## Sunday, 2 January 2011

### Mathematics : Cyclic Equations

Consider this (suspected ancient CE question)
Find value(s) of x, given a, b, c are real number satisfying
x = a/(b+c) = b/(a+c) = c/(a+b)
In my opinion this is quite hard in CE, even in senior olympiad level....
First of all, 3 equations, 4 unknowns, what you should think of is eliminating one of the variables.
Since x is what to be solved, we should not eliminate x, and the other three variables are in cyclic form so that they can't be eliminated. What should you do?

Consider the degree of the equations, they are degree zero, so that they are a proportion only.
Take an example, if (x,y,z) is a solution of (a,b,c), then (2x,2y,2z) can also be a solution.
Write k(a0,b0,c0) as the general solution of it. Then there exist a k such that k(a0,b0,c0) = (1,y,z).

Therefore, WLOG assume a=1, or you can assume a is a fixed constant.
Transform the equation like this: (a=1)
1+c = b^2+bc
b^2+b=c^2+c
The second equation shows the cyclic nature again. Consider x^2+x=y as a parabola, having y=-0.5 as the symmetrical axis, we should consider these two case:
1) b = c, which is obvious,
2) Consider the symmetrical axis, b+c=-1

For case 1, put b=c into 1+c=b^2+bc gives b = -0.5 or 1, -0.5 refers to b=c AND b+c=-1, so overlapped. Therefore b=c=1, a=b=c=1, x=1/(1+1) = 1/2

For case 2, 1/(b+c) = -1=x
Therefore x = 1/2 or -1, the corresponding (a,b,c) are: k(1,1,1), k(1,x,-1-x) AND their permutations. We should never forget the permutations in cyclic equations!

Here's another beautiful method to reach x = 1/2.
Rearrange the terms:
b+c = a/x
a+c = b/x
a+b = c/x
Sum them up: 2(a+b+c) = (a+b+c)/x, x = 1/2.
Why this method doesn't give x = -1? It's because both equations are divided by (a+b+c) is the last step, the dividing is correct ONLY IF a+b+c is non-zero. When x=-1, a+b+c = k(1+x-1-x) = 0 so it's wrong.

EDIT: Now, we may take a+b+c = 0 to sub. into the equation.
b+c = (-a) = a/x, x is obviously -1.
It seems that this solution satisfies the CE level, but it's still quite hard.

Another example from prelim 09':
t = u+1/v = v+1/w = w+1/x = x+1/u, where u,v,w,x, are distinct real and t is positive. Find t.
In the first thoughtyou may found the question extremely complicated. Also 5 unknowns with 4 eqautions, so you would like to assume u=1. However you should notice that there's more than one set of (u,v,w,x) satisfying t, but they are not linearly related. i.e., if (u0,v0,w0,x0) is one of the solution, k(u0,v0,w0,x0) (k is not 1) is not nessacarily be a solution. So a better way is to fix u as a constant, in another way, express v,w and x is the two fixed variables, u and t.
x = t-1/u = (ut-1)/u
w = t-1/x = (ut^2-t-u)/(ut-1)
v = t-1/w = (ut^3-t^2-2ut+1)/(ut^2-t-x)
Now sub. v into t = u+1/v, and rearrange the terms by a super hard factorization, we will get
t(t^2-2)(ut^2-u^2-1)=0
Note that the last term is non-zero, since the polynomial is parlindromic (refer to BAS note), it gives two roots of u which is reciprocal to each other, implies the existance of u+1/u = t = u+1/v, hence u=v which violates the given rule.
Therefore t = rtsq 2.

More to explore:
1) Find x if x = ab/(ac+bc) = bc/(ac+ab) = ac/(ab+bc)
2) Find x if x = a/(b+c+d) = b/(a+c+d) = c/(a+b+d) = d/(a+b+c)  (modified past IMO)
3) Find x if x = a^2/(b^2+c^2) = b^2/(a^2+c^2) = c^2/(a^2+b^2)
4) Find x if x = (a+b)/(c+d) + (a+c)/(b+d) + (a+d)/(b+c) + (b+c)/(a+d) + (b+d)/(a+c) + (c+d)/(a+b)
5) Find all integer solutions for (1) x+y+z = 3, (2) x^3+y^3+z^3 = 3  (HKMO 10, classical question)

fin.