6

For positive real numbers $a$ and $b,$ let $$a \diamond b = \frac{\sqrt{a^2 + 4ab + b^2 - 2a - 2b + 9}}{ab + 6}.$$Find $$( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1.$$

I can't find any quick way to do this. Can anyone help? Thanks in advance!

Mike Smith
  • 1,122
  • 6
  • 26

1 Answers1

16

Notice for all positive number $a$, we have

$$a \diamond 2 = \frac{\sqrt{a^2 + 8a + 4 - 2a - 4 + 9}}{2a+6} = \frac{\sqrt{a^2+6a+9}}{2a+6} = \frac12$$

The mess at hand equals to $$( (\cdots) \diamond 2) \diamond 1 = \frac12 \diamond 1 = \frac{\sqrt{37}}{13}$$

achille hui
  • 122,701
  • does it follow from the fact the $((a \diamond b) \diamond c) = (a \diamond (b \diamond c))$? – Alex Jun 01 '20 at 22:25
  • 1
    @Alex, nope. The first thing I do is check whether the $\diamond$ operation is associative and it isn't. – achille hui Jun 01 '20 at 22:26
  • In your calculation, $a \diamond 2$, what is $a$? All terms between 2017 and 3 or 3? – Alex Jun 01 '20 at 22:33
  • @Alex $a$ can be any positive number. if you substitute it by all the terms between 2017 and 3, you get the mess between 2017 and 2. – achille hui Jun 01 '20 at 22:36
  • OK, so in you construct $a=(\ldots)$, i.e. $\diamond$ between all terms between 3 and 2017. Is it guaranteed to be positive? – Alex Jun 01 '20 at 22:47
  • I tried solving for $a \diamond (a-1)$, didn't get anything useful – Alex Jun 01 '20 at 22:51
  • Since $a^2 + 4ab + b^2 - 2(a+b) + 9 = (a+b - 1)^2 + 2ab + 8$. If you start from any $a, b > 0$, this will be $> 8$, so $a \circ b$ is positive. Repeat applying this, one can deduce all terms between 2017 and 3 is positive. – achille hui Jun 01 '20 at 23:07
  • OK I think I got it, so essentially $a=(\ldots)$ – Alex Jun 02 '20 at 00:04