You are given a tetrahedron $T=ABCD$. Average opposing edges to create a second tetrahedron $T'=A'B'C'D'$ with $\overline{A'B'}=\overline{C'D'}=\frac 1 2[\overline{AB}+\overline{CD}]$ etc. Place $T$ and $T'$ in space; they may overlap if you like. For each of the three pairs $(\overline{AB'},\overline{A'B})$, $(\overline{AC'},\overline{D'B})$ and $(\overline{AD'},\overline{C'B})$ take its average, square it and, finally, sum the three to get $x$. Now relabel $T$ and $T'$ back-to-front, i.e. rename $A\rightarrow D, B\rightarrow C$ etc. And do the same calculation once more to get $y$. Compare $xy$ to the product $z=\overline{AA'}\,\overline{BB'}\,\overline{CC'}\,\overline{DD'}$.

Can $z$ ever be larger or at least (non-trivially) equal $xy$? If so which of the two?

(Conceived by me. Elegant---easily fit back of envelope---solution exists.)

Picture (may be considered a spoiler if you are really strict on that) spoileriness $\frac 1 2 / 10$: may give you ideas

enter image description here One possible arrangement. But note that, for example, $AC'BD'$ are not required to be in the same plane.

Maybe it's time for a first hint/spoiler don't read if you want the full (quite satisfying even if I say so myself) experience! spoileriness $3/10$: cryptic but not very

The average herring is maybe not red but a deep shade of pink.

Second hint/spoiler don't read unless you are positively desperate spoileriness $5/10$: won't feel like you solved yourself

Like someone trying to hide their mediocrity by pretending to be just an average lad/lass.

Third spoiler-not-hint don't read spoileriness $7/10$: enjoy being spoon-fed?

Not all mediocrities are equal.

Fourth wrecking-ball-not-spoiler impervious to advice aren't you?! spoileriness $10/10$: might as well come down to your place and solve it for you

Even if they hide behind a ptosh name.

... one ... last ... mega-spoiler only read if you are one of the I-like-geometry-but-have-zero-background tribe! How would you know? If you barely remember the sum of angles in a triangle and the name Pythagoras rings a bell but you've never heard of, say, inscribed angles or a chap called Heron then you may qualify---none of the names I just dropped have any bearing on the problem at hand, I just used them to gauge your general level of geometry. You need to know one result from elementary geometry which is a classic but not quite as famous as $a^2+b^2=c^2$ If and only if you think that's what's holding you back read the last spoiler spoileriness $20/10$

Look up Ptolemy's inequality.

If you answer, please state whether you looked at the hints!


1 Answer 1


Let's first get through the daunting task of actually writing out the expressions to be compared. Our $x$ is $$\left(\frac{AB'+BA'}{2}\right)^2+\left(\frac{AC'+BD'}{2}\right)^2+\left(\frac{AD'+BC'}{2}\right)^2.$$Now $y$ is basically this, except with a relabelling of variables, and writing out the entire product $xy$ is quite painful, so we will defer that to when we really need it.

Now we will start the estimates.

AM-GM inequality says $\left(\frac{a+b}2\right)^2\ge ab$, so $x$ is lower-bounded by $$AB'\cdot BA'+AC'\cdot BD'+AD'\cdot BC'.$$ Ptolemy's inequality now lets us say $$x\ge AB'\cdot BA'+\boxed{AC'\cdot BD'+AD'\cdot BC'}\ge AB'\cdot A'B+\boxed{AB\cdot C'D'}.$$ This seems to resist further simplication/estimation; but thankfully, the question says $C'D'$ is equal to $A'B'$! This and another bout of Ptolemy gets us to $$x\ge AB'\cdot A'B+AB\cdot C'D'=AB'\cdot A'B+AB\cdot A'B'\ge AA'\cdot BB'.$$

Looks like we are almost done! Since $y$ was just $x$ relabelled (and the problem conditions are all stable under that relabelling), the same logic applies to it, giving

$$y\ge CC'\cdot DD'.$$And so $$xy\ge AA'\cdot BB'\cdot CC'\cdot DD'\ge z.$$The conclusion follows immediately.

Puzzle feedback time!

Cute puzzle! There are some 'aha!' moments as to what piece of information feeds into where, and it's satisfying to finish.

I believe the main reason this sat unattempted so far is that the expressions of $x$ and $y$ are really scary. The problem statement describes these quantities in (a lot of) words, so any progress must begin with converting this to math expressions, and that kind of feels like a menial chore.

I'll admit I looked at all the hints, but that's primarily because I was hoping to find something that would cut down on the work involved. I thought of using the main result the later hints are cluing a while back (it's one of the very few ways of handing products of lengths I know) but felt "eh I'll wait till someone else works out the expression for $xy$". I feel the question would be better if that expression was given to begin with.

Another minor beef I have is that the fact that $A'B'$ is $\tfrac 12[AB+CD]$ seems to be inconsequential. It's not uncommon to include extra information in puzzles as red herrings; I tend not to like that. This is really a personal opinion, I don't know how many people share this view.

Other than these, I like the idea; I haven't used you-know-what inequality is 3-D situations before, so this feels fresh! Hope to see more of your creations. :)

  • 2
    $\begingroup$ Thank you and Well done! Re the $A'B'=...$ is inconsequential. I was indeed looking for a way of burying the $A'B'=C'D'$ because I was (needlessly with the benefit of hindsight) concerned otherwise it would be too obviouos. In the same way as the AM-GM owes its appearance mostly to the fact that otherwise our ptosh friend would be visually jumping out of the expressions you described as scary. $\endgroup$ Aug 12, 2020 at 7:01
  • 1
    $\begingroup$ Also, I felt the ugliness of writing out $xy$ would actually add to thet satisfaction that comes with the model solution since being a bit clever allowed you to completely bypass ever having to do that in the first place. $\endgroup$ Aug 12, 2020 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.