Problem Solving Skills for Mathcounts Competitions

Mathcounts => Mathcounts => : spotula May 21, 2013, 08:48:04 PM

: Matcounts 2011 National Target round problem 6
: spotula May 21, 2013, 08:48:04 PM

If a  and b are unequal real numbers  and (a-b)/a=b/(a-b), what is the sum of all possible values?

I am thankful for any insight.

: Re: Matcounts 2011 National Target round problem 6
: spotula May 21, 2013, 08:58:30 PM

Is this a possible answer

(a-b)^2 = ab

if we take b = 1;

(a-1)^2 = a; this is a^2-3a+1=0

quadratic solutions (3+sqrt(5))/2 and (3-sqrt(5))/2  and adding these two solutions gives an answer of 3.

--- This is not soulful but this is correct answer. Is there a better way to solve?

: Re: Matcounts 2011 National Target round problem 6
: yongcheng3315 May 22, 2013, 07:39:14 AM

We do this:
From (a-b)^2=ab, we get: a^2-3ab+b^2=0   (1)
We know that none of a and b is 0.  
Let a/b = x, (1) becomes x^2-3x+1=0 by dividing both sides by b^2.
This is a quadratic equation. By the Vieta's Theorem, the sum of the roots is -(-3) = 3.
3 is the answer.

Hope this will  help.

: Re: Matcounts 2011 National Target round problem 6
: spotula May 22, 2013, 08:14:00 AM

Thanks, you have the best :)