Re: equilateral tringles

In article <mn.08487d5c306dddd6.22155@xxxxxxx>,
Philippe 92 <nospam@xxxxxxxxxxxx> wrote:
>Dave Rusin wrote :
>> In article <3950172.1133377985893.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
>> eugene <jane1806@xxxxxxx> wrote:
>>
>>> Let ABC be an equilateral triangle. Points A_1,B_1,C_1 are chosen
>> inside the triangle in such a way that A_1 \in CC_1, B_1 \in AA_1,
>> C_1 \in BB_1 and AB_1=B_1A_1, BC_1=C_1B_1, CA_1=C_1A_1. Prove that
>> the triangle A_1B_1C_1 is also equilateral.

>> The word "chosen" is misleading here: the points A1, B1, C1 are
>> uniquely specified by the condition that each of them is a midpoint
>> of one of these three segments (and an endpoint of another).
>> But then, that means that a one-third turn of the figure preserves
>> all the interior line segments,
>
>Why that ? Yes if AA_1 = BB_1 = CC_1, or if angles are equal...
>that is if A_1B_1C_1 equilateral, you proove that it is equilateral !

No -- actually I think your question is inserted in the wrong space.

It is clear that the conditions describing the points {A1, B1, C1}
is invariant under a one-third turn of the triangle. (Rotate one such
triple and you obtain another triple of endpoints-which-are-midpoints.)
So then the fact that e.g AA_1 = BB_1 is a consequence of the
_uniqueness_ of the solution. So really your question "Why that?"
should be two lines earlier, when I asserted uniqueness.

As for the answer to that, I will tell you I approached it the same
way you did: the triangle lives in an affine space where we can
assert "XYZ are collinear with XY=YZ" as "Y = (1/2) X + (1/2) Y",
giving three equations in three unknown points, which have a
unique solution ( A1 = (1/7)A + (2/7)B + (4/7)C , and likewise for B1, C1 ).
You can express this fact in complex coordinates as you did, or in
real coordinates, or think of it as barycentric coordinates in the triangle.
If you actually DO the computations, you can SEE the three-fold
symmetry in the answers, but I opted simply to note that there was
a unique answer, and then see that the symmetry in the answer was
a consequence of symmetry in the question.


Here is a variant: What happens if we try this trick starting with
a regular tetrahedron P_1 P_2 P_3 P_4 ? Construct four more
points Q_i with Q_{i+1} being the midpoints of Q_i P_i.
Is there a unique such set of four points? Do they form a regular
tetrahedron? How does this shed light on what happened in the plane?

dave
.

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