Russian Math Olympiad Question #maths #mamtamaam

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[josephhoward]

You can get this done much faster. Once you have ab = -(1/2) and the a+b =1 you've got a system of equations. Sub them into each other and get a = (1+sqrt(3))/2 and b = (1-sqrt(3))/2. And then you are basically done.

[DokterrDanger]

As a calculus student i thought we had to do d²/dx²(a)+d²/dx²(b)...
a and b being conatants, the answer would be 0.

[MrGeralt]

For those people who found "short" solution: the author found an expression which allows her to calculate the answer with any desired accuracy. She found this expression by analytic way and she doesn't need any calculator - she just uses one of elementary arithmetic operation (division). When you use calculator and try to calculate square root you use some algorithm that is part of calculator firmware. The accuracy depends on this algorithm - it can be different for different calculators. This is difference between "short" solution and this one.

[hnautiyal]

You can figure out that both a and b satisfy
x2 = x + 1/2
It is easy to see that in a recurrence defined by
Tn = Tn-1 + Tn-2 / 2
Tn = a^n + b^n
If you just set up T1 = 1 and T2 = 2

Successively applying the recurrence gives you the following terms and you are looking for the 11th one.

1
2
5/2
14/4
38/8
52/8
142/16
194/16
530/32
724/32
989/32 as the video also arrives at

[MadScientyst]

For all comments pointing to an 'easy' solution, yes it can be approached in that manner...BUT, that kills the joy of looking deeper into the problem & discovering other elegant solutions.
Take for example zero (0), which can be also written as 0+0i in the Complex Plane.
That's the sheer beauty of Math, there's always more than one approach particularly at the Olympiad level.
IMO...if u aren't interested in exploring a problem from alternative perspectives, then u have no business being in Math.
I've always found the Indian (Vedic) approach to Math to be fascinating, so once again I'm subbing to another valuable Channel on the pedagogy of its principles....🤩🤩

[jimwinchester]

I just re-wrote equation #1 as b = (1 - a), and then used that substitution for 'b' in equation 2, to arrive at -
a^2 + (1 - a)^2 = 2. (1 - a)^2 is a^2 - 2a + 1, so the sub'd form of eq. (2) is 2a^2 - 2a + 1 = 2, or 2a^2 - 2a -1 = 0. Continue via quadratic formula (a==2, b==-2, c==-1), to get (a), then sub-back (1 - a) to get (b). Then get a calculator to compute each raised to 11th power and sum.

[pandagoeshome]

I’d appreciate it if you didn’t use X to signify multiplication. It threw me off (I was watching without sound). Thank you

[deananderson]

Nice solution personally I got to the step that AB=-1/2 then subbed A=-1/(2B) into A+B=1 to get B-1/(2B)=1 which simplified to (2B^2-1)/(2B)=1 => 2B^2-1=2B which means 2B^2-2B-1=0 use the quadratic formula to find that B equals (1+-sqrt(3))/2 and use the original equation to find that a=1-(1+-sqrt(3))/2 (something nice to notice whichever route you choose for b, a will be equal to the other one and the equations are symmetric for a and b so whichever you choose we can see we will only have one answer for a^11+b^11 after that just use whichever method you prefer to find these numbers raised to the 11th power and add them up.

[quakers]

Instead of just solving the problem wouldn't it be helpful to explain why you are doing these manipulations. At one point you cube both sides of the equation. There must be a reason for this. It is not intuitive.

[abeladamu]

It is self explanatory… beautifully done !!!

[nabilsleiman]

A nice python function can solve it for any power
fun(n) = a^n + b^ n
fun(0) = a^0 + b^0 = 1 + 1 = 2
fun(1) = a + b = 1 (given)
fun(2) = a^2 + b^2 = 2 (given)

def fun(n):
k = 0
if (n == 0):
k = 2
elif (n == 1):
k = 1
elif (n == 2):
k = 2
else:
k = fun(n - 1) + fun(n - 2)/2
return k

print(fun(3), fun(5), fun(6), fun(11))
fun(3) -> 2.5
fun(5) -> 4.75
fun(6) -> 6.5
fun(11) -> 30.90625

[Nikioko]

(a + b)² = 1²
a² + 2ab + b² = 1
2 + 2ab = 1
2ab = - 1
ab = - 1/2

a² - 2ab + b² = 1 + 2
(a - b)² = 3
a - b = ±√3

a² - b² = (a + b) ⋅ (a - b)
a² - b² = 1 ⋅ (±√3)

a² + b² + a² - b² = 2 ± √3
2a² = 2 ± √3
a² = 1 ± √3/2
a = √(1 ± √3/2)

b = 1 - a
= 1 - √(1 ± √3/2)

[SeaScoutDan]

How about b=1-a. Subsitute in to get a^2+ (1-a)^2=2
Solve for a, then solve for b.

[UberHummus]

Great video — very clear solution.

I’m getting pretty good at doing this mentally. Fractions are easy

[someguyification]

... I don't understand the notation for the third line of the question. I've only seen a'' to be used for derivatives, and those make no sense here (because a and b are unknown constants)

edit: ooh those are exponents. helps to actually watch the video.

[nigelwilliams]

a+b=1
a^2+b^2=2
Solve a^11+b^11

Given: b=1-a
so
a^2+(1-a)^2=2
a^2+(1-2a+a^2)=2
2a^2-2a+1=2
2a^2-2a-1=0

Via the quadratic formula the resulting roots for a are:
a= 1.366025 or -0.36603
and so b is
b = -0.36603 or 1.366025
Then either pair of roots raised to power of 11 and summed gives 30.9065 which is 989/32.
Seemed to me to be a simpler journey, tho your exercise of discovering a^11 was wonderful to watch!

[radhakrishnamohanty]

Kindly don't use X symbol for multiplication in any algebraic expressions.

[joseeduardomachado]

GOSTEI MUITO DO EXERCÍCIO. PARABENS A QUEM PENSOU O EXERCÍCIO E A QUEM RESOLVEU TAMBÉM

[fio]

2 pow 7 = 128. 2pow a + 2 pow b = 20. 4 + 16 =20. a =2, b =4, c=7.

[sidgar]

I thought the 11's were actually derivative symbols '' at first, like how tf are you supposed to get the 2nd derivative of a and b from the first two equations?