Don't get trapped in a loop
a b + a + 1 2 a x + b c + b + 1 2 b x + c a + c + 1 2 c x = 1
If there are distinct positive numbers a , b , c such that a b c = 1 and they satisfy the above equation , Find the value of x .
Source : A book
The answer is 0.5.
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7 solutions
Moderator note:
A common error made in solving this problem is to calculate it for a particular set of x , y , z . That allows you to determine the numerical answer to the problem. However, it is not a proof for all possible sets x y z = 1 .
a, b, c can be substituted as the cube roots of unity.
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Can you explain how that works?
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a = 1 , b = ω , c = ω 2
⇒ 2 a x ( ω + 2 1 + ω + 2 ω 2 1 + 1 ) = 1
⇒ 2 x ( 2 ω 2 + 1 + 2 + ω + 1 + 2 ω 2 ) = ( ω + 2 ) ( 2 ω 2 + 1 )
⇒ 2 x ( 4 ω 2 + ω + 4 ) = 4 ω 2 + ω + 4
⇒ 2 x = 1
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@Shubhendra Singh – As stated, you have only shown this for one particular case of a , b , c , and not for all possible cases of a b c = 1 .
@shubhendra singh Can you post it as an independent solution?
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I've posted it.
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@Shubhendra Singh – I saw it and upvoted :)Thanks for following me.Congrats for being my 800th follower. :P
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@Nihar Mahajan – Oh ! I just didn't noticed that !! Congrats .
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@Shubhendra Singh – Thanks. LOL we are congratulating each other. Cheers!
Substituting a = q p , b = r q , a n d c = p r in c y l ∑ a b + a + 1 2 a x ⇒ c y l ∑ b + 1 + a 1 2 x ⇒ c y l ∑ r q + 1 + p q 2 x ⇒ c y l ∑ ∑ p q 2 p r x which can be seen as 2x = 1, hence x = 0.5
Put a = b = c = 1 and you will get x = 2 1
Moderator note:
Even if you have a = 2 1 , b = 1 , c = 2 , how would you know that it's true that x = 0 . 5 for all a , b , c such that a b c = 1 .
Nice .But now I have edited the question to distinct. How would you proceed now? @Samrit Pramanik
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Ok!, then I will put a = 2 1 , b = 1 , c = 2 ,
now we get x = 2 1
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Nice.... !but incomplete. :P Change your solution accordingly.
@Calvin Lin There is a typo mistake in the challenge master's note. a b c = 1 .
Hey i too did the same way :p
a , b , c can be substituted as the roots of the equation x 3 = 1
a = 1 , b = ω , c = ω 2
⇒ 2 a x ( ω + 2 1 + ω + 2 ω 2 1 + 1 ) = 1
⇒ 2 x ( 2 ω 2 + 1 + 2 + ω + 1 + 2 ω 2 ) = ( ω + 2 ) ( 2 ω 2 + 1 )
⇒ 2 x ( 4 ω 2 + ω + 4 ) = 4 ω 2 + ω + 4
⇒ 2 x = 1
NOTE- ω 3 = 1
Same issue as before, how do you know that this will hold for any set of distinct positive numbers, a , b , c ? You have shown that it holds for a special set that is not restricted to positive numbers.
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My guess is that the logic used is that this puzzle wouldn't have been set unless it has a constant solution regardless of the particular values of a , b , c as long as they satisfy a b c = 1 .
So while the rigorous way round is to manipulate the equation to eliminate a , b , c and therefore solve for x , plugging specific values in is a convenient shortcut - cutting the Gordian knot, so to speak.
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Right, there's a difference between "finding the answer to the problem" and "proving that the answer is correct". So this solution would only get the "answer mark", but not the "method marks".
a b + a + 1 2 a x + b c + b + 1 2 b x + a c + c + 1 2 c x = 1
and a b c = 1 so a c = b 1 and a 1 = b c
Simplify
a 1 a 1 ⋅ a b + a + 1 2 a x + b c + b + 1 2 b x + a c + c + 1 2 c x = 1
b + 1 + a 1 2 x + b c + b + 1 2 b x + b 1 + c + 1 2 c x = 1
substitute in a 1 = b c
b + 1 + b c 2 x + b c + b + 1 2 b x + b b ⋅ b 1 + c + 1 2 c x = 1
b + 1 + b c 2 x + b c + b + 1 2 b x + 1 + b c + b 2 b c x = 1
b c + b + 1 2 b c x + 2 b x + 2 x = 1
2 x = 1
x = 2 1
Ignore everything except the fact the abc = 1. In each numberator, we see the distincitive letters "a" "b" and "c" being modified by x and 2. Since we KNOW these modifiers must still allow abc to = 1, x must cancel out the 2 such that 2x = 1. Therefore, x = 0.5
Could you please elaborate on that?
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Sure. From number theory: if abc=1, and we assume a and b and c are positive integers, then a=1, b=1, and c=1. Now we can see that the three fractions are 2x/3 + 2x/b + 2x/c = 1. We need thirds so as to add up to one. Therefore, x= 1/2 or 0.5. No complicated algebra necessary. Just observational skills and common sense.
I should note that this solution applies only to this particular set of a, b and c. There are other solutions possible of course, and solving them as the cube roots of unity allows for this fact. I will leave the proof-writing to those with more interest in the topic :)
a b + a + 1 2 a x + b c + b + 1 2 b x + c a + c + 1 2 c x = 1 ⇒ ( b c ) [ a b + a + 1 ] ( b c ) 2 a x + b c + b + 1 2 b x + ( b ) [ c a + c + 1 ] ( b ) 2 c x = 1 ⇒ a b c ( b ) + a b c + b c 2 a b c x + b c + b + 1 2 b x + a b c + b c + b 2 b c x = 1 ⇒ b + 1 + b c 2 x + b c + b + 1 2 b x + 1 + b c + b 2 b c x = 1 ⇒ 1 + b + b c 2 x ( 1 + b + b c ) = 1 ⇒ 2 x = 1 ⇒ x = 2 1
Note: Appreciate the trick used above.This is not my solution.