Proving a,b,c are in an AP: Solving 25(9a^2+b^2)+9c^2-15(5ab+bc+3ca)=0

[utkarshakash]

Homework Statement

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.

Homework Equations

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?

 

[Mandelbroth]

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 Arithmetic Progression

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

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.

 

[Ray Vickson]

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

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.

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

RGV

 

[utkarshakash]

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

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.

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.

 

[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.

 

[utkarshakash]

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.

Give me an example.

 

[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

 

[haruspex]

Rewrite the equation as (15a)²+(5b)²+9c²-(15a)(5b)-3(5b)c-3(15a)c=0

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.

 

[utkarshakash]

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

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

 

[ehild]

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

ehild