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Problem 494

AMC 10/12, early questions

Algebra Difficulty 3.0 Prove it CEMC Hypatia · Canada · 2014

For real numbers $a$ and $b$ with $a\geq 0$ and $b\geq 0$ , the operation $\odot$ is defined by $a \odot b=\sqrt{a+4b}.$ For example, $5 \odot 1=\sqrt{5+4(1)}=\sqrt{9}=3$ .

What is the value of $8\odot 7$ ?
If $16\odot n=10$ , what is the value of $n$ ?
Determine the value of $(9\odot 18)\odot 10$ .
With justification, determine all possible values of $k$ such that $k\odot k=k$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Using the given definition, $8 \odot 7=\sqrt{8+4(7)}=\sqrt{36}=6$ .
Since $16 \odot n=10$ , then $\sqrt{16+4n}=10$ or $16+4n=100$ (by squaring both sides) or $4n=84$ and so $n=21$ .

We check that indeed $16 \odot n=16 \odot 21=\sqrt{16+4(21)}=\sqrt{100}=10$ .
We first determine the value inside the brackets: $9 \odot 18=\sqrt{9+4(18)}=\sqrt{81}=9$ .

So then $(9 \odot 18)\odot 10=9 \odot 10=\sqrt{9+4(10)}=\sqrt{49}=7$ .
Using the definition, $k \odot k=\sqrt{k+4k}=\sqrt{5k}$ .

So we are asked to solve the equation $\sqrt{5k}=k$ .

Squaring both sides we get, $5k=k^2$ and so $k^2-5k=0$ or $k(k-5)=0$ , and so $k=0$ or $k=5$ .

Checking $k=0$ , we obtain $k \odot k=0 \odot 0=\sqrt{0+4(0)}=\sqrt{0}=0=k$ , as required.

Checking $k=5$ , we obtain $k \odot k=5 \odot 5=\sqrt{5+4(5)}=\sqrt{25}=5=k$ , as required.

Thus, the only possible solutions are $k=0$ and $k=5$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.