One of the HARDEST International Math Olympiad Question

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A great exponential equation! What do you think about this problem?

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@higher_mathematics

#maths #algebra

[mokouf]

I just remember that this is the golden ratio number, and it satisfies the quadratic equation, and it is easy.

[dalesmart]

An easy way of computing this is square 1/2(1 + root 5), simplify to get 1/2(3 + root 5), square that result again to get 1/2(7 + root 3), then find the product of the two to give x^6 =(9 + 4 root 5). x^12 is obtained by squaring this result to get 161 + 72 root 5. Alternatively, Binomial theorem could have been used to evaluate (1/2 + root 5/3)^12, but that method may not be all that sexy!

[theelectro]

another interesting approach that I find easier (and very worthy to understand) is if ur familiar with the fibonacci series, we denote fn as the nth term of the series (eg: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... so f3 = 2 f6 = 8 and so on) we can notice that our x is the solution to the ecuation x+1 = 1/x so it's actually the golden ratio, therefore we will denote it with Φ and if you know some properties you would know that Φ to the nth power is equal to Φ times the nth term of the fibonacci's series plus the (n-1)th term of the fibonacci's series: Φ^n = Φ*fn + f(n-1) for all n>=1.
In our case:
n = 12, fn = 144, f(n-1) = 89, so;
x^12 = Φ^12 = 144*Φ + 89 (= 144x + 89)
just from here you can see that the following steps are the exact same.
also his formula for x^2 = x + 1 also is explained this way:
n = 2, fn = 1, f(n-1) = 1, so;
x^2 = Φ^2 = Φ*1 + 1 (= x + 1)

Anyways your approach is optimal as almost nobody knows these properties of the golden ratio so it makes yours more universal than mine, I just wanted to show to the interested people another way of computing it instantly with some basic knowledge. Thanks for reading!!

EXTRA fact :) : the limit fn/f(n-1) when n approaches infinity is exactly equal to Φ.

[Notthatkindofdr]

Once you realize that x^2=x+1 (which is pretty easy to do if you recognize the golden ratio, or just the fact that it looks like the solution of a quadratic equation, it is simpler to do the following computations: x^3 = x(x+1) = x^2+x = (x+1)+x = 2x+1. Then x^6 = (2x+1)^2 = 4x^2+4x+1 = 4(x+1)+4x+1 = 8x+5, and finally x^12 = (8x+5)^2 = 64x^2+80x+25 = 64(x+1)+80x+25 = 144x+89. Then just plug back in the definition of x and get the answer.

[FaneBenMezd]

It's a piece of cake when you realize that the powers of the golden ratio are actually a Fibonacci sequence where every new term is the sum the the 2 previous ones.
φ^0=1
φ^1=φ
φ^2=φ+1
φ^3=2φ+1
φ^4=3φ+2

[AlexanderPatrakov]

Not simpler than the brute force approach, illustrated below.

x^2 = (1 + sqrt(5))^2 / 4 = (1 + 2 sqrt(5) + 5) / 4 = (3 + sqrt(5))/2
x^3 = x * x^2 = (1 + sqrt(5)) * (3 + sqrt(5)) / 4 = (3 + sqrt(5) + 3*sqrt(5) + 5)/4 = 2 + sqrt(5)
x^6 = (x^3)^2 = (2 + sqrt(5))^2 = 4 + 4*sqrt(5) + 5 = 9 + 4*sqrt(5)
x^12 = (x^6)^2 = (9 + 4*sqrt(5))^2 = 81 + 2 * 36 * sqrt(5) + 16 * 5 = 161 + 72 * sqrt(5)

[niccolocervelli]

A cool way to approximate it, even tho i dont find the exact value: knowing that the nth fibonacci number is (((1+√5)/2)^n-((1-√5)/2)^n)/√5, you notice that the term ((1-√5)/2)^n is really close to 0 if n=12, so the 12th fibonacci number multiplied by √5, that is 144√5, is incredibly near the solution. Try it yourself, their difference should be 0.001 or something like this.

[olivergroning]

Having worked in quasicrystals, I recognized the golden ratio tau=(1+sqrt(5))/2=tau, which has the property that tau^2=1+tau. From this it is easy to show that tau^n=f(n)*tau+f(n-1). Here f(n) is the nth Fibonacci number. So tau^12=f(12)*tau+f(11). With f(12)=144 and f(11)=89 thus, tau^12=144*tau+89=321.9969...

[subhamchaterjeeguitar]

Another interesting approach could be solving the problem by using binomial expansion.

[nuclearwarhead]

This is more like the easiest IMO problem

[jivitesh]

Or maybe use binomial theorem? It's much easier in this case

[higher_mathematics]

What do You think about this approach? Much love and respect. Have a great day! ❤️❤️❤️

[robfrohwein]

Very smart ... I will try later to reproduce this.... thanks ❤

[AaronDude-rc]

this was a fun question. i had no idea how to solve it, so i did it a dumb way, by just having one thing lead to another. and i got the answer. here's what i did:

x = [1+sqr(5)]/2 y = [1-sqr(5)]/2

xy = -1

x+y = 1
x2+y2 = 3
x3+y3 = 4

x3+y3 = 4
(x3+y3)^4 = 4^4
(x3+y3)^2 * (x3+y3)^2 = 256
[x6+2(x3)(y3)+y6] * [x6+2(x3)(y3)+y6] = 256
[x6+2(x3)(y3)+y6] = 16
[x6+2(xy)^3+y6] = 16
[x6+2(-1)^3+y6] = 16
[x6 -2 + y6] = 16
[x6 + y6] = 18

[x6 + y6]^2 = 18^2
[x6 + y6]^2 = 18^2
[x12 + 2(x6)(y6) + y12] = 18^2
[x12 + 2(xy)^6 + y12] = 18^2
[x12 + 2(-1)^6 + y12] = 18^2
[x12 + 2 + y12] = 18^2
[x12 + y12] = 18^2 - 2

previously:
xy = -1
x+y = 1
x2+y2 = 3
x3+y3 = 4

now:
x6+y6 = 18
x12+y12 = 18^2 - 2

if x6+y6 = 18
then y6 = 18-x6
then y12 = 18^2 - 36(x6) + (x12)

[x12 + y12] = 18^2 - 2
x12 + y12 = 18^2 - 2
x12 + [y12] = 18^2 - 2
x12 + [18^2 - 36(x6) + (x12)] = 18^2 - 2
x12 - 36(x6) + (x12) = -2
(x12) - 36(x6) + (x12) = -2
2(x12) - 36(x6) = -2
2(x12) - 36(x6) + 2 = 0
(x12) - 18(x6) + 1 = 0


let m=(x6)
m2 - 18m + 1 = 0

quadratic formula
m = 9+4sqr(5)
m = (x6)
x6 = 9+4sqr(5)

therefore

(x6)^2 = x^12
x^12 = 161+72sqr(5)

[alscents]

It is much simpler:
(1) Notice that x = f the Fibonacci constant, which has the property that x^2 = x + 1
(2) x^4 = (x + 1)^2 = 3x + 2
(3) x^8 = (3x + 2)^2 = 21x + 13
(4) x^12 = x^8 * x^4 = (21x + 13)(3x + 2) = 63 (x+1) + 42x + 39x + 26 = 144x + 89
(5) x^12 = 144 (1 + sqrt(5))/2 + 89 = 161 + 72 sqrt(5)

[itachishisuiuchiha]

this is in fact one of the easiest IMO problems

[aleattorium]

I am sure there are better approaches using the favt of phi being involved.

[threety-three]

Easiest way is to just start multiplying the term by itself, and at x^3, you'll notice it is 2+ root 5. then you can just easily do the rest, takes like a minute

[Nzargnalphabet]

It should be very close to the twelfth Lucas number

[nissimlevy]

This is an easy problem. Just calculate the square. Then multiply the two squares to get the 4th power. then multiply three fourth powers