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The numbers $2020,2019,2018,...,1$ are written on the blackboard from left to right.

John repeats the following process until there is only one number left:

John chooses the first two numbers from the left, namely $a,b$, and replaces them by $$\frac{\sqrt{a^2+3ab+b^2-2a-2b+4}}{ab+4}$$, which is written on the left of all other numbers.

What is the remaining number?


Source: HK Prelim 2010 Q19(Numbers changed)

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    $\begingroup$ Should this have the [no-computers] tag? $\endgroup$ – Gareth McCaughan Jun 20 at 14:09
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Let $$f(a,b) := \frac{\sqrt{a^2+3ab+b^2-2a-2b+4}}{ab+4}.$$

John's procedure is now to repeatedly replace the leftmost two values $a,b$ on the blackboard by the single value $f(a,b)$.

I claim that John never writes a negative value onto the blackboard.

Proof: the blackboard always begins with positive values. Suppose $a$ and $b$ are positive values John erases. Then $\sqrt{a^2+3ab+b^2-2a-2b+4} = \sqrt{(a-1)^2+(b-1)^2 + 3ab + 2}$ is also positive, as well as $ab+4$. So the new blackboard value $f(a,b)$ is also positive.

Suppose John does his thing until there are only three values left on the blackboard: $x, 2, 1$.

Then after the next step, the two remaining numbers will be $f(x,2)$ and $1$.

Let's compute $f(x,2)$:

$$f(x,2) = \frac{\sqrt{x^2+6x+4-2x-4+4}}{2x+4} = \frac{\sqrt{x^2+4x+4}}{2x+4} =\frac{|x+2|}{2x+4} = \frac 12.$$

After that step, the board contains $\frac 12$ and $1$.

The final number on the board is therefore $$f\left(\frac 12,1\right) = \frac{\sqrt{\frac 14 + 3\cdot\frac 12 + 1 - 1 - 2 + 4}}{\frac 12 + 4} = \frac{\sqrt {\frac{15}4}}{\frac 92} = \frac{\sqrt {15}}9.$$

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  • 3
    $\begingroup$ That's quite neat. $\endgroup$ – hexomino Jun 20 at 17:51
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    $\begingroup$ Damn I missed the wording that the result is placed on the left and reused, so I didn't know what to do with f(a,2) = 1/2. Enjoy your rep :) $\endgroup$ – ghosts_in_the_code Jun 20 at 18:08
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    $\begingroup$ Correct! This is the solution! $\endgroup$ – Culver Kwan Jun 21 at 0:11

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