A math olympiad problem that looks complicated, but it's easy to solve

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@higher_mathematics

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[jessewallis]

I used a=100/b; c=200/b; then substituted. Ended up with the right answer.

[srmpenedo]

Its easier to calculate a, b and c and then sum them up

[Urgleflogue]

There's no way this is math olympiad problem, I almost solved it in my head.

[МихаилКоган-чн]

2ab=bc => c=2a => 2a^2=300 => a=+-5sqrt(6), b=+-10sqrt(6)/3, c=+-10sqrt(6)
a+b+c=+-55sqrt(6)/3

[허공]

simple solution
multiplying 3 equations,
(abc)^2 = 6*10^6
abc = +-sqrt(6)*10^3 =>
a = +-sqrt(6)/2*10,
b = +-sqrt(6)/3*10,
c = +-sqrt(6)/1*10,
a+b+c = +-(sqrt(6)*11/6*10)

[wolfgangmaichen]

you missed the negative solution! Very easy to see. Just enter -a, -b, and -c in the original equations (instead of a, b, c) and the equations still remain valid. Thus, -a-b-c = -(a+b+c) is aso a valid solution.

[hongphuc]

*a/c=1/2=>c=2a*
*ac=a×2a=2a^2=300*
*a=√150=√(6×5^2) = 5√6*

[zoezulma]

It certainly is easy to solve, if you are a math whiz. For an ordinary mortal it is rather gnarly. My approach which I find to be simpler but uses more steps uses substitutions to solve for each variable: c times a equals 300 so c equals 300 over a. Replacing c with 300 over a in b times c equals 200 you wind up with b equals 2a over 3. Substituting that for b in a times b equals 100 you get a squared equals 150. Continuing this approach to solve for the other variables you get the same answers for a squared, b squared, and c squared as you but I find transforming the sum of the squares of the three variables to the sum of the three variables to be a bit daunting. It is straight forward but 'easy' is not the adjective I would use to describe the process.

[shawngayner]

I had a geometric solution. Draw a right trapezoid with the top being a, the side being b, and the bottom being c. The diagonal is the hypotenuse of a right triangle with sides b and (c-a) which fortunately for us is just a since c = 2a. This triangle also has an area of 50 from (200 - 100)/2. Using the formula for the area of a triangle, 1/2 b*a = 50, and since a = (3/2)b, we can substitute and solve, b = +/- sqrt(200/3). Since a=(3/2)b and c=3b, a+b+c is 5.5b. Take out the sqrt(100) to get +/-55*sqrt2/sqrt3 and simplify. I enjoyed this very much, thank you for the content!

[MuTcyXuDe]

ac = 3ab => c = 3a
thus 3a^2 = 300.
a = 5*sqrt(6)
rest is simple substitution

[duckwantbreads]

This method seems overly complicated considering that the ratios are so nice to work with, it's easy to see 2a = c immediately.

From there you can work out (by subbing into the third equation) that

a = sqrt(150) = 5*sqrt(6)

Therefore c = 10*sqrt(6).

Then subbing into the first equation you have

b = 20/sqrt(6)

Then it's pretty easy (by multiplying top and bottom of a, b and c by sqrt(6)) to add the three numbers up to the correct answer.

[mikmak]

easiest way: abc=1000*route(6) then abc/ab=c=10*route(6) and so route(6) c=10 route(6)....the add them up

[kkokaki]

6:44 you forget to add the negatif solution here

[XfeeXg]

ab=100 (*); bc=200(**) ; ca=300(***).
Nếu a, b, c khác 0 thì
bc=2ab
Chia 2 vế cho b ta được
c=2a
Thay kết quả trên vào(***), ta có: 2a*2=300
a*2=150
a1=+ 5×6^2 ;b1=10×6^2/3; c1=10×6^2
a2= -5×6^2; b2= -10×6^2 /3 ;
c2= - 10×6^2.

[EvanEscher]

We can figure out that a/b = 3/2, c/b = 3/1, and c/a = 2/1. Given that, let a = x. Since we know ca=300, we know that 2x^2=300, then x=√150, which can be reduced to a=5√6. Since we know c is 2 times a, c=10√6. Since we know c is 3 times b, b = (10√6)/3

Therefore, a+b+c=(55√6)/3

[GIANFRANCOZOLA]

It was quicker for me noticing that all the 3 multiplied are which means abc=1000sqrt(6). Then from the 1st you find c= 10sqrt(6), and the other two swiftly follow consequently, and then sum the 3. basically 6 steps.

[TheOnlyOpie]

3 pages later... super easy, barely an inconvenience

[davidhowe]

For practical reasons, as each product is a multiple of 100, you can take the multiples of 10 of a, b, c so AB=1, BC=2, CA=3 (but you have to remember to convert back at the end!)

[Gnowop]

Apply method of elimination and or substitution to get a b c is standard. Then it is easy enough to add a b c results together. Don't need to do (a+b+c)^2. Don't forget the negative answer though.

[jharvick]

This is a very long way to solve this. Here's a faster solution:

multiply all three equations together, you get (abc)^2 = 6, 000, 000, or, abc=1000rad(6)

Divide abc = 1000rad(6) by each of the equations to get each value. Two variables cancel out each time, leaving you with one left.

c= 10rad6
a=5rad6
b=10rad(6)/3

c= 30rad(6)/3
a=15rad(6)/3
b=10rad(6)/3

Add them up, 55rad(6)/3