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hie..my problem is in topic ratio & proportion

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pratik
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hie..my problem is in topic ratio & proportion [#permalink] New post   24 Jan 2010, 02:28
if (ab+bc+ac)whole square=ac(a+b+c)whole square
prove that:a & b are in continued proportion
i.e:a/b=b/c



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Re: hie..my problem is in topic ratio & proportion [#permalink] New post   24 Jan 2010, 12:14
There's no way you'd see this on the GMAT, but here goes.

Prove: a/b = b/c ---> ac = b^2
Given: (ab+bc+ac)^2 = ac(a+b+c)^2

Distribute:
(ab+bc+ac)^2 = ac(a+b+c)^2
ab(ab+bc+ac)+ bc(ab+bc+ac) + ac(ab+bc+ac) = ac(a^2 +b^2+c^2+2ab+2ac+2bc)
(ab)^2 + 2abbc + 2aabc + (bc)^2 + 2abcc +(ac)^2 = aaac + abbc + accc +2aabc+2(ac)^2+2abcc

Cancel identical terms on right and left:
(ab)^2 + abbc + (bc)^2 = aaac + accc + (ac)^2

Factor out a b^2 on the left, and an ac on the right:
(b^2)(a^2 + ac + c^2) = (ac)(a^2 + c^2 + ac)

Cancel a^2 + c^2 + ac from each side:
b^2 = ac
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Re: hie..my problem is in topic ratio & proportion [#permalink] New post   25 Jan 2010, 04:24
esledge wrote:
There's no way you'd see this on the GMAT, but here goes.

Prove: a/b = b/c ---> ac = b^2
Given: (ab+bc+ac)^2 = ac(a+b+c)^2

Distribute:
(ab+bc+ac)^2 = ac(a+b+c)^2
ab(ab+bc+ac)+ bc(ab+bc+ac) + ac(ab+bc+ac) = ac(a^2 +b^2+c^2+2ab+2ac+2bc)
(ab)^2 + 2abbc + 2aabc + (bc)^2 + 2abcc +(ac)^2 = aaac + abbc + accc +2aabc+2(ac)^2+2abcc

Cancel identical terms on right and left:
(ab)^2 + abbc + (bc)^2 = aaac + accc + (ac)^2

Factor out a b^2 on the left, and an ac on the right:
(b^2)(a^2 + ac + c^2) = (ac)(a^2 + c^2 + ac)

Cancel a^2 + c^2 + ac from each side:
b^2 = ac

hey....
thank you very much.....
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Re: hie..my problem is in topic ratio & proportion [#permalink] New post   12 Mar 2010, 12:47
Is there any way we can solve this problem without expanding the squares both thew sides?
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Re: hie..my problem is in topic ratio & proportion [#permalink] New post   14 Mar 2010, 15:02
sidhu4u wrote:
Is there any way we can solve this problem without expanding the squares both thew sides?

I don't think so, just because of order-of-operations. On the left you have a square, on the right a term times something in parentheses squared. You have to deal with the parentheses first, meaning you must square to be able to compare the right and left sides.

However, there is probably some strategic term grouping and/or cancelling that could be done at this step:

ab(ab+bc+ac)+ bc(ab+bc+ac) + ac(ab+bc+ac) = ac(a^2 +b^2+c^2+2ab+2ac+2bc)

I just showed the full squaring and distribution for clarity.



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Re: hie..my problem is in topic ratio & proportion [#permalink]
14 Mar 2010, 15:02
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