# Prove a,b,c are in AP

1. Nov 22, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
If a,b,c are real numbers satisfying the equation $25(9a^2+b^2)+9c^2-15(5ab+bc+3ca)=0$, then prove that a,b,c are in A.P.

2. Relevant equations

3. The attempt at a solution
Rearranging the question I can say that I have to prove this
2b=a+c

I tried simplifying the terms and I arrived at this
$(15a+5b+3c)^2=45(5ab+bc+3ac)$

Any ideas how should I proceed?

2. Nov 22, 2012

### Mandelbroth

$25(9a^2+b^2)+9c^2-15(5ab+bc+3ca)=0$

Distribute.
$225a^{2} + 25b^{2} + 9c^{2} - 75ab - 15bc - 45ac = 0$

Now comes the weird part. You don't need to know the values of a, b, and c if you want to prove that $2b = a + c$. You just need to know the ratios between them.

Solve for a, b, and c in terms of each other. I decided to solve b and c in terms of a.

This way, we get $b = 3a$ and $c = 5a - \frac{5\sqrt{-9 a^2 + 18a^{2} - (3a)^{2}}}{2\sqrt{3}} = 5a - \frac{5\sqrt{9a^{2} - 9a^{2}}}{2\sqrt{3}} = 5a$

Thus, $2(3a) = a + 5a$, ∴ $2b = a + c$. QED.

3. Nov 22, 2012

### Ray Vickson

You are not supposed to do a person's homework for them; hints only.

RGV

4. Nov 22, 2012

### utkarshakash

I simply can't understand what you're up to. Are you making a quadratic in b and then solving it? If yes then how did you get such a nice value? Mine becomes very long and complicated. The other possible situation I can think of is this.

Are you expanding it like this

$15(15a^2-bc)+25b(b-3a)+9c(c-5a)=0$

and setting each of term equal to zero (Since sum of +ve numbers can't be 0 each term must be zero) and then obtaining values of b and c in terms of a. If this is the case then I've got it.

Last edited: Nov 22, 2012
5. Nov 22, 2012

### haruspex

You can greatly simplify the expressions by making some simple substitutions for a, b and c.
Notice how wherever a appears there's a factor of 15 around? Similarly the other two variables.

6. Nov 23, 2012

### utkarshakash

Give me an example.

7. Nov 23, 2012

### ehild

Rewrite the equation as (15a)2+(5b)2+9c2-(15a)(5b)-3(5b)c-3(15a)c=0
You can choose x=15a/c and y=5b/c as new variables.

The new equation is

x2+y2+9-xy-3y-3x=0

or x2-x(y+3)+y2-3y+9=0

It is a quadratic equation in x. The discriminant is D=(y+3)2-4(y2-3y+9)=-3(y2-6y+9), negative or zero. To get a solution, D must be zero, so y=3. That involves x= 3.

ehild

8. Nov 23, 2012

### haruspex

I had in mind something very similar: t = 15a, u =5b, v = 3c producing t2 + u2 + v2 = uv + vt + tu.
This can be turned into (t-u)2 + (u-v)2 + (v-t)2 = 0, which has only the solution t = u = v.

9. Nov 24, 2012

### utkarshakash

I always find your solutions to be better than others. You are a life-saver. Thank you so much for helping me!

10. Nov 24, 2012

### ehild

To be fair, it was Haruspex's idea and his solution is the best.

ehild