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Consider a hexagon which is equiangular but not equilateral: all angles equal to 120 degrees but four consecutive sides of length $a,b,c,d$ not necessarily equal. What is the perimeter of the hexagon?


A very simple problem with a very neat solution. Inspired by (but more general than) a Dutch Maths Olympiad question.

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  • $\begingroup$ I can't quite explain how I got to the result, but is it 3(a+d)? Never been good at writing proofs, but my (rusty) intuition makes me suspect that the sum of opposite sides (a+d, b+e, c+f) is always the same, and since a,b,c,d omits e and f, a+d is the only pair of opposite sides that we can make. $\endgroup$ – Flater Mar 2 '18 at 15:58
  • $\begingroup$ This was the first problem in the Italian Math Olympiad in 2001 (with prescribed values of $a,b,c,d$, and the question "find $e,f$") --- see olimpiadi.dm.unibo.it/le-gare/gara-nazionale if you can read Italian. $\endgroup$ – Federico Poloni Mar 2 '18 at 17:42
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    $\begingroup$ It's the sum of the length of its sides. $\endgroup$ – Paul Mar 2 '18 at 17:44
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    $\begingroup$ @FedericoPoloni Right, this is such a simple and elegant puzzle that it must have appeared in many different places. I just included the source where I found it, to avoid the risk of having my question deleted by overzealous mods cruising for plagiarism ;-) $\endgroup$ – Rand al'Thor Mar 3 '18 at 14:58
  • $\begingroup$ Claiming rep for posting known puzzles, tsts. :c) Actually: thanks. I think this site can do with some really good puzzles (plus their correct source citation). Thanks for showcasing that to me. +1 $\endgroup$ – BmyGuest Mar 3 '18 at 18:45
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Image of solution:

Image credits due to Geogebra and MS Paint

The neat step is that

by erecting equilateral triangles on every other side, we make a larger equilateral triangle. Thus, we can determine the other two sides in terms of $a,b,c,d$. Thus we get the answer $a+2b+2c+d$.

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  • $\begingroup$ Excellent answer. As a bonus, it shows for which values the perimeter is defined. a shouldn't be larger than c+d and d shouldn't be larger than a+b. $\endgroup$ – Eric Duminil Mar 4 '18 at 14:54
  • $\begingroup$ Fun fact: you don't even need to find the lengths of the final two sides individually. It's enough to know that their sum plus $b$ equals $b+c+d$, which is clear from examining the two slanted sides of the big equilateral triangle. $\endgroup$ – Rand al'Thor Mar 16 '18 at 22:38
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This yields the same answer (of course!) as phenomist's, which was posted earlier; I think it's an interestingly different way of seeing the same thing.

Any two opposite sides are parallel, as in the diagram here where I've drawn them in black.

enter image description here

The other sides, drawn in red in the diagram, are all at the same angle (120 degrees) to those black lines, which means that when we project them onto the vertical blue line in the diagram the ratio (length of projection) / (length of original line) is the same for all. (It happens to equal $\frac{\sqrt3}2$, but we don't need to know that here.) But the sum of projections is the same on the left as on the right: it's just the distance between the parallel black lines. Hence the sum of lengths is the same on the left and on the right too.

In other words: if you take two "opposite" pairs of adjacent sides, their total lengths are the same.

In particular, the missing sides e,f have the same total length as the known sides b,c, and we're done.

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  • $\begingroup$ Yup, this is a common math contest trick. $\endgroup$ – qwr Mar 3 '18 at 18:10
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If we assume there is such a function $P(a,b,c,d)$, how could it look like?

$P$ should be Linear

The perimeter should be $0$ if every length is $0$, and the perimeter should be twice as long if every side is twice as long. It means that $P$ is linear, and can be written:

$$P(a,b,c,d) = ma + nb + oc + pd$$

$P$ should be symmetrical

The perimeter shouldn't change if we consider the mirror image of the hexagon. It means that :

$$P(a,b,c,d) = P(d,c,b,a)\\ \implies \begin{cases} p = m\\ o = n \end{cases}\\ \implies P(a,b,c,d) = m(a + d) + n(b + c)$$

$P$ should be valid for any equiangular hexagon

In particular, $P$ should be valid for a regular hexagon:

$$P(a,a,a,a) = 6a\\ \implies 2ma + 2na = 6a\\ \implies m + n = 3\\ \implies P(a,b,c,d) = (3-n)(a + d) + n(b+c)$$

$P$ should be valid for very small $a$ and $d$

enter image description here

If we let $a$ and $d$ tend to $0$ and keep $b=c$, the hexagon will tend to a rhombus. It means that

$$P(0,b,b,0) = 4b\\ \implies n(b+b) = 4b\\ \implies n = 2\\ \implies m = 1\\ \implies P(a,b,c,d) = a + 2(b+c) + d \\ $$

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    $\begingroup$ Love it that it's a totally different approach. As usual, there is no "best" answer and only "one accepted", but +1 from me for adding this for sure! $\endgroup$ – BmyGuest Mar 3 '18 at 18:48
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    $\begingroup$ This is a neat analytical approach to a geometric problem! $\endgroup$ – Lawrence Mar 4 '18 at 15:23
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Based on Phenmist's answer: embed the hexagon into an equilateral triangle, with side b on the bottom. Label the two other edges e and f.

The base of the triangle is divided into three parts; the first and third are the bases of equilateral triangles of lengths $a$ and $c$ respectively (the middle section is side b).

The right side of the triangle is divided into three parts, of length c, d, and e. The left side is divided into three parts: a, f, and e.

Thus, we have:

$$ a + b + c = c + d + e = a + e + f $$

From there, we get: $e = a + b - d$, $f = b + c - e = d + c - a$.

Thus, add them all together and do a bit of math:

Perimeter = $a + 2b + 2c + d$.

Interestingly, the sum of the lengths of any two consecutive sides is equal to the sum of the lengths of the two opposite sides.

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    $\begingroup$ Isn't this just the same as phenomist's existing answer, only without the diagram and with slightly different words? $\endgroup$ – Gareth McCaughan Mar 2 '18 at 3:59
  • $\begingroup$ @Gareth phenomist's answer was originally slightly wrong: right method, but assuming the sides were in arithmetic progression. Perhaps this answer was written/posted before phenomist corrected theirs? $\endgroup$ – Rand al'Thor Mar 2 '18 at 12:19
  • $\begingroup$ Nope. It was written after mine, and mine was written after Phenomist fixed his bug. $\endgroup$ – Gareth McCaughan Mar 2 '18 at 13:36
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    $\begingroup$ Not long after, though. So perhaps user3294068 started writing while Phenomist's answer was still buggy, but posted after Phenomist fixed it. $\endgroup$ – Gareth McCaughan Mar 2 '18 at 13:38
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    $\begingroup$ Yeah, sorry, my answer came in about five minutes after Gareth's answer, and I didn't notice at the time. $\endgroup$ – user3294068 Mar 2 '18 at 17:23
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Again the same answer but more logical and simple explanation. No need for diagrams it's based on just based facts. Can't comment yet so posting as an answer only.

Equiangular hexagon (a polygon with 120° angle) since the user diminished the rule for not having 4 consecutive sides as equal it can be speculated as all sides are different)

As an equiangular hexagon has an angle of 120, it can be inscribed into 2 equilateral triangles of sides equal to 3 consecutive sides taken into a pair. (Since an equilateral triangle has an angle of 60°)

So considering it has 6 sides from (a to f such as a b c d e and f) the pair of would be any 3 consecutive sides along with the remaining triplet. So in this case would be ( a b c) and (d e f) the triplets would be as such as the sum of a pair of consecutive sides will be equal yo the parallel pair of consecutive sides.

This gives as:

(a+b = d+e) (b+c = e+f) (c+d = f+a). Since the pairs are swappable the perimeter of the hexagon would be the sum of 2 of any two consecutive sides along with the sum of the respective consecutive sides to the selected sides or in simple words, it would be the sum of any two parallel sides and twice of the sum of the sides trapped in between them.

As such we can speculate it as [a + 2(b+c) + d] or if the remaining e and f sides are taken into consideration [d + 2(e+f) + a] since it makes them 4 consecutive sides still, the answer remains the same.

This is possible because of the fact that the 2 equilateral triangles entrapping the hexagon will have sides (a+b+c) and (d+e+f) having b+c=e+f so the total perimeter of the hexagon instead of being a+b+c+d+e+f becomes a+2(b+c)+d

For an example of the theory see this link. https://mvtrinh.wordpress.com/2016/12/20/equiangular-hexagon/

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Here is the neatest solution I could find:

Consider two horizontal lines, A and B, at a distance of h from each other.

Any (non-backtracking) path connecting the two lines, that consists of line segments at angle $\alpha$ from the vertical, will have the same length, namely $\frac{h}{cos(\alpha)}$.

enter image description here

An equiangular hexagon's opposite sides are parallel, so you can align both a and d horizontally. If you do so, the other sides' angle from the vertical is a constant 30°, which means that the paths connecting the endpoints of a to the endpoints of d must be equal in length, that is, $\mathbf{c+b = e+f}$. The perimeter of the hexagon then becomes $\mathbf{a+2b+2c+d}$

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  • $\begingroup$ Interesting way to look at it! $\endgroup$ – Eric Duminil May 16 '18 at 11:25
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I'm pretty surprised that nobody has yet mentioned the standard and thoughtless way of solving problems like this. Toss the figure into the complex plane, and invoke the following well known lemma:

The complex numbers $z_1, z_2, z_3, \cdots, z_{n-1}, z_n$ (treat them as vectors) can be translated to form a closed $n$-sided polygon (possibly intersecting) iff $z_1 + z_2 + \cdots + z_n = 0$

Now, starting form a side, let the sides be $a, b, c, d, e, f$, and $\omega \in \mathbb{C} - \mathbb{R}$ be a primitive third root of unity.

The equiangular condition is equivalent to the side vectors being $a, -b\omega^2, \omega c, -d, \omega^2 e, - \omega f$, so they form a hexagon iff (it's easy to see the case when the polygon is interesecting is impossible) $a - b\omega^2 + \omega c - d + \omega^2 e - \omega f = 0 \Rightarrow (a-d) + \omega (e-b) + \omega^2 (c-f) = 0$ Now using $\omega^2 = -\omega-1$, the condition reduces to $(a+f-d-c) + \omega (e-b-c+f) = 0$. Now if either of the numbers $(a+f-d-c)$ or $(e-b-c+f)$ are nonzero, we then find a polynomial with degree at max one and real coefficients having a complex root, which is obviously nonsense, so both of them are zero.

So $f= c+d-a$ and $e = b+c-f = b+a-d$, and we get the desired perimeter to be $a+b+c+d+e+f = a+2b+2c+d$, as desired.

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