Math Olympiad Problem | A Nice Algebra Challenge

Math Olympiad Problem | A Nice Algebra Challenge

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[anson]

just substitute b for (1-a) and use quadratic equation to find (a) in the second equation, then calculate a^11 + (1-a)^11, and the answer will be 989/32

[marizacabral]

You can use the original 2 equations to solve for a and b directly. Two equations with 2 unknowns...

[lusax]

a+b=1 … set T1
a²+b²=2 … set T2 (ex. a^n+b^n=Tn)
∴ ab=-1/2
equation whose root is a, b
t²-t-1/2=0
it's identity so we can substitute t to x
x²-x-1/2=0 … ×x^n
x^(n+2)=x^(n+1)+(1/2)x^n
same method,
y^(n+2)=y^(n+1)+(1/2)x^n
add two equations

then
Tn+2=Tn+1+(1/2)Tn
∴ T3=T2+(1/2)T1
T4…
and finally, solve T11. that's the answer.

[claudiosaccon]

Let c_n:=a^n+b^n.
Then c_1=1 and c_2=2.
Moreover 1=(a+b)^2=a^2+2ab+b^2=2+2ab hence ab=-1/2.
Let's find a recursive formula for c_n :
c_n=1 × c_n=c_1 c_n=(a+b)(a^n+b^n)=
a^(n+1)+a^n b+a b^n+b^(n+1)=
c_(n+1)+ab c_(n-1)=c_(n+1)-(1/2)c_(n-1).
It follows:
c_(n+1)=c_n+(1/2)c_(n-1).

[GodbornNoven]

a+b=1
b=1-a
a²+(1-a)²=2
a²+1-2a+a²=2
2a²-2a=1
a²-a- 1/2 = 0
Solve for a by using quadratic formula, solve for b by using a. Then raise a and b individually to the 11th power and add them.
No need to overcomplicate it

[davidash]

Another way of approaching this is to let S_n=a^n+b^n for any n. Then S_0=2 (trivially) and S_1=1, S_2=2 (we are given this). a and b must both be roots of some quadratic x^2+ux+v=0. So a^2+ua+v=0 and b^2+ub+v=0, and hence for any n, a^(n+2)+ua^(n+1)+va^n=0 and b^(n+2)+ub^(n+1)+vb^n=0. Adding the recurrence S_(n+2)=-uS_(n+1)-vS_n holds for all n. u must be the negation of (a+b) so u=-1, and S_(n+2)=S_(n+1)-vS_n. Given S_2=2, S_1=1, and S_0=2, we can solve for v: v=-1/2.

The recurrence then becomes S_(n+2)=S_(n+1)+S_n/2, and we can then evaluate S_11 by repeated application of this recurrence.

[vishalmishra]

*Simpler Solution*
a + b =1 and a^2 + b^2 = 2 Therefore, 2ab = 1^2 - 2 = -1. So, ab = -1/2. Therefore, a and b = (1 ± √3) / 2.
Using Pascal's triangle, the co-efficients of (a + b)^11 = [ 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1 ]
Due to ± the odd terms will cancel out and the even terms will double. So √3 terms will cancel out and powers of 3 will double. Denominator = 2^11 = 2048.
Powers of 3 are [ 1, 3, 9, 27, 81, 243 ] and (even) binomial coefficients are [ 1, 55, 330, 462, 165, 11 ]
Hence, a^11 + b^11 = 2 x [ 1, 3, 9, 27, 81, 243 ] . [ 1, 55, 330, 462, 165, 11 ] / 2^11 = 2 x 31648 / 2048 = 989 / 32

[MeNowDealWIthIt]

My solution:

a+b=1 -> b=1-a
So a^2+b^2=2 -> a^2+(a-1)^2=2
2a^2-2a+1=2
a^2-a-.5=0
a=(1+-sqrt(3))/2
Checking the solutions for the first to equations means a=(1+sqrt(3))/2 and b=(1-sqrt(3))/2 or vice versa, it doesn't change what we're trying to solve.

Pascal's triangle starting at the part I have memorized:
1 6 15 20
1 7 21 35 35
1 8 28 56 70
1 9 36 94 126 126
1 10 45 132 220 252
1 11 55 177 352 472 472 352 177 55 11 1

a^11+b^11
=((1+sqrt(3))/2)^11 + ((1-sqrt(3))/2)^11


Pascal's triangle starting at the part I have memorized:
1 6 15 20
1 7 21 35 35
1 8 28 56 70
1 9 36 94 126 126
1 10 45 132 220 252
1 11 55 177 352 472 472 352 177 55 11 1
I know the odd degree terms will be opposites, and the even degree powers will be the same, so just doubling the even degree powers will be equivalent.





=31088/1024
=16544/512
=8272/256
=4136/128
=2068/64
=1034/32
which means I did some arithmetic wrong somewhere. damn.

[channelsixtysix]

That was an enjoyable nine minutes and fifty seconds. You had me laughing as it all resolved. Marvellous.

[jcb]

If f(n) = a^n + b^n then f(n) satisfies the recurrence f(n+1)=f(n)+ f(n-1)/2. Given f(1)=1 and f(2)=2 it is pretty quick to compute f(11)

[vincenzofonseca]

Guys, I think that if we expand (1-a)^11 using binomial theorem, we would take less time as the video's resolution took. Almost 10 minutes is a lot of time...

[atikshchawla]

Another simpler method for this question is as follows
first get the value of ab by using the formula
(a+b)^2 = a^2 +b^2 +2ab

then use the formula
(a-b)^2 = a^2 +b^2 -2ab
to get the value of a-b

then simply equate a+b and a-b together to get the value of a and b
and then simply substitute them to the required equation i.e. a^11 +b^11

[l.w.paradis]

Your approach brought back my many winter nights of solving definite integrals. :)

[owlsmath]

Good problem. Was hoping for a shortcut but I guess you just have to keep multiplying in this case

[MrLidless]

Once you get ab = -0.5, ie b = -1/2a, slot it into the first equation.
a - 1/2a = 1

Rearranging:
a^2 = a + 1/2
a^3 = a^2 + a/2 = a + 1/2 + a/2 = 3a/2 + 1/2. So a^3 is a multiple of a + constant. Multiply this by a again to get
a^4 = 2a + 3/4
Etc etc. You’ll soon see the pattern for a^n = f(n)a + g(n) and can therefore jump to a^11

Ditto with b.

You’ll get to the answer more elegantly.

[michroz]

I got 30.90625 straightforward with Desmos graphing calculator to skip calculations.
a = 1-b; b=1/2-(sqrt(3))/2, a=1/2+(sqrt(3))/2, etc.
Not sure I'd pass the Olympiad this way - they prefer tricky solutions.

[filippocontiberas]

This because there are only 2 solutions for A and two for B such that all of them are shared valid solutions for the 3 equations :
Case 1: A = [ 1 - sqrt( 3 ) ] / 2 and B = [ 1 + sqrt( 3 ) ] / 2
Case 2: A = [ 1 + sqrt( 3 ) ] / 2 and B = [ 1 - sqrt( 3 ) ] / 2
The case 1 and case 2 are two points where the line ( first linear equation ), the circle ( second quadratic equation ) and the curve ( third 11th equation ) meets all at the same time.

[rameshchandra-qgsw]

If you had a calculator then you could have solved a and b by using the property: a^2+b^2 = (a+b)^2+(a-b)^2 => a-b= sqrt (3). a=(1 + sqrt(3))/2 and b=(1 - sqrt(3))/2. a= 1.366, b= -0.366. Now substitute back in a^11 + b^11. Approx. 30.89. Off course, if no calculator your elegant method is needed

[hetanshpatel]

A faster method:

a + b = 1
a = b-1

a² + b² = 2
(b-1)² + b² = 2
b² - 2b + 1 + b² = 2
2b² - 2b + 1 = 2

with quadratic formula:
b = (1 ± √3)/2
a= (-1 ± √3)/2

We disregard negative since it gives negative output

b = (1 + √3)/2
a= (-1 + √3)/2

Now we substitute in a¹¹+b¹¹ and we get a value close to 30.60925, which is 19/4.

[erkinjonmeliboyev]

Respect, you are very clever person.
Thank you for this problem which is solved by you