Dear Uncle Colin,

I’m struggling with this STEP question. The first two parts are fine – equality holds when there is some constant $k$ for which $a = kx$, $b = ky$ and $c=kz$, and part (i) follows directly from the original inequality.

I can get an answer to part (ii) – $\colvecthree {p}{q}{r} = \colvecthree{24}{6}{1}$ – but that’s from luck, not from using the inequality. (I tried using $\colvecthree {1}{4}{9}$, but didn’t get anywhere. How would you go about it?

-- Students Can’t Always Look At Rationales

That would have been my first thought as well, SCALAR! However, there’s a clue (under the devious STEP-setter definition of “something that’s not really a clue unless you know what you’re looking for”) in that all of the elements on the left of the first equation are squares. That suggests, to a certain kind of mind, that $\colvecthree{p}{2q}{3r}$ might be the relevant one. This suggestion is strengthened by noticing that the second equation can be written as $\colvecthree {8}{4}{1} \cdot \colvecthree {p}{2q}{3r} = 243$.

Let’s let $\bb a = \colvecthree {p}{2q}{3r}$ and $\bb b = \colvecthree {8}{4}{1}$, and look at the original inequality, which says that $\left( \bb a \cdot \bb b \right)^2 \le \left| \bb a \right|^2 \left| \bb b \right|^2$.

Here, the left hand side is $(243)^2$, which I’m not going to work out because a) I can see it’s $3^{10}$ and that’s all I care about and b) I’m going to divide it by stuff shortly.

The right hand side is $\left| \bb a \right|^2 \left| \bb b \right|^2$; however, the first factor is 729 (from what we’re told) and the second is 81 (by working out the modulus of $\colvecthree {8}{4}{1}$).

That tells us that $3^{10} \le 3^6 \times 3^4$. Well, we sorta knew that already. What we didn’t know was that the two were equal – which means $\bb a$ is a multiple of $\bb b$, as established before.

The easiest way I can see to find out what multiple is to notice that we already know $\left| \bb{a} \right|^2 = 729$ and $\left| \bb{b} \right|^2 = 81$, so if $\bb{a} = k \bb{b}$, then $729 = k^2 \times 81$ and $k=\pm 3$. (It has to be positive, since $\bb a \cdot \bb b = k \bb b \cdot \bb b = k \left| \bb b \right|^2$ is.)

So, since we know $k=3$, we know that $\colvecthree {p}{2q}{3r} = \colvecthree {24}{12}{3}$, so $p=24$, $q=6$ and $r=1$. A quick check in the original equations shows that these do, indeed work.

Boom!

-- Uncle Colin

* Post edited 2016-07-20 to put the question in! Thanks to Andy for pointing out that it was missing.