Two from Africa!

[@onealjerry]

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]

Teacher: what is 1+2? Students: 4(sqrt(4-2sqrt(3)))+sqrt(97-56sqrt(3))

[@freepimaths9698]

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]

@@freepimaths9698 then define 3 as an infinite nested square root

[@freepimaths9698]

@@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]

​@@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]

@@JordHaj Oh, no actually, looking back I see I clearly just misinterpreted when I shouldn't have 😅 the sorry is from my direction

[@OCinTexile]

Your reasoning in that second question especially is amazing to me. I couldn't imagine even how to start answering it. Thanks!

[@andreivila7607]

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]

Oh cool. Thanks Stalin!

[@shatishankaryadav8428]

First is easy make combination of 2ab then it become (a-b)^2

[@andreivila7607]

@@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]

@@madhukushwaha4578 the first video that appears is literally why odd x odd is even

[@V-for-Vendetta01]

@@evanev7 lmfao

[@NavyBlueMan]

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]

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]

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

[@mach2570]

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]

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]

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.

[@kozokosa9289]

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]

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]

I agree with gameguy

[@카르비젤]

@@gameguy8101 hey gameguy, are u genius?

[@gfest1119]

Yeah man, i watched it very fast, and dont understand how the author hasnt watched it

[@pshr2447]

@@gameguy8101 love you game guy

[@CaradhrasAiguo49]

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

[@V-for-Vendetta01]

many have already pointed this out, but the first problem can easily be solved by converting the terms into (a+b)^2 form

[@chhabisarkar9057]

I did like that lmao and took me almost 4 steps

[@lambdcalculus]

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.

[@songguo2243]

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]

If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@caiotmz]

I swear the I read 'Toto from Africa' in the title

[@xevira]

Dammit, now I got that song stuck in my head again... again. :P

[@doctorb9264]

Love playing that tune on bass !

[@DanielWalvin]

I bless the math!

[@jamirimaj6880]

Legendary song!

[@ellenmarch3095]

I think everybody did, lol.

[@mathwithjanine]

Awesome video! :) Enjoyed these fun math problems!

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@franckquessette9198]

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.

[@shadowsky]

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]

If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@jeanf6295]

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

[@vishalmishra3046]

*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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@burhan8795]

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.

[@edusoto91]

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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@tuffleader2448]

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.

[@pierineri]

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* .

[@siulibasak3804]

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....

[@laokratis55]

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]

correction: 7-4 sqrt 3

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@matthieumoussiegt]

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

[@Utesfan100]

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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@cesaremarcolazzarini3289]

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.

[@himanshigoyal7471]

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

[@Chess-ks8lk]

You're great mathematician!

[@anirudh67]

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

[@kelvinella]

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)

[@CM63_France]

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".

[@tomatrix7525]

Great video, as usual.

[@tomatrix7525]

@@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

[@recklessroges]

Those are both beautiful. Thanks.

[@VideoFusco]

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.

[@Pablo360able]

Second question seemed straightforward to me, but that's because I've used all the logic involved to solve cryptic sudoku.

[@malawigw]

These problems were quite simple but fun!

[@ankitbhattacharjee280]

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.

[@HagenvonEitzen]

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

[@pfeilspitze]

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.

[@ericzgrey]

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

[@f5673-t1h]

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.

[@rmschad5234]

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.

[@alihaydar728]

I used : sqrt [ a - sqrt ( b ) ] = sqrt [ a + sqrt ( a^2 - b ) ] / sqrt ( 2 ) - sqrt [ a - sqrt ( a^2 - b ) ] / sqrt ( 2 )

[@alihaydar728]

or sqrt [ a - sqrt ( b ) ] = sqrt (x) - sqrt (y) such that a = x+y , b=4*x*y

[@damascus21]

Thanks for showing us something from Africa. This helps contradict stereotypes about the continent

[@aviralsood8141]

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]

It in no way contradicts stereotypes about Africa - none of which deal with mathematics. All the stereotypes are accurate.

[@johnmccrae2932]

@@garymclaughlin9559 What?

[@dandan-gf4jk]

@@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

[@eliaskar5084]

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.

[@jlhjlh]

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.

[@spartacus8875]

Thank you for the movies ...the best

[@eminvahid3976]

Very smart solutions. 🤓

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@jeanlismonde8718]

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

[@randomlife7935]

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]

this is the good stuff right here, thanks

[@Downloader77]

You did it your way and I like that.

[@shivansh668]

When I saw the thumbnail,I thought that it's an Integer Partition in a irrational way !😅

[@madhukushwaha4578]

If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@kristianwichmann9996]

#2 was a really nice little problem :-)

[@orchestrain88keys]

The first problem is fairly straightforward, but solved in a convoluted way. The second problem is interesting.

[@somanathdash8143]

Loved the problem

[@somanathdash8143]

@@madhukushwaha4578 thanks : )

[@jesusthroughmary]

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]

@@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]

@@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]

@@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]

@@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"

[@jonathanbonicel1654]

That time skip at 18:28 was PERFECT

[@akshatjangra4167]

Can do the first one easily by decomposing 4 into 3 and 1

[@akshatjangra4167]

Which can be written as root 3 squared and 1 squared

[@akshatjangra4167]

Then we will get (a-b)^2

[@akshatjangra4167]

Square and root will cancel out

[@akshatjangra4167]

Same for solving the next radical

[@mehdimarashi1736]

#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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@KingstonCzajkowski]

Does an equilateral triangle not work? Use the Pythagorean/Gougu Theorem for the height.

[@mehdimarashi1736]

@@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]

@@mehdimarashi1736 In an equilateral triangle, the bisector, median, and altitude all have the same length.

[@mehdimarashi1736]

@@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.

[@bndrcr82a08e349g]

I do not expect negative value in the second problem

[@jindagibura7427]

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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@akshatjangra4167]

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]

@@akshatjangra4167 Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@hamdiel-sissi7760]

Super complicated when the solution is trivial!!

[@michaelhall5801]

Yay! Representation

[@Archik4]

At school, they solve such problems. You can just select a full square under the root.

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@alphalunamare]

well that just blew my marbles :-)

[@alimokadem1055]

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 👍👍

[@goodplacetostop2973]

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]

This video is 2 Minutes old. What are you doing at 20:56? haha

[@goodplacetostop2973]

@@tobiasgorgen7592 I’m the time wizard

[@hydraitm5248]

Yes !

[@manishtripathy5156]

Perhaps this guy is related to Micheal Penn or is he Micheal penn??

[@goodplacetostop2973]

@@manishtripathy5156 I wish

[@rafael7696]

Beautiful problems

[@robertruta687]

Honestly... easiest thing to do is approximate sqrt(3) as 1.7 or something. It works out to certainly be 3

[@IvanisIvan]

the second problem reminded me of a software problem

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@rhythmmandal3377]

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.

[@victormd1100]

2:00 thank you so much

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@gfest1119]

Answer 3. Cause 4√(√3-1)^2+√(7-4√3)^2=-4+7=3

[@marcusdecarvalho1354]

The 14-th Pan-African Mathematical Olympiad (Tunis, Tunisia, 2004 ) - First Day : Questions 2 and 3.

[@wallaceferreiradasilva68]

I wish I was able to fully understand problem 3 solution.

[@NileshKumar-gu2ti]

Can we use complex number for the second question root of unity concept?

[@ProjetoEquilatero]

Usa-se duas vezes radical duplo na primeira questão. Foi isso que ele fez, mas de forma diferente.

[@raymond2814]

Didn’t you miss out the 4 coefficient

[@Namo_Amitabha2024]

These 2 q's are remarkably easier than the others...

[@hydraitm5248]

1st, Love from India ❤️ ❤️

[@madhukushwaha4578]

If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@ryanleung6753]

4:00 doesn’t understand the progression of this step

[@varunmishra7556]

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]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@adelinniculae]

In the first exercise, a is wrong... that quadratic ecuation can't be written as you did!! You used x^2-y^2=(x+y)(x-y) where x is a^2 and y is 2a-2 so this will give us a^4-(2a-2)^2 which is not equal to a^4-8a^2+4!!! Did I miss something?!

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@SachinSachin-gr2fu]

Sir i didn't understood the 3rd line of answer no. 3

[@skipugh]

Impressive

[@targetiitbcse1761]

a sub(n)= a sub(n+4) shouldn't be be true for all integers rather than natural numbers

[@pedrokrause7553]

√(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

[@roberttelarket4934]

I have no audio. What's going on?

[@shreyathebest100]

Can anyone tell me how we factor such expressions

[@user-A168]

Good

[@legendarydisciplegaming4316]

Is there a soln. For this equation

[@madhukushwaha4578]

Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun

[@baniksden7481]

√[97-56√3] = √[49-2.7.4√3+48] = 7-4√3 Isn't it?

[@denisphelipon4695]

I was not in the engrenages of the elephants .

[@johnloony68]

The answer to question 3 could be anything you like; it is unknowable and uncalculable until and unless you insert some extra assumptions into the question. You did that by interpreting "the sum of 20 consecutive numbers is 75" as meaning "the sum of any consecutive numbers is 75" - which is not what the question asks.

[@keithmasumoto9698]

194b^2 eqn not factored right

[@cepatwaras]

I thought so too at first. But then I realize that 196 is reduced by 2 from the other multiplication terms with b².

[@keithmasumoto9698]

@@cepatwaras Oh right! 14^2-2=196-2 ty!

[@speedsterh]

Thought the same at first glance, but Michael is right (of course)

[@geraldomelo2751]

4(√3 - 1)+7- 4√3 = 3

[@hamidkh5488]

Thanks Michael . I suggest this way Sqr(A-Sqr(B)) = Sqr((A+C)/2) - Sqr((A-C)/2) A^2 - B = C^2

[@brunoparra7204]

I know I'm 2 months late, but how can i investigate about it? Or does this thing has name?

[@hamidkh5488]

@@brunoparra7204 Sorry I don't know if this formula has a special name or not. I learned it from my algebra teacher in high school about 35 years ago.

[@mathsamtube2741]

wow

[@alnitaka]

Instead of trying to guess a factorization, the quartic formula may help get what we want. For a^4-8a^2+4=0, the resolvent cubic is y^3-16y^2+48y=0, which has roots 0, 4, and 12. So a root is given by (1/2)(sqrt(0)+sqrt(4)+sqrt(12)) = 1+sqrt(3). There are three other roots so that gives us +-1+-sqrt(3), so as in above we find sqrt(3)-1. For the second term, b^4-194b^2+1=0. The resolvent cubic is z^3-388z^2+37632z, which has solutions 0, 192, and 196. So a root is (1/2)(sqrt(0)+sqrt(192)+sqrt(196)) = 7+4sqrt(3) and 3 other roots - the one that works is 7-4sqrt(3). I note that 7-4sqrt(3) = (2-sqrt(3))^2, but that has little to do with the solution - apparently a red herring that distracts.

[@skyper8779]

Q2: Answer: Yes and No. Not sure why you eliminated the negative roots, all 8 values of a and b are legit, giving us 12 possible outcomes for the sum, 4 of them are integers: + - 3 and + - 11 5:02 square root of something bigger than 0 can pretty much be less than 0