Algebra Set

Hi! This week's set consists in three problems in algebra that seem nice. I'm using the word "seem" because I've been working on them and haven't been able to solve them. Anyway, enjoy the problems, and, if you find them nice and/or fun, please reshare the note, so more people will be able to see the problems and try to tackle these hard problems. Just as a reference, problem 1 was taken from the Colombian book "Un recorrido por el álgebra" and problems 2 and 3 were taken from the first Costa Rican IMO TST 2014.

  1. Find all functions f:Z+Rf:\mathbb{Z}^+\rightarrow\mathbb{R} such that f(n+m)+f(nm)=f(3n)f(n+m)+f(n-m)=f(3n) for every n,mZ+,n>mn,m\in{\mathbb{Z}^+}, n>m.

  2. Find all real solutions (x1,...,xn)(x_1,...,x_n) of the equation (1x1)2+(x1x2)2+...+(xn1xn)2+xn2=1n+1(1-x_1)^2+(x_1-x_2)^2+...+(x_{n-1}-x_n)^2+x_n^2=\dfrac{1}{n+1}

  3. Let x,yx, y be real numbers such that y>0y>0 and 7x2+3xy+3y2=17x^2+3xy+3y^2=1. Show that x2+y2y12\dfrac{x^2+y^2}{y}\ge{\dfrac{1}{2}}

Note by José Marín Guzmán
5 years, 4 months ago

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For the second one:

Using the Cauchy-Schwarz Inequality we can write, ((1x1)2+(x1x2)2++(xn1xn)2+xn2)(12+12++12+12)((1x1)+(x1x2)++(xn1xn)+xn)2 \displaystyle \begin{array}{c}\\ \bigg( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \bigg) \bigg( 1^2 + 1^2 + \dotsb + 1^2 + 1^2 \bigg) \geq & \\ \bigg( (1-x_1) + (x_1 - x_2 ) +\dotsb + (x_{n-1} - x_n)+ x_n \bigg)^2 & \\ \end{array}

((1x1)2+(x1x2)2++(xn1xn)2+xn2)(n+1)(1)2 \displaystyle \Rightarrow \Big( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \Big) (n+1) \geq ( 1)^2 (1x1)2+(x1x2)2++(xn1xn)2+xn21n+1\displaystyle \Rightarrow (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \geq \frac{1}{n+1}

But it is given that the equality condition holds true. Therefore, from the condition of equality in Cauchy-Schwarz, we obtain,
1x11=x1x21==xn1xn1=xn1 \displaystyle \frac{1-x_1}{1} = \frac{x_1 - x_2}{1} = \dots = \frac{x_{n-1} - x_n}{1} = \frac{x_n}{1} xn1=2xn , xn2=3xn,, x2=(n1)xn and x1=nxn\displaystyle \Rightarrow x_{n-1} = 2x_n \text{ , } x_{n-2} = 3x_n ,\dotsc, \text{ } x_2 = (n-1)x_n \text{ and } x_1 = nx_n .
Also, 1=x1+xn(n+1)xn=1xn=1n+1\displaystyle 1 = x_1 + x_n \Rightarrow (n+1)x_n = 1 \Rightarrow x_n = \frac{1}{n+1} .

Hence, xi=n+1in+1 ; i=1,2,3,,n1,n\displaystyle x_i = \frac{n+1-i}{n+1} \text{ ; } i = 1,2,3, \dotsc , n-1, n

Sudeep Salgia - 5 years, 4 months ago

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May I know how you arrive at a conclusion for xix_{i}?

Shaun Loong - 5 years, 4 months ago

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From third last line, xi=(n+1i)xnx_{i} = (n+1-i)x_{n} and xn=1n+1x_{n} = \frac{1}{n+1}.

Samuraiwarm Tsunayoshi - 5 years, 4 months ago

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@Samuraiwarm Tsunayoshi Thanks, I see it now :)

Shaun Loong - 5 years, 4 months ago

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I know of a "crude" solution to 3, which only uses basic knowledge of conic sections. It has a nice interpretation that way.

Calvin Lin Staff - 5 years, 4 months ago

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I'm afraid I don't know much about conic sections, but share the solution and I'll try to look up the stuff I don't understand :). Besides, it would be useful for other people who may know about conic sections.

José Marín Guzmán - 5 years, 4 months ago

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Show that the circle 2x2+2y2y=0 2 x^2 + 2 y^2 - y = 0 is contained within the ellipse 7x2+3xy+3y2=1 7x^2 + 3xy + 3y^2 = 1 . In fact, these conic sections are tangential at the point (15,25) ( \frac{ 1}{5}, \frac{2}{5}) .

Hence, it follows that for any point on the ellipse which satisfy y0 y \geq 0 , we have

2x2+2y2y0x2+y2y12. 2 x^2 + 2y^2 - y \geq 0 \Rightarrow \frac{ x^2+y^2} { y} \geq \frac{1}{2}.

The tangential point is the equality case.

Calvin Lin Staff - 5 years, 4 months ago

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@Calvin Lin awesome man!!........... nice approach....... thanks:)

Abhinav Raichur - 5 years, 4 months ago

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For no.1 I have a weird start, this might helps.

Substituting m1,2,3,...,n1m \rightarrow 1, 2, 3, ..., n-1 we get

f(2ni)+f(i)=f(3n)=f(2nj)+f(j)f(2n-i) + f(i) = f(3n) = f(2n-j) + f(j) for every i,j=1,2,...,n1i,j = 1, 2, ..., n-1

We see that if a+b=c+da+b = c+d, then f(a)+f(b)=f(c)+f(d)f(a)+f(b)=f(c)+f(d) for positive integers a,b,c,da,b,c,d.

I have no idea to do anything right meow. I suck at this thing XD

Samuraiwarm Tsunayoshi - 5 years, 4 months ago

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Feel free to post ideas and solutions! :D

José Marín Guzmán - 5 years, 4 months ago

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For question 22, does the solution only occur when n=0n=0 with x0=1x_{0}=1?

Shaun Loong - 5 years, 4 months ago

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