Interesting little problem?

[The Highlander]

I thought this might be an interesting exercise for some to practise their algebra...

ab = 100
bc = 200
ca = 300

Find the (exact) value of: a + b + c​

I'm not asking for any help or suggestions (I already have two separate procedures to get to the answer) I just thought it would be interesting to see how others approach it.

If you can be bothered to tackle it then please post your solution(s) in full.

NB: The answer should be an exact value, ie: left as a (fully simplified) surd multiple not a decimal approximation.

 

[mrtwhs]

[imath]\dfrac{55 \sqrt{6}}{3}[/imath] or [imath]\dfrac{-55 \sqrt{6}}{3}[/imath]

 

[The Highlander]

[imath]\dfrac{55 \sqrt{6}}{3}[/imath] or [imath]\dfrac{-55 \sqrt{6}}{3}[/imath]

You need to show the solution (ie: the working) not just the answer (which is readily available through any Maths calculator App).

 

[mrtwhs]

From the third equation [imath]c = \dfrac{300}{a}[/imath]

Plug into the second equation and simplify to get [imath]\dfrac{b}{a}=\dfrac{2}{3}[/imath] so [imath]b=\dfrac{2a}{3}[/imath]

Plug into the first equation and simplify to get [imath]a^2=150[/imath]

Square root and back substitute.

 

[mrtwhs]

... or ... multiply all the equations together to get [imath]a^2b^2c^2=6000000[/imath]

Square root and then divide by each equation.

 

[The Highlander]

I've since discovered there are dozens of solutions already posted online, lol.
Wasn't as original as I first thought.
(eg: Googling it,)

 

[Aion]

I find this approach appealing because it avoids almost all algebra and works directly with the symmetry of the given products.


Given:
[math]ab=100,\quad bc=200,\quad ca=300[/math]
We reconstruct each variable symmetrically:
[math]a = \sqrt{\frac{ab \cdot ac}{bc}}, \quad b = \sqrt{\frac{ab \cdot bc}{ca}}, \quad c = \sqrt{\frac{bc \cdot ca}{ab}}[/math]
Then (up to a global sign),
[math]a = \pm 5\sqrt{6}, \quad b = \pm \frac{10\sqrt{6}}{3}, \quad c = \pm 10\sqrt{6}[/math]
Hence,
[math]a+b+c = \pm \frac{55\sqrt{6}}{3}.[/math]

 

[The Highlander]

I find this approach appealing because it avoids almost all algebra and works directly with the symmetry of the given products.

Given: \(\displaystyle \quad ab=100,\quad bc=200,\quad ca=300\)
We reconstruct each variable symmetrically:\(\displaystyle \quad a = \sqrt{\frac{ab\cdot ac}{bc}}, \quad b = \sqrt{\frac{ab \cdot bc}{ca}}, \quad c = \sqrt{\frac{bc \cdot ca}{ab}}\)
Then (up to a global sign),\(\displaystyle \quad a = \pm 5\sqrt{6}, \quad b = \pm \frac{10\sqrt{6}}{3}, \quad c = \pm 10\sqrt{6}\)
Hence,\(\displaystyle \; a+b+c = \pm \frac{55\sqrt{6}}{3}.\)​

Brilliant! I love it.
Some readers might struggle to see how you get from one line to the next but the efficiency & clarity of your post is stunning!
Bravo!