. 4√(4-2√3) + √(97-56√3) 4-2√3 = 3+1-2√3 = (√3)²+1²-2(√3)(1) = (√3-1)² 97-56√3 = 48+49-56√3 = (4√3)²+7²-2(4√3)(7) = (7-4√3)² 4(√3-1)+ 7-4√3 = 4√3-4 + 7-4√3 = 3
@eminvahid3976 3 жыл бұрын
Very smart solutions. 🤓
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@cesaremarcolazzarini3289 3 жыл бұрын
I solved problem 2 thinking that 2*sqrt(3) could be the double product in (1 - sqrt(3))² [and realizing 4 = 1 + 3, sum of the squares], then developing 97 - 56*sqrt(3) as (7 - 4*sqrt(3))². I got 4(sqrt(3) - 1) + (7 - 4*sqrt(3)), that gives the integer you were looking for.
@laokratis55 3 жыл бұрын
I did the first one in a slightly different way, not using the bi-quadratic equation. Each term under the root can be expressed in the form (A+B sgrt 3)=(a+b sqrt 3)^2 . This leads to the equations a^2+3b^2=A and B=2ab. For the second term which is less obvious we get for all the possible splitting of 97 the only possibility that works 97=49+3*16 =7^2+3 * 4^2 and for -56=-2*4*7 which finally give (4 sqrt 3-7).
@laokratis55 3 жыл бұрын
correction: 7-4 sqrt 3
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@user-jc2lz6jb2e 3 жыл бұрын
First is easy if you assume straight away that both a and b are of the form x+y*sqrt(3). For a, you get (x^2+3y^2)+(2xy)sqrt(3), but the only solutions for x^2+3y^2 = 4 is either x^2 = 1 and y^2 = 1, OR x^2 = 4 and y^2 = 0. Second doesn't work because we need 2xy = -2, and this second equation tells us x and y have opposite signs, so a is sqrt(3)-1 or its negative, but we take the positive root which is sqrt(3)-1. The same assumption applis to b, but now also assume b's y*sqrt(3) will cancel out with that of y, so b = x-4sqrt(3), square that to get (x^2+48)+x8sqrt(3), and we get x = 7, so b = 7-4sqrt(3). Final answer is 3.
@NileshKumar-gu2ti 3 жыл бұрын
Can we use complex number for the second question root of unity concept?
@somanathdash8143 3 жыл бұрын
Loved the problem
@somanathdash8143 3 жыл бұрын
@@madhukushwaha4578 thanks : )
@anirudh67 3 жыл бұрын
Well for the first one we can just go on writing the numbers as a perfect square. Like for sqrt(4-2✓3) we look it as 1^2 +(✓3)^2 -2(1)(✓3) and so on
@alphalunamare 3 жыл бұрын
well that just blew my marbles :-)
@jeanlismonde8718 3 жыл бұрын
on peut trouver le résultat égal à 3 en remarquant que : 4.racinecarrée[4 - 2rac(3)] = 4.racinecarrée[(rac(3) - 1)²]=4.rac(3) - 4 d'autre part : racinecarrée[97 - 56.rac(3)] = racinecarrée[(7 - 4.rac(3))²} = 7 - 4.rac(3) en faisant la somme des deux expressions on trouve 7 - 4 = 3
@kelvinella 3 жыл бұрын
This is not an efficient way to do problem 2, just let (a+b*sqrt3)^2 = 97-56sqrt3 for some natural numbers a and b. (sqrt3-1)^2 = 4-2sqrt3 is obvious. Now, you have a^2+3b^2=97 and ab=28 and then by eyeballing, its not hard to find that 49 + 3*16 = 97 and hence 7-4sqrt3 = sqrt(97-56sqrt3)
@bndrcr82a08e349g 2 жыл бұрын
I do not expect negative value in the second problem
@varunmishra7556 3 жыл бұрын
Hey I have some problems in first ques where can I show u my solutions so as to get ans where I am wrong
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@jonathanbonicel1654 3 жыл бұрын
That time skip at 18:28 was PERFECT
@rafael7696 3 жыл бұрын
Beautiful problems
@shreyathebest100 3 жыл бұрын
Can anyone tell me how we factor such expressions
@jesusthroughmary 3 жыл бұрын
Can't we just say that the sum of any four consecutive numbers is 15, so 9 + 3 + 4 + x = 15, therefore x = -1?
@jesusthroughmary 3 жыл бұрын
@@angelmendez-rivera351 He went all the way out to 20, basically multiplying everything by 5 just to divide everything again by 5 at the end. It was redundant.
@surem8319 3 жыл бұрын
@@jesusthroughmary He also said (18:35) "and I should point out that we only really care about what's happening in the first four positions, but I am gonna /explicitly/ use the fact that the sum of any 20 consecutive numbers is 75, just so that everything is a bit more clear". So yeah, you could just do that. He did it the long way to make it more clear what is going on.
@jesusthroughmary 3 жыл бұрын
@@surem8319 I just figured that since we went through the trouble of proving that there are only four different numbers and that they appear in a 4-cycle, it would have been justifiable from there to reduce 75/20 to 15/4 and save time.
@RedRad1990 3 жыл бұрын
@@jesusthroughmary What *Sure m8* meant is that Michael said it himself. He basically did: "I'm going to do it the long way so that everyone understands"
@ProjetoEquilatero 3 жыл бұрын
Usa-se duas vezes radical duplo na primeira questão. Foi isso que ele fez, mas de forma diferente.
@shivansh668 3 жыл бұрын
When I saw the thumbnail,I thought that it's an Integer Partition in a irrational way !😅
@madhukushwaha4578 3 жыл бұрын
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@hybmnzz2658 3 жыл бұрын
@@madhukushwaha4578 😐
@wallaceferreiradasilva68 3 жыл бұрын
I wish I was able to fully understand problem 3 solution.
@michaelhall5801 3 жыл бұрын
Yay! Representation
@IvanisIvan 3 жыл бұрын
the second problem reminded me of a software problem
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@marcusdecarvalho1354 3 жыл бұрын
The 14-th Pan-African Mathematical Olympiad (Tunis, Tunisia, 2004 ) - First Day : Questions 2 and 3.
@Utesfan100 3 жыл бұрын
For A it seems easier to guess and check that (sqrt(3)-1)^2 = 4-2sqrt(3). Since sqrt(3) is irrational, 4A+B = n tells us B = n+4-4sqrt(3) = u - 4sqrt(3). B^2 = u^2 - 8u sqrt(3) + 48 = 97 - 56sqrt(3). Since sqrt(3) is irrational we have u^2+48 = 97 and 8u = 56. The second equation has the single solution of 7, which works in the other. Thus B = 7-4sqrt(3). Now 4A+B = 4(-1+sqrt(3)) + (7-4sqrt(3)) = 3.
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@damascus21 3 жыл бұрын
Thanks for showing us something from Africa. This helps contradict stereotypes about the continent
@aviralsood8141 3 жыл бұрын
Yeah I always thought Africa was too poor to put 268 numbers around a table but apparently they can do it. My eyes have been opened.
@garymclaughlin9559 3 жыл бұрын
It in no way contradicts stereotypes about Africa - none of which deal with mathematics. All the stereotypes are accurate.
@johnmccrae2932 3 жыл бұрын
@@garymclaughlin9559 What?
@hybmnzz2658 3 жыл бұрын
I know you are trying to say nice things but this comment is so stupidly political. If I didn't know better I could even turn this around and say "oh wow! African math olympiads have such EASY problems... must be an uneducated place! The african stereotypes are real". Like how does showcasing math problems from a country say anything about stereotypes about said country? Your comment seems to be supporting Africa but it looks more like pity for no reason. By the way, Africa is a huge continent and thus has huge educational disparities. The learning crisis is real and there is a shortage of teachers in the world compared to students. A math olympiad is not changing this reality.
@dandan-gf4jk 3 жыл бұрын
@@angelmendez-rivera351 "Have you been to every town in Africa?" It doesn't help the cause if you show you don't have basic understanding of what it means for something to be a stereotype
@pfeilspitze 2 жыл бұрын
20:18 You distributed the 5, but didn't just divide both sides by 5. Much easier to go to x + 16 = 15 rather than going through 5x + 80.
@Namo_Amitabha2024 3 жыл бұрын
These 2 q's are remarkably easier than the others...
@alihaydar728 3 жыл бұрын
I used : sqrt [ a - sqrt ( b ) ] = sqrt [ a + sqrt ( a^2 - b ) ] / sqrt ( 2 ) - sqrt [ a - sqrt ( a^2 - b ) ] / sqrt ( 2 )
@alihaydar728 3 жыл бұрын
or sqrt [ a - sqrt ( b ) ] = sqrt (x) - sqrt (y) such that a = x+y , b=4*x*y
@SachinSachin-gr2fu 3 жыл бұрын
Sir i didn't understood the 3rd line of answer no. 3
@victormd1100 3 жыл бұрын
2:00 thank you so much
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@skipugh 11 ай бұрын
Impressive
@vishalmishra3046 3 жыл бұрын
*Simpler* Assume 4 integers a, b, M, N, such that (a - b✓3)^2 = (a^2 + 3b^2) - ab(2✓3) = M - N✓3. Therefore, N = ab = (2/2=1=1x1=ab and 56/2=28=7x4=ab) and M = a^2 + 3b^2 = (4=1^2+3x1^2=a^2 + 3b^2 and 97 = 48 + 48 = 49 + 3 x 16 = 7^2 + 3 x 4^2 = a^2 + 3b^2). So the expressions (a - b✓3)^2 are (✓3-1)^2 and (7-4✓3)^2 found by integer factoring of N/2 in all such class of expressions containing N✓3 inside a radical. Finally, 4(✓3-1) + (7-4✓3) = 7 - 4 = 3, since 4✓3 cancels off. This approach can be further generalized to any prime square root. (a + b✓P)^2 = (a^2 + P x b^2) + 2(ab)(✓P) where P is any prime positive integer (3 in this case), turning this class of radical simplification problems into integer factorization problems.
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@legendarydisciplegaming4316 3 жыл бұрын
Is there a soln. For this equation
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@jindagibura7427 3 жыл бұрын
Say, if I don't square anything...but divide both sides by 2.... I get a/2=√1-sin2.30°=sin30°-cos30°=> a = 1-√3 directly is this approach correct?
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@akshatjangra4167 3 жыл бұрын
I will give another easy version Break 4 into 3 and 1 , it is just an identity now of a minus b whole squared to get it in seconds
@madhukushwaha4578 3 жыл бұрын
@@akshatjangra4167 Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@gfest1119 3 жыл бұрын
Answer 3. Cause 4√(√3-1)^2+√(7-4√3)^2=-4+7=3
@Archik4 3 жыл бұрын
At school, they solve such problems. You can just select a full square under the root.
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@robertruta687 3 жыл бұрын
Honestly... easiest thing to do is approximate sqrt(3) as 1.7 or something. It works out to certainly be 3
@raymond2814 3 жыл бұрын
Didn’t you miss out the 4 coefficient
@randomlife7935 3 жыл бұрын
For Problem 2, a formula is available: mathforum.org/library/drmath/view/65409.html. Suppose we want to write sqrt( a + b * sqrt(c) ) as a sum of square roots of rational numbers. Then we calculate a^2 - b^2*c and check it it is the square of a rational number. If its square root is not rational, then you can't write your square root as a sum of square roots of rational numbers. On the other hand, if its square root is m, so that m^2 = a^2 - b^2*c then if b is positive, sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) + sqrt((a-m)/2) and if b is negative, sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) - sqrt((a-m)/2)
@jesusthroughmary 3 жыл бұрын
this is the good stuff right here, thanks
@rhythmmandal3377 3 жыл бұрын
Honestly for the 1st question I saw 4*sqrt(1+3-2sqrt(3))+(-2*4*7*sqrt(3)+49+48) and went ohhh... 4*(sqrt(3)-1)+(7-4*sqrt(3)) Did not even cross my mind to go full quad equation.
@alimokadem1055 3 жыл бұрын
Pr Michel Pen vous aurez pu essayé certaines valeurs et déduire de façon assez simple et rapide que 4-2sqrt3=(sqrt3-1)^2 & 97-56sqrt3=(4sqrt3-7)^2 ! Malgré que certaines idées simples vous échappe parfois Mais vos vidéos sont fabuleuses et très enrichissantes ! Vous faites un travail remarquable 👍👍
@roberttelarket4934 3 жыл бұрын
I have no audio. What's going on?
@akshatjangra4167 3 жыл бұрын
Can do the first one easily by decomposing 4 into 3 and 1
@akshatjangra4167 3 жыл бұрын
Which can be written as root 3 squared and 1 squared
@akshatjangra4167 3 жыл бұрын
Then we will get (a-b)^2
@akshatjangra4167 3 жыл бұрын
Square and root will cancel out
@akshatjangra4167 3 жыл бұрын
Same for solving the next radical
@starshipx1282 3 жыл бұрын
The second problem was a beauty.
@user-A168 3 жыл бұрын
Good
@goodplacetostop2973 3 жыл бұрын
20:57 LMAO, the homework I have today is from the same contest but the year 2008. I like this homework but I have yet to find an official solution for it so here we go... Determine all functions f (R -> R) satisfying f(x + y)
@tobiasgorgen7592 3 жыл бұрын
This video is 2 Minutes old. What are you doing at 20:56? haha
@goodplacetostop2973 3 жыл бұрын
@@tobiasgorgen7592 I’m the time wizard
@hydraitm5248 3 жыл бұрын
Yes !
@manishtripathy5156 3 жыл бұрын
Perhaps this guy is related to Micheal Penn or is he Micheal penn??
@goodplacetostop2973 3 жыл бұрын
@@manishtripathy5156 I wish
@mehdimarashi1736 3 жыл бұрын
#suggestion Hi Michael, I love your channel and your problem solving style. I wish your youtube channel was around in the old days when I was studying for the olympiad. I wanna suggest a geometry problem that you might find interesting for your channel. This was in our planar geometry exam in the 10th grade, but to me, it looks like olympiad material. I've been going back to it once and again for the last 20 years, but I never had any success with it. The problem goes: Draw a triangle in which the lengths of the altitude, the bisector, and the median passing through a single vertex are known. I don't consider myself and my classmates geometry prodigies, but nevertheless, we found it an impossible problem. We, 50 of us, collectively, got a zero on this problem. We had no idea how to approach it. It's been bothering me since, and I never found a solution for it. Is it hard, or do I suck?
@madhukushwaha4578 3 жыл бұрын
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@KingstonCzajkowski 3 жыл бұрын
Does an equilateral triangle not work? Use the Pythagorean/Gougu Theorem for the height.
@mehdimarashi1736 3 жыл бұрын
@@KingstonCzajkowski I don't know what you mean. I think that we had to use the angle bisector theorem that relates the length of the sides to the lengths of the base segments, but I cannot imagine how. And angle bisector theorem is not something that every high school student knows.
@KingstonCzajkowski 3 жыл бұрын
@@mehdimarashi1736 In an equilateral triangle, the bisector, median, and altitude all have the same length.
@mehdimarashi1736 3 жыл бұрын
@@KingstonCzajkowski Yep, but unfortunately the three given lengths are arbitrary and not necessarily equal, and if they are equal, the problem has infinite number of answers, any equilateral tr with the given altitude is the answer.
@geraldomelo2751 3 жыл бұрын
4(√3 - 1)+7- 4√3 = 3
@mathsamtube2741 3 жыл бұрын
wow
@PokerPlayerJames 2 жыл бұрын
Couldn't you have just written 4-2rt3 as 1 - 2rt3 +3 and factorised as (1- rt3)^2? Dealing with a Quartic seems unnecessarily complicated.
@denisphelipon4695 3 жыл бұрын
I was not in the engrenages of the elephants .
@youssefallani6407 3 жыл бұрын
Heroooo
@rbdgr8370 3 жыл бұрын
pappa flammy will go nuts if he saw the guessing you did for the first bi-quadratic factorization in Q1
@hydraitm5248 3 жыл бұрын
1st, Love from India ❤️ ❤️
@madhukushwaha4578 3 жыл бұрын
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@youssefallani6407 3 жыл бұрын
I hope you have 200000000 like in this video
@djvalentedochp 3 жыл бұрын
Nice
@targetiitbcse1761 3 жыл бұрын
a sub(n)= a sub(n+4) shouldn't be be true for all integers rather than natural numbers
@keithmasumoto9698 3 жыл бұрын
194b^2 eqn not factored right
@cepatwaras 3 жыл бұрын
I thought so too at first. But then I realize that 196 is reduced by 2 from the other multiplication terms with b².
@keithmasumoto9698 3 жыл бұрын
@@cepatwaras Oh right! 14^2-2=196-2 ty!
@speedsterh 3 жыл бұрын
Thought the same at first glance, but Michael is right (of course)
@pedrokrause7553 3 жыл бұрын
√(4-2√3) = √(1-2√3+3) = √((√3-1)²)= √3-1 So, x = 4√(4-2√3) + √(97-56√3) x = 4√3 + √(97-56√3) - 4 If x is an integer, then 4√3 + √(97-56√3) is also one. Let's call it q q = 4√3 + √(97-56√3) √(97-56√3) = q - 4√3 97 - 56√3 = q² - 8√3 + 48 q² = 49 - 8√3(7 - q) There is no way for the RHS to be rational or integer due to the second term, unless it gets cancelled, which actually happens when q = 7, which is a solution. So, x = q - 4 x = 3 Integer
@peelysl Жыл бұрын
bruh if that's the olympiad problem in the US then I would make IMO :skull:
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