PseudonomousDaughterTwo learned vector-addition at school and

important formulas such as the _Chasles-Moebius_ equation

$\forall A,B,C \in \mathbb{R}^2~:~\vec{AB}+\vec{BC} = \vec{AC} $

Last evening I helped her a bit with her homework and there was one

problem she could not do immediately (but it was a starred exercise so

you didn't have to do it, but…) :

consider a regular pentagon

with center $\vec{0} $. Prove that

$\vec{0A} + \vec{0B} +

\vec{0C} + \vec{0D} + \vec{0E} = \vec{0} $

PD2 : How would

_you_ do this? (with a tone like : I bet even you can't do

it)

Me : Symmetry!

PD2 : Huh?

Me : Rotate the plane

1/5 turn, then $A \mapsto B $, $B \mapsto C $ and so on. So the vector

giving the sum of all five terms must be mapped to itself under this

rotation and the only vector doing this is the zero vector.

PD2 :

That cannot be the solution, you didn't take sums of vectors and all

other exercises did that.

Me : I don't care, it is an elegant

solution, you don't have to compute a thing!

But clearly

she was not convinced and I had to admit there was nothing in her

textbook preparing her for such an argument. I was about to explain that

there was even more symmetry : reflecting along a line through a vertex

giving dihedral symmetry when I saw what the _intended solution_

of the exercise was :

Me : Okay, if you _have_ to do

sums let us try this. Fix a vertex, say A. Then the sum

$\vec{0E}+\vec{0B} $ must lie on the line 0A by the parallellogram-rule

(always good to drop in a word from the textbook to gain some

trust…), similarly the sum $\vec{0C}+\vec{0D} $ must lie on the

line 0A. So you now have to do a sum of three vectors lying on the

line 0A so the result must lie on 0A

PD2 : Yes, and???

Me : But there was nothing special about $A$. I could have started with

B and do the whole argument all over again and then I would get that

the sum is a vector on the line 0B

PD2 : And the only vector

lying on both 0A and 0B is $\vec{0} $

Me : Right! But

all we did now was just redoing the symmetry argument because the line

0A is mapped to 0B

PD2 : Don't you get started on

_that symmetry_ again!

I wonder which of the two

solutions she will sell today as her own. I would love to see the face

of a teacher when a 15yr old says “Clearly that is trivial because

the zero vector is the only one left invariant under

pentagon-symmetry!”

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